Properties of Gases
0/3 completed

Gas Laws
0/21 completed

- Boyle's Law
- Exercise 1
- Exercise 2
- Charles's Law and STP
- Exercise 2
- Avogadro's Law
- Exercise 4
- Exercise 5
- Ideal Gas Equation
- Exercise 6
- Exercise 7
- Determining Molar Mass from Ideal Gas Equation
- Exercise 8
- Determining Gas Density from Ideal Gas Equation
- Exercise 9
- Exercise 10
- Law of Combining Volumes
- Exercise 11
- Exercise 11 - Alternative Solution
- Exercise 12
- Exercise 13

Mixtures of Gases
0/8 completed

Kinetic Theory of Gases
0/7 completed

Nonideal Gases
0/2 completed

{"Free":0,"Sample":1,"Paid":2}

[{"Name":"Properties of Gases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Liquid Pressure","Duration":"10m 45s","ChapterTopicVideoID":17410,"CourseChapterTopicPlaylistID":80094,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.830 ","End":"00:04.785","Text":"In this video, we\u0027ll discuss the pressure of a liquid."},{"Start":"00:04.785 ","End":"00:06.240","Text":"In the next video,"},{"Start":"00:06.240 ","End":"00:10.995","Text":"we\u0027ll use this liquid pressure to measure the pressure of a gas."},{"Start":"00:10.995 ","End":"00:14.860","Text":"First, let\u0027s begin with discussing pressure."},{"Start":"00:23.090 ","End":"00:28.120","Text":"Pressure is defined as the force per unit area."},{"Start":"00:31.310 ","End":"00:35.309","Text":"For example, if you don\u0027t want to sink into snow,"},{"Start":"00:35.309 ","End":"00:42.070","Text":"you use skis which have a larger area than shoes so that the pressure is lower."},{"Start":"00:43.250 ","End":"00:48.589","Text":"Here\u0027s a ski, there\u0027s a long area,"},{"Start":"00:48.589 ","End":"00:51.515","Text":"whereas the shoe, here\u0027s the sole of a shoe,"},{"Start":"00:51.515 ","End":"00:54.170","Text":"just has a smaller area."},{"Start":"00:54.170 ","End":"00:58.640","Text":"Here, the pressure will be less than with the shoe."},{"Start":"00:58.640 ","End":"01:02.150","Text":"Unfortunately, in this video,"},{"Start":"01:02.150 ","End":"01:06.290","Text":"we have to discuss a lot of units and the first unit we\u0027re going to talk"},{"Start":"01:06.290 ","End":"01:13.020","Text":"about is the SI unit of force because pressure is proportional to force."},{"Start":"01:14.080 ","End":"01:17.930","Text":"The SI unit of force is called the Newton,"},{"Start":"01:17.930 ","End":"01:22.595","Text":"N, after the famous English physicist Isaac Newton."},{"Start":"01:22.595 ","End":"01:25.080","Text":"I\u0027m sure you\u0027ve all heard of him."},{"Start":"01:27.110 ","End":"01:32.380","Text":"Newton is defined as the force required to produce an acceleration of"},{"Start":"01:32.380 ","End":"01:37.460","Text":"1 meter per second square of a kilogram mass."},{"Start":"01:37.460 ","End":"01:44.650","Text":"If you want 1 kilogram mass to move with an acceleration of 1 meter per second squared,"},{"Start":"01:44.650 ","End":"01:49.580","Text":"you will need to apply a force of 1 newton."},{"Start":"01:49.970 ","End":"01:52.665","Text":"You may have learned some physics,"},{"Start":"01:52.665 ","End":"01:55.840","Text":"and then remember that Newton\u0027s second law says that"},{"Start":"01:55.840 ","End":"01:59.065","Text":"the force is equal to the mass times the acceleration."},{"Start":"01:59.065 ","End":"02:02.365","Text":"M is the mass and a is acceleration,"},{"Start":"02:02.365 ","End":"02:09.665","Text":"so 1 newton is 1 kilogram times 1 meter per second squared."},{"Start":"02:09.665 ","End":"02:13.735","Text":"Now we can discuss the SI unit of pressure."},{"Start":"02:13.735 ","End":"02:17.210","Text":"The SI unit of pressure is called the Pascal,"},{"Start":"02:17.210 ","End":"02:23.210","Text":"Pa, after the famous French polymath Blaise Pascal."},{"Start":"02:23.210 ","End":"02:27.290","Text":"You\u0027ve probably met the Pascal triangle in your mathematics."},{"Start":"02:27.290 ","End":"02:36.145","Text":"A pascal is defined as the pressure exerted by 1 newton on 1 meter squared."},{"Start":"02:36.145 ","End":"02:38.740","Text":"1 newton is the force,"},{"Start":"02:38.740 ","End":"02:41.360","Text":"1 meter squared is the area."},{"Start":"02:41.360 ","End":"02:44.450","Text":"The pressure in pascal units is"},{"Start":"02:44.450 ","End":"02:48.125","Text":"the force in Newtons divided by the area in meter squared."},{"Start":"02:48.125 ","End":"02:53.285","Text":"1 pascal is 1 Newton divided by meter squared,"},{"Start":"02:53.285 ","End":"02:55.660","Text":"so 1 newton per meter squared,"},{"Start":"02:55.660 ","End":"02:57.640","Text":"and that\u0027s equal to 1."},{"Start":"02:57.640 ","End":"03:03.340","Text":"Now, we can substitute for Newton kilogram times meters per second squared"},{"Start":"03:03.340 ","End":"03:09.395","Text":"times and the area is meters power minus 2 and if we multiply that out,"},{"Start":"03:09.395 ","End":"03:11.900","Text":"meter times meter minus 2,"},{"Start":"03:11.900 ","End":"03:19.190","Text":"is meters to the power of minus 1 so we\u0027re left with 1 kilogram per meter,"},{"Start":"03:19.190 ","End":"03:25.130","Text":"meter to the power minus 1 per second squared seconds to the power minus 2."},{"Start":"03:25.130 ","End":"03:30.965","Text":"Now we have, we know what a pascal is and that\u0027s the official SI unit."},{"Start":"03:30.965 ","End":"03:34.715","Text":"We\u0027re going to discuss the pressure exerted by a liquid."},{"Start":"03:34.715 ","End":"03:45.030","Text":"Suppose we have a column of liquid and here is its base and the height\u0027s going to be h,"},{"Start":"03:45.030 ","End":"03:48.710","Text":"and the area of the base is going to be A."},{"Start":"03:48.710 ","End":"03:54.035","Text":"Now we want to know what the pressure exerted by this column of liquid is."},{"Start":"03:54.035 ","End":"03:56.780","Text":"First thing to notice is that weight is"},{"Start":"03:56.780 ","End":"04:00.860","Text":"a force and if we go back to Newton\u0027s second law, weight,"},{"Start":"04:00.860 ","End":"04:07.500","Text":"which is a force is equal to the mass times g. G is acceleration due to gravity,"},{"Start":"04:07.500 ","End":"04:12.470","Text":"so that\u0027s 9.81 meters per second squared at sea level."},{"Start":"04:12.470 ","End":"04:15.830","Text":"That\u0027s the relationship between the weight and the mass."},{"Start":"04:15.830 ","End":"04:20.405","Text":"Now we can write that the pressure is equal to force divided by area."},{"Start":"04:20.405 ","End":"04:25.180","Text":"This case the force is the weight w divided by A."},{"Start":"04:25.180 ","End":"04:31.955","Text":"We can write w mass times acceleration due to gravity divided by A."},{"Start":"04:31.955 ","End":"04:39.140","Text":"Now we know the relationship between the mass of a liquid and its density."},{"Start":"04:39.140 ","End":"04:43.490","Text":"Remember that density is equal to mass divided by"},{"Start":"04:43.490 ","End":"04:48.530","Text":"volume so we know that the mass is equal to the density times the volume."},{"Start":"04:48.530 ","End":"04:50.930","Text":"We can write here instead of mass,"},{"Start":"04:50.930 ","End":"04:53.285","Text":"we can write density times volume."},{"Start":"04:53.285 ","End":"04:57.380","Text":"Now we have g times d times v divided by A."},{"Start":"04:57.380 ","End":"05:00.799","Text":"Now, for this column of liquid,"},{"Start":"05:00.799 ","End":"05:06.995","Text":"the volume is equal to the area times the height."},{"Start":"05:06.995 ","End":"05:09.200","Text":"We can now substitute here,"},{"Start":"05:09.200 ","End":"05:12.455","Text":"volume is equal to area times height."},{"Start":"05:12.455 ","End":"05:16.369","Text":"The area appears in both numerator and denominator."},{"Start":"05:16.369 ","End":"05:21.495","Text":"We can cancel them and we\u0027re left with g times d times"},{"Start":"05:21.495 ","End":"05:27.440","Text":"h so that\u0027s the acceleration due to gravity times the density, times the height."},{"Start":"05:27.440 ","End":"05:31.790","Text":"We can say that the pressure of a liquid column is directly"},{"Start":"05:31.790 ","End":"05:36.890","Text":"proportional to the height of the liquid column and its density."},{"Start":"05:36.890 ","End":"05:39.695","Text":"Now, what liquids can we use?"},{"Start":"05:39.695 ","End":"05:42.920","Text":"Well, the most common are water,"},{"Start":"05:42.920 ","End":"05:47.000","Text":"which has a density of 1 gram per milliliter, and mercury,"},{"Start":"05:47.000 ","End":"05:51.484","Text":"which has a density of 13.6 grams per milliliter."},{"Start":"05:51.484 ","End":"05:58.385","Text":"Mercury is extremely dense and is often used to measure pressure."},{"Start":"05:58.385 ","End":"06:02.210","Text":"Now we\u0027re going to measure the barometric pressure."},{"Start":"06:02.210 ","End":"06:06.290","Text":"The barometric pressure is the atmospheric pressure."},{"Start":"06:06.290 ","End":"06:08.840","Text":"It\u0027s the same thing as the atmospheric pressure."},{"Start":"06:08.840 ","End":"06:11.495","Text":"Now we\u0027re going to measure the barometric pressure."},{"Start":"06:11.495 ","End":"06:14.750","Text":"The barometric pressure is the same as the atmospheric pressure."},{"Start":"06:14.750 ","End":"06:16.640","Text":"It\u0027s just another name for it."},{"Start":"06:16.640 ","End":"06:19.305","Text":"Let\u0027s look at the picture."},{"Start":"06:19.305 ","End":"06:21.285","Text":"On the left-hand side,"},{"Start":"06:21.285 ","End":"06:27.080","Text":"we have a container of mercury and a glass tube in it."},{"Start":"06:27.080 ","End":"06:30.755","Text":"The important thing to notice is that the tube is open at the end."},{"Start":"06:30.755 ","End":"06:35.510","Text":"Now the atmospheric pressure pushes down on the mercury,"},{"Start":"06:35.510 ","End":"06:39.370","Text":"but it also pushes down inside the tube."},{"Start":"06:39.370 ","End":"06:42.860","Text":"The level of mercury inside the tube is the same as"},{"Start":"06:42.860 ","End":"06:46.250","Text":"the level of mercury in the beaker in the container."},{"Start":"06:46.250 ","End":"06:47.750","Text":"We look at the next picture,"},{"Start":"06:47.750 ","End":"06:50.155","Text":"we\u0027ll see that here it\u0027s closed."},{"Start":"06:50.155 ","End":"06:52.820","Text":"The glass tube is closed."},{"Start":"06:52.820 ","End":"06:55.835","Text":"If we start off with a vacuum with no air in it,"},{"Start":"06:55.835 ","End":"07:00.530","Text":"the atmospheric pressure will push down and push the mercury up"},{"Start":"07:00.530 ","End":"07:06.200","Text":"the column so that the pressure exerted by the mercury in the column,"},{"Start":"07:06.200 ","End":"07:10.655","Text":"the liquid pressure, is exactly equal to the atmospheric pressure."},{"Start":"07:10.655 ","End":"07:14.480","Text":"The mercury barometer was invented by somebody called"},{"Start":"07:14.480 ","End":"07:20.060","Text":"Torricelli in 1643 to measure atmospheric pressure."},{"Start":"07:20.060 ","End":"07:24.055","Text":"We\u0027ll see later he gives his name to a unit called the torr."},{"Start":"07:24.055 ","End":"07:27.515","Text":"The atmospheric pressure supported by column of mercury,"},{"Start":"07:27.515 ","End":"07:32.915","Text":"which is exactly 760 millimeters high at 0 degrees Celsius,"},{"Start":"07:32.915 ","End":"07:37.880","Text":"is given in pascals by 101,325 pascals."},{"Start":"07:37.880 ","End":"07:40.819","Text":"This is too pretty high accuracy,"},{"Start":"07:40.819 ","End":"07:43.745","Text":"but it\u0027s not absolutely precise."},{"Start":"07:43.745 ","End":"07:47.720","Text":"This, you get it by using a very precise measure of"},{"Start":"07:47.720 ","End":"07:51.635","Text":"g and a very precise measure of the density."},{"Start":"07:51.635 ","End":"07:55.024","Text":"If we were to do the same thing with water,"},{"Start":"07:55.024 ","End":"07:59.390","Text":"we\u0027d see that the column of water that is pushed"},{"Start":"07:59.390 ","End":"08:05.132","Text":"up by atmospheric pressure is 10.3 meters."},{"Start":"08:05.132 ","End":"08:09.425","Text":"That\u0027s the highest atmospheric pressure can push water."},{"Start":"08:09.425 ","End":"08:11.120","Text":"If you\u0027re building is higher,"},{"Start":"08:11.120 ","End":"08:17.690","Text":"you will need some sort of mechanical pump to pump the water to the top of the building."},{"Start":"08:17.690 ","End":"08:21.500","Text":"Unfortunately, there are even more units of pressure used"},{"Start":"08:21.500 ","End":"08:26.395","Text":"in this subject of gas pressure."},{"Start":"08:26.395 ","End":"08:30.080","Text":"Chemists very much like a standard atmosphere,"},{"Start":"08:30.080 ","End":"08:32.230","Text":"which they write as atm,"},{"Start":"08:32.230 ","End":"08:36.790","Text":"and it\u0027s defined to be exactly 101,325"},{"Start":"08:36.790 ","End":"08:43.235","Text":"pascals or you can give it in kilopascals 101.325 kilopascals."},{"Start":"08:43.235 ","End":"08:49.630","Text":"Now, while atmosphere is also taken to be equivalent to 760 torrs,"},{"Start":"08:49.630 ","End":"08:51.350","Text":"that\u0027s how you define a torr."},{"Start":"08:51.350 ","End":"08:54.355","Text":"I\u0027ve told you that comes from the name Torricelli,"},{"Start":"08:54.355 ","End":"08:59.750","Text":"1 atmosphere is precisely equal to 760 torr,"},{"Start":"08:59.750 ","End":"09:05.960","Text":"so 1 torr is 1/760 atmospheres."},{"Start":"09:05.960 ","End":"09:09.860","Text":"Remember, it\u0027s almost precisely the same as 1 millimeter of mercury."},{"Start":"09:09.860 ","End":"09:13.265","Text":"In this course will take the two to be the same."},{"Start":"09:13.265 ","End":"09:17.420","Text":"We use torr and millimeters of mercury interchangeably."},{"Start":"09:17.420 ","End":"09:20.870","Text":"Now there\u0027s one more unit that the physicists like very"},{"Start":"09:20.870 ","End":"09:24.740","Text":"much and sometimes chemists use it too,"},{"Start":"09:24.740 ","End":"09:30.360","Text":"1 bar is equal to 10^5 pascals,"},{"Start":"09:30.360 ","End":"09:32.910","Text":"that\u0027s 100,000 pascals,"},{"Start":"09:32.910 ","End":"09:35.420","Text":"and we can convert that to atmospheres."},{"Start":"09:35.420 ","End":"09:43.240","Text":"Because remember that 1 atmosphere we defined as exactly 101,325 pascals."},{"Start":"09:43.240 ","End":"09:50.420","Text":"If pascals cancel and if we multiply or divide 10^5 by this big number,"},{"Start":"09:50.420 ","End":"09:54.425","Text":"we get 0.986923 atmospheres,"},{"Start":"09:54.425 ","End":"09:57.289","Text":"which is very close to 1 atmosphere."},{"Start":"09:57.289 ","End":"09:59.780","Text":"If we wanted to do it the other way round,"},{"Start":"09:59.780 ","End":"10:02.510","Text":"you get 1 atmosphere can be"},{"Start":"10:02.510 ","End":"10:05.900","Text":"converted to bars because we know this is our conversion unit."},{"Start":"10:05.900 ","End":"10:11.450","Text":"1 bar is the same as 0.986923 atmospheres,"},{"Start":"10:11.450 ","End":"10:13.100","Text":"we just calculated that,"},{"Start":"10:13.100 ","End":"10:16.755","Text":"and then it comes here, refused it here."},{"Start":"10:16.755 ","End":"10:18.965","Text":"If we do the calculation,"},{"Start":"10:18.965 ","End":"10:23.890","Text":"that will be 1.01325 bars so it\u0027s almost exactly 1 bar,"},{"Start":"10:23.890 ","End":"10:27.200","Text":"so 1 atmosphere, 1 bar are very very similar."},{"Start":"10:27.200 ","End":"10:29.180","Text":"We\u0027ll see it makes a little difference,"},{"Start":"10:29.180 ","End":"10:30.410","Text":"but not a great deal."},{"Start":"10:30.410 ","End":"10:34.400","Text":"Just to note that weather forecasters"},{"Start":"10:34.400 ","End":"10:39.545","Text":"use millibars as unit of pressure so that\u0027s where you may have seen it before."},{"Start":"10:39.545 ","End":"10:41.600","Text":"In this video, we discussed"},{"Start":"10:41.600 ","End":"10:45.960","Text":"liquid pressure and some of the units used to measure pressure."}],"ID":18157},{"Watched":false,"Name":"Gas Pressure","Duration":"6m 39s","ChapterTopicVideoID":17411,"CourseChapterTopicPlaylistID":80094,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"In the previous video,"},{"Start":"00:01.845 ","End":"00:04.950","Text":"we discussed the pressure exerted by a column of liquid."},{"Start":"00:04.950 ","End":"00:09.600","Text":"In this video, we will use liquid pressure to measure gas pressure."},{"Start":"00:09.600 ","End":"00:14.580","Text":"We\u0027re going to measure the gas pressure with an open-ended manometer."},{"Start":"00:14.580 ","End":"00:18.540","Text":"Here we have open-ended manometers, 3 pictures."},{"Start":"00:18.540 ","End":"00:20.175","Text":"Let\u0027s look at it."},{"Start":"00:20.175 ","End":"00:28.260","Text":"We see that there is a glass U-tube which is open at the top and a gas in a glass bulb,"},{"Start":"00:28.260 ","End":"00:32.210","Text":"and inside the U-tube there is mercury."},{"Start":"00:32.210 ","End":"00:40.205","Text":"There\u0027s also a stopcock which allows us to release more or less gas."},{"Start":"00:40.205 ","End":"00:43.090","Text":"In this picture, the first picture A,"},{"Start":"00:43.090 ","End":"00:46.670","Text":"we see that the level is the same in"},{"Start":"00:46.670 ","End":"00:51.740","Text":"the open tube here and in the other side which is connected to the gas,"},{"Start":"00:51.740 ","End":"00:55.610","Text":"so the atmospheric pressure pressing down on"},{"Start":"00:55.610 ","End":"00:59.825","Text":"one side is exactly equal to the gas pressure on the other side."},{"Start":"00:59.825 ","End":"01:01.520","Text":"In the second picture,"},{"Start":"01:01.520 ","End":"01:05.360","Text":"we\u0027ve let out more gas and we see that the gas is"},{"Start":"01:05.360 ","End":"01:10.835","Text":"pushing the mercury down in the column and up on the other side."},{"Start":"01:10.835 ","End":"01:19.110","Text":"Now, the pressure of the gas we\u0027ll see is proportional to this h,"},{"Start":"01:19.110 ","End":"01:22.189","Text":"h is the difference in height between"},{"Start":"01:22.189 ","End":"01:26.164","Text":"the left-hand side and the right-hand side of the manometer."},{"Start":"01:26.164 ","End":"01:30.380","Text":"In the third picture C we have less gas,"},{"Start":"01:30.380 ","End":"01:36.920","Text":"so the pressure of the gas is less than the atmospheric pressure."},{"Start":"01:36.920 ","End":"01:44.585","Text":"It isn\u0027t sufficient to push the mercury right up to the same height in both sides."},{"Start":"01:44.585 ","End":"01:47.270","Text":"What\u0027s a manometer that we\u0027ve just seen?"},{"Start":"01:47.270 ","End":"01:49.730","Text":"Let\u0027s see what a manometer is."},{"Start":"01:49.730 ","End":"01:53.435","Text":"A manometer is a device for measuring pressure."},{"Start":"01:53.435 ","End":"01:57.500","Text":"There is mercury in the U-tube and gas in the glass bulb,"},{"Start":"01:57.500 ","End":"01:58.835","Text":"in the glass sphere."},{"Start":"01:58.835 ","End":"02:01.100","Text":"Let\u0027s look at the 3 cases."},{"Start":"02:01.100 ","End":"02:02.630","Text":"In the first case,"},{"Start":"02:02.630 ","End":"02:06.620","Text":"remember the mercury was the same height in both sides,"},{"Start":"02:06.620 ","End":"02:08.780","Text":"so the pressure of the gas is equal to"},{"Start":"02:08.780 ","End":"02:12.290","Text":"the pressure of the atmosphere, the barometric pressure."},{"Start":"02:12.290 ","End":"02:15.005","Text":"In the second case B,"},{"Start":"02:15.005 ","End":"02:18.745","Text":"the pressure of the gas was greater than the atmospheric pressure."},{"Start":"02:18.745 ","End":"02:21.350","Text":"We can write this, the pressure of the gas"},{"Start":"02:21.350 ","End":"02:24.440","Text":"is equal to the pressure of the atmosphere plus"},{"Start":"02:24.440 ","End":"02:27.425","Text":"Delta P. Delta P is a difference"},{"Start":"02:27.425 ","End":"02:31.795","Text":"between the gas pressure and the pressure of the atmosphere."},{"Start":"02:31.795 ","End":"02:37.339","Text":"Delta P, we can write the pressure is g,"},{"Start":"02:37.339 ","End":"02:39.710","Text":"acceleration due to gravity,"},{"Start":"02:39.710 ","End":"02:44.150","Text":"times the density of the mercury times the h, the height,"},{"Start":"02:44.150 ","End":"02:49.010","Text":"the difference in height between the 2 sides of the manometer."},{"Start":"02:49.010 ","End":"02:54.350","Text":"Then C, the pressure of the gas was less than the atmospheric pressure."},{"Start":"02:54.350 ","End":"02:56.060","Text":"We can write the pressure gases,"},{"Start":"02:56.060 ","End":"02:58.880","Text":"the atmospheric pressure plus Delta P,"},{"Start":"02:58.880 ","End":"03:00.485","Text":"same equation as before,"},{"Start":"03:00.485 ","End":"03:04.160","Text":"but in this case Delta P is negative,"},{"Start":"03:04.160 ","End":"03:08.250","Text":"it\u0027s minus g times d times"},{"Start":"03:08.250 ","End":"03:14.630","Text":"h. If we use millimeters of mercury to measure the pressure,"},{"Start":"03:14.630 ","End":"03:19.774","Text":"then we can see that Delta P is equal to plus or minus"},{"Start":"03:19.774 ","End":"03:23.180","Text":"each millimeters of mercury depending on"},{"Start":"03:23.180 ","End":"03:26.720","Text":"whether we\u0027ll have to add it when it\u0027s positive or negative."},{"Start":"03:26.720 ","End":"03:28.490","Text":"Let\u0027s take an example."},{"Start":"03:28.490 ","End":"03:32.855","Text":"Supposing Delta P is minus 10 millimeters of mercury."},{"Start":"03:32.855 ","End":"03:36.665","Text":"What is P gas in millimeters of mercury?"},{"Start":"03:36.665 ","End":"03:40.660","Text":"Then, what is P gas in pascals,"},{"Start":"03:40.660 ","End":"03:44.845","Text":"in units of atmosphere and in bars."},{"Start":"03:44.845 ","End":"03:51.640","Text":"We saw the equation was P gas is equal to P atmosphere plus Delta P. In this case,"},{"Start":"03:51.640 ","End":"03:53.380","Text":"Delta P is negative."},{"Start":"03:53.380 ","End":"03:55.780","Text":"So P atmosphere is 760,"},{"Start":"03:55.780 ","End":"03:57.850","Text":"Delta P is minus 10,"},{"Start":"03:57.850 ","End":"04:01.855","Text":"so that\u0027s 760 minus 10 millimeters of mercury,"},{"Start":"04:01.855 ","End":"04:05.170","Text":"which is 750 millimeters of mercury."},{"Start":"04:05.170 ","End":"04:11.335","Text":"In this case, the gas pressure is less than the atmospheric pressure."},{"Start":"04:11.335 ","End":"04:15.310","Text":"Now we have to convert the pressure of gas,"},{"Start":"04:15.310 ","End":"04:20.745","Text":"the gas pressure from millimeters of mercury to pascal,"},{"Start":"04:20.745 ","End":"04:25.240","Text":"so 750 millimeters of mercury is equals to 750 millimeters of"},{"Start":"04:25.240 ","End":"04:31.285","Text":"mercury with a conversion factor between pascals and millimeters of mercury."},{"Start":"04:31.285 ","End":"04:36.670","Text":"Here it is 101,325 pascals,"},{"Start":"04:36.670 ","End":"04:42.325","Text":"it\u0027s almost precisely equal to 760 millimeters of mercury."},{"Start":"04:42.325 ","End":"04:45.400","Text":"The millimeters of mercury cancel,"},{"Start":"04:45.400 ","End":"04:52.850","Text":"and when we multiply 750 times 101,325 and divide by 760,"},{"Start":"04:52.850 ","End":"04:58.630","Text":"we get 99,991.8 pascals."},{"Start":"04:58.630 ","End":"05:04.845","Text":"Now we want to convert the pascals to atmospheres."},{"Start":"05:04.845 ","End":"05:09.350","Text":"We can write 99,991.8 pascals,"},{"Start":"05:09.350 ","End":"05:11.030","Text":"that\u0027s what we got before,"},{"Start":"05:11.030 ","End":"05:13.220","Text":"is equal to the same number."},{"Start":"05:13.220 ","End":"05:18.245","Text":"Now we need the conversion factor between atmospheres and pascals."},{"Start":"05:18.245 ","End":"05:25.925","Text":"Remember 1 atmosphere was defined as precisely equal to 101,325 pascals."},{"Start":"05:25.925 ","End":"05:30.210","Text":"So pascals cancel and we\u0027re left with atmospheres,"},{"Start":"05:30.210 ","End":"05:40.075","Text":"and if we do the division on our calculators we get out 0.986842 atmospheres,"},{"Start":"05:40.075 ","End":"05:43.205","Text":"which of course is less than 1 atmosphere."},{"Start":"05:43.205 ","End":"05:45.740","Text":"Now we have to convert it to bars,"},{"Start":"05:45.740 ","End":"05:48.530","Text":"so we take the number of atmospheres,"},{"Start":"05:48.530 ","End":"05:51.530","Text":"again it\u0027s equal to the same number and we need"},{"Start":"05:51.530 ","End":"05:55.055","Text":"our conversion unit between bars and atmospheres."},{"Start":"05:55.055 ","End":"06:01.175","Text":"Remember, we calculated that 1 atmosphere is 1.01325 bars."},{"Start":"06:01.175 ","End":"06:04.475","Text":"If we do, atmosphere cancels with atmosphere,"},{"Start":"06:04.475 ","End":"06:09.500","Text":"we multiply 0.986842 times"},{"Start":"06:09.500 ","End":"06:16.925","Text":"1.01325 and we get 0.999918 bars."},{"Start":"06:16.925 ","End":"06:22.345","Text":"In this video, we learned about gas pressure and how it\u0027s measured."},{"Start":"06:22.345 ","End":"06:25.625","Text":"We learned about lots of different units for pressure."},{"Start":"06:25.625 ","End":"06:30.920","Text":"In the next few videos we\u0027ll talk about the gas laws that relate to pressure,"},{"Start":"06:30.920 ","End":"06:35.345","Text":"to the volume, to the amount of gas, and the temperature."},{"Start":"06:35.345 ","End":"06:39.149","Text":"That\u0027s the subject of the next few videos."}],"ID":18158},{"Watched":false,"Name":"Exercise 1","Duration":"3m 53s","ChapterTopicVideoID":22961,"CourseChapterTopicPlaylistID":80094,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.970","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.970 ","End":"00:07.764","Text":"Convert each pressure to an equivalent pressure in standard atmospheres, a,"},{"Start":"00:07.764 ","End":"00:10.020","Text":"654 millimeters of Hg,"},{"Start":"00:10.020 ","End":"00:12.045","Text":"b, 723 Torr,"},{"Start":"00:12.045 ","End":"00:14.715","Text":"c, 46.2 centimeters of Hg,"},{"Start":"00:14.715 ","End":"00:17.115","Text":"and d, 435 kilo pascal."},{"Start":"00:17.115 ","End":"00:22.455","Text":"In a, we have 654 millimeters of Hg, which is Mercury."},{"Start":"00:22.455 ","End":"00:25.050","Text":"We have to convert these into atmospheres."},{"Start":"00:25.050 ","End":"00:33.195","Text":"Recall that 1 atmosphere equals 760 millimeters of Hg."},{"Start":"00:33.195 ","End":"00:38.590","Text":"We\u0027re just going to multiply our 654 millimeters of Hg by a conversion factor."},{"Start":"00:38.720 ","End":"00:41.250","Text":"In 1 atmosphere,"},{"Start":"00:41.250 ","End":"00:45.900","Text":"we have 760 millimeters Hg,"},{"Start":"00:45.900 ","End":"00:48.950","Text":"and we\u0027re going to put the 760 millimeters of Hg in our denominator,"},{"Start":"00:48.950 ","End":"00:51.170","Text":"so the millimeters of Hg will cancel out."},{"Start":"00:51.170 ","End":"00:53.180","Text":"Milliliters of Hg cancel out,"},{"Start":"00:53.180 ","End":"00:57.729","Text":"and this equals 0.861 atmosphere."},{"Start":"00:57.729 ","End":"01:01.905","Text":"That\u0027s our answer for a, 0.861 atmospheres."},{"Start":"01:01.905 ","End":"01:05.800","Text":"We\u0027re going to go on to b and b we have 723 Torr."},{"Start":"01:08.360 ","End":"01:11.910","Text":"We have to convert the Torr into atmosphere."},{"Start":"01:11.910 ","End":"01:14.970","Text":"Recall that in 1 atmosphere,"},{"Start":"01:14.970 ","End":"01:18.820","Text":"we have 760 Torr."},{"Start":"01:19.430 ","End":"01:23.195","Text":"Again, we\u0027re going to multiply by a conversion factor,"},{"Start":"01:23.195 ","End":"01:30.210","Text":"1 atmosphere equals 760 Torr and again,"},{"Start":"01:30.210 ","End":"01:34.010","Text":"our Torr will be in the denominator since we want our Torr to cancel out,"},{"Start":"01:34.010 ","End":"01:38.915","Text":"and this equals 0.951 atmosphere."},{"Start":"01:38.915 ","End":"01:41.370","Text":"That\u0027s our answer for b."},{"Start":"01:41.570 ","End":"01:48.480","Text":"In c we want to convert 46.2 centimeters of Hg into atmospheres."},{"Start":"01:50.720 ","End":"01:53.300","Text":"Again we\u0027re going to use the relationship,"},{"Start":"01:53.300 ","End":"02:00.535","Text":"1 atmosphere equals 760 millimeters of Hg, mercury."},{"Start":"02:00.535 ","End":"02:05.870","Text":"In order to get a relationship between atmospheres and centimeters of mercury,"},{"Start":"02:05.870 ","End":"02:08.435","Text":"we\u0027re going to multiply by a conversion factor."},{"Start":"02:08.435 ","End":"02:13.975","Text":"Now we know that 1 centimeter equals 10 millimeters."},{"Start":"02:13.975 ","End":"02:22.515","Text":"We\u0027re going to multiply this by 1 centimeter of Hg for every 10 millimeters of Hg."},{"Start":"02:22.515 ","End":"02:27.000","Text":"The millimeters of Hg will cancel out and we have 760 divided by 10,"},{"Start":"02:27.000 ","End":"02:31.895","Text":"so this equals 76 centimeters of Hg."},{"Start":"02:31.895 ","End":"02:35.653","Text":"Again, we\u0027ll just write the relationship between atmospheres and centimeters Hg,"},{"Start":"02:35.653 ","End":"02:42.720","Text":"so 1 atmosphere equals 76 centimeters of Hg."},{"Start":"02:42.720 ","End":"02:45.185","Text":"Again, we\u0027re going to multiply by a conversion factor."},{"Start":"02:45.185 ","End":"02:53.700","Text":"For every 1 atmosphere we have 76 centimeters of Hg."},{"Start":"02:53.700 ","End":"02:55.575","Text":"Centimeters of Hg cancel out,"},{"Start":"02:55.575 ","End":"03:01.930","Text":"and this equals 0.608 atmosphere."},{"Start":"03:03.680 ","End":"03:11.160","Text":"Now we\u0027re going to go on to d. We want to convert 435 kilopascal into atmospheres."},{"Start":"03:15.520 ","End":"03:24.100","Text":"1 atmosphere equals 101.325 kilopascal."},{"Start":"03:24.100 ","End":"03:27.110","Text":"Therefore, to convert 435 kilopascal"},{"Start":"03:27.110 ","End":"03:30.215","Text":"into atmospheres we\u0027re just going to multiply again by conversion factor."},{"Start":"03:30.215 ","End":"03:32.305","Text":"In every 1 atmosphere,"},{"Start":"03:32.305 ","End":"03:38.260","Text":"we have 101.325 kilopascal."},{"Start":"03:38.330 ","End":"03:47.080","Text":"Kilopascal cancel out and our answer for d is 4.29 atmosphere."},{"Start":"03:47.390 ","End":"03:53.110","Text":"That\u0027s our answer for d. Thank you very much for watching."}],"ID":23807}],"Thumbnail":null,"ID":80094},{"Name":"Gas Laws","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Boyle\u0027s Law","Duration":"5m 2s","ChapterTopicVideoID":17412,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.865","Text":"In this and the next few videos,"},{"Start":"00:02.865 ","End":"00:05.595","Text":"we will study the simple gas laws."},{"Start":"00:05.595 ","End":"00:07.695","Text":"We will start with Boyle\u0027s Law,"},{"Start":"00:07.695 ","End":"00:10.065","Text":"which is the first and oldest law."},{"Start":"00:10.065 ","End":"00:14.085","Text":"It was discovered in 1662 by Robert Boyle."},{"Start":"00:14.085 ","End":"00:18.974","Text":"Boyle\u0027s Law states that for a fixed amount of gas at fixed temperature,"},{"Start":"00:18.974 ","End":"00:24.135","Text":"the gas volume is inversely proportional to the gas pressure."},{"Start":"00:24.135 ","End":"00:27.480","Text":"In other words, as the pressure increases,"},{"Start":"00:27.480 ","End":"00:30.150","Text":"the volume will decrease."},{"Start":"00:30.150 ","End":"00:32.160","Text":"We can illustrate that."},{"Start":"00:32.160 ","End":"00:35.505","Text":"Supposing we have a container containing gas,"},{"Start":"00:35.505 ","End":"00:37.530","Text":"and here we have a piston."},{"Start":"00:37.530 ","End":"00:39.810","Text":"Suddenly it can go up and down."},{"Start":"00:39.810 ","End":"00:45.955","Text":"Supposing we have 1 weight pressing down,"},{"Start":"00:45.955 ","End":"00:51.080","Text":"then the pressure will be P_1 and the volume will be V_1."},{"Start":"00:51.080 ","End":"00:54.990","Text":"Supposing we now increase the number of weights,"},{"Start":"00:54.990 ","End":"00:58.685","Text":"so we now we have 2 weights pressing downwards,"},{"Start":"00:58.685 ","End":"01:00.875","Text":"so the pressure is P_2."},{"Start":"01:00.875 ","End":"01:04.670","Text":"Then the volume will be V_2 and we\u0027ll see that"},{"Start":"01:04.670 ","End":"01:09.650","Text":"the volume is smaller in the second situation than in the first."},{"Start":"01:09.650 ","End":"01:14.780","Text":"In other words, the more we press down on this piston,"},{"Start":"01:14.780 ","End":"01:17.450","Text":"the more we compress the gas,"},{"Start":"01:17.450 ","End":"01:19.295","Text":"the gas has a smaller volume."},{"Start":"01:19.295 ","End":"01:21.730","Text":"Let\u0027s write that mathematically."},{"Start":"01:21.730 ","End":"01:25.040","Text":"The pressure is proportional to 1 over the volume."},{"Start":"01:25.040 ","End":"01:28.730","Text":"That\u0027s pressures inversely proportional to the volume."},{"Start":"01:28.730 ","End":"01:32.930","Text":"We can write that the pressure is equal to a constant,"},{"Start":"01:32.930 ","End":"01:34.865","Text":"which we can determine later,"},{"Start":"01:34.865 ","End":"01:37.100","Text":"divided by the volume."},{"Start":"01:37.100 ","End":"01:40.910","Text":"If we now multiply both sides of this equation by V,"},{"Start":"01:40.910 ","End":"01:47.225","Text":"we\u0027ll get PV the problem of the pressure and the volume equal to k, a constant."},{"Start":"01:47.225 ","End":"01:52.015","Text":"In other words, the pressure times the volume is a constant."},{"Start":"01:52.015 ","End":"01:55.860","Text":"We can write that as P_1 times V_1."},{"Start":"01:55.860 ","End":"01:57.485","Text":"That\u0027s our first situation."},{"Start":"01:57.485 ","End":"02:02.000","Text":"The pressure times the volume is equal to P_2V_2,"},{"Start":"02:02.000 ","End":"02:05.510","Text":"the pressure and the volume in the second situation."},{"Start":"02:05.510 ","End":"02:09.425","Text":"This is a good statement of Boyle\u0027s law."},{"Start":"02:09.425 ","End":"02:13.715","Text":"P_1V_1=P_2 V_2."},{"Start":"02:13.715 ","End":"02:20.510","Text":"Now we can plot the pressure as a function of the volume, pressure versus volume."},{"Start":"02:20.510 ","End":"02:25.355","Text":"Here\u0027s our graph, so here we have the pressure of written as an atmospheres,"},{"Start":"02:25.355 ","End":"02:28.160","Text":"and here\u0027s the volume in liters."},{"Start":"02:28.160 ","End":"02:31.565","Text":"Now, as the pressure decreases,"},{"Start":"02:31.565 ","End":"02:34.090","Text":"we\u0027re going to lower pressures here."},{"Start":"02:34.090 ","End":"02:37.550","Text":"We see that the volume increases,"},{"Start":"02:37.550 ","End":"02:39.245","Text":"or we could go the other way,"},{"Start":"02:39.245 ","End":"02:41.765","Text":"going to higher pressures,"},{"Start":"02:41.765 ","End":"02:44.630","Text":"the volume is decreasing."},{"Start":"02:44.630 ","End":"02:50.270","Text":"A better way of plotting this is to plot the pressure versus 1 over"},{"Start":"02:50.270 ","End":"02:53.180","Text":"the volume because we saw that Boyle\u0027s law tells"},{"Start":"02:53.180 ","End":"02:57.170","Text":"us that the pressure is proportional to 1 over the volume."},{"Start":"02:57.170 ","End":"03:04.880","Text":"Now we get a straight line is pressure on the y-axis and 1 over volume on"},{"Start":"03:04.880 ","End":"03:08.810","Text":"the x-axis and we see we have a nice straight line with"},{"Start":"03:08.810 ","End":"03:13.760","Text":"slope k. Let\u0027s show an example of this."},{"Start":"03:13.760 ","End":"03:18.965","Text":"If the volume of gas is 1 liter and the pressure is 2 atmospheres,"},{"Start":"03:18.965 ","End":"03:23.090","Text":"what will be the volume if we double the pressure?"},{"Start":"03:23.090 ","End":"03:26.960","Text":"The first thing to do is to decide what\u0027s P_1,"},{"Start":"03:26.960 ","End":"03:29.950","Text":"V_1, P_2, and V_2."},{"Start":"03:29.950 ","End":"03:32.930","Text":"At the beginning, the pressure is 2 atmospheres,"},{"Start":"03:32.930 ","End":"03:34.730","Text":"so P_1 is 2 atmospheres."},{"Start":"03:34.730 ","End":"03:38.255","Text":"The volume V_1 is 1 liter."},{"Start":"03:38.255 ","End":"03:40.850","Text":"Now we\u0027re going to double the pressure."},{"Start":"03:40.850 ","End":"03:45.080","Text":"P_2 is 4 atmospheres, twice 2 atmospheres."},{"Start":"03:45.080 ","End":"03:47.855","Text":"We need to work out what V_2 is,"},{"Start":"03:47.855 ","End":"03:51.530","Text":"which is the volume in the second circumstance."},{"Start":"03:51.530 ","End":"03:57.590","Text":"Now, V_2=P_1, V_1 divided by P_2."},{"Start":"03:57.590 ","End":"03:58.910","Text":"How do I know that?"},{"Start":"03:58.910 ","End":"04:06.290","Text":"Because I know that P_2V_2=P_1V_1."},{"Start":"04:06.290 ","End":"04:08.990","Text":"If I divide by P_2,"},{"Start":"04:08.990 ","End":"04:15.050","Text":"I get V_2=P_1 times V_1 divided by P_2."},{"Start":"04:15.050 ","End":"04:17.090","Text":"That\u0027s the equation we have here."},{"Start":"04:17.090 ","End":"04:19.340","Text":"Let\u0027s substitute the numbers."},{"Start":"04:19.340 ","End":"04:22.640","Text":"P_1 is 2 atmospheres,"},{"Start":"04:22.640 ","End":"04:27.155","Text":"V_1 is 1 liter and P_2 is 4 atmospheres."},{"Start":"04:27.155 ","End":"04:31.570","Text":"Atmospheres cancels with atmospheres and we\u0027re left with 2 divided by 4,"},{"Start":"04:31.570 ","End":"04:35.590","Text":"which is 0.5 and the units are liters,"},{"Start":"04:35.590 ","End":"04:39.900","Text":"so V_2 is 0.5 liters."},{"Start":"04:39.900 ","End":"04:43.655","Text":"We can conclude that if we double the pressure,"},{"Start":"04:43.655 ","End":"04:45.650","Text":"we half the volume."},{"Start":"04:45.650 ","End":"04:51.555","Text":"We\u0027ve got from pressure of 2-4 and a volume of 1 liter to 1/2 liter,"},{"Start":"04:51.555 ","End":"04:53.270","Text":"so if we double the pressure,"},{"Start":"04:53.270 ","End":"04:55.115","Text":"we have the volume."},{"Start":"04:55.115 ","End":"04:58.130","Text":"In this video, we discussed Boyle\u0027s law,"},{"Start":"04:58.130 ","End":"05:02.130","Text":"which relates the pressure and volume of a gas."}],"ID":18162},{"Watched":false,"Name":"Exercise 1","Duration":"3m 36s","ChapterTopicVideoID":22968,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.640","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.640 ","End":"00:08.430","Text":"Sample of oxygen gas has a volume of 25.4 liters at 735 Torr."},{"Start":"00:08.430 ","End":"00:11.370","Text":"What is the new volume if the pressure is a,"},{"Start":"00:11.370 ","End":"00:13.800","Text":"lowered to 367 Torr,"},{"Start":"00:13.800 ","End":"00:16.935","Text":"b increased to 4.02 atmospheres."},{"Start":"00:16.935 ","End":"00:20.070","Text":"The temperature and amount of gas are held constant."},{"Start":"00:20.070 ","End":"00:25.920","Text":"We have oxygen gas and we know that the volume is 25.4 liters."},{"Start":"00:25.920 ","End":"00:30.280","Text":"Our initial volume is 25.4 liters."},{"Start":"00:30.830 ","End":"00:35.745","Text":"We know that the initial pressure is 735 Torr."},{"Start":"00:35.745 ","End":"00:39.855","Text":"Equals 735 Torr."},{"Start":"00:39.855 ","End":"00:43.849","Text":"Now the pressure has changed and we have to find out the volume."},{"Start":"00:43.849 ","End":"00:48.020","Text":"For this purpose, we know that the initial pressure times"},{"Start":"00:48.020 ","End":"00:53.750","Text":"the initial volume equals the final pressure times the final volume. Let\u0027s begin with a."},{"Start":"00:53.750 ","End":"00:57.560","Text":"In a, we know that our pressure is lowered to 367 Torr."},{"Start":"00:57.560 ","End":"01:04.270","Text":"Our final pressure equals 367 Torr."},{"Start":"01:04.520 ","End":"01:07.215","Text":"We\u0027re looking for our new volume."},{"Start":"01:07.215 ","End":"01:09.370","Text":"We\u0027re actually looking for the final volume."},{"Start":"01:09.370 ","End":"01:12.980","Text":"We can see here that if we divide both sides by the final pressure,"},{"Start":"01:12.980 ","End":"01:15.425","Text":"we\u0027re going to get the final volume, Vf,"},{"Start":"01:15.425 ","End":"01:24.995","Text":"our final volume equals Pi times Vi divided by Pf."},{"Start":"01:24.995 ","End":"01:27.905","Text":"Again, our initial pressure is"},{"Start":"01:27.905 ","End":"01:36.230","Text":"735 Torr times our initial volume,"},{"Start":"01:36.230 ","End":"01:39.120","Text":"which is 25.4 liters."},{"Start":"01:42.100 ","End":"01:45.320","Text":"This is divided by our final pressure,"},{"Start":"01:45.320 ","End":"01:48.510","Text":"which equals 367 Torr."},{"Start":"01:51.410 ","End":"01:57.945","Text":"The Torr canceled out and we get 50.87 liters."},{"Start":"01:57.945 ","End":"02:00.035","Text":"That\u0027s our final volume for a."},{"Start":"02:00.035 ","End":"02:01.775","Text":"Now we\u0027re going to go on to b."},{"Start":"02:01.775 ","End":"02:06.215","Text":"In b, our final pressure equals 4.02 atmospheres."},{"Start":"02:06.215 ","End":"02:12.470","Text":"In b, the final pressure equals 4.02 atmospheres."},{"Start":"02:12.470 ","End":"02:15.080","Text":"Again, we\u0027re going to use the same equation."},{"Start":"02:15.080 ","End":"02:23.195","Text":"So Vf final volume equals Pi times Vi divided by Pf."},{"Start":"02:23.195 ","End":"02:25.100","Text":"Pi again is the same,"},{"Start":"02:25.100 ","End":"02:30.530","Text":"735 Torr times our initial volume,"},{"Start":"02:30.530 ","End":"02:34.050","Text":"which is the same, 25.4 liters."},{"Start":"02:35.420 ","End":"02:40.140","Text":"The only change is our final pressure."},{"Start":"02:40.140 ","End":"02:43.579","Text":"Our final pressure is 4.02 atmospheres."},{"Start":"02:45.470 ","End":"02:48.555","Text":"This is why we have Torr atmospheres."},{"Start":"02:48.555 ","End":"02:50.220","Text":"We\u0027re going to convert one of them."},{"Start":"02:50.220 ","End":"02:52.130","Text":"Let\u0027s convert atmospheres to Torr."},{"Start":"02:52.130 ","End":"02:58.150","Text":"Now remember that one atmosphere equals 760 Torr."},{"Start":"02:58.220 ","End":"03:01.945","Text":"We\u0027re going to multiply this by a conversion factor."},{"Start":"03:01.945 ","End":"03:07.985","Text":"We have 760 Torr in one atmosphere."},{"Start":"03:07.985 ","End":"03:10.915","Text":"The atmosphere is in the denominator."},{"Start":"03:10.915 ","End":"03:13.030","Text":"That way our atmospheres cancel out."},{"Start":"03:13.030 ","End":"03:15.235","Text":"Our Torr also cancel out."},{"Start":"03:15.235 ","End":"03:19.285","Text":"This equals 6.11 liters."},{"Start":"03:19.285 ","End":"03:21.560","Text":"That\u0027s our volume for b."},{"Start":"03:21.560 ","End":"03:24.190","Text":"That\u0027s 6.11 liters."},{"Start":"03:24.190 ","End":"03:26.930","Text":"Now I just want to remind you that the equation we use"},{"Start":"03:26.930 ","End":"03:33.695","Text":"PiVi=PiVf is correct because the temperature and the amount of gas are held constant."},{"Start":"03:33.695 ","End":"03:35.210","Text":"That\u0027s our final answer."},{"Start":"03:35.210 ","End":"03:37.860","Text":"Thank you very much for watching."}],"ID":23822},{"Watched":false,"Name":"Exercise 2","Duration":"3m 29s","ChapterTopicVideoID":22970,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.315","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.315 ","End":"00:07.230","Text":"A 1900 liter tank is first evacuated and then"},{"Start":"00:07.230 ","End":"00:12.315","Text":"connected to a 32.9 liters cylinder of compressed argon gas."},{"Start":"00:12.315 ","End":"00:17.580","Text":"If the temperature is held constant and the final pressure is 763 mmHg,"},{"Start":"00:17.580 ","End":"00:22.110","Text":"what must have been the original guess pressure in the cylinder in atmospheres?"},{"Start":"00:22.110 ","End":"00:24.750","Text":"First of all, we can see that we\u0027re given the volume of a tank."},{"Start":"00:24.750 ","End":"00:27.580","Text":"The tank is evacuated and then connected to a cylinder,"},{"Start":"00:27.580 ","End":"00:29.820","Text":"and we\u0027re also given the volume of the cylinder."},{"Start":"00:29.820 ","End":"00:32.985","Text":"We have an initial volume and a final volume."},{"Start":"00:32.985 ","End":"00:37.635","Text":"Our initial volume is a cylinder volume,"},{"Start":"00:37.635 ","End":"00:40.890","Text":"which is 32.9 liters and"},{"Start":"00:40.890 ","End":"00:44.390","Text":"our final volume is the sum of the volume of"},{"Start":"00:44.390 ","End":"00:48.425","Text":"the tank and the volume of the cylinder since they are connected."},{"Start":"00:48.425 ","End":"00:51.185","Text":"Our final volume is 1900 liters,"},{"Start":"00:51.185 ","End":"00:53.015","Text":"which is the volume of the tank,"},{"Start":"00:53.015 ","End":"00:57.980","Text":"plus 32.9 liters, which is the volume of the cylinder."},{"Start":"00:57.980 ","End":"01:02.175","Text":"This equals 1,932.9 liters."},{"Start":"01:02.175 ","End":"01:03.650","Text":"That\u0027s our final volume."},{"Start":"01:03.650 ","End":"01:06.215","Text":"Now we\u0027re also given a final pressure here."},{"Start":"01:06.215 ","End":"01:15.000","Text":"The final pressure equals 763 millimeters of mercury,"},{"Start":"01:15.000 ","End":"01:17.300","Text":"and we\u0027re asked to find the original gas pressure,"},{"Start":"01:17.300 ","End":"01:18.905","Text":"meaning the initial pressure."},{"Start":"01:18.905 ","End":"01:20.600","Text":"That\u0027s the question. Now we know that"},{"Start":"01:20.600 ","End":"01:22.990","Text":"the temperature is held constant because it\u0027s given,"},{"Start":"01:22.990 ","End":"01:26.960","Text":"and we also know that the amount of the gas is constant because in the beginning,"},{"Start":"01:26.960 ","End":"01:30.030","Text":"it was in the cylinder and then it was connected to the tank,"},{"Start":"01:30.030 ","End":"01:32.300","Text":"but the amount of gas is kept constant."},{"Start":"01:32.300 ","End":"01:35.495","Text":"Now since the temperature and the amount of gas are kept constant,"},{"Start":"01:35.495 ","End":"01:38.090","Text":"we know that the initial pressure times"},{"Start":"01:38.090 ","End":"01:43.370","Text":"the initial volume equals the final pressure times the final volume."},{"Start":"01:43.370 ","End":"01:46.910","Text":"As we said, we\u0027re looking for the initial pressure."},{"Start":"01:46.910 ","End":"01:49.159","Text":"The initial pressure equals,"},{"Start":"01:49.159 ","End":"01:52.025","Text":"we divide both sides by the initial volume,"},{"Start":"01:52.025 ","End":"01:57.850","Text":"so it\u0027s the final pressure times the final volume divided by the initial volume."},{"Start":"01:57.850 ","End":"02:02.270","Text":"This equals again, the final pressure is 763 mmHg,"},{"Start":"02:02.270 ","End":"02:08.330","Text":"so it\u0027s 763 mmHg times the final volume,"},{"Start":"02:08.330 ","End":"02:10.970","Text":"which we said is 1,932.9"},{"Start":"02:10.970 ","End":"02:18.510","Text":"liters divided by the initial volume,"},{"Start":"02:18.510 ","End":"02:20.930","Text":"and the initial volume equals 32.9 liters,"},{"Start":"02:20.930 ","End":"02:23.580","Text":"which is the volume of the cylinder."},{"Start":"02:25.000 ","End":"02:35.990","Text":"Liters cancel out and this equals 44,826.83 millimeters of mercury."},{"Start":"02:35.990 ","End":"02:39.830","Text":"Now we\u0027re asked to find the gas pressure in atmospheres so now we"},{"Start":"02:39.830 ","End":"02:43.670","Text":"have to convert mmHg to atmospheres."},{"Start":"02:43.670 ","End":"02:51.135","Text":"Now recall that 1 atmosphere equals 760 mmHg."},{"Start":"02:51.135 ","End":"02:54.080","Text":"We\u0027re going to take the gas pressure that we found,"},{"Start":"02:54.080 ","End":"03:01.335","Text":"which is 44,826.83 mmHg,"},{"Start":"03:01.335 ","End":"03:03.800","Text":"and we\u0027re going to multiply this by a conversion factor"},{"Start":"03:03.800 ","End":"03:06.260","Text":"in order to convert it into atmospheres."},{"Start":"03:06.260 ","End":"03:08.749","Text":"In every 1 atmosphere,"},{"Start":"03:08.749 ","End":"03:14.210","Text":"we have 760 mmHg."},{"Start":"03:14.210 ","End":"03:16.520","Text":"The millimeters of Hg cancel out,"},{"Start":"03:16.520 ","End":"03:20.690","Text":"and this equals 59 atmospheres."},{"Start":"03:20.690 ","End":"03:25.400","Text":"The initial pressure in atmospheres is 59 atmospheres."},{"Start":"03:25.400 ","End":"03:27.425","Text":"That is our final answer."},{"Start":"03:27.425 ","End":"03:30.150","Text":"Thank you very much for watching."}],"ID":23824},{"Watched":false,"Name":"Charles\u0027s Law and STP","Duration":"7m 58s","ChapterTopicVideoID":18226,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.395","Text":"In the previous video,"},{"Start":"00:01.395 ","End":"00:04.950","Text":"we discussed the first gas law, Boyle\u0027s law."},{"Start":"00:04.950 ","End":"00:09.975","Text":"In this video, we\u0027ll discuss the second of the gas laws, Charles\u0027s law."},{"Start":"00:09.975 ","End":"00:15.165","Text":"In 1787, well over 100 years after Boyle\u0027s law,"},{"Start":"00:15.165 ","End":"00:18.344","Text":"the French physicist, Jacques Charles,"},{"Start":"00:18.344 ","End":"00:23.205","Text":"discovered the relationship between the volume and temperature of a gas."},{"Start":"00:23.205 ","End":"00:27.185","Text":"He found out that as the temperature of the gas increases,"},{"Start":"00:27.185 ","End":"00:29.450","Text":"so does its volume."},{"Start":"00:29.450 ","End":"00:31.865","Text":"Now, we can plot this."},{"Start":"00:31.865 ","End":"00:37.115","Text":"I\u0027m going to plot the volume versus the temperature in degrees Celsius."},{"Start":"00:37.115 ","End":"00:43.160","Text":"I plotted the volume in liters against the temperature in degrees Celsius,"},{"Start":"00:43.160 ","End":"00:48.590","Text":"from 0 degrees Celsius up to 400 degrees Celsius."},{"Start":"00:48.590 ","End":"00:52.415","Text":"We can see that as the temperature increases,"},{"Start":"00:52.415 ","End":"00:54.050","Text":"so does the volume."},{"Start":"00:54.050 ","End":"00:55.715","Text":"The volume is also going up."},{"Start":"00:55.715 ","End":"00:58.834","Text":"This is at line with a positive slope."},{"Start":"00:58.834 ","End":"01:03.945","Text":"If I were to extrapolate this line downwards,"},{"Start":"01:03.945 ","End":"01:10.800","Text":"I would find that it reached 0 volume at -273 degrees Celsius,"},{"Start":"01:10.800 ","End":"01:14.500","Text":"-273,"},{"Start":"01:17.480 ","End":"01:22.490","Text":"exactly, 0.15 degrees Celsius."},{"Start":"01:22.490 ","End":"01:24.460","Text":"Here we\u0027ve written it."},{"Start":"01:24.460 ","End":"01:26.620","Text":"When we plot the volume versus temperature,"},{"Start":"01:26.620 ","End":"01:27.955","Text":"we get a straight line."},{"Start":"01:27.955 ","End":"01:29.410","Text":"If we extrapolate the line,"},{"Start":"01:29.410 ","End":"01:36.250","Text":"we find the volume becomes 0 at -273.15 degrees Celsius."},{"Start":"01:36.250 ","End":"01:40.079","Text":"This is often called absolute 0,"},{"Start":"01:40.079 ","End":"01:42.730","Text":"because you cannot get a lower temperature."},{"Start":"01:42.730 ","End":"01:46.625","Text":"In fact, you can\u0027t get as quite as low as that."},{"Start":"01:46.625 ","End":"01:48.680","Text":"But in modern times,"},{"Start":"01:48.680 ","End":"01:51.995","Text":"with the help of sophisticated laser techniques,"},{"Start":"01:51.995 ","End":"01:55.235","Text":"people have got very close to 0."},{"Start":"01:55.235 ","End":"01:59.510","Text":"Now, it\u0027s usual, especially in gas laws,"},{"Start":"01:59.510 ","End":"02:02.390","Text":"to use a different temperature scale,"},{"Start":"02:02.390 ","End":"02:05.330","Text":"it\u0027s called the Kelvin scale of temperature,"},{"Start":"02:05.330 ","End":"02:11.045","Text":"after Lord Kelvin, who was a Scottish physicist."},{"Start":"02:11.045 ","End":"02:13.820","Text":"Instead of using degrees Celsius,"},{"Start":"02:13.820 ","End":"02:18.255","Text":"which is an arbitrary scale based on the freezing point,"},{"Start":"02:18.255 ","End":"02:22.710","Text":"0 degrees Celsius and boiling point 100 degrees Celsius water,"},{"Start":"02:22.710 ","End":"02:25.010","Text":"we use the Kelvin scale."},{"Start":"02:25.010 ","End":"02:29.195","Text":"How do we convert degrees Celsius to Kelvin?"},{"Start":"02:29.195 ","End":"02:31.385","Text":"You don\u0027t write degrees Kelvin,"},{"Start":"02:31.385 ","End":"02:39.305","Text":"just K. You add 273.15 to the number in Celsius."},{"Start":"02:39.305 ","End":"02:42.740","Text":"The temperature in Kelvin is equal to"},{"Start":"02:42.740 ","End":"02:49.085","Text":"the temperature in degrees Celsius with plus 273.15."},{"Start":"02:49.085 ","End":"02:50.930","Text":"Let\u0027s take an example."},{"Start":"02:50.930 ","End":"02:56.045","Text":"Supposing we have minus 273.15 degrees Celsius,"},{"Start":"02:56.045 ","End":"02:58.945","Text":"which we said was absolute 0."},{"Start":"02:58.945 ","End":"03:05.070","Text":"It\u0027s minus 273.15 and we have to add 273.15."},{"Start":"03:05.070 ","End":"03:08.834","Text":"When we add these two we\u0027ll get 0 Kelvin."},{"Start":"03:08.834 ","End":"03:10.020","Text":"On the other hand,"},{"Start":"03:10.020 ","End":"03:12.220","Text":"if we take 0 degrees Celsius,"},{"Start":"03:12.220 ","End":"03:15.510","Text":"which is the freezing point of water, so that\u0027s 0,"},{"Start":"03:15.510 ","End":"03:21.245","Text":"and we add to 273.15 to get the temperature in Kelvin,"},{"Start":"03:21.245 ","End":"03:25.850","Text":"the final temperature will be 273.15 Kelvin."},{"Start":"03:25.850 ","End":"03:30.950","Text":"Now, we can plot the volume versus the temperature in Kelvin."},{"Start":"03:30.950 ","End":"03:34.295","Text":"Here\u0027s the same graph as we had before,"},{"Start":"03:34.295 ","End":"03:37.715","Text":"except we\u0027re plotting in temperature, in Kelvin."},{"Start":"03:37.715 ","End":"03:39.320","Text":"You see it\u0027s going from 0,"},{"Start":"03:39.320 ","End":"03:41.090","Text":"it starts at 0."},{"Start":"03:41.090 ","End":"03:44.365","Text":"Once again, we\u0027ve stopped in the same place."},{"Start":"03:44.365 ","End":"03:48.229","Text":"We\u0027re going to extrapolate this back to 0."},{"Start":"03:48.229 ","End":"03:53.900","Text":"We see that now we\u0027ve got 0 volume at 0 Kelvin."},{"Start":"03:53.900 ","End":"03:59.555","Text":"Now we can write Charles\u0027s law in a more mathematical way."},{"Start":"03:59.555 ","End":"04:04.400","Text":"The volume is proportional to the temperature, capital T,"},{"Start":"04:04.400 ","End":"04:11.015","Text":"in Kelvin, V is equal to k sub constant times T,"},{"Start":"04:11.015 ","End":"04:15.920","Text":"capital T, which is temperature in Kelvin."},{"Start":"04:15.920 ","End":"04:22.235","Text":"If we divide both sides of this equation by capital T, we\u0027ll get V/T=k."},{"Start":"04:22.235 ","End":"04:27.725","Text":"In other words, volume divided by temperature is a constant."},{"Start":"04:27.725 ","End":"04:32.630","Text":"In other words, we can write that if we compare two circumstances,"},{"Start":"04:32.630 ","End":"04:39.575","Text":"V_1 divided by T_1 with another one, V_2/ T_2."},{"Start":"04:39.575 ","End":"04:42.410","Text":"At any two points,"},{"Start":"04:42.410 ","End":"04:47.250","Text":"V_1/T_1 will be exactly equal to V_2/T_2."},{"Start":"04:47.250 ","End":"04:49.350","Text":"Let\u0027s take an example."},{"Start":"04:49.350 ","End":"04:53.420","Text":"If the volume of a gas is 1 liter at 20 degrees Celsius,"},{"Start":"04:53.420 ","End":"04:57.590","Text":"what would be its volume at 60 degrees Celsius?"},{"Start":"04:57.590 ","End":"05:00.230","Text":"We have to decide what\u0027s V_1,"},{"Start":"05:00.230 ","End":"05:03.140","Text":"T_1, V_2, and T_2."},{"Start":"05:03.140 ","End":"05:07.235","Text":"At the beginning, V_1 is 1 liter,"},{"Start":"05:07.235 ","End":"05:12.260","Text":"and temperature is 20 plus 273."},{"Start":"05:12.260 ","End":"05:13.750","Text":"I\u0027ve just written it for short,"},{"Start":"05:13.750 ","End":"05:18.485","Text":"273 Kelvin, that\u0027s 293 Kelvin."},{"Start":"05:18.485 ","End":"05:22.895","Text":"Now, after we\u0027ve heated the gas,"},{"Start":"05:22.895 ","End":"05:25.815","Text":"the temperature is 60 degrees Celsius,"},{"Start":"05:25.815 ","End":"05:27.710","Text":"but we need the temperature in Kelvin."},{"Start":"05:27.710 ","End":"05:34.160","Text":"T_2 is 60 plus 273 Kelvin, that\u0027s 333 Kelvin."},{"Start":"05:34.160 ","End":"05:40.950","Text":"The volume, V_2 is V_1T_2/T_1."},{"Start":"05:40.950 ","End":"05:42.180","Text":"How did I know that?"},{"Start":"05:42.180 ","End":"05:50.930","Text":"I know that V_2/T_2 is V_1/T_1."},{"Start":"05:50.930 ","End":"05:52.190","Text":"If I want V_2,"},{"Start":"05:52.190 ","End":"05:55.655","Text":"I have to multiply both sides of the equation by T_2,"},{"Start":"05:55.655 ","End":"06:00.840","Text":"so then I get V_2 is T_2V_1/T_1."},{"Start":"06:02.230 ","End":"06:07.395","Text":"Here it is. Now we can substitute."},{"Start":"06:07.395 ","End":"06:09.135","Text":"V_1 is 1 liter,"},{"Start":"06:09.135 ","End":"06:12.030","Text":"T_2 is 333 Kelvin,"},{"Start":"06:12.030 ","End":"06:14.710","Text":"T_1 is 293 Kelvin."},{"Start":"06:14.710 ","End":"06:16.760","Text":"Kelvins cancel."},{"Start":"06:16.760 ","End":"06:22.643","Text":"Now I can divide 333 by 293 and I get 1.14,"},{"Start":"06:22.643 ","End":"06:24.990","Text":"and units are still liter."},{"Start":"06:24.990 ","End":"06:29.910","Text":"Now the volume is 1.14 liters."},{"Start":"06:29.910 ","End":"06:33.440","Text":"In other words, after heating the gas,"},{"Start":"06:33.440 ","End":"06:38.285","Text":"its volume increased from 1 liter to 1.14 liters."},{"Start":"06:38.285 ","End":"06:43.670","Text":"Now, often you\u0027ll see that there are problems where you\u0027re asked to"},{"Start":"06:43.670 ","End":"06:48.620","Text":"calculate something at the standard temperature and pressure, STP."},{"Start":"06:48.620 ","End":"06:51.280","Text":"What is meant by this?"},{"Start":"06:51.280 ","End":"06:55.810","Text":"There\u0027s an old definition which says that that"},{"Start":"06:55.810 ","End":"07:00.180","Text":"is 0 degrees Celsius and pressure of 1 atmosphere."},{"Start":"07:00.180 ","End":"07:02.540","Text":"Temperature of 0 degrees Celsius,"},{"Start":"07:02.540 ","End":"07:04.645","Text":"and a pressure of 1 atmosphere."},{"Start":"07:04.645 ","End":"07:12.880","Text":"There\u0027s a new definition which says 0 degrees Celsius and 1 bar pressure."},{"Start":"07:12.880 ","End":"07:15.040","Text":"Instead of using 1 atmosphere,"},{"Start":"07:15.040 ","End":"07:16.720","Text":"they use 1 bar."},{"Start":"07:16.720 ","End":"07:21.850","Text":"Now, we know that there\u0027s very little difference between 1 atmosphere and 1 bar."},{"Start":"07:21.850 ","End":"07:32.175","Text":"We calculated in a previous video that 1 atmosphere is equal to 1.01325 bars."},{"Start":"07:32.175 ","End":"07:38.885","Text":"In practice, using atmosphere or bars doesn\u0027t make a great deal of difference."},{"Start":"07:38.885 ","End":"07:41.900","Text":"You have to be very careful to see whether in"},{"Start":"07:41.900 ","End":"07:45.139","Text":"a particular book or a particular set of problems,"},{"Start":"07:45.139 ","End":"07:48.110","Text":"they mean 0 degrees Celsius and 1 atmosphere,"},{"Start":"07:48.110 ","End":"07:50.900","Text":"or 0 degrees Celsius and 1 bar,"},{"Start":"07:50.900 ","End":"07:53.810","Text":"but there\u0027s really very little difference between them."},{"Start":"07:53.810 ","End":"07:57.990","Text":"In this video, we discussed Charles\u0027s law."}],"ID":18968},{"Watched":false,"Name":"Exercise 2","Duration":"2m 42s","ChapterTopicVideoID":22969,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.655","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.655 ","End":"00:06.750","Text":"We want to increase the volume of a fixed amount of gas from 54.5"},{"Start":"00:06.750 ","End":"00:11.140","Text":"milliliters to 147 milliliters while holding the pressure constant."},{"Start":"00:11.140 ","End":"00:13.470","Text":"To what value must we change the temperature if"},{"Start":"00:13.470 ","End":"00:16.110","Text":"the initial temperature is 27-degrees Celsius?"},{"Start":"00:16.110 ","End":"00:19.455","Text":"We can see that we\u0027re given an initial and final volume."},{"Start":"00:19.455 ","End":"00:24.640","Text":"The initial volume equals 54.5 milliliters,"},{"Start":"00:25.430 ","End":"00:31.719","Text":"and the final volume equals 147 milliliters."},{"Start":"00:32.570 ","End":"00:40.090","Text":"We can also see that we\u0027re given an initial temperature which equals 27-degrees Celsius."},{"Start":"00:40.090 ","End":"00:42.140","Text":"Soon we\u0027re going to convert this into kelvin."},{"Start":"00:42.140 ","End":"00:44.345","Text":"But first of all, we can also see that we have"},{"Start":"00:44.345 ","End":"00:47.270","Text":"a constant pressure and a fixed amount of gas."},{"Start":"00:47.270 ","End":"00:51.920","Text":"Meaning we can use the equation initial volume divided by"},{"Start":"00:51.920 ","End":"00:57.935","Text":"initial temperature equals final volume divided by the final temperature."},{"Start":"00:57.935 ","End":"00:59.338","Text":"Now let\u0027s go back to the temperature."},{"Start":"00:59.338 ","End":"01:00.380","Text":"Our initial temperature is"},{"Start":"01:00.380 ","End":"01:04.550","Text":"27-degrees Celsius and we want to change this to convert this into kelvin."},{"Start":"01:04.550 ","End":"01:12.775","Text":"Therefore we\u0027re going to add 273 and this is going to give us 300 kelvin."},{"Start":"01:12.775 ","End":"01:15.830","Text":"Initial temperature is 300 kelvin."},{"Start":"01:15.830 ","End":"01:17.855","Text":"Then we\u0027re going to go back to our equation."},{"Start":"01:17.855 ","End":"01:20.171","Text":"We can see that we have the initial volume,"},{"Start":"01:20.171 ","End":"01:23.525","Text":"the final volume, and the initial temperature."},{"Start":"01:23.525 ","End":"01:25.879","Text":"We\u0027re looking for the final temperature."},{"Start":"01:25.879 ","End":"01:27.515","Text":"The final temperature is here."},{"Start":"01:27.515 ","End":"01:31.580","Text":"The final temperature equals the final volume"},{"Start":"01:31.580 ","End":"01:36.815","Text":"times the initial temperature divided by the initial volume."},{"Start":"01:36.815 ","End":"01:40.475","Text":"We\u0027re just doing that with a simple math manipulation here."},{"Start":"01:40.475 ","End":"01:43.220","Text":"As we said this is the final volume equals"},{"Start":"01:43.220 ","End":"01:50.040","Text":"147 milliliters times the initial temperature which we said is"},{"Start":"01:50.040 ","End":"01:59.535","Text":"300 kelvin divided by the initial volume which equals 54.5 milliliters."},{"Start":"01:59.535 ","End":"02:09.875","Text":"The milliliters cancel out and this equals 809 kelvin."},{"Start":"02:09.875 ","End":"02:11.570","Text":"Now we can leave it this way,"},{"Start":"02:11.570 ","End":"02:17.135","Text":"the temperature is 809 kelvin or we could also convert this temperature into Celsius."},{"Start":"02:17.135 ","End":"02:20.525","Text":"If we want our temperature in degrees Celsius we\u0027re just going to take"},{"Start":"02:20.525 ","End":"02:28.430","Text":"our 809 K minus 273."},{"Start":"02:28.430 ","End":"02:34.360","Text":"This is going to give us 536-degrees Celsius."},{"Start":"02:34.360 ","End":"02:39.500","Text":"Temperature is either 809 or 536-degrees Celsius."},{"Start":"02:39.500 ","End":"02:40.730","Text":"That is our final answer."},{"Start":"02:40.730 ","End":"02:43.260","Text":"Thank you very much for watching."}],"ID":23823},{"Watched":false,"Name":"Avogadro\u0027s Law","Duration":"3m 5s","ChapterTopicVideoID":18228,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.710","Text":"In previous videos, we talked about Boyle\u0027s and Charles\u0027s law."},{"Start":"00:04.710 ","End":"00:08.910","Text":"In this video, we\u0027re going to discuss Avogadro\u0027s Law."},{"Start":"00:08.910 ","End":"00:13.170","Text":"Avogadro\u0027s Law tells us that at a fixed temperature and pressure,"},{"Start":"00:13.170 ","End":"00:18.510","Text":"the volume of a gas is proportional to the number of moles of gas."},{"Start":"00:18.510 ","End":"00:21.623","Text":"So if we have a container,"},{"Start":"00:21.623 ","End":"00:25.610","Text":"and in it there\u0027s 1 mole of gas,"},{"Start":"00:25.610 ","End":"00:28.085","Text":"it will have a particular volume."},{"Start":"00:28.085 ","End":"00:35.740","Text":"If we now take the same container and put 2 moles of gas in,"},{"Start":"00:37.030 ","End":"00:42.125","Text":"the new volume will be twice as much as the old formula."},{"Start":"00:42.125 ","End":"00:43.880","Text":"Now, in previous videos,"},{"Start":"00:43.880 ","End":"00:49.595","Text":"we learned that 1 mole of any gas contains Avogadro\u0027s number of molecules."},{"Start":"00:49.595 ","End":"00:54.785","Text":"In fact, 1 mole of anything contains Avogadro\u0027s number of molecules."},{"Start":"00:54.785 ","End":"00:58.310","Text":"Just to remind you that Avogadro\u0027s number,"},{"Start":"00:58.310 ","End":"01:00.950","Text":"which we indicated as N_A,"},{"Start":"01:00.950 ","End":"01:07.190","Text":"is approximately 6.022 times 10^23."},{"Start":"01:07.190 ","End":"01:13.075","Text":"So that the volume is not only proportional to the number of moles,"},{"Start":"01:13.075 ","End":"01:17.889","Text":"but since each mole contains Avogadro\u0027s number of molecules,"},{"Start":"01:17.889 ","End":"01:21.430","Text":"it\u0027s also proportional to the number of molecules."},{"Start":"01:21.430 ","End":"01:25.060","Text":"Let\u0027s write Avogadro\u0027s law in a more mathematical way."},{"Start":"01:25.060 ","End":"01:27.260","Text":"The volume, that\u0027s V,"},{"Start":"01:27.260 ","End":"01:28.570","Text":"is proportional to n,"},{"Start":"01:28.570 ","End":"01:30.310","Text":"the number of moles."},{"Start":"01:30.310 ","End":"01:33.025","Text":"The V is equal to c,"},{"Start":"01:33.025 ","End":"01:34.975","Text":"a constant, times n,"},{"Start":"01:34.975 ","End":"01:36.355","Text":"the number of moles."},{"Start":"01:36.355 ","End":"01:38.470","Text":"So what\u0027s this constant?"},{"Start":"01:38.470 ","End":"01:44.890","Text":"C is equal to V divided by n. That\u0027s called the molar volume,"},{"Start":"01:44.890 ","End":"01:47.125","Text":"the volume per mole,"},{"Start":"01:47.125 ","End":"01:49.990","Text":"now we\u0027re going to talk about molar volume."},{"Start":"01:49.990 ","End":"01:53.935","Text":"The molar volume, that\u0027s the volume of 1 mole of gas,"},{"Start":"01:53.935 ","End":"01:56.800","Text":"varies to some extent from gas to gas."},{"Start":"01:56.800 ","End":"01:59.470","Text":"We can find this out experimentally."},{"Start":"01:59.470 ","End":"02:02.500","Text":"However, for an ideal gas,"},{"Start":"02:02.500 ","End":"02:03.640","Text":"what\u0027s an ideal gas?"},{"Start":"02:03.640 ","End":"02:06.895","Text":"One that obeys all the simple gas laws."},{"Start":"02:06.895 ","End":"02:15.175","Text":"C is equal to 22.414 liters at 0-degrees Celsius and 1 atmosphere."},{"Start":"02:15.175 ","End":"02:18.370","Text":"That\u0027s the old definition of STP."},{"Start":"02:18.370 ","End":"02:20.710","Text":"Let\u0027s say STP,"},{"Start":"02:20.710 ","End":"02:23.915","Text":"standard temperature and pressure."},{"Start":"02:23.915 ","End":"02:30.985","Text":"Equal to 22.717 liters at 0-degrees Celsius and 1 bar,"},{"Start":"02:30.985 ","End":"02:35.020","Text":"that\u0027s the new STP."},{"Start":"02:35.120 ","End":"02:43.265","Text":"This is often written as 1 mole of gas occupies 22.414 liters"},{"Start":"02:43.265 ","End":"02:51.665","Text":"at the old STP and 22.717 liters at the new STP."},{"Start":"02:51.665 ","End":"02:56.165","Text":"We\u0027re going to use this information in the next video,"},{"Start":"02:56.165 ","End":"03:00.140","Text":"which we\u0027re going to talk about the ideal gas law."},{"Start":"03:00.140 ","End":"03:05.220","Text":"So in this video we discussed Avogadro\u0027s Law."}],"ID":18970},{"Watched":false,"Name":"Exercise 4","Duration":"2m 41s","ChapterTopicVideoID":22971,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.490","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.490 ","End":"00:07.440","Text":"What is the mass of argon gas in a 63 milliliter volume at STP?"},{"Start":"00:07.440 ","End":"00:12.320","Text":"STP or standard conditions of temperature and pressure and we know that at STP,"},{"Start":"00:12.320 ","End":"00:21.310","Text":"1 mole of gas equals 22.4 liters of gas."},{"Start":"00:21.310 ","End":"00:23.250","Text":"We know that the volume,"},{"Start":"00:23.250 ","End":"00:26.710","Text":"which is given in the question as 63 milliliters,"},{"Start":"00:29.480 ","End":"00:33.525","Text":"and we\u0027re asked to find the mass of the argon gas."},{"Start":"00:33.525 ","End":"00:35.880","Text":"First of all, we\u0027re going to calculate the number of moles"},{"Start":"00:35.880 ","End":"00:38.300","Text":"of the gas using this relationship,"},{"Start":"00:38.300 ","End":"00:41.065","Text":"and then we\u0027ll convert the number of moles into mass."},{"Start":"00:41.065 ","End":"00:42.500","Text":"Let\u0027s begin, as we said,"},{"Start":"00:42.500 ","End":"00:46.130","Text":"we have a volume of 63 milliliters and we know that 1 mole of gas equals"},{"Start":"00:46.130 ","End":"00:50.930","Text":"22.4 liters of gas since we have STP conditions."},{"Start":"00:50.930 ","End":"00:56.134","Text":"The number of moles of our argon gas equals our volume"},{"Start":"00:56.134 ","End":"01:02.565","Text":"times 1 mole per every 22.4 liters of gas,"},{"Start":"01:02.565 ","End":"01:11.000","Text":"so volume is 63 milliliters times 1 mole for 22.4 liters."},{"Start":"01:11.000 ","End":"01:12.845","Text":"Since we have milliliters and liters,"},{"Start":"01:12.845 ","End":"01:15.335","Text":"what we\u0027re going to do is convert our milliliters into liters."},{"Start":"01:15.335 ","End":"01:19.850","Text":"So this equals 63 milliliters times,"},{"Start":"01:19.850 ","End":"01:21.200","Text":"we\u0027re going to use a conversion factor,"},{"Start":"01:21.200 ","End":"01:23.165","Text":"remember that in every 1 liter,"},{"Start":"01:23.165 ","End":"01:25.805","Text":"we have 1000 milliliters."},{"Start":"01:25.805 ","End":"01:27.875","Text":"Milliliters cancel out,"},{"Start":"01:27.875 ","End":"01:32.550","Text":"times what we have leftover here, 1 mole,"},{"Start":"01:32.750 ","End":"01:37.410","Text":"22.4 liters, so liters will also cancel out,"},{"Start":"01:37.410 ","End":"01:44.205","Text":"and this equals 2.81 times 10^negative 3 moles."},{"Start":"01:44.205 ","End":"01:47.990","Text":"Now we have the number of moles and we want to convert these into the mass."},{"Start":"01:47.990 ","End":"01:50.720","Text":"For this first part, we\u0027re going to use the equation n,"},{"Start":"01:50.720 ","End":"01:52.520","Text":"the number of moles equals m,"},{"Start":"01:52.520 ","End":"01:55.270","Text":"which is the mass divided by the molar mass,"},{"Start":"01:55.270 ","End":"01:57.080","Text":"and since we want to find the mass,"},{"Start":"01:57.080 ","End":"01:59.230","Text":"we\u0027re going to multiply both sides by the molar mass."},{"Start":"01:59.230 ","End":"02:05.050","Text":"The mass equals the number of moles times the molar mass."},{"Start":"02:05.050 ","End":"02:08.150","Text":"Now the number of moles is what we calculated,"},{"Start":"02:08.150 ","End":"02:16.470","Text":"and it equals 2.81 times 10^negative 3 moles times the molar mass,"},{"Start":"02:16.470 ","End":"02:18.470","Text":"and since we\u0027re talking about argon gas,"},{"Start":"02:18.470 ","End":"02:24.685","Text":"the molar mass equals 39.95 grams per mole."},{"Start":"02:24.685 ","End":"02:28.520","Text":"The molar mass of argon is taken from the periodic table of elements,"},{"Start":"02:28.520 ","End":"02:33.695","Text":"so the moles cancel out and this equals 0.112 grams,"},{"Start":"02:33.695 ","End":"02:38.870","Text":"the mass of the argon gas at STP equals 0.112 grams."},{"Start":"02:38.870 ","End":"02:39.995","Text":"That is our final answer."},{"Start":"02:39.995 ","End":"02:42.540","Text":"Thank you very much for watching."}],"ID":23825},{"Watched":false,"Name":"Exercise 5","Duration":"3m 2s","ChapterTopicVideoID":22972,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.030 ","End":"00:05.130","Text":"If 33.2 milliliter sample,"},{"Start":"00:05.130 ","End":"00:07.510","Text":"of phosphine gas is obtained at STP,"},{"Start":"00:07.510 ","End":"00:10.635","Text":"how many molecules of phosphine are present?"},{"Start":"00:10.635 ","End":"00:11.970","Text":"We know that STP,"},{"Start":"00:11.970 ","End":"00:16.155","Text":"are standard conditions of temperature and pressure and at STP,"},{"Start":"00:16.155 ","End":"00:18.370","Text":"1 mole of gas,"},{"Start":"00:19.040 ","End":"00:24.540","Text":"equals 22.4 liters of gas."},{"Start":"00:24.540 ","End":"00:29.740","Text":"We also know that the volume of the phosphine is 33.2 milliliters."},{"Start":"00:34.520 ","End":"00:39.350","Text":"We have to calculate the number of molecules of phosphine which are present."},{"Start":"00:39.350 ","End":"00:42.980","Text":"In order to do this, we will first calculate the number of moles of phosphine,"},{"Start":"00:42.980 ","End":"00:45.005","Text":"and then calculate the number of molecules."},{"Start":"00:45.005 ","End":"00:47.500","Text":"The number of moles of phosphine,"},{"Start":"00:47.500 ","End":"00:50.745","Text":"equal the volume of phosphine,"},{"Start":"00:50.745 ","End":"00:55.480","Text":"times again, in every 1 mole of gas,"},{"Start":"00:55.490 ","End":"01:00.070","Text":"we have 22.4 liters of gas."},{"Start":"01:00.490 ","End":"01:05.640","Text":"The volume of phosphine, we said is 33.2 milliliters."},{"Start":"01:06.520 ","End":"01:10.025","Text":"Since we have milliliters and liters in our equation,"},{"Start":"01:10.025 ","End":"01:12.665","Text":"we\u0027re going to multiply by a conversion factor."},{"Start":"01:12.665 ","End":"01:17.209","Text":"Recall that, in every 1 liter we have 1000 milliliters."},{"Start":"01:17.209 ","End":"01:20.555","Text":"We\u0027re going to multiply by 1 mole,"},{"Start":"01:20.555 ","End":"01:24.365","Text":"divided by 22.4 liters,"},{"Start":"01:24.365 ","End":"01:26.165","Text":"so the liter\u0027s cancel out,"},{"Start":"01:26.165 ","End":"01:28.700","Text":"and milliliters also cancel out."},{"Start":"01:28.700 ","End":"01:31.835","Text":"This equals 1.48,"},{"Start":"01:31.835 ","End":"01:36.520","Text":"times 10^negative 3 mole, of phosphine."},{"Start":"01:36.520 ","End":"01:38.660","Text":"Now we know the number of moles of phosphine."},{"Start":"01:38.660 ","End":"01:41.615","Text":"We can calculate the number of molecules or phosphine."},{"Start":"01:41.615 ","End":"01:43.910","Text":"In order to calculate the number of molecules of a phosphine,"},{"Start":"01:43.910 ","End":"01:45.095","Text":"we\u0027ll use the equation n,"},{"Start":"01:45.095 ","End":"01:46.385","Text":"number of moles equals,"},{"Start":"01:46.385 ","End":"01:49.150","Text":"N divided by NA."},{"Start":"01:49.150 ","End":"01:52.210","Text":"NA is Avogadro\u0027s number,"},{"Start":"01:56.300 ","End":"01:59.270","Text":"and N recall, could be atoms,"},{"Start":"01:59.270 ","End":"02:00.635","Text":"molecules and so on."},{"Start":"02:00.635 ","End":"02:03.510","Text":"In this case it\u0027s molecules."},{"Start":"02:04.180 ","End":"02:07.300","Text":"I now want to calculate the number of molecules,"},{"Start":"02:07.300 ","End":"02:09.400","Text":"so N equals,"},{"Start":"02:09.400 ","End":"02:15.680","Text":"n the number of moles times, NA Avogadro\u0027s number."},{"Start":"02:15.680 ","End":"02:17.750","Text":"The number of moles is what we calculated here,"},{"Start":"02:17.750 ","End":"02:22.500","Text":"1.48, times 10^negative 3 mole,"},{"Start":"02:22.570 ","End":"02:28.145","Text":"times Avogadro\u0027s number, which equals 6.022,"},{"Start":"02:28.145 ","End":"02:33.960","Text":"times 10^23, molecules per mole."},{"Start":"02:37.220 ","End":"02:39.585","Text":"The moles cancel out,"},{"Start":"02:39.585 ","End":"02:41.970","Text":"and we get 8.91,"},{"Start":"02:41.970 ","End":"02:51.490","Text":"times 10^20 molecules, of phosphine."},{"Start":"02:52.510 ","End":"02:56.120","Text":"The number of molecules of phosphine are 8.91,"},{"Start":"02:56.120 ","End":"02:58.789","Text":"times 10^20, molecules."},{"Start":"02:58.789 ","End":"03:00.350","Text":"That is our final answer."},{"Start":"03:00.350 ","End":"03:02.850","Text":"Thank you very much for watching."}],"ID":23826},{"Watched":false,"Name":"Ideal Gas Equation","Duration":"10m 31s","ChapterTopicVideoID":18227,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.220","Text":"In the last few videos,"},{"Start":"00:02.220 ","End":"00:06.945","Text":"we learned about three simple gas laws: Boyle\u0027s law,"},{"Start":"00:06.945 ","End":"00:13.095","Text":"where the volume of the gas is proportional to 1 over its pressure; Charles\u0027s law,"},{"Start":"00:13.095 ","End":"00:18.435","Text":"where the volume of the gas is proportional to its temperature in Kelvin;"},{"Start":"00:18.435 ","End":"00:23.355","Text":"and Avogadro\u0027s Law, where the volume is proportional to the number of moles."},{"Start":"00:23.355 ","End":"00:28.415","Text":"We can combine these three laws into one single equation."},{"Start":"00:28.415 ","End":"00:32.600","Text":"We can write that the volume is proportional to the number of moles,"},{"Start":"00:32.600 ","End":"00:37.190","Text":"n times the temperature divided by the pressure."},{"Start":"00:37.190 ","End":"00:41.029","Text":"If we multiply both sides of this equation by the pressure,"},{"Start":"00:41.029 ","End":"00:43.760","Text":"we get the pressure times the volume is"},{"Start":"00:43.760 ","End":"00:47.620","Text":"proportional to the number of moles times the temperature."},{"Start":"00:47.620 ","End":"00:54.050","Text":"If we add the constant we can write this as PV=nRT,"},{"Start":"00:54.050 ","End":"00:58.985","Text":"where the constant is R that\u0027s called the gas constant."},{"Start":"00:58.985 ","End":"01:05.740","Text":"This is called the ideal gas equation, PV=nRT."},{"Start":"01:05.740 ","End":"01:12.035","Text":"All three simple gas laws can be derived from this equation, for example,"},{"Start":"01:12.035 ","End":"01:14.840","Text":"here we have pressure times volume,"},{"Start":"01:14.840 ","End":"01:16.460","Text":"and we look at the right-hand side,"},{"Start":"01:16.460 ","End":"01:19.040","Text":"if we keep the number of moles constant,"},{"Start":"01:19.040 ","End":"01:22.030","Text":"then R is in any case a constant."},{"Start":"01:22.030 ","End":"01:24.245","Text":"We keep the temperature constant."},{"Start":"01:24.245 ","End":"01:29.135","Text":"Then we can see that PV is equal to a constant."},{"Start":"01:29.135 ","End":"01:35.660","Text":"Boyle\u0027s law is included in this general equation."},{"Start":"01:35.660 ","End":"01:38.870","Text":"We can similarly find out all the laws."},{"Start":"01:38.870 ","End":"01:47.900","Text":"If you only want to remember one equation in connection with gases, just remember PV=nRT."},{"Start":"01:47.900 ","End":"01:52.384","Text":"All three simple gas laws can be derived from this equation."},{"Start":"01:52.384 ","End":"01:57.050","Text":"Now if we want to find out what the value of R is,"},{"Start":"01:57.050 ","End":"02:02.015","Text":"we can first note that R=PV divided by nT,"},{"Start":"02:02.015 ","End":"02:04.880","Text":"this is the gas constant."},{"Start":"02:04.880 ","End":"02:08.450","Text":"Now in the next few minutes,"},{"Start":"02:08.450 ","End":"02:12.185","Text":"we\u0027re going to talk about the values of the gas constant."},{"Start":"02:12.185 ","End":"02:16.455","Text":"It all depends on which units you use for the pressure,"},{"Start":"02:16.455 ","End":"02:18.875","Text":"and which units you use for the volume."},{"Start":"02:18.875 ","End":"02:27.095","Text":"Then you will get different values of R and different units of R. Let\u0027s examine this."},{"Start":"02:27.095 ","End":"02:30.365","Text":"We\u0027re going to talk about the gas constant."},{"Start":"02:30.365 ","End":"02:32.255","Text":"Now in the previous video,"},{"Start":"02:32.255 ","End":"02:34.760","Text":"we learned about the molar volume,"},{"Start":"02:34.760 ","End":"02:39.470","Text":"that\u0027s the volume of 1 mole at STP and we can use this to"},{"Start":"02:39.470 ","End":"02:45.440","Text":"calculate the gas constant R at 0 degrees Celsius and 1 atmosphere,"},{"Start":"02:45.440 ","End":"02:49.910","Text":"which you remember is the old definition of STP,"},{"Start":"02:49.910 ","End":"02:52.730","Text":"but one can still like to use."},{"Start":"02:52.730 ","End":"02:54.455","Text":"We can write the R,"},{"Start":"02:54.455 ","End":"03:00.235","Text":"which is PV divided by nT is 1 atmosphere,"},{"Start":"03:00.235 ","End":"03:02.750","Text":"the volume at 1 atmosphere,"},{"Start":"03:02.750 ","End":"03:08.630","Text":"22.414 liters, n is 1 mole and the temperature,"},{"Start":"03:08.630 ","End":"03:10.819","Text":"of course is 0 degrees Celsius,"},{"Start":"03:10.819 ","End":"03:15.125","Text":"which is 273.15 Kelvin."},{"Start":"03:15.125 ","End":"03:17.150","Text":"If we divide that,"},{"Start":"03:17.150 ","End":"03:23.910","Text":"divide 22.414 by 273.15 we get 0.08206,"},{"Start":"03:26.270 ","End":"03:32.745","Text":"and the units are atmosphere times liter divided by mole,"},{"Start":"03:32.745 ","End":"03:34.030","Text":"divided by K,"},{"Start":"03:34.030 ","End":"03:41.215","Text":"which I\u0027ve written as atmosphere times liter mole^-1 times K^-1."},{"Start":"03:41.215 ","End":"03:45.075","Text":"This is the first way of writing R,"},{"Start":"03:45.075 ","End":"03:47.345","Text":"and we have to remember the units."},{"Start":"03:47.345 ","End":"03:51.619","Text":"Now if we use the new definition of STP,"},{"Start":"03:51.619 ","End":"03:53.945","Text":"0 degrees Celsius and 1 bar,"},{"Start":"03:53.945 ","End":"03:55.840","Text":"that\u0027s the new STP."},{"Start":"03:55.840 ","End":"04:01.280","Text":"We can work out that R now we have for the pressure 1 bar,"},{"Start":"04:01.280 ","End":"04:08.610","Text":"we have the value of the molar volume at 0 degrees Celsius and 1 bar,"},{"Start":"04:08.610 ","End":"04:13.560","Text":"that\u0027s 22.711 liters and of course,"},{"Start":"04:13.560 ","End":"04:21.290","Text":"n is 1 mole and the temperature is 273.15 Kelvin that\u0027s 0 degrees Celsius."},{"Start":"04:21.290 ","End":"04:31.760","Text":"Now when we divide 22.711 by 273.15, we get 0.08315."},{"Start":"04:31.760 ","End":"04:34.190","Text":"The units are now bar,"},{"Start":"04:34.190 ","End":"04:37.550","Text":"because we use bar for pressure, liter,"},{"Start":"04:37.550 ","End":"04:44.800","Text":"because we use liter for the volume times mole^-1 and K^-1."},{"Start":"04:44.800 ","End":"04:48.880","Text":"This is the definition for the new STP."},{"Start":"04:48.880 ","End":"04:54.370","Text":"Now, we see that the value is not much different than the value with old STP."},{"Start":"04:54.370 ","End":"05:00.330","Text":"It\u0027s 0.08315 instead of 0.08206."},{"Start":"05:01.970 ","End":"05:07.370","Text":"Now which one you use will depend on the question you\u0027re asked"},{"Start":"05:07.370 ","End":"05:12.500","Text":"and what units of pressure and volume are used in the question,"},{"Start":"05:12.500 ","End":"05:16.775","Text":"now we can convert the atmosphere to Pascal."},{"Start":"05:16.775 ","End":"05:19.130","Text":"Remember, in a previous video,"},{"Start":"05:19.130 ","End":"05:24.325","Text":"we showed how many Pascals there are in 1 atmosphere."},{"Start":"05:24.325 ","End":"05:32.160","Text":"Now writing 1 atmosphere in Pascals is, 101,325 Pascals."},{"Start":"05:32.160 ","End":"05:34.665","Text":"The same volume,"},{"Start":"05:34.665 ","End":"05:44.810","Text":"as the molar volume at 1 atmosphere and again divide by 1 mole and 273.15 Kelvin."},{"Start":"05:44.810 ","End":"05:46.460","Text":"If we work this out,"},{"Start":"05:46.460 ","End":"05:56.245","Text":"we get 8,314.5 and the units are Pascal times liter"},{"Start":"05:56.245 ","End":"06:03.650","Text":"per mole per K. So this number is also very important because we can show that"},{"Start":"06:03.650 ","End":"06:13.380","Text":"8,314.5 Pascals times liter is equal to 8.3145 Joules,"},{"Start":"06:13.380 ","End":"06:17.475","Text":"where the Joule that\u0027s J is the SI unit"},{"Start":"06:17.475 ","End":"06:22.265","Text":"of energy and we\u0027re going to use this quite a lot in the future."},{"Start":"06:22.265 ","End":"06:31.550","Text":"Other units for R is R = 8.3145 Joules per"},{"Start":"06:31.550 ","End":"06:41.510","Text":"mole per K and we\u0027re going to use it whenever we\u0027re talking about energy."},{"Start":"06:41.510 ","End":"06:46.609","Text":"When we\u0027re not talking about pressures and volumes, but energy,"},{"Start":"06:46.609 ","End":"06:49.765","Text":"we use this definition of R,"},{"Start":"06:49.765 ","End":"06:54.035","Text":"this value of R. Now what\u0027s an ideal gas?"},{"Start":"06:54.035 ","End":"06:59.030","Text":"An ideal gas is a gas that obeys the ideal gas equation."},{"Start":"06:59.030 ","End":"07:04.790","Text":"It obeys the equation, P times V=nRT."},{"Start":"07:04.790 ","End":"07:10.730","Text":"We\u0027ll see later that there are quite a lot of exceptions to this simple law,"},{"Start":"07:10.730 ","End":"07:12.860","Text":"but it\u0027s much, much in use."},{"Start":"07:12.860 ","End":"07:18.635","Text":"It\u0027s called the ideal gas equation and the ideal gas obeys it."},{"Start":"07:18.635 ","End":"07:23.000","Text":"Let\u0027s take an example of how we can use this equation."},{"Start":"07:23.000 ","End":"07:24.965","Text":"Here\u0027s a simple question."},{"Start":"07:24.965 ","End":"07:30.170","Text":"What volumes occupied by 32 grams of O_2 at"},{"Start":"07:30.170 ","End":"07:36.140","Text":"a pressure of 380 millimeters of mercury and the temperature of 60 degrees Celsius?"},{"Start":"07:36.140 ","End":"07:39.800","Text":"We can use any value of R that we want but in"},{"Start":"07:39.800 ","End":"07:43.880","Text":"this case I\u0027m going to use the value of R that has"},{"Start":"07:43.880 ","End":"07:51.875","Text":"the units atmosphere times liter times mole^-1 times K^-1."},{"Start":"07:51.875 ","End":"07:55.100","Text":"In order to use this value of R,"},{"Start":"07:55.100 ","End":"07:59.270","Text":"the pressure must be in atmospheres,"},{"Start":"07:59.270 ","End":"08:02.195","Text":"the volume must be in liters,"},{"Start":"08:02.195 ","End":"08:04.820","Text":"the amount must be in moles,"},{"Start":"08:04.820 ","End":"08:07.370","Text":"and the temperature must be in K,"},{"Start":"08:07.370 ","End":"08:08.975","Text":"because if we look at R,"},{"Start":"08:08.975 ","End":"08:14.360","Text":"it has the units atmosphere liter mole^-1 K^-1."},{"Start":"08:14.360 ","End":"08:19.340","Text":"The pressure is given in millimeters of mercury,"},{"Start":"08:19.340 ","End":"08:25.615","Text":"380 millimeters of mercury and we need to remember this conversion factor,"},{"Start":"08:25.615 ","End":"08:32.295","Text":"that 1 atmosphere is equivalent to 760 millimeters of mercury,"},{"Start":"08:32.295 ","End":"08:34.955","Text":"380 millimeters of mercury,"},{"Start":"08:34.955 ","End":"08:38.840","Text":"we can divide by 760 and millimeters of"},{"Start":"08:38.840 ","End":"08:44.270","Text":"mercury cancels and we\u0027re left with 0.5 atmospheres."},{"Start":"08:44.270 ","End":"08:45.935","Text":"That\u0027s the pressure."},{"Start":"08:45.935 ","End":"08:47.630","Text":"Now we need the temperature."},{"Start":"08:47.630 ","End":"08:51.440","Text":"We have to convert the temperature to the temperature in Kelvin."},{"Start":"08:51.440 ","End":"08:56.135","Text":"We take 60 degrees Celsius and add 273"},{"Start":"08:56.135 ","End":"09:01.475","Text":"to get the temperature in Kelvin, that\u0027s 333 kelvin."},{"Start":"09:01.475 ","End":"09:06.840","Text":"Now the number of moles is just 1 mole because it was 32 grams and"},{"Start":"09:06.840 ","End":"09:12.615","Text":"32 grams is the molar mass of O_2."},{"Start":"09:12.615 ","End":"09:15.165","Text":"Now we can complete the calculation,"},{"Start":"09:15.165 ","End":"09:18.225","Text":"V=nRT divided by P,"},{"Start":"09:18.225 ","End":"09:19.665","Text":"n is 1 mole,"},{"Start":"09:19.665 ","End":"09:24.140","Text":"R is 0.082 atmospheres times liters per"},{"Start":"09:24.140 ","End":"09:29.720","Text":"mole per Kelvin and we multiply the temperature 333 Kelvin,"},{"Start":"09:29.720 ","End":"09:34.295","Text":"divide by the pressure, 0.5 atmospheres."},{"Start":"09:34.295 ","End":"09:39.620","Text":"When we multiply out the numbers, we get 54.6."},{"Start":"09:39.620 ","End":"09:43.540","Text":"Now we have to look at the units."},{"Start":"09:43.540 ","End":"09:50.455","Text":"We have mole times mole^-1, so that\u0027s 1."},{"Start":"09:50.455 ","End":"09:53.995","Text":"Then we have atmosphere divide by atmosphere."},{"Start":"09:53.995 ","End":"09:57.765","Text":"Then we have K^-1 times K,"},{"Start":"09:57.765 ","End":"09:59.715","Text":"which is again is just 1."},{"Start":"09:59.715 ","End":"10:03.135","Text":"So we\u0027re left only with the liter,"},{"Start":"10:03.135 ","End":"10:10.745","Text":"so we have 54.6 liters that\u0027s something you have to do all the time."},{"Start":"10:10.745 ","End":"10:14.765","Text":"You always have to check the units if you\u0027ve chosen"},{"Start":"10:14.765 ","End":"10:19.730","Text":"the right value of R and the right units for the pressure,"},{"Start":"10:19.730 ","End":"10:22.579","Text":"temperature and number of moles."},{"Start":"10:22.579 ","End":"10:27.004","Text":"Then you will get out with the right units."},{"Start":"10:27.004 ","End":"10:31.530","Text":"In this video, we learned about the ideal gas equation."}],"ID":18969},{"Watched":false,"Name":"Exercise 6","Duration":"6m 33s","ChapterTopicVideoID":22973,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.585","Text":"Hi, We\u0027re going to solve the following exercise."},{"Start":"00:03.585 ","End":"00:10.530","Text":"What is the volume in milliliters occupied by 92.3 grams of carbon dioxide gas at"},{"Start":"00:10.530 ","End":"00:18.885","Text":"42-degrees Celsius and 754 millimeters of mercury mmHg?"},{"Start":"00:18.885 ","End":"00:20.730","Text":"To calculate the volume,"},{"Start":"00:20.730 ","End":"00:23.290","Text":"we\u0027re going to use the equation PV=nRT,"},{"Start":"00:24.890 ","End":"00:30.970","Text":"this is called the ideal gas equation and P is the pressure,"},{"Start":"00:32.270 ","End":"00:35.530","Text":"V is the volume,"},{"Start":"00:36.140 ","End":"00:39.555","Text":"N is the number of moles,"},{"Start":"00:39.555 ","End":"00:48.550","Text":"R is the gas constant and T is the temperature."},{"Start":"00:51.760 ","End":"00:54.350","Text":"Now we\u0027re looking for the volume."},{"Start":"00:54.350 ","End":"00:56.690","Text":"Therefore, we\u0027re going to divide both sides by the pressure."},{"Start":"00:56.690 ","End":"00:58.670","Text":"Then we\u0027re going to get V,"},{"Start":"00:58.670 ","End":"01:08.390","Text":"the volume=nRT divided by P. Final question,"},{"Start":"01:08.390 ","End":"01:11.420","Text":"we can see that we\u0027re given the mass of the carbon dioxide."},{"Start":"01:11.420 ","End":"01:18.980","Text":"The mass of the carbon dioxide equals 92.3 grams."},{"Start":"01:18.980 ","End":"01:22.940","Text":"We also know that the temperature is 42-degrees Celsius."},{"Start":"01:25.880 ","End":"01:31.780","Text":"We are given the pressure, which is 754 mmHg."},{"Start":"01:36.630 ","End":"01:40.135","Text":"From the mass, we\u0027re going to calculate the number of moles."},{"Start":"01:40.135 ","End":"01:41.650","Text":"Just to remind you, n,"},{"Start":"01:41.650 ","End":"01:45.835","Text":"the number of moles equals m divided by Mw,"},{"Start":"01:45.835 ","End":"01:48.805","Text":"which equals the mass divided by the molar mass."},{"Start":"01:48.805 ","End":"01:50.590","Text":"In order to calculate the number of moles,"},{"Start":"01:50.590 ","End":"01:53.155","Text":"we need to calculate the molar mass of carbon dioxide."},{"Start":"01:53.155 ","End":"01:58.540","Text":"The molar mass of carbon dioxide equals the molar mass of carbon."},{"Start":"01:58.540 ","End":"02:00.820","Text":"We can see that we have 1 carbon plus"},{"Start":"02:00.820 ","End":"02:06.140","Text":"2 times the molar mass of oxygen since we have 2 oxygens,"},{"Start":"02:06.570 ","End":"02:10.010","Text":"this equals 12.01 grams per mole,"},{"Start":"02:10.410 ","End":"02:14.900","Text":"plus 2 times 16 grams per mole,"},{"Start":"02:16.230 ","End":"02:22.820","Text":"this equals 44.01 grams per mole,"},{"Start":"02:22.820 ","End":"02:25.910","Text":"that\u0027s the molar mass of carbon dioxide."},{"Start":"02:25.910 ","End":"02:29.375","Text":"Now I can go on to calculating the number of moles."},{"Start":"02:29.375 ","End":"02:35.270","Text":"The number of moles of carbon dioxide equals the mass divided by the molar mass."},{"Start":"02:35.270 ","End":"02:42.230","Text":"The mass of carbon dioxide is 92.3 grams divided by the molar mass,"},{"Start":"02:42.230 ","End":"02:45.720","Text":"which is 44.01 grams per mole,"},{"Start":"02:47.790 ","End":"02:51.130","Text":"this equals 2.1 mole."},{"Start":"02:51.130 ","End":"02:53.650","Text":"Just a small reminder, look at the units."},{"Start":"02:53.650 ","End":"02:57.620","Text":"We can see that we\u0027re dividing gram by gram per mole."},{"Start":"02:58.130 ","End":"03:02.315","Text":"We\u0027re dividing by a fraction and dividing by a fraction equals,"},{"Start":"03:02.315 ","End":"03:05.410","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction,"},{"Start":"03:05.410 ","End":"03:09.250","Text":"so it equals grams times mole per grams."},{"Start":"03:09.250 ","End":"03:13.780","Text":"The grams cancel out and we\u0027re left with mole in our units."},{"Start":"03:13.780 ","End":"03:16.135","Text":"Now we know the number of moles."},{"Start":"03:16.135 ","End":"03:18.160","Text":"If we look at our temperature,"},{"Start":"03:18.160 ","End":"03:21.055","Text":"we can see that we\u0027re given 42-degrees Celsius."},{"Start":"03:21.055 ","End":"03:23.470","Text":"Now we need our temperature in Kelvin."},{"Start":"03:23.470 ","End":"03:33.625","Text":"Temperature in kelvin equals the temperature in degrees Celsius plus 273."},{"Start":"03:33.625 ","End":"03:44.600","Text":"In our case, we have 42-degrees Celsius plus 273 and our temperature equals 315 Kelvin."},{"Start":"03:44.600 ","End":"03:50.415","Text":"Now we\u0027re going to get back to our equation to calculate our volume and again,"},{"Start":"03:50.415 ","End":"03:57.255","Text":"the volume V=nRT divided by P, the pressure."},{"Start":"03:57.255 ","End":"03:59.760","Text":"This equals number of moles,"},{"Start":"03:59.760 ","End":"04:06.160","Text":"which is 2.1 mole times R,"},{"Start":"04:06.160 ","End":"04:07.580","Text":"R is the gas constant,"},{"Start":"04:07.580 ","End":"04:09.679","Text":"as I said, and it has a number of values."},{"Start":"04:09.679 ","End":"04:16.580","Text":"In our case, we\u0027re going to use R=0.082 and"},{"Start":"04:16.580 ","End":"04:24.530","Text":"its units are liters times atmospheres divided by mole times Kelvin."},{"Start":"04:24.530 ","End":"04:29.585","Text":"Okay? This is times 0.082"},{"Start":"04:29.585 ","End":"04:38.150","Text":"liters times atmospheres divided by moles times Kelvin times the temperature,"},{"Start":"04:38.150 ","End":"04:40.970","Text":"which we said is 315 K,"},{"Start":"04:43.730 ","End":"04:46.320","Text":"divided by the pressure,"},{"Start":"04:46.320 ","End":"04:50.440","Text":"which is 754 mmHg."},{"Start":"04:53.950 ","End":"04:56.375","Text":"Now if we look at our units,"},{"Start":"04:56.375 ","End":"04:59.470","Text":"we see the moles cancels out with the moles."},{"Start":"04:59.470 ","End":"05:06.230","Text":"Kelvin cancels out with Kelvin and we\u0027re left with liters times atmosphere."},{"Start":"05:06.230 ","End":"05:08.570","Text":"That\u0027s in our numerator and in our denominator,"},{"Start":"05:08.570 ","End":"05:10.790","Text":"we have milliliters of Hg."},{"Start":"05:10.790 ","End":"05:16.820","Text":"We\u0027re going to multiply the 754 millimeters of Hg, by conversion factor."},{"Start":"05:16.820 ","End":"05:25.455","Text":"Remember that 1 atmosphere equals 760 millimeters of mercury."},{"Start":"05:25.455 ","End":"05:27.200","Text":"Therefore, we\u0027re going to multiply this by"},{"Start":"05:27.200 ","End":"05:35.630","Text":"1 atmosphere divided by 760 millimeters of mercury."},{"Start":"05:35.630 ","End":"05:37.760","Text":"That way the millimeters of mercury are going to cancel"},{"Start":"05:37.760 ","End":"05:41.285","Text":"out and the atmospheres will also cancel out."},{"Start":"05:41.285 ","End":"05:42.740","Text":"If we look at our units,"},{"Start":"05:42.740 ","End":"05:44.510","Text":"all that\u0027s left is liters,"},{"Start":"05:44.510 ","End":"05:52.855","Text":"so this equals 54.67 liters,"},{"Start":"05:52.855 ","End":"05:55.395","Text":"that makes sense because it\u0027s the volume,"},{"Start":"05:55.395 ","End":"05:57.250","Text":"so the volume came out as liters."},{"Start":"05:57.250 ","End":"06:00.770","Text":"Now in the question, we were asked to find the volume in milliliters."},{"Start":"06:00.770 ","End":"06:02.840","Text":"We\u0027re going to multiply by a conversion factor."},{"Start":"06:02.840 ","End":"06:05.240","Text":"In every liter we have 1000 milliliters,"},{"Start":"06:05.240 ","End":"06:10.610","Text":"so it\u0027s times 1000 milliliters for every 1 liter."},{"Start":"06:10.610 ","End":"06:13.760","Text":"Liters cancel out and"},{"Start":"06:13.760 ","End":"06:22.250","Text":"this equals 54,670 milliliters."},{"Start":"06:22.250 ","End":"06:29.025","Text":"Volume equals 54,670 milliliters,"},{"Start":"06:29.025 ","End":"06:30.725","Text":"that is our final answer."},{"Start":"06:30.725 ","End":"06:33.480","Text":"Thank you very much for watching."}],"ID":23827},{"Watched":false,"Name":"Exercise 7","Duration":"5m 22s","ChapterTopicVideoID":22974,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:03.285","Text":"We\u0027re going to solve the following exercise."},{"Start":"00:03.285 ","End":"00:07.320","Text":"Krypton gas in a 16.7 liters cylinder exerts"},{"Start":"00:07.320 ","End":"00:11.340","Text":"a pressure of 13.4 atmosphere at 25.9-degrees Celsius."},{"Start":"00:11.340 ","End":"00:13.335","Text":"How many grams of gas are present?"},{"Start":"00:13.335 ","End":"00:14.730","Text":"In order to solve this question,"},{"Start":"00:14.730 ","End":"00:17.290","Text":"we\u0027re going to use the equation PV=nRT."},{"Start":"00:18.560 ","End":"00:21.180","Text":"This is the ideal gas equation."},{"Start":"00:21.180 ","End":"00:27.160","Text":"P is the pressure, V is the volume,"},{"Start":"00:27.830 ","End":"00:30.780","Text":"n is the number of moles,"},{"Start":"00:30.780 ","End":"00:34.030","Text":"R is the gas constant,"},{"Start":"00:36.100 ","End":"00:39.360","Text":"and T is the temperature."},{"Start":"00:41.800 ","End":"00:45.050","Text":"Since we want to find the mass of the gas,"},{"Start":"00:45.050 ","End":"00:47.240","Text":"we\u0027re going to first find the number of moles of the gas"},{"Start":"00:47.240 ","End":"00:49.895","Text":"and then find the mass of the gas."},{"Start":"00:49.895 ","End":"00:54.995","Text":"We\u0027re going to isolate n, which is the number of moles by dividing both sides by RT,"},{"Start":"00:54.995 ","End":"01:00.770","Text":"so n which is the number of moles equals PV, divided by RT."},{"Start":"01:00.770 ","End":"01:03.340","Text":"Now we\u0027re going to look at what we know from the question,"},{"Start":"01:03.340 ","End":"01:07.710","Text":"so we know the volume which equals 16.7 liters."},{"Start":"01:08.860 ","End":"01:13.650","Text":"We also know the pressure, which equals 13.4 atmosphere,"},{"Start":"01:16.460 ","End":"01:20.645","Text":"and we know the temperature equals 25.9-degrees Celsius."},{"Start":"01:20.645 ","End":"01:22.150","Text":"Now we need the temperature in Kelvin,"},{"Start":"01:22.150 ","End":"01:28.845","Text":"so it\u0027s temperature in Kelvin equals the temperature in degrees Celsius plus 273,"},{"Start":"01:28.845 ","End":"01:35.145","Text":"so this equals 25.9-degrees Celsius plus 273,"},{"Start":"01:35.145 ","End":"01:40.050","Text":"and this equals 298.9 Kelvin,"},{"Start":"01:40.050 ","End":"01:41.775","Text":"so that\u0027s our temperature."},{"Start":"01:41.775 ","End":"01:50.070","Text":"PV said is 13.4 atmosphere times the volume,"},{"Start":"01:50.070 ","End":"01:58.185","Text":"which is 16.7 liters divided by R which is a gas constant."},{"Start":"01:58.185 ","End":"02:00.650","Text":"The gas constant has a number of values."},{"Start":"02:00.650 ","End":"02:05.360","Text":"We\u0027re going to use R=0.082"},{"Start":"02:05.360 ","End":"02:13.150","Text":"liters times atmospheres divided by mole times Kelvin."},{"Start":"02:13.150 ","End":"02:22.400","Text":"Again, 0.082 liters times atmospheres divided by mole times Kelvin,"},{"Start":"02:22.400 ","End":"02:25.385","Text":"and this is times that temperature,"},{"Start":"02:25.385 ","End":"02:28.010","Text":"so our temperature we found is 298.9"},{"Start":"02:28.010 ","End":"02:37.415","Text":"K. Now we\u0027re just going to divide the numbers and the units so it\u0027ll be easier."},{"Start":"02:37.415 ","End":"02:48.640","Text":"The numbers we have are 13.4 times 16.7 divided by 0.082,"},{"Start":"02:48.640 ","End":"02:55.665","Text":"which is over here, times 298.9, that\u0027s our numbers."},{"Start":"02:55.665 ","End":"02:57.350","Text":"Now we\u0027re going to take our units."},{"Start":"02:57.350 ","End":"03:00.665","Text":"Here just be more comfortable to do that."},{"Start":"03:00.665 ","End":"03:08.120","Text":"We can see that we have atmosphere times liter divided by liter times"},{"Start":"03:08.120 ","End":"03:18.500","Text":"atmosphere times Kelvin divided by mole times Kelvin."},{"Start":"03:18.500 ","End":"03:20.675","Text":"Now I just want to remind you something."},{"Start":"03:20.675 ","End":"03:22.595","Text":"When we\u0027re dividing by a fraction,"},{"Start":"03:22.595 ","End":"03:25.520","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"03:25.520 ","End":"03:27.270","Text":"Again, we\u0027re going to look first at the numbers,"},{"Start":"03:27.270 ","End":"03:30.135","Text":"they come to 9.13,"},{"Start":"03:30.135 ","End":"03:32.735","Text":"and now we\u0027re going to look at the units."},{"Start":"03:32.735 ","End":"03:35.420","Text":"Remember that dividing by a fraction is the same"},{"Start":"03:35.420 ","End":"03:38.165","Text":"as multiplying by the reciprocal of the fraction,"},{"Start":"03:38.165 ","End":"03:44.520","Text":"so this over here is the same as atmosphere times liter times,"},{"Start":"03:44.520 ","End":"03:46.973","Text":"we consider reciprocal is"},{"Start":"03:46.973 ","End":"03:55.745","Text":"mole times Kelvin divided by liters times atmosphere times Kelvin."},{"Start":"03:55.745 ","End":"03:59.270","Text":"The Kelvin we can see cancels out right away,"},{"Start":"03:59.270 ","End":"04:03.365","Text":"and we can also see that we have atmosphere here divided by atmosphere,"},{"Start":"04:03.365 ","End":"04:05.045","Text":"liters divided by liters,"},{"Start":"04:05.045 ","End":"04:09.055","Text":"and we\u0027re left with mole so this equals mole,"},{"Start":"04:09.055 ","End":"04:13.735","Text":"and our answer comes to 9.13 mole."},{"Start":"04:13.735 ","End":"04:16.370","Text":"We know that we have 9.13 mole of krypton,"},{"Start":"04:16.370 ","End":"04:19.970","Text":"and now we have to calculate the mass of the krypton."},{"Start":"04:19.970 ","End":"04:21.530","Text":"In order to calculate the mass,"},{"Start":"04:21.530 ","End":"04:26.650","Text":"we\u0027re going to use the equation n number of moles equals mass divided by the molar mass."},{"Start":"04:26.650 ","End":"04:29.410","Text":"For the mass, we\u0027re going to multiply both sides by the molar mass,"},{"Start":"04:29.410 ","End":"04:35.310","Text":"so the mass equals n number of moles times the molar mass."},{"Start":"04:37.670 ","End":"04:41.590","Text":"We\u0027re looking for the mass of the krypton."},{"Start":"04:41.600 ","End":"04:43.880","Text":"We\u0027re going to use the moles of krypton,"},{"Start":"04:43.880 ","End":"04:52.370","Text":"which we calculated equal to 9.13 mole times the molar mass of the krypton,"},{"Start":"04:52.370 ","End":"05:02.290","Text":"which is taken from the periodic table of elements and equals 83.80 grams per mole."},{"Start":"05:02.680 ","End":"05:04.955","Text":"We can see that the moles cancel out,"},{"Start":"05:04.955 ","End":"05:12.100","Text":"and our answer comes to 765.09 grams."},{"Start":"05:12.100 ","End":"05:13.970","Text":"So the mass of the krypton,"},{"Start":"05:13.970 ","End":"05:18.290","Text":"which we were asked to find equals 765.09 grams."},{"Start":"05:18.290 ","End":"05:20.195","Text":"That is our final answer."},{"Start":"05:20.195 ","End":"05:22.860","Text":"Thank you very much for watching."}],"ID":23828},{"Watched":false,"Name":"Determining Molar Mass from Ideal Gas Equation","Duration":"3m 46s","ChapterTopicVideoID":17414,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.530","Text":"In the previous video,"},{"Start":"00:01.530 ","End":"00:04.290","Text":"we studied the ideal gas equation."},{"Start":"00:04.290 ","End":"00:09.030","Text":"In this video, we will use it to find the molar mass of a gas."},{"Start":"00:09.030 ","End":"00:12.825","Text":"Let\u0027s begin by recalling the ideal gas equation."},{"Start":"00:12.825 ","End":"00:16.890","Text":"The ideal gas equation is PV=nRT,"},{"Start":"00:16.890 ","End":"00:18.840","Text":"P is the pressure, V,"},{"Start":"00:18.840 ","End":"00:20.610","Text":"the volume, n,"},{"Start":"00:20.610 ","End":"00:22.500","Text":"the number of moles, R,"},{"Start":"00:22.500 ","End":"00:24.570","Text":"the gas constant, and T,"},{"Start":"00:24.570 ","End":"00:26.505","Text":"the temperature in kelvin."},{"Start":"00:26.505 ","End":"00:31.395","Text":"Now we\u0027re going to use the ideal gas equation to find the molar mass."},{"Start":"00:31.395 ","End":"00:34.650","Text":"We begin by recalling that the number of moles is"},{"Start":"00:34.650 ","End":"00:38.100","Text":"equal to the mass divided by the molar mass,"},{"Start":"00:38.100 ","End":"00:41.055","Text":"m divided by Mw."},{"Start":"00:41.055 ","End":"00:43.920","Text":"Here\u0027s our ideal gas equation,"},{"Start":"00:43.920 ","End":"00:49.474","Text":"PV=nRT, and we can substitute the expression we got before,"},{"Start":"00:49.474 ","End":"00:52.950","Text":"m over Mw for n,"},{"Start":"00:52.950 ","End":"00:54.120","Text":"the number of moles,"},{"Start":"00:54.120 ","End":"01:01.180","Text":"so we have PV=m/Mw times RT and now,"},{"Start":"01:01.180 ","End":"01:05.690","Text":"we can multiply both sides of equation by Mw,"},{"Start":"01:05.690 ","End":"01:10.860","Text":"so we have Mw times PV=mRT."},{"Start":"01:10.860 ","End":"01:16.085","Text":"Now, in order to isolate the molar mass that says Mw,"},{"Start":"01:16.085 ","End":"01:20.410","Text":"we can divide both sides of the equation by PV,"},{"Start":"01:20.410 ","End":"01:22.035","Text":"so we have Mw,"},{"Start":"01:22.035 ","End":"01:24.770","Text":"the molar mass, equal to m,"},{"Start":"01:24.770 ","End":"01:26.480","Text":"the mass times R,"},{"Start":"01:26.480 ","End":"01:28.400","Text":"the gas constant times T,"},{"Start":"01:28.400 ","End":"01:30.125","Text":"the temperature in kelvin,"},{"Start":"01:30.125 ","End":"01:32.345","Text":"divided by PV,"},{"Start":"01:32.345 ","End":"01:34.780","Text":"which is the pressure times the volume."},{"Start":"01:34.780 ","End":"01:39.605","Text":"That\u0027s our expression for the molar mass."},{"Start":"01:39.605 ","End":"01:41.960","Text":"Now let\u0027s take an example."},{"Start":"01:41.960 ","End":"01:52.275","Text":"What is the molar mass of a gas of mass 0.365 grams and volume 0.224 liters at STP."},{"Start":"01:52.275 ","End":"01:55.275","Text":"We\u0027re going to take the old STP here,"},{"Start":"01:55.275 ","End":"01:59.420","Text":"that\u0027s 1 atmosphere and 0-degrees Celsius."},{"Start":"01:59.420 ","End":"02:02.885","Text":"Then suggest the identity of the gas."},{"Start":"02:02.885 ","End":"02:05.570","Text":"Here\u0027s our expression for the molar mass."},{"Start":"02:05.570 ","End":"02:09.335","Text":"Molar mass is equal to mRT divided by PV,"},{"Start":"02:09.335 ","End":"02:11.885","Text":"and now we can insert the numbers."},{"Start":"02:11.885 ","End":"02:14.900","Text":"The mass is 0.365 grams,"},{"Start":"02:14.900 ","End":"02:21.394","Text":"the gas constant is 0.082 atmospheres times liter,"},{"Start":"02:21.394 ","End":"02:24.440","Text":"mole^minus 1, K^minus 1,"},{"Start":"02:24.440 ","End":"02:30.395","Text":"the temperature in kelvin is 273 kelvin, that\u0027s 0-degrees Celsius."},{"Start":"02:30.395 ","End":"02:32.675","Text":"The pressure is 1 atmosphere,"},{"Start":"02:32.675 ","End":"02:38.675","Text":"and V, the volume 0.224 liters."},{"Start":"02:38.675 ","End":"02:41.225","Text":"First, let\u0027s look at the units."},{"Start":"02:41.225 ","End":"02:44.785","Text":"Atmosphere goes with atmosphere,"},{"Start":"02:44.785 ","End":"02:47.905","Text":"liter goes with liter,"},{"Start":"02:47.905 ","End":"02:52.195","Text":"K to power minus 1 times K is just 1,"},{"Start":"02:52.195 ","End":"02:59.695","Text":"and we\u0027re left with grams times mole to the power minus 1."},{"Start":"02:59.695 ","End":"03:02.765","Text":"If we work out the numbers,"},{"Start":"03:02.765 ","End":"03:07.235","Text":"we get 36.5 grams per mole,"},{"Start":"03:07.235 ","End":"03:12.500","Text":"36.5 grams times mole to the power of minus 1,"},{"Start":"03:12.500 ","End":"03:14.480","Text":"so that\u0027s our molar mass,"},{"Start":"03:14.480 ","End":"03:18.200","Text":"36.5 grams per mole."},{"Start":"03:18.200 ","End":"03:20.450","Text":"Now what about the identity?"},{"Start":"03:20.450 ","End":"03:25.109","Text":"Now possible gas is HCl, is that possible?"},{"Start":"03:25.109 ","End":"03:28.555","Text":"Where the molar mass of each is 1,"},{"Start":"03:28.555 ","End":"03:32.450","Text":"the molar mass of HCl is 35.5,"},{"Start":"03:32.450 ","End":"03:35.680","Text":"giving a total of 36.5."},{"Start":"03:35.680 ","End":"03:38.960","Text":"So that seems a reasonable possibility."},{"Start":"03:38.960 ","End":"03:42.650","Text":"In this video, we showed how to determine the molar mass"},{"Start":"03:42.650 ","End":"03:46.950","Text":"of a gas using the ideal gas equation."}],"ID":18164},{"Watched":false,"Name":"Exercise 8","Duration":"6m 10s","ChapterTopicVideoID":22975,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.509","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.509 ","End":"00:07.320","Text":"A 0.536 grams sample of a gas has a volume of"},{"Start":"00:07.320 ","End":"00:15.360","Text":"127 milliliters at 75.4 degrees Celsius, and 736 mmHg."},{"Start":"00:15.360 ","End":"00:18.135","Text":"What is the molar mass of this gas?"},{"Start":"00:18.135 ","End":"00:19.650","Text":"In order to solve this question,"},{"Start":"00:19.650 ","End":"00:24.950","Text":"we\u0027re going to use the equation PV=nRT."},{"Start":"00:24.950 ","End":"00:28.260","Text":"This is ideal gas equation, P is the pressure,"},{"Start":"00:31.760 ","End":"00:34.740","Text":"V is the volume,"},{"Start":"00:34.740 ","End":"00:37.395","Text":"N is the number of moles,"},{"Start":"00:37.395 ","End":"00:40.480","Text":"R is the gas constant,"},{"Start":"00:42.140 ","End":"00:45.460","Text":"and T is the temperature."},{"Start":"00:47.620 ","End":"00:51.470","Text":"Now we\u0027re asked to find the molar mass of the gas."},{"Start":"00:51.470 ","End":"00:54.190","Text":"First, we\u0027re going to find the number of moles of the gas,"},{"Start":"00:54.190 ","End":"00:57.110","Text":"and then we\u0027ll calculate the molar mass of the gas."},{"Start":"00:57.110 ","End":"00:58.730","Text":"In order to find the number of moles,"},{"Start":"00:58.730 ","End":"01:01.100","Text":"we\u0027re going to isolate n number of moles,"},{"Start":"01:01.100 ","End":"01:03.845","Text":"and in order to do this, we\u0027re going to divide both sides by RT."},{"Start":"01:03.845 ","End":"01:09.725","Text":"N, the number of moles equals PV divided by RT."},{"Start":"01:09.725 ","End":"01:11.900","Text":"Now if we look at our question,"},{"Start":"01:11.900 ","End":"01:14.150","Text":"we can see that we know the volume of the gas,"},{"Start":"01:14.150 ","End":"01:16.330","Text":"it equals 127 milliliters,"},{"Start":"01:16.330 ","End":"01:18.800","Text":"we also know the pressure of the gas,"},{"Start":"01:18.800 ","End":"01:22.500","Text":"which is 736 mmHg,"},{"Start":"01:26.180 ","End":"01:28.790","Text":"and we also know the temperature of the gas,"},{"Start":"01:28.790 ","End":"01:31.920","Text":"which is 75.4 degrees Celsius."},{"Start":"01:34.030 ","End":"01:36.710","Text":"We\u0027re also given the mass of the gas,"},{"Start":"01:36.710 ","End":"01:37.880","Text":"which we\u0027ll use later,"},{"Start":"01:37.880 ","End":"01:41.400","Text":"so the mass is 0.536 grams."},{"Start":"01:42.880 ","End":"01:46.354","Text":"Now, since we need our temperature in Kelvin,"},{"Start":"01:46.354 ","End":"01:47.600","Text":"we\u0027re just going to do that here,"},{"Start":"01:47.600 ","End":"01:49.670","Text":"so temperature in Kelvin,"},{"Start":"01:49.670 ","End":"01:55.500","Text":"equals the temperature in degrees Celsius plus 273,"},{"Start":"01:55.500 ","End":"02:06.360","Text":"so this equals 75.4 degrees Celsius plus 273,"},{"Start":"02:06.360 ","End":"02:13.750","Text":"and this equals 348.4 Kelvin."},{"Start":"02:13.910 ","End":"02:16.445","Text":"That\u0027s our temperature in Kelvin."},{"Start":"02:16.445 ","End":"02:18.860","Text":"Now if we go back to our equation, again,"},{"Start":"02:18.860 ","End":"02:20.770","Text":"we have PV divided by RT,"},{"Start":"02:20.770 ","End":"02:23.750","Text":"now R is the gas constant and it has a number of values,"},{"Start":"02:23.750 ","End":"02:27.865","Text":"we\u0027re going to use the value R equals 0.082,"},{"Start":"02:27.865 ","End":"02:30.075","Text":"in the unit R,"},{"Start":"02:30.075 ","End":"02:32.025","Text":"atmosphere times liters,"},{"Start":"02:32.025 ","End":"02:35.735","Text":"divided by mole times Kelvin."},{"Start":"02:35.735 ","End":"02:38.540","Text":"Now since we\u0027re using atmosphere and liters,"},{"Start":"02:38.540 ","End":"02:42.560","Text":"we have to convert our pressure and our volume."},{"Start":"02:42.560 ","End":"02:49.650","Text":"We\u0027re going to start with the pressure because we have P, it\u0027s 736 mmHg."},{"Start":"02:49.650 ","End":"02:54.710","Text":"But, since our gas constant R has atmosphere in it,"},{"Start":"02:54.710 ","End":"02:58.775","Text":"we\u0027re just going to go and convert the mmHg to atmosphere right now."},{"Start":"02:58.775 ","End":"03:06.660","Text":"Remember that 1 atmosphere, equals 760 mmHg."},{"Start":"03:06.660 ","End":"03:10.790","Text":"The mmHg will cancel out and what will be left with atmosphere, so that\u0027s our pressure."},{"Start":"03:10.790 ","End":"03:13.025","Text":"Now we have times the volume."},{"Start":"03:13.025 ","End":"03:16.270","Text":"Volume is 127 milliliters,"},{"Start":"03:16.270 ","End":"03:20.060","Text":"but again, if we look at our gas constant,"},{"Start":"03:20.060 ","End":"03:21.910","Text":"we can see we have liters in our gas constant,"},{"Start":"03:21.910 ","End":"03:24.620","Text":"so we can multiply this by a conversion factor."},{"Start":"03:24.620 ","End":"03:28.650","Text":"We know that 1 liter equals 1,000 milliliters,"},{"Start":"03:29.240 ","End":"03:34.130","Text":"so the milliliters cancel out and we\u0027ll be left with liters."},{"Start":"03:34.130 ","End":"03:37.040","Text":"That\u0027s our numerator, and I will go to our denominator."},{"Start":"03:37.040 ","End":"03:47.800","Text":"We have R equals 0.082, atmosphere times liter,"},{"Start":"03:48.560 ","End":"03:53.670","Text":"divided by mole times K Kelvin,"},{"Start":"03:53.670 ","End":"03:56.030","Text":"and R is multiplied by T,"},{"Start":"03:56.030 ","End":"03:59.540","Text":"which is the temperature which we found is 348.4K,"},{"Start":"03:59.540 ","End":"04:04.025","Text":"so this is times 348.4K."},{"Start":"04:04.025 ","End":"04:07.150","Text":"There we can see right away that the Kelvin cancel out."},{"Start":"04:07.150 ","End":"04:16.340","Text":"This equals 4.31 times 10 to the negative 3 in the numbers."},{"Start":"04:16.340 ","End":"04:17.960","Text":"Now if we look at our units,"},{"Start":"04:17.960 ","End":"04:22.590","Text":"we can see that what we have left in the numerators atmosphere times liters,"},{"Start":"04:26.320 ","End":"04:31.040","Text":"we can see here we have atmospheres times liters divided by mole."},{"Start":"04:31.040 ","End":"04:34.040","Text":"We\u0027re just going to write that down."},{"Start":"04:34.040 ","End":"04:37.729","Text":"Now if you look, we can see we\u0027re dividing here by a fraction."},{"Start":"04:37.729 ","End":"04:40.055","Text":"Now, remember when you divide by a fraction,"},{"Start":"04:40.055 ","End":"04:43.220","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"04:43.220 ","End":"04:49.385","Text":"Atmosphere times liter divided by mole is the same as,"},{"Start":"04:49.385 ","End":"04:53.150","Text":"atmosphere times liter times mole,"},{"Start":"04:53.150 ","End":"04:56.230","Text":"divided by atmosphere times liter,"},{"Start":"04:56.230 ","End":"04:59.360","Text":"because that\u0027s the reciprocal of the fraction."},{"Start":"04:59.360 ","End":"05:03.020","Text":"Now atmosphere times liters cancel out,"},{"Start":"05:03.020 ","End":"05:05.150","Text":"and we\u0027re left with mole."},{"Start":"05:05.150 ","End":"05:11.880","Text":"This equals 4.31 times 10 to the negative 3 mole."},{"Start":"05:11.880 ","End":"05:13.590","Text":"That\u0027s the number of moles of the gas."},{"Start":"05:13.590 ","End":"05:15.755","Text":"Now, you were asked to find the molar mass of the gas."},{"Start":"05:15.755 ","End":"05:17.360","Text":"Just a reminder, n,"},{"Start":"05:17.360 ","End":"05:21.365","Text":"the number of moles equals mass divided by the molar mass,"},{"Start":"05:21.365 ","End":"05:24.360","Text":"and you want the molar mass,"},{"Start":"05:24.410 ","End":"05:28.350","Text":"so molar mass equals the mass,"},{"Start":"05:28.350 ","End":"05:31.800","Text":"divided by the number of moles."},{"Start":"05:31.800 ","End":"05:34.415","Text":"We just did a simple math manipulation here."},{"Start":"05:34.415 ","End":"05:36.080","Text":"The mass was given in the question,"},{"Start":"05:36.080 ","End":"05:40.495","Text":"and the mass equals 0.536 grams,"},{"Start":"05:40.495 ","End":"05:44.000","Text":"divided by the number of moles which we calculated,"},{"Start":"05:44.000 ","End":"05:50.465","Text":"which equals 4.31 times 10 to the negative 3 mole,"},{"Start":"05:50.465 ","End":"05:59.130","Text":"and this equals 124.36 grams per mole."},{"Start":"05:59.450 ","End":"06:06.140","Text":"Again, the molar mass equals 124.36 grams per mole."},{"Start":"06:06.140 ","End":"06:07.805","Text":"That is our final answer."},{"Start":"06:07.805 ","End":"06:10.680","Text":"Thank you very much for watching."}],"ID":23829},{"Watched":false,"Name":"Determining Gas Density from Ideal Gas Equation","Duration":"4m 14s","ChapterTopicVideoID":17415,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"In the previous video,"},{"Start":"00:01.935 ","End":"00:06.044","Text":"we determine the molar mass of a gas from the ideal gas equation."},{"Start":"00:06.044 ","End":"00:11.580","Text":"In this video, we\u0027ll use the ideal gas equation to find the density of a gas."},{"Start":"00:11.580 ","End":"00:14.535","Text":"Here\u0027s our ideal gas equation."},{"Start":"00:14.535 ","End":"00:17.220","Text":"PV=nRT."},{"Start":"00:17.220 ","End":"00:19.725","Text":"P is the pressure, V is the volume,"},{"Start":"00:19.725 ","End":"00:21.375","Text":"n is the number of moles,"},{"Start":"00:21.375 ","End":"00:23.220","Text":"R is the gas constant,"},{"Start":"00:23.220 ","End":"00:25.785","Text":"and T is the temperature in kelvin."},{"Start":"00:25.785 ","End":"00:30.555","Text":"We\u0027re going to work out the gas density from the ideal gas equation."},{"Start":"00:30.555 ","End":"00:36.655","Text":"The first thing to recall is that the density is equal to the mass divided by the volume."},{"Start":"00:36.655 ","End":"00:42.365","Text":"Let\u0027s start with a gas equation, PV=nRT."},{"Start":"00:42.365 ","End":"00:44.510","Text":"As in the previous video,"},{"Start":"00:44.510 ","End":"00:46.340","Text":"we\u0027ll substitute for n,"},{"Start":"00:46.340 ","End":"00:49.285","Text":"the mass divided by the molar mass,"},{"Start":"00:49.285 ","End":"00:56.880","Text":"so n is equal to the mass divided by the molar mass which we called Mw."},{"Start":"00:58.970 ","End":"01:09.265","Text":"We\u0027ve substituted m over Mw for n. We have PV=mRT divided by Mw."},{"Start":"01:09.265 ","End":"01:14.670","Text":"We\u0027re going to divide both sides of the equation by V. We have"},{"Start":"01:14.670 ","End":"01:20.185","Text":"P is equal to mRT divided by V times Mw."},{"Start":"01:20.185 ","End":"01:24.470","Text":"Recall that d, the density,"},{"Start":"01:24.470 ","End":"01:26.270","Text":"is mass divide by volume."},{"Start":"01:26.270 ","End":"01:28.210","Text":"Here we have the density."},{"Start":"01:28.210 ","End":"01:33.875","Text":"We can write that as density times RT divided by Mw."},{"Start":"01:33.875 ","End":"01:37.280","Text":"We want to isolate the density."},{"Start":"01:37.280 ","End":"01:44.200","Text":"To do that, we multiply both sides of the equation by Mw and divide by RT."},{"Start":"01:44.200 ","End":"01:51.560","Text":"When we\u0027ve done that, we get the density is equal to Mw times the pressure divided by RT."},{"Start":"01:51.560 ","End":"01:54.770","Text":"Here\u0027s our equation for the density,"},{"Start":"01:54.770 ","End":"01:59.435","Text":"d is equal to Mw times P divided by RT."},{"Start":"01:59.435 ","End":"02:01.310","Text":"Let\u0027s take an example."},{"Start":"02:01.310 ","End":"02:03.710","Text":"What is the density of HCl,"},{"Start":"02:03.710 ","End":"02:07.195","Text":"that\u0027s the gas we guessed the last video,"},{"Start":"02:07.195 ","End":"02:12.995","Text":"with the molar mass of 36.5 grams per mole at STP?"},{"Start":"02:12.995 ","End":"02:16.280","Text":"We\u0027re taking the old STP,"},{"Start":"02:16.280 ","End":"02:21.070","Text":"that\u0027s 1 atmosphere and 0 degrees Celsius."},{"Start":"02:21.070 ","End":"02:23.090","Text":"Here\u0027s our expression for the density."},{"Start":"02:23.090 ","End":"02:27.799","Text":"The density is equal to the molar mass times the pressure divided by RT."},{"Start":"02:27.799 ","End":"02:32.075","Text":"The molar mass is 36.5 grams per mole."},{"Start":"02:32.075 ","End":"02:35.765","Text":"The pressure at STP is 1 atmosphere,"},{"Start":"02:35.765 ","End":"02:44.698","Text":"R is 0.082 atmospheres times liter mole^minus 1 times k^minus 1,"},{"Start":"02:44.698 ","End":"02:49.460","Text":"and the temperature at 0 degrees Celsius is 273 kelvin."},{"Start":"02:49.460 ","End":"02:52.025","Text":"Let\u0027s look at the units first."},{"Start":"02:52.025 ","End":"02:56.430","Text":"We have atmosphere, atmosphere, those cancel."},{"Start":"02:56.430 ","End":"03:00.485","Text":"We have mole^minus 1, mole^minus 1."},{"Start":"03:00.485 ","End":"03:05.185","Text":"We have k^minus 1 times k and that\u0027s what? Just 1."},{"Start":"03:05.185 ","End":"03:11.420","Text":"What we\u0027re left with is grams divided by liters, grams per liter."},{"Start":"03:11.420 ","End":"03:19.010","Text":"We can divide 36.5 by 0.082 and divide also by 273,"},{"Start":"03:19.010 ","End":"03:23.510","Text":"and we get 1.63 grams per liter."},{"Start":"03:23.510 ","End":"03:28.580","Text":"The density is 1.63 grams per liter."},{"Start":"03:28.580 ","End":"03:31.720","Text":"We could have solved this problem another way."},{"Start":"03:31.720 ","End":"03:39.590","Text":"We recall that 1 mole of any gas at STP occupies 22.4 liters."},{"Start":"03:39.590 ","End":"03:44.670","Text":"We know that 1 mole of HCl is 36.5 grams so"},{"Start":"03:44.670 ","End":"03:51.249","Text":"36.5 grams of HCl occupies 22.4 liters."},{"Start":"03:51.249 ","End":"03:54.260","Text":"The density is simply the mass,"},{"Start":"03:54.260 ","End":"03:57.410","Text":"36.5 grams, divided by the volume,"},{"Start":"03:57.410 ","End":"04:00.500","Text":"22.4 liters, and divide that,"},{"Start":"04:00.500 ","End":"04:04.258","Text":"we get 1.63 grams per liter,"},{"Start":"04:04.258 ","End":"04:07.280","Text":"the same answer as we got before."},{"Start":"04:07.280 ","End":"04:10.130","Text":"In this video, we showed how to determine"},{"Start":"04:10.130 ","End":"04:14.880","Text":"the molar mass of a gas using the ideal equation."}],"ID":18165},{"Watched":false,"Name":"Exercise 9","Duration":"5m 8s","ChapterTopicVideoID":22976,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.090","Text":"Hi. We\u0027re going to solve the following exercise: a particular application calls for"},{"Start":"00:06.090 ","End":"00:12.360","Text":"N_2 gas with a density of 1.72 grams per liter at 28-degrees Celsius."},{"Start":"00:12.360 ","End":"00:16.875","Text":"What must be the pressure of the N_2 gas in milliliters of mercury?"},{"Start":"00:16.875 ","End":"00:20.580","Text":"We\u0027re going to use the equation PV=nRT,"},{"Start":"00:20.580 ","End":"00:22.455","Text":"it\u0027s the ideal gas equation."},{"Start":"00:22.455 ","End":"00:24.550","Text":"P is the pressure,"},{"Start":"00:24.890 ","End":"00:28.120","Text":"V is the volume,"},{"Start":"00:28.610 ","End":"00:31.280","Text":"n is the number of moles,"},{"Start":"00:31.280 ","End":"00:34.140","Text":"R is the gas constant,"},{"Start":"00:36.940 ","End":"00:39.440","Text":"and T is the temperature."},{"Start":"00:39.440 ","End":"00:43.250","Text":"Now, first of all,"},{"Start":"00:43.250 ","End":"00:46.160","Text":"we\u0027re looking for the pressure of the N_2 gas, therefore,"},{"Start":"00:46.160 ","End":"00:47.300","Text":"we\u0027re going to isolate the pressure,"},{"Start":"00:47.300 ","End":"00:49.685","Text":"so we\u0027re going to divide both sides by the volume."},{"Start":"00:49.685 ","End":"00:57.110","Text":"Pressure=nRT divided by the volume."},{"Start":"00:57.110 ","End":"00:58.490","Text":"Now, if we look at our question,"},{"Start":"00:58.490 ","End":"01:00.500","Text":"we can see that we\u0027re given the density of the gas,"},{"Start":"01:00.500 ","End":"01:05.220","Text":"so the density equals 1.72 grams per liter."},{"Start":"01:05.320 ","End":"01:08.000","Text":"We\u0027re also given the temperature of the gas,"},{"Start":"01:08.000 ","End":"01:10.680","Text":"which is 28-degrees Celsius."},{"Start":"01:12.250 ","End":"01:15.425","Text":"Now again, we need our temperature in kelvin."},{"Start":"01:15.425 ","End":"01:22.805","Text":"Temperature in kelvin equals temperature in degrees Celsius plus 273,"},{"Start":"01:22.805 ","End":"01:28.415","Text":"which equals 28-degrees Celsius plus 273,"},{"Start":"01:28.415 ","End":"01:31.655","Text":"and this equals 301 kelvin."},{"Start":"01:31.655 ","End":"01:34.700","Text":"Now again, we have P=nRT divided by V,"},{"Start":"01:34.700 ","End":"01:36.410","Text":"and we know our density."},{"Start":"01:36.410 ","End":"01:38.960","Text":"Now, I want to remind you,"},{"Start":"01:38.960 ","End":"01:44.345","Text":"d=m divided by V. If we look at our equation,"},{"Start":"01:44.345 ","End":"01:45.980","Text":"we have n number of moles,"},{"Start":"01:45.980 ","End":"01:48.170","Text":"and I want to remind you again that N,"},{"Start":"01:48.170 ","End":"01:52.235","Text":"the number of moles equals the mass divided by the molar mass."},{"Start":"01:52.235 ","End":"01:57.065","Text":"Now, if we substitute mass divided by molar mass instead of the n,"},{"Start":"01:57.065 ","End":"02:05.990","Text":"we get mass times R times T divided by the molar mass times V. Now,"},{"Start":"02:05.990 ","End":"02:09.440","Text":"as you can see, in our equation now we have m divided by V,"},{"Start":"02:09.440 ","End":"02:11.460","Text":"which is the density."},{"Start":"02:13.120 ","End":"02:22.695","Text":"Our equation equals m divided by V times RT divided by the molar mass."},{"Start":"02:22.695 ","End":"02:25.130","Text":"We had to do all of this since we\u0027re given"},{"Start":"02:25.130 ","End":"02:29.120","Text":"the density in the question. Now, we can substitute."},{"Start":"02:29.120 ","End":"02:37.710","Text":"The density equals 1.72 grams per liter, times RT,"},{"Start":"02:37.710 ","End":"02:46.835","Text":"which is the gas constant 0.082 atmosphere times liter divided by mole times kelvin,"},{"Start":"02:46.835 ","End":"02:49.078","Text":"times the temperature, which is 301"},{"Start":"02:49.078 ","End":"02:57.030","Text":"K. We can already cancel out the kelvin divided by the molar mass."},{"Start":"02:57.710 ","End":"03:05.880","Text":"The molar mass of N_2=2 times the molar mass of nitrogen,"},{"Start":"03:05.880 ","End":"03:13.150","Text":"which equals 2 times 14.01 grams per mole,"},{"Start":"03:13.640 ","End":"03:18.970","Text":"which equals 28.02 grams per mole."},{"Start":"03:18.970 ","End":"03:21.710","Text":"Again, this is divided by the molar mass,"},{"Start":"03:21.710 ","End":"03:26.100","Text":"which is 28.02 grams per mole."},{"Start":"03:26.720 ","End":"03:29.835","Text":"This equals 1.52,"},{"Start":"03:29.835 ","End":"03:31.130","Text":"and if we look at our units,"},{"Start":"03:31.130 ","End":"03:36.200","Text":"we can see that what we have left is grams per liter, which is here,"},{"Start":"03:36.200 ","End":"03:42.685","Text":"times atmosphere times liter, divided by mole."},{"Start":"03:42.685 ","End":"03:49.820","Text":"Here we can see the liters cancel out and all of this is divided by grams per mole."},{"Start":"03:49.820 ","End":"03:53.600","Text":"Now, I want to remind you that"},{"Start":"03:53.600 ","End":"03:55.700","Text":"when we were dividing by a fraction it\u0027s the same as"},{"Start":"03:55.700 ","End":"03:58.565","Text":"multiplying by the reciprocal of the fraction."},{"Start":"03:58.565 ","End":"04:00.740","Text":"We\u0027ll just do that over here."},{"Start":"04:00.740 ","End":"04:10.389","Text":"We can see that we have grams times atmosphere divided by mole."},{"Start":"04:11.330 ","End":"04:15.120","Text":"This is divided by a fraction grams divided by moles."},{"Start":"04:15.120 ","End":"04:21.170","Text":"We\u0027re going to multiply by the reciprocal of the fraction which is mole per gram."},{"Start":"04:21.170 ","End":"04:23.810","Text":"This way we can see that our moles cancel out,"},{"Start":"04:23.810 ","End":"04:25.130","Text":"our grams cancel out,"},{"Start":"04:25.130 ","End":"04:27.780","Text":"and we\u0027re left with atmosphere."},{"Start":"04:27.880 ","End":"04:34.020","Text":"This equals 1.52 atmosphere."},{"Start":"04:34.220 ","End":"04:38.395","Text":"Now, we were asked to find the pressure in millimeters of mercury."},{"Start":"04:38.395 ","End":"04:41.600","Text":"We\u0027re going to use a conversion factor and we\u0027re going to multiply this by"},{"Start":"04:41.600 ","End":"04:48.340","Text":"760 millimeters of Hg for every 1 atmosphere."},{"Start":"04:48.440 ","End":"04:50.805","Text":"The atmospheres cancel out,"},{"Start":"04:50.805 ","End":"05:00.260","Text":"and this equals 1,155.2 millimeters of mercury."},{"Start":"05:00.260 ","End":"05:05.510","Text":"Again, the pressure equals 1,155.2 millimeters of mercury."},{"Start":"05:05.510 ","End":"05:06.800","Text":"That is our final answer."},{"Start":"05:06.800 ","End":"05:09.390","Text":"Thank you very much for watching."}],"ID":23830},{"Watched":false,"Name":"Exercise 10","Duration":"4m 33s","ChapterTopicVideoID":22977,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Hi, We\u0027re going to solve the following exercise."},{"Start":"00:03.030 ","End":"00:07.500","Text":"Consider air to have a molar mass of 29.57 grams per mole."},{"Start":"00:07.500 ","End":"00:13.005","Text":"Determine the density of air at 30-degrees Celsius in 1.26 atmosphere in grams per liter."},{"Start":"00:13.005 ","End":"00:15.180","Text":"We need to determine the density of air."},{"Start":"00:15.180 ","End":"00:17.730","Text":"Remember that d, the density equals m,"},{"Start":"00:17.730 ","End":"00:19.350","Text":"which is the mass divided by V,"},{"Start":"00:19.350 ","End":"00:23.975","Text":"which is the volume we know that PV equals nRT."},{"Start":"00:23.975 ","End":"00:31.015","Text":"P is the pressure, V is the volume,"},{"Start":"00:31.015 ","End":"00:33.765","Text":"n is the number of moles,"},{"Start":"00:33.765 ","End":"00:36.790","Text":"R is the gas constant,"},{"Start":"00:38.870 ","End":"00:42.340","Text":"and T is the temperature."},{"Start":"00:43.460 ","End":"00:45.680","Text":"Now we also know that n,"},{"Start":"00:45.680 ","End":"00:47.360","Text":"the number of moles, equals m,"},{"Start":"00:47.360 ","End":"00:49.670","Text":"which is the mass divided by Mw,"},{"Start":"00:49.670 ","End":"00:51.095","Text":"which is the molar mass."},{"Start":"00:51.095 ","End":"00:55.160","Text":"Now since we need d, the density which equals n divided by V,"},{"Start":"00:55.160 ","End":"01:00.440","Text":"we\u0027re going to substitute m divided by the molar mass for n. This gives"},{"Start":"01:00.440 ","End":"01:06.335","Text":"us PV equals mRT divided by the molar mass."},{"Start":"01:06.335 ","End":"01:09.770","Text":"We\u0027re looking for the density which equals m divided by V. So we\u0027re going to"},{"Start":"01:09.770 ","End":"01:13.100","Text":"isolate m divided by V from our equation."},{"Start":"01:13.100 ","End":"01:15.920","Text":"First of all, we see that we have our mass here."},{"Start":"01:15.920 ","End":"01:17.300","Text":"We have our volume here."},{"Start":"01:17.300 ","End":"01:19.550","Text":"We\u0027re going to divide both sides by the volume."},{"Start":"01:19.550 ","End":"01:29.090","Text":"We\u0027re going to get P, the pressure equals mRT divided by V times Mw, the molar mass."},{"Start":"01:29.090 ","End":"01:30.470","Text":"Since we want the density,"},{"Start":"01:30.470 ","End":"01:33.020","Text":"we\u0027re going to isolate m divided by V. We\u0027re going to move"},{"Start":"01:33.020 ","End":"01:35.615","Text":"everything else to the other side of the equation."},{"Start":"01:35.615 ","End":"01:38.705","Text":"We\u0027re going to multiply both sides by the molar mass at first."},{"Start":"01:38.705 ","End":"01:40.830","Text":"Let\u0027s do this slowly. We have m divided by"},{"Start":"01:40.830 ","End":"01:45.560","Text":"V on one side equals if we look at the other side,"},{"Start":"01:45.560 ","End":"01:54.290","Text":"we have the pressure times the molar mass divided by what\u0027s left RT,"},{"Start":"01:54.290 ","End":"01:56.510","Text":"which goes to the other side."},{"Start":"01:56.510 ","End":"01:59.149","Text":"If we go back to the question,"},{"Start":"01:59.149 ","End":"02:08.150","Text":"we know that the pressure equals 1.26 atmosphere times the molar mass,"},{"Start":"02:08.150 ","End":"02:12.000","Text":"which equals 29.57 grams per mole."},{"Start":"02:15.400 ","End":"02:19.370","Text":"All of this is divided by R,"},{"Start":"02:19.370 ","End":"02:20.720","Text":"which is the gas constant."},{"Start":"02:20.720 ","End":"02:22.700","Text":"Now the gas constant has a number of values,"},{"Start":"02:22.700 ","End":"02:26.910","Text":"and we\u0027re going to choose 0.082"},{"Start":"02:26.910 ","End":"02:33.470","Text":"and the units are atmospheres times liters divided by mole times kelvin."},{"Start":"02:33.470 ","End":"02:36.406","Text":"This we\u0027re going to multiply it by the temperature."},{"Start":"02:36.406 ","End":"02:38.300","Text":"Now the temperature we need in kelvin,"},{"Start":"02:38.300 ","End":"02:40.460","Text":"and we have the temperature in degrees Celsius."},{"Start":"02:40.460 ","End":"02:42.620","Text":"You see here 30-degrees Celsius."},{"Start":"02:42.620 ","End":"02:51.395","Text":"The temperature in kelvin equals the temperature in degrees Celsius plus 273."},{"Start":"02:51.395 ","End":"02:59.495","Text":"Temperature equals 30-degrees Celsius plus 273."},{"Start":"02:59.495 ","End":"03:08.945","Text":"This comes out to 303 kelvin times the temperature times 303 kelvin."},{"Start":"03:08.945 ","End":"03:11.645","Text":"Now this equals 1.5."},{"Start":"03:11.645 ","End":"03:15.950","Text":"If we look at the numbers and if we look at our units,"},{"Start":"03:15.950 ","End":"03:20.760","Text":"we have atmospheres times gram divided by mole."},{"Start":"03:24.520 ","End":"03:33.715","Text":"All of this is divided by atmospheres times liter divided by,"},{"Start":"03:33.715 ","End":"03:36.710","Text":"all of this is divided by atmospheres times liters times"},{"Start":"03:36.710 ","End":"03:42.660","Text":"kelvin divided by mole times kelvin."},{"Start":"03:42.660 ","End":"03:45.170","Text":"We can see that the kelvin cancels out right away."},{"Start":"03:45.170 ","End":"03:47.080","Text":"We could have canceled it out here too."},{"Start":"03:47.080 ","End":"03:49.040","Text":"Now we\u0027re dividing by a fraction."},{"Start":"03:49.040 ","End":"03:50.870","Text":"Remember that when you divide by a fraction,"},{"Start":"03:50.870 ","End":"03:54.080","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"03:54.080 ","End":"03:57.740","Text":"This equals again 1.5 atmosphere times"},{"Start":"03:57.740 ","End":"04:03.455","Text":"grams divided by mole times the reciprocal of the fraction."},{"Start":"04:03.455 ","End":"04:09.025","Text":"It\u0027s times mole divided by atmospheres times liters."},{"Start":"04:09.025 ","End":"04:15.450","Text":"You see that the atmosphere cancels out, so it is the mole."},{"Start":"04:15.450 ","End":"04:22.470","Text":"What we get here is 1.5 grams divided by liter."},{"Start":"04:23.230 ","End":"04:29.495","Text":"So d, the density of air equals 1.5 grams per liter."},{"Start":"04:29.495 ","End":"04:31.235","Text":"That is our final answer."},{"Start":"04:31.235 ","End":"04:33.810","Text":"Thank you very much for watching."}],"ID":23831},{"Watched":false,"Name":"Law of Combining Volumes","Duration":"7m 13s","ChapterTopicVideoID":17425,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"In a previous video,"},{"Start":"00:01.770 ","End":"00:03.990","Text":"we studied Avogadro\u0027s law."},{"Start":"00:03.990 ","End":"00:08.880","Text":"In this video, we\u0027ll use it to derive the law of combining volumes."},{"Start":"00:08.880 ","End":"00:12.270","Text":"Let\u0027s first consider Avogadro\u0027s law."},{"Start":"00:12.270 ","End":"00:15.990","Text":"Avogadro\u0027s law says that at a fixed temperature and pressure,"},{"Start":"00:15.990 ","End":"00:20.205","Text":"the volume of a gas is proportional to the number of moles of gas."},{"Start":"00:20.205 ","End":"00:22.230","Text":"That is V, the volume,"},{"Start":"00:22.230 ","End":"00:23.550","Text":"is proportional to n,"},{"Start":"00:23.550 ","End":"00:25.065","Text":"the number of moles."},{"Start":"00:25.065 ","End":"00:28.830","Text":"We can use Avogadro\u0027s law to derive the law of"},{"Start":"00:28.830 ","End":"00:33.510","Text":"combining volumes for a chemical reaction involving gases."},{"Start":"00:33.510 ","End":"00:37.240","Text":"This is only true for the gases involved in the reaction."},{"Start":"00:37.240 ","End":"00:38.810","Text":"Let\u0027s take an example."},{"Start":"00:38.810 ","End":"00:43.250","Text":"The reaction I\u0027m going to consider is called the Haber process."},{"Start":"00:43.250 ","End":"00:50.330","Text":"In this reaction, nitrogen gas reacts with hydrogen gas to form ammonia."},{"Start":"00:50.330 ","End":"00:57.229","Text":"This is an industrial process used to take nitrogen from the atmosphere to ammonia,"},{"Start":"00:57.229 ","End":"01:02.180","Text":"which can then be used in other to make for example fertilizers."},{"Start":"01:02.180 ","End":"01:06.330","Text":"This equation tells us that 1 mole of nitrogen reacts"},{"Start":"01:06.330 ","End":"01:10.635","Text":"with 3 moles of hydrogen to give 2 moles of ammonia."},{"Start":"01:10.635 ","End":"01:15.175","Text":"Now, if we measure all these gases at the same temperature and pressure,"},{"Start":"01:15.175 ","End":"01:19.790","Text":"and if the volume of 1 mole of gas is V liters."},{"Start":"01:19.790 ","End":"01:25.400","Text":"Then we can write the equation in the following form; 1 volume,"},{"Start":"01:25.400 ","End":"01:30.660","Text":"1 times V liters of nitrogen reacts with"},{"Start":"01:30.660 ","End":"01:39.185","Text":"3 times V liters of hydrogen to form 2 times V liters of ammonia."},{"Start":"01:39.185 ","End":"01:43.970","Text":"Because we know that 1 mole occupies V liters."},{"Start":"01:43.970 ","End":"01:45.680","Text":"Now, if we look at this equation,"},{"Start":"01:45.680 ","End":"01:49.055","Text":"we see that V appears 3 times."},{"Start":"01:49.055 ","End":"01:55.050","Text":"We can divide the equation by V. Then we get 1 liter"},{"Start":"01:55.050 ","End":"02:01.220","Text":"of nitrogen reacts with 3 liters of hydrogen to give 2 liters of ammonia."},{"Start":"02:01.220 ","End":"02:04.855","Text":"Now we can state the law of combining volumes."},{"Start":"02:04.855 ","End":"02:09.524","Text":"Before stating it, we should notice that we had 1 mole at the top,"},{"Start":"02:09.524 ","End":"02:13.155","Text":"1 mole in this reaction of nitrogen,"},{"Start":"02:13.155 ","End":"02:15.465","Text":"and here we have 1 liter of nitrogen,"},{"Start":"02:15.465 ","End":"02:17.700","Text":"we had 3 moles of hydrogen,"},{"Start":"02:17.700 ","End":"02:19.995","Text":"here we have 3 liters of hydrogen,"},{"Start":"02:19.995 ","End":"02:22.305","Text":"we had 2 moles of ammonia,"},{"Start":"02:22.305 ","End":"02:24.824","Text":"now we have 2 liters of ammonia."},{"Start":"02:24.824 ","End":"02:26.775","Text":"So it\u0027s the same numbers."},{"Start":"02:26.775 ","End":"02:29.734","Text":"We can state the law of combining volumes."},{"Start":"02:29.734 ","End":"02:35.915","Text":"The ratio of the volumes of gases that react and are produced in a chemical reaction,"},{"Start":"02:35.915 ","End":"02:38.794","Text":"is the same as the mole ratio,"},{"Start":"02:38.794 ","End":"02:43.310","Text":"provided all the gases are measured at the same pressure and temperature."},{"Start":"02:43.310 ","End":"02:45.440","Text":"Instead of using the mole ratio,"},{"Start":"02:45.440 ","End":"02:48.725","Text":"we can use the ratio of the volumes."},{"Start":"02:48.725 ","End":"02:50.495","Text":"Now, just as an aside,"},{"Start":"02:50.495 ","End":"02:56.570","Text":"you may recall that you learned once about the Gay-Lussac\u0027s law of combining volumes."},{"Start":"02:56.570 ","End":"03:03.005","Text":"This was a previous version of this better law of combining volumes."},{"Start":"03:03.005 ","End":"03:09.030","Text":"This is an improvement of Gay-Lussac\u0027s law of combining volumes."},{"Start":"03:09.030 ","End":"03:12.770","Text":"Because there they said that the volumes were in a simple ratio,"},{"Start":"03:12.770 ","End":"03:15.050","Text":"but didn\u0027t specify which ratio."},{"Start":"03:15.050 ","End":"03:19.950","Text":"Now we know that the ratio is the mole ratio."},{"Start":"03:23.540 ","End":"03:26.609","Text":"Let\u0027s consider an example."},{"Start":"03:26.609 ","End":"03:32.885","Text":"0.5 liters of nitrogen gas reacts with 3 liters of hydrogen gas."},{"Start":"03:32.885 ","End":"03:37.198","Text":"First of all, what is the limiting reagent or reactant."},{"Start":"03:37.198 ","End":"03:38.780","Text":"Another way of calling it,"},{"Start":"03:38.780 ","End":"03:41.360","Text":"limited reagent or limiting reactant."},{"Start":"03:41.360 ","End":"03:44.675","Text":"What volume of ammonia is produced?"},{"Start":"03:44.675 ","End":"03:48.395","Text":"We\u0027re assuming that everything is the same pressure and temperature."},{"Start":"03:48.395 ","End":"03:51.100","Text":"Let\u0027s go back to our original equation."},{"Start":"03:51.100 ","End":"03:54.510","Text":"We saw that 1 mole of nitrogen reacts with 3 moles of"},{"Start":"03:54.510 ","End":"03:58.230","Text":"hydrogen to give 2 moles of ammonia."},{"Start":"03:58.230 ","End":"03:59.775","Text":"Or in other words,"},{"Start":"03:59.775 ","End":"04:07.010","Text":"1 liter of nitrogen reacts with 3 liters of hydrogen to give 2 liters of ammonia."},{"Start":"04:07.010 ","End":"04:11.510","Text":"We can write that the volume of hydrogen divided by"},{"Start":"04:11.510 ","End":"04:16.190","Text":"the volume of nitrogen is 3 liters to 1 liter."},{"Start":"04:16.190 ","End":"04:18.110","Text":"The liters cancel,"},{"Start":"04:18.110 ","End":"04:20.020","Text":"and we get 3."},{"Start":"04:20.020 ","End":"04:22.325","Text":"Let\u0027s look at the problem."},{"Start":"04:22.325 ","End":"04:26.419","Text":"Where if the volume of nitrogen is 0.5 liters,"},{"Start":"04:26.419 ","End":"04:29.435","Text":"and we want all of that to react,"},{"Start":"04:29.435 ","End":"04:34.894","Text":"then we require the volume of hydrogen to be equal to the volume of nitrogen,"},{"Start":"04:34.894 ","End":"04:39.120","Text":"times this ratio that we\u0027ve just discussed,"},{"Start":"04:39.120 ","End":"04:42.840","Text":"and decided that it\u0027s 3 to 1, 3."},{"Start":"04:42.840 ","End":"04:45.170","Text":"To get the volume of hydrogen,"},{"Start":"04:45.170 ","End":"04:50.490","Text":"we need the volume of nitrogen multiplied by the ratio which is 3."},{"Start":"04:50.490 ","End":"04:55.340","Text":"In the problem we saw that the volume of nitrogen was 0.5 liters,"},{"Start":"04:55.340 ","End":"04:57.170","Text":"multiply that by 3,"},{"Start":"04:57.170 ","End":"04:59.975","Text":"and we get 1.5 liters."},{"Start":"04:59.975 ","End":"05:06.475","Text":"Now, we\u0027re told that the volume of hydrogen is 3 liters."},{"Start":"05:06.475 ","End":"05:08.915","Text":"We certainly have enough hydrogen."},{"Start":"05:08.915 ","End":"05:11.765","Text":"Now if we want all the hydrogen to react,"},{"Start":"05:11.765 ","End":"05:14.015","Text":"the 3 liters of hydrogen to react,"},{"Start":"05:14.015 ","End":"05:17.585","Text":"what volume of nitrogen do we require?"},{"Start":"05:17.585 ","End":"05:20.960","Text":"The volume of nitrogen is the volume of hydrogen"},{"Start":"05:20.960 ","End":"05:24.850","Text":"multiplied by 1 over the ratio we had before,"},{"Start":"05:24.850 ","End":"05:27.855","Text":"and that means that it\u0027s 3 liters,"},{"Start":"05:27.855 ","End":"05:29.760","Text":"we have 3 liters of hydrogen,"},{"Start":"05:29.760 ","End":"05:32.960","Text":"times 1 over 3 in this case because this is"},{"Start":"05:32.960 ","End":"05:38.665","Text":"the inverse of the original ratio that we calculated."},{"Start":"05:38.665 ","End":"05:41.475","Text":"So 3 liters times 1/3,"},{"Start":"05:41.475 ","End":"05:43.920","Text":"which is of course 1 liter."},{"Start":"05:43.920 ","End":"05:47.250","Text":"This tells us that if we want all the hydrogen to react,"},{"Start":"05:47.250 ","End":"05:50.685","Text":"we require 1 liter of nitrogen."},{"Start":"05:50.685 ","End":"05:53.430","Text":"But we don\u0027t have 1 liter of nitrogen,"},{"Start":"05:53.430 ","End":"05:57.530","Text":"we only have 0.5 liters of nitrogen."},{"Start":"05:57.530 ","End":"06:02.540","Text":"We can conclude that nitrogen is the limiting reagent."},{"Start":"06:02.540 ","End":"06:05.105","Text":"All of the nitrogen will react,"},{"Start":"06:05.105 ","End":"06:08.675","Text":"and we will be left over with some hydrogen."},{"Start":"06:08.675 ","End":"06:11.950","Text":"Hydrogen is in excess."},{"Start":"06:11.950 ","End":"06:15.005","Text":"Assuming that all the nitrogen reacts,"},{"Start":"06:15.005 ","End":"06:19.235","Text":"we can calculate how much ammonia we will produce."},{"Start":"06:19.235 ","End":"06:25.025","Text":"The volume of ammonia compared to the volume of nitrogen is the same as the mole ratio."},{"Start":"06:25.025 ","End":"06:28.115","Text":"That\u0027s 2 liters to 1 liter,"},{"Start":"06:28.115 ","End":"06:29.675","Text":"which is equal to 2."},{"Start":"06:29.675 ","End":"06:32.155","Text":"That\u0027s just like the mole ratio."},{"Start":"06:32.155 ","End":"06:35.970","Text":"You look at the equation, it\u0027s 2, 2, 1."},{"Start":"06:35.970 ","End":"06:40.595","Text":"2 moles of ammonia are formed from 1 mole of nitrogen."},{"Start":"06:40.595 ","End":"06:43.220","Text":"From that, we can conclude that"},{"Start":"06:43.220 ","End":"06:47.795","Text":"the volume of ammonia produced will be the volume of nitrogen,"},{"Start":"06:47.795 ","End":"06:53.315","Text":"times the ratio volume of ammonia compared to the volume of nitrogen."},{"Start":"06:53.315 ","End":"06:56.485","Text":"The volume of nitrogen 0.5 liters,"},{"Start":"06:56.485 ","End":"06:58.635","Text":"and multiply it by 2,"},{"Start":"06:58.635 ","End":"07:01.185","Text":"we get 1 liter."},{"Start":"07:01.185 ","End":"07:03.505","Text":"That\u0027s the answer to our question."},{"Start":"07:03.505 ","End":"07:08.840","Text":"The amount of ammonia that will be produced is 1 liter."},{"Start":"07:08.840 ","End":"07:14.220","Text":"In this video, we learned about the law of combining volumes."}],"ID":18172},{"Watched":false,"Name":"Exercise 11","Duration":"5m 17s","ChapterTopicVideoID":22979,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.970","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.970 ","End":"00:06.330","Text":"What volume of O_2 gas is consumed in the combustion of"},{"Start":"00:06.330 ","End":"00:11.460","Text":"72.8 liters of propane gas if both gases are measured at STP."},{"Start":"00:11.460 ","End":"00:15.345","Text":"First of all, we\u0027re going to write the equation for the combustion of propane gas."},{"Start":"00:15.345 ","End":"00:22.150","Text":"We have C_3H_8 plus oxygen,"},{"Start":"00:24.260 ","End":"00:31.780","Text":"gives us carbon dioxide plus water."},{"Start":"00:33.200 ","End":"00:36.230","Text":"We\u0027re going to balance this equation and we\u0027re going to divide"},{"Start":"00:36.230 ","End":"00:40.385","Text":"the equation into our reactant side and our product side."},{"Start":"00:40.385 ","End":"00:44.729","Text":"On the reactant side, we can see we have 3 carbons,"},{"Start":"00:44.920 ","End":"00:48.360","Text":"we have 8 hydrogens,"},{"Start":"00:48.490 ","End":"00:51.414","Text":"we have 2 oxygens."},{"Start":"00:51.414 ","End":"00:55.425","Text":"On our product side, we have 1 carbon,"},{"Start":"00:55.425 ","End":"01:00.750","Text":"2 hydrogens, and we have 3 oxygens,"},{"Start":"01:00.750 ","End":"01:02.040","Text":"1 oxygen from the water,"},{"Start":"01:02.040 ","End":"01:04.020","Text":"and 2 oxygens from the carbon dioxide,"},{"Start":"01:04.020 ","End":"01:05.750","Text":"so that\u0027s 3 oxygens in all."},{"Start":"01:05.750 ","End":"01:07.310","Text":"First, we\u0027re going to balance the carbons."},{"Start":"01:07.310 ","End":"01:08.660","Text":"We can see that on our reactant side,"},{"Start":"01:08.660 ","End":"01:10.950","Text":"we have 3 carbons and on our product side we have 1."},{"Start":"01:10.950 ","End":"01:13.825","Text":"We\u0027re going to multiply our carbon dioxide by 3."},{"Start":"01:13.825 ","End":"01:17.895","Text":"This is going to change our number of carbons into 3 instead of 1."},{"Start":"01:17.895 ","End":"01:19.640","Text":"This is also going to change the oxygens,"},{"Start":"01:19.640 ","End":"01:22.615","Text":"because now we have 6 oxygens from our carbon dioxide,"},{"Start":"01:22.615 ","End":"01:23.630","Text":"it\u0027s 2 times 3,"},{"Start":"01:23.630 ","End":"01:27.010","Text":"so that\u0027s 6 oxygens plus 1 from the water."},{"Start":"01:27.010 ","End":"01:30.430","Text":"We have 7 oxygens at all instead of 3."},{"Start":"01:30.430 ","End":"01:32.570","Text":"Now we\u0027re going to balance our hydrogens."},{"Start":"01:32.570 ","End":"01:34.760","Text":"We can see on the reactant side we have 8 hydrogens,"},{"Start":"01:34.760 ","End":"01:36.350","Text":"and on our product side, we have 2."},{"Start":"01:36.350 ","End":"01:39.470","Text":"We\u0027re going to multiply our water by 4."},{"Start":"01:39.470 ","End":"01:42.035","Text":"That way we\u0027re going to get 8 hydrogens, 2 times 4."},{"Start":"01:42.035 ","End":"01:45.435","Text":"We\u0027re going to change the number of hydrogens to 8."},{"Start":"01:45.435 ","End":"01:48.043","Text":"We can see that we also changed the number of oxygens,"},{"Start":"01:48.043 ","End":"01:50.210","Text":"because now in the water we have 4,"},{"Start":"01:50.210 ","End":"01:55.340","Text":"4 times 1, 4 oxygens plus 6 from the carbon dioxide,"},{"Start":"01:55.340 ","End":"01:56.870","Text":"that\u0027s 10 in all."},{"Start":"01:56.870 ","End":"01:59.870","Text":"The number of oxygens now is 10."},{"Start":"01:59.870 ","End":"02:01.790","Text":"Now we have to balance the oxygens."},{"Start":"02:01.790 ","End":"02:04.010","Text":"We can see that on our product side we have 10 oxygens,"},{"Start":"02:04.010 ","End":"02:05.500","Text":"and on the reactant side, we have 2,"},{"Start":"02:05.500 ","End":"02:10.200","Text":"we\u0027re going to balance the oxygens by multiplying our O_2 by 5."},{"Start":"02:10.200 ","End":"02:13.330","Text":"We\u0027re going to change the number of oxygens now to 10,"},{"Start":"02:13.330 ","End":"02:16.090","Text":"because we have 2 times 5 equals 10."},{"Start":"02:16.090 ","End":"02:18.500","Text":"Now we can see that we have a balanced equation."},{"Start":"02:18.500 ","End":"02:20.720","Text":"Now that we\u0027d have a balanced equation,"},{"Start":"02:20.720 ","End":"02:24.065","Text":"we can calculate the number of moles of propane gas."},{"Start":"02:24.065 ","End":"02:28.955","Text":"Now we know that we have a volume of 72.8 liters of profane,"},{"Start":"02:28.955 ","End":"02:31.280","Text":"and we know that both gases,"},{"Start":"02:31.280 ","End":"02:34.085","Text":"the oxygen and the propane are measured at STP."},{"Start":"02:34.085 ","End":"02:36.480","Text":"Remember that at STP,"},{"Start":"02:38.920 ","End":"02:47.175","Text":"1 mole of gas occupies 22.4 liters."},{"Start":"02:47.175 ","End":"02:52.840","Text":"The number of moles of propane gas equal the volume of"},{"Start":"02:52.840 ","End":"03:01.070","Text":"the propane gas times 1 mole of gas for every 22.4 liters."},{"Start":"03:03.060 ","End":"03:05.110","Text":"The volume of the propane,"},{"Start":"03:05.110 ","End":"03:12.220","Text":"which is given as 72.8 liters times 1 mole,"},{"Start":"03:12.220 ","End":"03:15.500","Text":"divided by 22.4 liters,"},{"Start":"03:15.500 ","End":"03:17.625","Text":"so liters cancel out,"},{"Start":"03:17.625 ","End":"03:24.190","Text":"and this equals 3.25 moles of propane."},{"Start":"03:24.420 ","End":"03:27.805","Text":"Now that we know the number of moles of propane,"},{"Start":"03:27.805 ","End":"03:33.565","Text":"we\u0027re going to calculate the number of moles of the oxygen using our equation."},{"Start":"03:33.565 ","End":"03:38.440","Text":"Next, we\u0027ll calculate the volume of the oxygen."},{"Start":"03:38.440 ","End":"03:40.405","Text":"If we look at our reaction,"},{"Start":"03:40.405 ","End":"03:42.940","Text":"we can see that for every 1 mole of propane gas,"},{"Start":"03:42.940 ","End":"03:45.715","Text":"which reacts, 5 moles of oxygen react."},{"Start":"03:45.715 ","End":"03:49.690","Text":"The number of moles of oxygen gas equal the number of moles of"},{"Start":"03:49.690 ","End":"03:58.390","Text":"propane times 5 moles of oxygen for every 1 mole of propane."},{"Start":"03:59.100 ","End":"04:04.460","Text":"The number of moles of propane equal 3.25 mole."},{"Start":"04:08.940 ","End":"04:18.385","Text":"Again, this is times 5 moles of oxygen for every 1 mole of propane."},{"Start":"04:18.385 ","End":"04:20.675","Text":"The moles of propane cancel out,"},{"Start":"04:20.675 ","End":"04:27.155","Text":"and this equals 16.25 moles of oxygen."},{"Start":"04:27.155 ","End":"04:31.610","Text":"Now we know the number of moles of oxygen and we have to calculate the volume of oxygen."},{"Start":"04:31.610 ","End":"04:34.490","Text":"But remember the oxygen is also measured at STP."},{"Start":"04:34.490 ","End":"04:39.770","Text":"This is the volume of oxygen equals the number of moles of oxygen times"},{"Start":"04:39.770 ","End":"04:46.575","Text":"22.4 liters for every 1 mole of oxygen."},{"Start":"04:46.575 ","End":"04:57.030","Text":"Moles of oxygen, which we calculated our 16.25 times 22.4 liters,"},{"Start":"04:57.030 ","End":"05:02.400","Text":"divided by 1 mole of oxygen."},{"Start":"05:02.400 ","End":"05:04.550","Text":"The moles of oxygen cancel out,"},{"Start":"05:04.550 ","End":"05:09.005","Text":"and this equals 364 liters."},{"Start":"05:09.005 ","End":"05:14.630","Text":"The volume of the oxygen is found to be 364 liters."},{"Start":"05:14.630 ","End":"05:15.965","Text":"That is our final answer."},{"Start":"05:15.965 ","End":"05:18.690","Text":"Thank you very much for watching."}],"ID":23833},{"Watched":false,"Name":"Exercise 11 - Alternative Solution","Duration":"2m 7s","ChapterTopicVideoID":22978,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.030 ","End":"00:07.200","Text":"What volume of oxygen gas is consumed in the combustion of 72.8"},{"Start":"00:07.200 ","End":"00:11.760","Text":"liters of propane gas if both gases are measured at STP?"},{"Start":"00:11.760 ","End":"00:16.740","Text":"To one of our previous video we calculated the volume of the oxygen by"},{"Start":"00:16.740 ","End":"00:21.600","Text":"converting the volume of the propane gas to number of moles,"},{"Start":"00:21.600 ","End":"00:23.490","Text":"and after calculating the number of moles of"},{"Start":"00:23.490 ","End":"00:27.870","Text":"propane gas we calculated the number of moles of the oxygen,"},{"Start":"00:27.870 ","End":"00:31.290","Text":"and then converted the number of moles of the oxygen to volume."},{"Start":"00:31.290 ","End":"00:34.455","Text":"That\u0027s what we did in the previous video."},{"Start":"00:34.455 ","End":"00:38.820","Text":"There is another way that we can calculate the volume of the oxygen,"},{"Start":"00:38.820 ","End":"00:41.325","Text":"and we\u0027re going to do that here. It\u0027s a very simple way."},{"Start":"00:41.325 ","End":"00:43.230","Text":"Now, since both gases,"},{"Start":"00:43.230 ","End":"00:46.620","Text":"the oxygen gas and the propane gas are both"},{"Start":"00:46.620 ","End":"00:51.290","Text":"measured at the same temperature and pressure since they\u0027re both measured at STP,"},{"Start":"00:51.290 ","End":"00:53.929","Text":"we can use the law of combining volumes."},{"Start":"00:53.929 ","End":"00:56.210","Text":"Now that means if we look at our reaction,"},{"Start":"00:56.210 ","End":"01:01.654","Text":"that 1 liter of propane gas reacts with 5 liters of oxygen."},{"Start":"01:01.654 ","End":"01:04.520","Text":"If we want to calculate the volume of the oxygen,"},{"Start":"01:04.520 ","End":"01:05.521","Text":"which is what we want to do, the volume of the oxygen gas,"},{"Start":"01:06.710 ","End":"01:11.760","Text":"it equals the volume of the propane gas,"},{"Start":"01:12.040 ","End":"01:17.310","Text":"times 5 liters of oxygen,"},{"Start":"01:17.780 ","End":"01:22.540","Text":"divided by 1 liter of propane gas."},{"Start":"01:23.780 ","End":"01:27.895","Text":"Again, we can use this equation and the law of combining volumes,"},{"Start":"01:27.895 ","End":"01:30.160","Text":"since both gases are measured at the same temperature and pressure."},{"Start":"01:30.160 ","End":"01:35.200","Text":"This equals the volume of the propane gas,"},{"Start":"01:35.200 ","End":"01:41.865","Text":"which equals 72.8 liters of propane,"},{"Start":"01:41.865 ","End":"01:48.700","Text":"times 5 liters of oxygen for every 1 liter of propane."},{"Start":"01:48.700 ","End":"01:52.008","Text":"The liters of propane cancel out,"},{"Start":"01:52.008 ","End":"01:58.480","Text":"and this equals 364 liters of oxygen."},{"Start":"01:58.480 ","End":"01:59.830","Text":"The volume of the oxygen,"},{"Start":"01:59.830 ","End":"02:03.775","Text":"which we calculated, equals 364 liters."},{"Start":"02:03.775 ","End":"02:05.200","Text":"That is our final answer."},{"Start":"02:05.200 ","End":"02:07.670","Text":"Thank you very much for watching."}],"ID":23832},{"Watched":false,"Name":"Exercise 12","Duration":"9m 37s","ChapterTopicVideoID":22980,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.910","Text":"Hey, we\u0027re going to solve the following exercise."},{"Start":"00:02.910 ","End":"00:07.320","Text":"A 2.49 grams sample of a potassium chloride,"},{"Start":"00:07.320 ","End":"00:13.470","Text":"potassium chlorate mixture is decomposed by heating and produces 212 milliliters of"},{"Start":"00:13.470 ","End":"00:19.890","Text":"oxygen gas measured at 24.5 degrees Celsius and 97 kilo pascal."},{"Start":"00:19.890 ","End":"00:23.445","Text":"What is the mass percent of potassium chloride in the mixture?"},{"Start":"00:23.445 ","End":"00:29.805","Text":"We have the equation, 2 potassium chlorate give us 2 potassium chloride plus 3 oxygen."},{"Start":"00:29.805 ","End":"00:33.585","Text":"The first step here is going to be to find the number of moles of oxygen."},{"Start":"00:33.585 ","End":"00:38.700","Text":"Then we will calculate the number of moles of the potassium chlorate from the oxygen,"},{"Start":"00:38.700 ","End":"00:43.565","Text":"and then we can calculate the mass percent of the potassium chlorate. Let\u0027s begin."},{"Start":"00:43.565 ","End":"00:45.770","Text":"In order to calculate the number of moles of the oxygen,"},{"Start":"00:45.770 ","End":"00:50.360","Text":"we\u0027re going to use the equation PV equals nRT."},{"Start":"00:50.360 ","End":"00:52.920","Text":"P is the pressure,"},{"Start":"00:54.220 ","End":"00:57.270","Text":"V is the volumes,"},{"Start":"00:57.800 ","End":"01:00.855","Text":"n is the number of moles,"},{"Start":"01:00.855 ","End":"01:04.120","Text":"R is the gas constant,"},{"Start":"01:05.350 ","End":"01:08.850","Text":"and T is the temperature."},{"Start":"01:11.380 ","End":"01:14.750","Text":"Now we\u0027re looking for the number of moles of the oxygen."},{"Start":"01:14.750 ","End":"01:17.330","Text":"Therefore, we\u0027re going to isolate the end the number of moles,"},{"Start":"01:17.330 ","End":"01:18.695","Text":"so n, the number of moles,"},{"Start":"01:18.695 ","End":"01:23.400","Text":"we can say that we\u0027re just going to divide both sides by RT,"},{"Start":"01:23.400 ","End":"01:25.400","Text":"and then we\u0027ll isolate the number of moles,"},{"Start":"01:25.400 ","End":"01:29.490","Text":"so equals PV divided by RT."},{"Start":"01:31.870 ","End":"01:37.190","Text":"We need the pressure and the pressure equals 97 kilopascal."},{"Start":"01:37.190 ","End":"01:39.960","Text":"I\u0027m just going to write that down,"},{"Start":"01:40.130 ","End":"01:42.390","Text":"times the volume,"},{"Start":"01:42.390 ","End":"01:50.630","Text":"and the volume is 212 milliliters divided by the gas constant,"},{"Start":"01:50.630 ","End":"01:51.890","Text":"which has a number of values."},{"Start":"01:51.890 ","End":"01:55.100","Text":"We\u0027re going to use 0.082,"},{"Start":"01:55.100 ","End":"02:01.010","Text":"and the units are atmospheres times liters divided by mole times kelvin."},{"Start":"02:01.010 ","End":"02:05.315","Text":"The temperature we have is 24.5 degrees Celsius."},{"Start":"02:05.315 ","End":"02:07.160","Text":"We need the temperature in Kelvin."},{"Start":"02:07.160 ","End":"02:15.090","Text":"Temperature in Kelvin equal temperature in degrees Celsius plus 273."},{"Start":"02:15.090 ","End":"02:23.080","Text":"Here we have 24.5 degrees Celsius plus 273,"},{"Start":"02:24.500 ","End":"02:31.080","Text":"and this equals 297.5 Kelvin."},{"Start":"02:31.080 ","End":"02:32.370","Text":"I\u0027m just going to write that down here,"},{"Start":"02:32.370 ","End":"02:37.955","Text":"the temperature is 297.5 Kelvin."},{"Start":"02:37.955 ","End":"02:40.295","Text":"Now if we look at our units,"},{"Start":"02:40.295 ","End":"02:44.700","Text":"we can see that in the numerator we have kilopascal and milliliters."},{"Start":"02:44.700 ","End":"02:48.590","Text":"We can see that in the denominator we have atmosphere and liters."},{"Start":"02:48.590 ","End":"02:52.385","Text":"We\u0027re just going to go ahead and multiply by conversion factors."},{"Start":"02:52.385 ","End":"02:55.660","Text":"Again, 97 kilopascal."},{"Start":"02:55.660 ","End":"02:59.075","Text":"We\u0027re going to convert the kilopascal into atmosphere."},{"Start":"02:59.075 ","End":"03:07.710","Text":"Just remember that 1 atmosphere equals 101.325 kilopascal."},{"Start":"03:07.710 ","End":"03:10.115","Text":"We\u0027re just going to multiply by a conversion factor,"},{"Start":"03:10.115 ","End":"03:17.245","Text":"1 atmosphere and 101.325 kilopascal."},{"Start":"03:17.245 ","End":"03:21.650","Text":"Now we\u0027re going to multiply again by the volume which is 212 milliliters."},{"Start":"03:21.650 ","End":"03:24.230","Text":"We\u0027re going to convert this to milliliters into"},{"Start":"03:24.230 ","End":"03:26.630","Text":"liters because we have leaders in the denominator."},{"Start":"03:26.630 ","End":"03:28.970","Text":"We\u0027re going to multiply by the conversion factor,"},{"Start":"03:28.970 ","End":"03:32.505","Text":"in 1 liter we have 1000 milliliters."},{"Start":"03:32.505 ","End":"03:34.820","Text":"We\u0027re going to go on, but before we go on you can see"},{"Start":"03:34.820 ","End":"03:37.140","Text":"here that we have kilopascal divided by kilopascal,"},{"Start":"03:37.140 ","End":"03:41.180","Text":"so we can just cancel those out and we can cancel our milliliters out."},{"Start":"03:41.180 ","End":"03:47.380","Text":"Everything here is divided by 0.082."},{"Start":"03:47.800 ","End":"03:55.280","Text":"Again, atmospheres times liters divided by mole times Kelvin times the temperature,"},{"Start":"03:55.280 ","End":"03:58.460","Text":"which is 297.5 Kelvin."},{"Start":"03:58.460 ","End":"04:02.510","Text":"Here we can also see in the denominator that we have times Kelvin divided by Kelvin,"},{"Start":"04:02.510 ","End":"04:05.150","Text":"so we can cancel that out too."},{"Start":"04:05.150 ","End":"04:11.350","Text":"This equals 8.32 times 10^negative 3."},{"Start":"04:11.350 ","End":"04:12.920","Text":"That\u0027s all the numbers."},{"Start":"04:12.920 ","End":"04:14.810","Text":"Now if we look at our units,"},{"Start":"04:14.810 ","End":"04:18.950","Text":"we can see that what we have left in the numerator is atmospheres times liters,"},{"Start":"04:18.950 ","End":"04:24.769","Text":"so atmospheres times liters divided by, in the denominator,"},{"Start":"04:24.769 ","End":"04:31.415","Text":"we have a fraction, we have atmospheres times liters divided by mole."},{"Start":"04:31.415 ","End":"04:34.440","Text":"Again, a quick reminder,"},{"Start":"04:35.930 ","End":"04:38.130","Text":"we divide by a fraction,"},{"Start":"04:38.130 ","End":"04:40.770","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"04:40.770 ","End":"04:50.020","Text":"We have atmosphere times liters times mole divided by atmospheres times liters."},{"Start":"04:50.020 ","End":"04:54.740","Text":"The atmosphere and the liters cancel out and we\u0027re left with moles."},{"Start":"04:54.740 ","End":"04:58.490","Text":"The number of moles of the oxygen equals"},{"Start":"04:58.490 ","End":"05:06.265","Text":"8.32 times 10 to the negative 3 moles."},{"Start":"05:06.265 ","End":"05:09.665","Text":"Now that we have the number of moles of the oxygen,"},{"Start":"05:09.665 ","End":"05:13.205","Text":"we\u0027re going to calculate the number of moles of the potassium chlorate."},{"Start":"05:13.205 ","End":"05:17.140","Text":"Now we can see that for every 2 moles of potassium chlorate which react,"},{"Start":"05:17.140 ","End":"05:19.965","Text":"you get 3 moles of oxygen."},{"Start":"05:19.965 ","End":"05:23.405","Text":"The number of moles of potassium chlorate"},{"Start":"05:23.405 ","End":"05:28.940","Text":"equal the number of moles of oxygen times 2 moles of"},{"Start":"05:28.940 ","End":"05:31.140","Text":"potassium chlorate"},{"Start":"05:35.570 ","End":"05:40.450","Text":"for every 3 moles of oxygen."},{"Start":"05:42.230 ","End":"05:50.346","Text":"The moles of oxygen equals 8.32 times 10^negative 3 mole,"},{"Start":"05:50.346 ","End":"05:54.630","Text":"again times 2 moles of"},{"Start":"05:54.630 ","End":"06:01.330","Text":"potassium chlorate divided by 3 moles of oxygen."},{"Start":"06:01.330 ","End":"06:05.660","Text":"The moles of oxygen cancel out and the number of moles of the potassium chlorate"},{"Start":"06:05.660 ","End":"06:14.500","Text":"equals 5.55 times 10^negative 3 moles."},{"Start":"06:16.040 ","End":"06:19.520","Text":"Now I\u0027m reminding you that we have a mixture."},{"Start":"06:19.520 ","End":"06:20.746","Text":"I can write that down here,"},{"Start":"06:20.746 ","End":"06:24.830","Text":"the potassium chloride and potassium chlorate,"},{"Start":"06:24.830 ","End":"06:30.785","Text":"and the mass of the mixture equals 2.49 grams."},{"Start":"06:30.785 ","End":"06:36.035","Text":"We are asked to find the mass percent of the potassium chlorate in this mixture."},{"Start":"06:36.035 ","End":"06:40.165","Text":"Now we calculated the number of moles of potassium chlorate."},{"Start":"06:40.165 ","End":"06:44.180","Text":"What we\u0027re going to do is calculate the mass of the potassium chlorate,"},{"Start":"06:44.180 ","End":"06:46.760","Text":"and then we\u0027ll calculate the mass percent of the potassium chlorate."},{"Start":"06:46.760 ","End":"06:49.130","Text":"In order to calculate the mass of the potassium chlorate,"},{"Start":"06:49.130 ","End":"06:50.869","Text":"we\u0027re going to use the equation,"},{"Start":"06:50.869 ","End":"06:55.970","Text":"the number of moles equals mass m divided by the molar mass."},{"Start":"06:55.970 ","End":"06:57.744","Text":"We\u0027re looking for the mass,"},{"Start":"06:57.744 ","End":"07:02.465","Text":"so the mass equals n times mw;"},{"Start":"07:02.465 ","End":"07:05.234","Text":"n times the molar mass."},{"Start":"07:05.234 ","End":"07:07.325","Text":"Number of moles times the molar mass."},{"Start":"07:07.325 ","End":"07:12.555","Text":"Now we know the number of moles of the potassium chlorate since we calculated them,"},{"Start":"07:12.555 ","End":"07:15.605","Text":"and now we have to calculate the molar mass of potassium chlorate."},{"Start":"07:15.605 ","End":"07:17.150","Text":"I\u0027m going to calculate that here,"},{"Start":"07:17.150 ","End":"07:22.430","Text":"the molar mass of potassium chlorate equals the molar mass of"},{"Start":"07:22.430 ","End":"07:30.799","Text":"potassium plus the molar mass of chlorine plus 3 times the molar mass of oxygen."},{"Start":"07:30.799 ","End":"07:32.750","Text":"We can see here there is 1 potassium,"},{"Start":"07:32.750 ","End":"07:34.580","Text":"1 chlorine, and 3 oxygens."},{"Start":"07:34.580 ","End":"07:40.370","Text":"This equals 39.1 grams per mole plus"},{"Start":"07:40.370 ","End":"07:49.650","Text":"35.45 grams per mole plus 3 times 16 grams per mole."},{"Start":"07:50.150 ","End":"07:56.150","Text":"This equals 122.55 grams per mole."},{"Start":"07:56.150 ","End":"07:58.930","Text":"That\u0027s the molar mass of the potassium chlorate."},{"Start":"07:58.930 ","End":"08:02.160","Text":"Now we know what the number of moles and the molar mass of potassium chloride,"},{"Start":"08:02.160 ","End":"08:05.225","Text":"and we\u0027re going to calculate the mass of potassium chlorate."},{"Start":"08:05.225 ","End":"08:11.430","Text":"Number of moles or 5.55 times 10^negative 3"},{"Start":"08:11.430 ","End":"08:19.895","Text":"mole times the molar mass which equals 122.55 grams per mole."},{"Start":"08:19.895 ","End":"08:25.250","Text":"We can see that the moles cancel out and this equals 0.68 grams."},{"Start":"08:25.250 ","End":"08:28.940","Text":"The mass of potassium chloride is 0.68 grams."},{"Start":"08:28.940 ","End":"08:31.820","Text":"Now we know that the mass of the mixture of the potassium chloride and"},{"Start":"08:31.820 ","End":"08:35.165","Text":"potassium chlorate equals 2.49 grams."},{"Start":"08:35.165 ","End":"08:43.460","Text":"The mass percent of the potassium chlorate in the mixture equals"},{"Start":"08:43.460 ","End":"08:51.950","Text":"the mass of potassium chlorate divided by the mass of the mixture,"},{"Start":"08:51.950 ","End":"08:59.830","Text":"which is potassium chloride plus potassium chlorate times 100 percent."},{"Start":"08:59.830 ","End":"09:03.830","Text":"This equals the mass of potassium chlorate,"},{"Start":"09:03.830 ","End":"09:08.930","Text":"which we found to be 0.68 grams,"},{"Start":"09:08.930 ","End":"09:18.810","Text":"divided by the mass of the mixture which is 2.49 grams, times 100 percent."},{"Start":"09:18.810 ","End":"09:20.460","Text":"The grams cancel out,"},{"Start":"09:20.460 ","End":"09:25.415","Text":"and this equals 27 percent."},{"Start":"09:25.415 ","End":"09:33.125","Text":"The mass percent of potassium chlorate in the mixture equals 27 percent."},{"Start":"09:33.125 ","End":"09:35.120","Text":"That is our final answer."},{"Start":"09:35.120 ","End":"09:37.830","Text":"Thank you very much for watching."}],"ID":23834},{"Watched":false,"Name":"Exercise 13","Duration":"6m 18s","ChapterTopicVideoID":22967,"CourseChapterTopicPlaylistID":80095,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.955","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.955 ","End":"00:06.780","Text":"Calculate the volume of hydrogen gas measured at 32-degrees Celsius in"},{"Start":"00:06.780 ","End":"00:12.540","Text":"722 Torr required to react with 32.7 liters of carbon monoxide gas,"},{"Start":"00:12.540 ","End":"00:16.005","Text":"measured at 5-degrees Celsius in 745 Torr,"},{"Start":"00:16.005 ","End":"00:19.320","Text":"3 carbon monoxide gas plus 7 hydrogen gas gives"},{"Start":"00:19.320 ","End":"00:22.530","Text":"us propane gas plus 3 water in the liquid state."},{"Start":"00:22.530 ","End":"00:24.720","Text":"In order to calculate the volume of the hydrogen,"},{"Start":"00:24.720 ","End":"00:28.140","Text":"we\u0027re first going to calculate the number of moles of the carbon monoxide gas,"},{"Start":"00:28.140 ","End":"00:29.790","Text":"then we\u0027ll calculate the number of moles of"},{"Start":"00:29.790 ","End":"00:33.090","Text":"hydrogen gas and finally the volume of the hydrogen gas."},{"Start":"00:33.090 ","End":"00:36.270","Text":"In order to calculate the number of moles of the carbon monoxide,"},{"Start":"00:36.270 ","End":"00:39.300","Text":"we\u0027re going to use the ideal gas equation, which is PV,"},{"Start":"00:39.300 ","End":"00:42.670","Text":"pressure times volume equals nRT,"},{"Start":"00:42.670 ","End":"00:44.030","Text":"n is the number of moles,"},{"Start":"00:44.030 ","End":"00:45.470","Text":"R is the gas constant,"},{"Start":"00:45.470 ","End":"00:46.625","Text":"and T is the temperature."},{"Start":"00:46.625 ","End":"00:48.110","Text":"Since we want to calculate n,"},{"Start":"00:48.110 ","End":"00:50.075","Text":"we\u0027re going to divide both sides by RT."},{"Start":"00:50.075 ","End":"00:56.405","Text":"The number of moles of the carbon monoxide equal PV divided by RT."},{"Start":"00:56.405 ","End":"01:01.530","Text":"The pressure of the carbon monoxide gas is 745 Torr."},{"Start":"01:04.570 ","End":"01:09.349","Text":"The volume of the carbon monoxide gas is 32.7 liters,"},{"Start":"01:09.349 ","End":"01:16.740","Text":"and this is divided by R,"},{"Start":"01:16.740 ","End":"01:17.925","Text":"which is the gas constant."},{"Start":"01:17.925 ","End":"01:22.640","Text":"We\u0027re going to use 0.082 atmospheric times"},{"Start":"01:22.640 ","End":"01:27.725","Text":"liters divided by mole times Kelvin times the temperature."},{"Start":"01:27.725 ","End":"01:29.390","Text":"The temperature is 5-degrees Celsius,"},{"Start":"01:29.390 ","End":"01:31.715","Text":"and we need the temperature in Kelvin."},{"Start":"01:31.715 ","End":"01:38.525","Text":"Temperature in Kelvin equals temperature in degrees Celsius plus 273."},{"Start":"01:38.525 ","End":"01:47.580","Text":"This equals 5-degrees Celsius plus 273, n=278 Kelvin."},{"Start":"01:47.580 ","End":"01:49.880","Text":"This is times 278 Kelvin."},{"Start":"01:49.880 ","End":"01:52.385","Text":"We can already see that our Kelvins cancel out."},{"Start":"01:52.385 ","End":"01:54.650","Text":"Also, since we have Torr and atmosphere,"},{"Start":"01:54.650 ","End":"01:57.500","Text":"we have Torr in the numerator and atmosphere in the denominator,"},{"Start":"01:57.500 ","End":"01:59.660","Text":"we\u0027re going to multiply by a conversion factor."},{"Start":"01:59.660 ","End":"02:06.935","Text":"We\u0027re just going to multiply this by 1 atmosphere divided by 760 Torr."},{"Start":"02:06.935 ","End":"02:09.580","Text":"The Torrs cancel out."},{"Start":"02:09.580 ","End":"02:12.930","Text":"This equals 1.41,"},{"Start":"02:12.930 ","End":"02:14.300","Text":"and if we look at our units,"},{"Start":"02:14.300 ","End":"02:23.490","Text":"we have liters times atmosphere divided by liters times atmosphere,"},{"Start":"02:23.490 ","End":"02:29.215","Text":"or atmospheres times liters divided by mole."},{"Start":"02:29.215 ","End":"02:31.910","Text":"Now remember that when you\u0027re dividing by a fraction,"},{"Start":"02:31.910 ","End":"02:34.895","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"02:34.895 ","End":"02:37.250","Text":"I\u0027m going to write this again,"},{"Start":"02:37.250 ","End":"02:45.370","Text":"atmosphere times liters divided by atmosphere times liters divided by mole."},{"Start":"02:45.740 ","End":"02:56.915","Text":"This equals atmosphere times liters times mole divided by atmosphere times liters."},{"Start":"02:56.915 ","End":"03:01.685","Text":"Atmosphere times liters cancel out and we\u0027re left with mole."},{"Start":"03:01.685 ","End":"03:05.635","Text":"This equals 1.41 mole."},{"Start":"03:05.635 ","End":"03:10.745","Text":"Again, the number of moles of the carbon monoxide equals 1.41 mole."},{"Start":"03:10.745 ","End":"03:12.740","Text":"The number of moles of carbon monoxide,"},{"Start":"03:12.740 ","End":"03:14.975","Text":"which we calculated equals 1.41 mole."},{"Start":"03:14.975 ","End":"03:17.960","Text":"Now we\u0027re going to calculate the number of moles of the hydrogen gas."},{"Start":"03:17.960 ","End":"03:20.390","Text":"The number of moles of"},{"Start":"03:20.390 ","End":"03:23.840","Text":"the hydrogen gas equals the number of moles of the carbon monoxide."},{"Start":"03:23.840 ","End":"03:25.290","Text":"If we look at our reaction,"},{"Start":"03:25.290 ","End":"03:29.045","Text":"we can see that for every 3 moles of carbon monoxide which react,"},{"Start":"03:29.045 ","End":"03:32.435","Text":"we have 7 moles of hydrogen which reacts."},{"Start":"03:32.435 ","End":"03:43.050","Text":"This is times 7 moles of hydrogen divided by 3 moles of carbon monoxide."},{"Start":"03:43.660 ","End":"03:46.429","Text":"Again, the carbon monoxide, which we calculated,"},{"Start":"03:46.429 ","End":"03:48.470","Text":"the number of moles is 1.41 mole."},{"Start":"03:48.470 ","End":"03:55.865","Text":"This is 1.41 mole of carbon monoxide times 7 moles of hydrogen,"},{"Start":"03:55.865 ","End":"03:59.790","Text":"divided by 3 moles of carbon monoxide."},{"Start":"03:59.790 ","End":"04:02.060","Text":"The moles of carbon monoxide cancel out."},{"Start":"04:02.060 ","End":"04:07.890","Text":"This equals 3.29 moles of hydrogen."},{"Start":"04:08.060 ","End":"04:12.520","Text":"Again, the number of moles of hydrogen are 3.29 moles."},{"Start":"04:12.520 ","End":"04:16.435","Text":"Now in order to calculate the volume of the hydrogen gas,"},{"Start":"04:16.435 ","End":"04:21.175","Text":"again, we\u0027re going to use the ideal gas equation, PV=nRT."},{"Start":"04:21.175 ","End":"04:23.230","Text":"However, this time we want the volume,"},{"Start":"04:23.230 ","End":"04:25.285","Text":"so we\u0027re going to divide both sides by the pressure."},{"Start":"04:25.285 ","End":"04:30.250","Text":"The volume of the hydrogen gas equals nRT divided by"},{"Start":"04:30.250 ","End":"04:36.380","Text":"P. The number of moles of the hydrogen is 3.29 mole."},{"Start":"04:37.910 ","End":"04:48.550","Text":"Again times r, which is 0.082 atmosphere times liters divided by mole Kelvin."},{"Start":"04:48.550 ","End":"04:52.565","Text":"We can already cancel out the moles times the temperature."},{"Start":"04:52.565 ","End":"04:56.030","Text":"The temperature of the hydrogen is 32-degrees Celsius."},{"Start":"04:56.030 ","End":"05:04.725","Text":"Again, the temperature in Kelvin equals the temperature in degrees Celsius plus 273."},{"Start":"05:04.725 ","End":"05:09.845","Text":"This equals 32 plus 273,"},{"Start":"05:09.845 ","End":"05:13.010","Text":"which equals 305 Kelvin."},{"Start":"05:13.010 ","End":"05:17.705","Text":"This is times the temperature times 305 Kelvin."},{"Start":"05:17.705 ","End":"05:20.450","Text":"All of this is divided by the pressure,"},{"Start":"05:20.450 ","End":"05:24.080","Text":"and the pressure of the hydrogen gas is 722 Torrs."},{"Start":"05:24.080 ","End":"05:31.800","Text":"This is divided by 722 Torr."},{"Start":"05:32.510 ","End":"05:36.040","Text":"Again, we have atmosphere and Torrs."},{"Start":"05:36.040 ","End":"05:38.345","Text":"We\u0027re going to multiply by a conversion factor."},{"Start":"05:38.345 ","End":"05:44.690","Text":"Multiply by 1 atmosphere for every 760 torr."},{"Start":"05:44.690 ","End":"05:46.965","Text":"Now, Torrs cancel out."},{"Start":"05:46.965 ","End":"05:50.730","Text":"This equals 86.6."},{"Start":"05:50.730 ","End":"05:52.310","Text":"If we look back at our units,"},{"Start":"05:52.310 ","End":"05:54.170","Text":"we can also see that the Kelvins cancel out,"},{"Start":"05:54.170 ","End":"05:55.625","Text":"so we\u0027re just going to cancel them out."},{"Start":"05:55.625 ","End":"06:03.450","Text":"In our units we have atmosphere times liters divided by atmosphere."},{"Start":"06:03.970 ","End":"06:09.920","Text":"The atmosphere cancel out and we\u0027re left with 86.6 liters."},{"Start":"06:09.920 ","End":"06:11.570","Text":"The volume of the hydrogen gas,"},{"Start":"06:11.570 ","End":"06:14.795","Text":"which we calculated equals 86.6 liters."},{"Start":"06:14.795 ","End":"06:16.430","Text":"That is our final answer."},{"Start":"06:16.430 ","End":"06:18.900","Text":"Thank you very much for watching."}],"ID":23821}],"Thumbnail":null,"ID":80095},{"Name":"Mixtures of Gases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Dalton\u0027s Law of Partial Pressures - Theory","Duration":"4m 11s","ChapterTopicVideoID":17416,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/17416.jpeg","UploadDate":"2019-03-05T05:59:06.6200000","DurationForVideoObject":"PT4M11S","Description":null,"MetaTitle":"Dalton\u0027s Law of Partial Pressures - Theory: Video + Workbook | Proprep","MetaDescription":"Gases - Mixtures of Gases. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/general-chemistry/gases/mixtures-of-gases/vid18159","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.110","Text":"In previous videos, we talked about"},{"Start":"00:02.110 ","End":"00:05.710","Text":"the gas laws for the case of an ideal gas in a container."},{"Start":"00:05.710 ","End":"00:08.365","Text":"In this video, we\u0027ll discuss a case when"},{"Start":"00:08.365 ","End":"00:11.845","Text":"more than one gas is present in the same container."},{"Start":"00:11.845 ","End":"00:16.075","Text":"We\u0027re going to talk about Dalton\u0027s law of partial pressures."},{"Start":"00:16.075 ","End":"00:20.170","Text":"The first thing to realize that when we have several gases in"},{"Start":"00:20.170 ","End":"00:25.555","Text":"the same container is that each gas behaves as if it is alone in the container."},{"Start":"00:25.555 ","End":"00:28.480","Text":"This is because the distances between the molecules are very"},{"Start":"00:28.480 ","End":"00:32.860","Text":"large so that 1 gas doesn\u0027t feel the presence of the other."},{"Start":"00:32.860 ","End":"00:37.510","Text":"Dalton\u0027s Law of partial pressures says the total pressure of a mixture of"},{"Start":"00:37.510 ","End":"00:42.845","Text":"gases is the sum of the partial pressures of each of the gases."},{"Start":"00:42.845 ","End":"00:45.020","Text":"Let\u0027s take an example."},{"Start":"00:45.020 ","End":"00:48.815","Text":"Let\u0027s consider 2 different gases in the same container."},{"Start":"00:48.815 ","End":"00:52.790","Text":"First, we write the gas laws for them."},{"Start":"00:52.790 ","End":"00:58.745","Text":"Remember that the ideal gas law is PV=nRT."},{"Start":"00:58.745 ","End":"01:00.980","Text":"For the first gas,"},{"Start":"01:00.980 ","End":"01:02.395","Text":"which we\u0027ll call A,"},{"Start":"01:02.395 ","End":"01:07.250","Text":"P_A is equal to n_A times RT divided by"},{"Start":"01:07.250 ","End":"01:12.380","Text":"V. We\u0027ve just taken PV=nRT and divided by V on both sides,"},{"Start":"01:12.380 ","End":"01:18.480","Text":"so we get P is equal to nRT divided by V. For the second gas,"},{"Start":"01:18.480 ","End":"01:21.780","Text":"which we\u0027ll call B, P_B is equal to N_B"},{"Start":"01:21.780 ","End":"01:26.405","Text":"times RT over V. These are called partial pressures."},{"Start":"01:26.405 ","End":"01:28.845","Text":"Now let\u0027s consider the total pressure."},{"Start":"01:28.845 ","End":"01:30.765","Text":"According to Dalton\u0027s law,"},{"Start":"01:30.765 ","End":"01:33.860","Text":"the total pressure is the sum of the partial pressures."},{"Start":"01:33.860 ","End":"01:37.145","Text":"P_total is P_A plus P_B."},{"Start":"01:37.145 ","End":"01:40.415","Text":"Now let\u0027s write the expressions for P_A and P_B."},{"Start":"01:40.415 ","End":"01:49.190","Text":"P_A is n_A times RT over V and PB is n_B times RT over V. We can take"},{"Start":"01:49.190 ","End":"01:58.045","Text":"the RT over V out and write n_A plus n_B times RT over V. N_A plus n_B,"},{"Start":"01:58.045 ","End":"02:00.205","Text":"we\u0027ll write as n_total."},{"Start":"02:00.205 ","End":"02:06.085","Text":"N_A plus n_B equal to n_total,"},{"Start":"02:06.085 ","End":"02:08.960","Text":"which I\u0027ve abbreviated as tot."},{"Start":"02:08.960 ","End":"02:13.960","Text":"Now we know that P_total is equal to n_total times RT over"},{"Start":"02:13.960 ","End":"02:19.510","Text":"V. Now let\u0027s express the partial pressure in terms of the total pressure."},{"Start":"02:19.510 ","End":"02:23.350","Text":"What\u0027s the relationship between the partial pressure and the total pressure?"},{"Start":"02:23.350 ","End":"02:28.065","Text":"In doing so, we\u0027ll introduce another concept called the mole fraction."},{"Start":"02:28.065 ","End":"02:36.507","Text":"We know that P_A is n_A times RT over V and P_total is n_total times RT over V,"},{"Start":"02:36.507 ","End":"02:38.800","Text":"average in n total distance."},{"Start":"02:38.800 ","End":"02:43.825","Text":"Now, we see the RT over V cancels with RT over V,"},{"Start":"02:43.825 ","End":"02:46.765","Text":"and we\u0027re left with n_A over n_total."},{"Start":"02:46.765 ","End":"02:49.075","Text":"This is called the mole fraction."},{"Start":"02:49.075 ","End":"02:52.030","Text":"N_A over n_total is equal to Chi,"},{"Start":"02:52.030 ","End":"02:54.420","Text":"the Greek letter Chi_A."},{"Start":"02:54.420 ","End":"02:58.030","Text":"This is the mole fraction of A."},{"Start":"02:58.030 ","End":"03:03.845","Text":"Now we can write P_A is equal to Chi_A times P_total."},{"Start":"03:03.845 ","End":"03:05.727","Text":"This is a fraction,"},{"Start":"03:05.727 ","End":"03:10.265","Text":"this is the partial pressure in terms of the total pressure."},{"Start":"03:10.265 ","End":"03:12.320","Text":"Now we can do the same thing for B,"},{"Start":"03:12.320 ","End":"03:16.355","Text":"P_B is equal to Chi_B times P_total."},{"Start":"03:16.355 ","End":"03:23.375","Text":"We have P_A equal to Chi_A times P_total and P_B equal to Chi_B times P_total."},{"Start":"03:23.375 ","End":"03:27.890","Text":"Chi_A and Chi_B are called the mole fractions and when we sum them,"},{"Start":"03:27.890 ","End":"03:29.775","Text":"we\u0027ll see that we get 1."},{"Start":"03:29.775 ","End":"03:33.015","Text":"Chi_A plus Chi_B,"},{"Start":"03:33.015 ","End":"03:35.985","Text":"Chi_A is n_A over n total."},{"Start":"03:35.985 ","End":"03:38.940","Text":"Chi_B is n_B over n_total."},{"Start":"03:38.940 ","End":"03:43.785","Text":"We sum the 2, we get n_A plus n_B over n_total."},{"Start":"03:43.785 ","End":"03:46.710","Text":"N_A plus n_B is n_total,"},{"Start":"03:46.710 ","End":"03:49.850","Text":"so n_total over n_total is equal to 1."},{"Start":"03:49.850 ","End":"03:53.060","Text":"The sum of the mole fractions is equal to 1."},{"Start":"03:53.060 ","End":"03:56.554","Text":"It doesn\u0027t matter how many we have if we have 3 gases,"},{"Start":"03:56.554 ","End":"04:02.360","Text":"the Chi_A plus Chi_B plus Chi_C equal to 1."},{"Start":"04:02.360 ","End":"04:06.935","Text":"In this video, we discuss Dalton\u0027s law of partial pressures."},{"Start":"04:06.935 ","End":"04:08.210","Text":"In the next video,"},{"Start":"04:08.210 ","End":"04:11.460","Text":"we\u0027ll solve a problem based on it."}],"ID":18159},{"Watched":false,"Name":"Dalton\u0027s Law of Partial Pressures - Example","Duration":"4m 41s","ChapterTopicVideoID":17417,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"In the previous video,"},{"Start":"00:01.830 ","End":"00:05.595","Text":"we discussed the theory of Dalton\u0027s Law of Partial Pressures."},{"Start":"00:05.595 ","End":"00:09.645","Text":"In this video, we will solve a problem based on it."},{"Start":"00:09.645 ","End":"00:11.115","Text":"Here\u0027s our problem."},{"Start":"00:11.115 ","End":"00:19.860","Text":"Suppose we have 1.00 moles of helium and 2.50 moles of hydrogen in the same container."},{"Start":"00:19.860 ","End":"00:21.450","Text":"If the volume of the container is"},{"Start":"00:21.450 ","End":"00:26.460","Text":"5.00 liters and the temperature is 100-degrees Celsius,"},{"Start":"00:26.460 ","End":"00:30.660","Text":"find the partial and total pressures in atmospheres."},{"Start":"00:30.660 ","End":"00:34.950","Text":"The number of moles of helium we\u0027re told is 1.00 moles,"},{"Start":"00:34.950 ","End":"00:41.145","Text":"and the number of moles of hydrogen is 2.50 moles so the total number of moles,"},{"Start":"00:41.145 ","End":"00:44.945","Text":"n_total is 3.50 moles."},{"Start":"00:44.945 ","End":"00:47.345","Text":"Now we can work out the mole fraction."},{"Start":"00:47.345 ","End":"00:50.105","Text":"Pi, which is the mole fraction of helium,"},{"Start":"00:50.105 ","End":"00:53.210","Text":"is 1 mole divided by 3.50 moles."},{"Start":"00:53.210 ","End":"00:57.425","Text":"So number of moles of helium divided by the total number of moles."},{"Start":"00:57.425 ","End":"01:03.700","Text":"The moles cancel, and that 1 over 3.50 is clearly 2 over 7."},{"Start":"01:03.700 ","End":"01:09.110","Text":"Now the number of moles of hydrogen is 1 because the total of"},{"Start":"01:09.110 ","End":"01:14.925","Text":"the mole fractions is 1 minus the whole fraction of helium is 2 over 7,"},{"Start":"01:14.925 ","End":"01:17.560","Text":"giving us 5 over 7."},{"Start":"01:17.560 ","End":"01:21.770","Text":"Now we know that the mole fraction of helium is 2 over 7,"},{"Start":"01:21.770 ","End":"01:25.340","Text":"and the mole fraction of hydrogen is 5 over 7."},{"Start":"01:25.340 ","End":"01:31.145","Text":"Now we will need the volume of RT divided by V so let\u0027s calculate it here."},{"Start":"01:31.145 ","End":"01:36.080","Text":"R is 0.082 atmospheres times liter per"},{"Start":"01:36.080 ","End":"01:41.495","Text":"mole per K and temperature is 100-degrees Celsius,"},{"Start":"01:41.495 ","End":"01:46.475","Text":"which is a 100 plus 273 kelvin."},{"Start":"01:46.475 ","End":"01:49.395","Text":"That\u0027s 373 kelvin,"},{"Start":"01:49.395 ","End":"01:52.930","Text":"so here we have 373 kelvin."},{"Start":"01:52.930 ","End":"01:57.860","Text":"The volume we\u0027re told in the problem is 5.00 liters."},{"Start":"01:57.860 ","End":"01:59.630","Text":"Now let\u0027s see what cancels."},{"Start":"01:59.630 ","End":"02:01.795","Text":"Liter cancels with liter."},{"Start":"02:01.795 ","End":"02:06.665","Text":"K^minus 1 times K is just 1 so we\u0027re left with"},{"Start":"02:06.665 ","End":"02:12.260","Text":"atmospheres per mole and if we work out the numbers,"},{"Start":"02:12.260 ","End":"02:16.100","Text":"we get 6.12 atmospheres per mole."},{"Start":"02:16.100 ","End":"02:18.890","Text":"Now we can calculate the partial pressure of helium."},{"Start":"02:18.890 ","End":"02:23.780","Text":"We know it\u0027s equal to the number of moles of helium times RT"},{"Start":"02:23.780 ","End":"02:28.550","Text":"over V. The number of moles of helium is 1.00 moles and"},{"Start":"02:28.550 ","End":"02:33.290","Text":"RT over V we\u0027ve just calculated to be 6.12 atmospheres per"},{"Start":"02:33.290 ","End":"02:38.710","Text":"mole so mole times mole^minus 1 is just 1."},{"Start":"02:38.710 ","End":"02:42.785","Text":"We\u0027re left with 6.12 atmospheres."},{"Start":"02:42.785 ","End":"02:48.720","Text":"Now we have the partial pressure of helium, 6.12 atmospheres."},{"Start":"02:48.720 ","End":"02:53.945","Text":"The partial pressure of hydrogen is the number of moles of hydrogen times RT"},{"Start":"02:53.945 ","End":"02:59.525","Text":"over V. The number of moles of hydrogen is 2.50 moles and again,"},{"Start":"02:59.525 ","End":"03:03.755","Text":"RT divided by V is 6.12 atmospheres per mole."},{"Start":"03:03.755 ","End":"03:07.315","Text":"Again, mole times mole^minus 1 is 1."},{"Start":"03:07.315 ","End":"03:09.530","Text":"We\u0027re left just with atmospheres."},{"Start":"03:09.530 ","End":"03:11.240","Text":"When we multiply out,"},{"Start":"03:11.240 ","End":"03:15.480","Text":"we get 15.3 atmospheres."},{"Start":"03:16.610 ","End":"03:19.925","Text":"Now we can calculate the total pressure."},{"Start":"03:19.925 ","End":"03:24.770","Text":"It\u0027s a sum of 6.12 and 15.3,"},{"Start":"03:24.770 ","End":"03:28.610","Text":"and that gives us 21.42 atmospheres."},{"Start":"03:28.610 ","End":"03:31.670","Text":"Now we have the partial pressures and the total pressure."},{"Start":"03:31.670 ","End":"03:33.770","Text":"Now we could have done this a different way."},{"Start":"03:33.770 ","End":"03:36.995","Text":"We could have calculated the total pressure first."},{"Start":"03:36.995 ","End":"03:40.550","Text":"The total pressure is equal total number of moles,"},{"Start":"03:40.550 ","End":"03:42.800","Text":"which is 3.50 moles."},{"Start":"03:42.800 ","End":"03:45.725","Text":"Again, times RT divided by V,"},{"Start":"03:45.725 ","End":"03:49.620","Text":"which we know is 6.12 atmospheres per mole."},{"Start":"03:49.620 ","End":"03:52.125","Text":"Mole times mole^minus 1 is 1,"},{"Start":"03:52.125 ","End":"03:57.425","Text":"3.50 times 6.12 is 21.42 atmospheres,"},{"Start":"03:57.425 ","End":"03:59.495","Text":"which is the same as we got before."},{"Start":"03:59.495 ","End":"04:01.805","Text":"Now we can calculate the partial pressures."},{"Start":"04:01.805 ","End":"04:07.865","Text":"The partial pressure of helium is the mole fraction of helium times P_total."},{"Start":"04:07.865 ","End":"04:09.540","Text":"That\u0027s 2/7."},{"Start":"04:09.540 ","End":"04:13.460","Text":"We calculated that the mole fraction of helium is 2 over 7,"},{"Start":"04:13.460 ","End":"04:18.830","Text":"2 over 7 times 21.42 atmospheres is 6.12 atmospheres,"},{"Start":"04:18.830 ","End":"04:21.590","Text":"just the same as we had before and the pressure of"},{"Start":"04:21.590 ","End":"04:26.345","Text":"hydrogen is the mole fraction of hydrogen times P_total."},{"Start":"04:26.345 ","End":"04:32.660","Text":"That\u0027s 5/7 times 21.42 atmospheres giving us 50.3 atmospheres,"},{"Start":"04:32.660 ","End":"04:35.375","Text":"which is again exactly what we got before."},{"Start":"04:35.375 ","End":"04:41.760","Text":"We found 2 ways of calculating the partial pressures and the total pressure."}],"ID":18160},{"Watched":false,"Name":"Collecting Gas over Water","Duration":"5m 52s","ChapterTopicVideoID":17418,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"In the 2 previous videos,"},{"Start":"00:01.935 ","End":"00:04.800","Text":"we learned about Dalton\u0027s law of partial pressures."},{"Start":"00:04.800 ","End":"00:09.195","Text":"In this video, we\u0027ll discuss the collection of gas over water,"},{"Start":"00:09.195 ","End":"00:12.270","Text":"which is an application of Dalton\u0027s Law."},{"Start":"00:12.270 ","End":"00:16.545","Text":"So we\u0027re going to talk about collected gas over water. Here\u0027s a picture."},{"Start":"00:16.545 ","End":"00:19.800","Text":"Here we have a container of water and a bottle"},{"Start":"00:19.800 ","End":"00:23.280","Text":"that was containing water that\u0027s been inverted."},{"Start":"00:23.280 ","End":"00:26.954","Text":"Of course, it\u0027ll be held still by some clamp."},{"Start":"00:26.954 ","End":"00:31.118","Text":"Now, oxygen gas is flowing in this tube,"},{"Start":"00:31.118 ","End":"00:36.335","Text":"and it goes through the water and accumulates at the top of the bottom."},{"Start":"00:36.335 ","End":"00:38.375","Text":"This is oxygen gas."},{"Start":"00:38.375 ","End":"00:42.800","Text":"However, the oxygen gas has been flowing through the water."},{"Start":"00:42.800 ","End":"00:44.635","Text":"So it\u0027s now wet,"},{"Start":"00:44.635 ","End":"00:47.435","Text":"and it contains water vapor."},{"Start":"00:47.435 ","End":"00:51.830","Text":"Here we have, the gas at the top consists of oxygen and water vapor."},{"Start":"00:51.830 ","End":"00:58.315","Text":"We want just to know about the pressure of the oxygen without the water vapor."},{"Start":"00:58.315 ","End":"01:02.360","Text":"The first thing to do is to adjust the position of the bottle,"},{"Start":"01:02.360 ","End":"01:05.975","Text":"so the height of the water inside and outside the bottle is the same."},{"Start":"01:05.975 ","End":"01:08.510","Text":"Now, the height of the water in"},{"Start":"01:08.510 ","End":"01:12.170","Text":"the container and the height of the water in the bottle should be the same."},{"Start":"01:12.170 ","End":"01:15.515","Text":"So the pressure of the atmosphere pushing down"},{"Start":"01:15.515 ","End":"01:19.250","Text":"should be exactly equal to the pressure of the gas pushing down."},{"Start":"01:19.250 ","End":"01:26.210","Text":"So P-bar is equal to the pressure of the oxygen and the pressure of the water vapor."},{"Start":"01:26.210 ","End":"01:30.469","Text":"The second thing we need to do is to measure the volume of the gas."},{"Start":"01:30.469 ","End":"01:34.190","Text":"Now we can apply Dalton\u0027s law of partial pressure."},{"Start":"01:34.190 ","End":"01:37.280","Text":"The atmospheric pressure P-bar is equal to"},{"Start":"01:37.280 ","End":"01:43.660","Text":"the partial pressure of the oxygen and the partial pressure of the water vapor."},{"Start":"01:43.660 ","End":"01:45.410","Text":"The pressure of the oxygen,"},{"Start":"01:45.410 ","End":"01:47.135","Text":"which is what we want to find,"},{"Start":"01:47.135 ","End":"01:53.045","Text":"is equal to the atmospheric pressure minus the pressure of the water,"},{"Start":"01:53.045 ","End":"01:55.865","Text":"which is called vapor pressure of water."},{"Start":"01:55.865 ","End":"01:57.649","Text":"Let\u0027s take an example."},{"Start":"01:57.649 ","End":"02:00.350","Text":"Oxygen gas is collected over water at"},{"Start":"02:00.350 ","End":"02:06.860","Text":"25-degrees Celsius and barometric pressure of 756 millimeters of mercury."},{"Start":"02:06.860 ","End":"02:13.640","Text":"The vapor pressure of water at that temperature is 23.8 millimeters of mercury."},{"Start":"02:13.640 ","End":"02:18.275","Text":"That\u0027s something we can find in tables and books."},{"Start":"02:18.275 ","End":"02:22.460","Text":"The volume of gas collected is a 100 milliliters."},{"Start":"02:22.460 ","End":"02:25.495","Text":"What is the mass of oxygen collected?"},{"Start":"02:25.495 ","End":"02:29.975","Text":"We\u0027re going to apply the ideal gas law."},{"Start":"02:29.975 ","End":"02:33.440","Text":"Let\u0027s see the data that must go into the law."},{"Start":"02:33.440 ","End":"02:35.780","Text":"First, the pressure of the oxygen is"},{"Start":"02:35.780 ","End":"02:39.140","Text":"the barometric pressure minus the pressure of the water."},{"Start":"02:39.140 ","End":"02:44.825","Text":"That\u0027s 756 minus 23.8 millimeters of mercury,"},{"Start":"02:44.825 ","End":"02:48.545","Text":"which gives us 732 millimeters of mercury."},{"Start":"02:48.545 ","End":"02:51.020","Text":"Now in order to apply the ideal gas law,"},{"Start":"02:51.020 ","End":"02:53.060","Text":"we need that in atmospheres."},{"Start":"02:53.060 ","End":"02:55.790","Text":"So 732 millimeters of mercury,"},{"Start":"02:55.790 ","End":"03:01.505","Text":"732 millimeters of mercury times this conversion factor,"},{"Start":"03:01.505 ","End":"03:07.835","Text":"which tells us that 1 atmosphere is equivalent to 760 millimeters of mercury."},{"Start":"03:07.835 ","End":"03:10.370","Text":"Millimeters of mercury cancels,"},{"Start":"03:10.370 ","End":"03:12.895","Text":"and we\u0027re left with atmospheres."},{"Start":"03:12.895 ","End":"03:17.325","Text":"When we divide 732 by 760,"},{"Start":"03:17.325 ","End":"03:19.860","Text":"we get 0.963,"},{"Start":"03:19.860 ","End":"03:22.625","Text":"and the units are atmospheres."},{"Start":"03:22.625 ","End":"03:28.070","Text":"Now we have the pressure of oxygen is 0.963 atmospheres."},{"Start":"03:28.070 ","End":"03:30.499","Text":"What about the rest of the data required?"},{"Start":"03:30.499 ","End":"03:34.310","Text":"The temperature is 25-degrees Celsius."},{"Start":"03:34.310 ","End":"03:35.510","Text":"We want that in Kelvin,"},{"Start":"03:35.510 ","End":"03:37.670","Text":"so we have to add 273,"},{"Start":"03:37.670 ","End":"03:39.905","Text":"we get 298 Kelvin."},{"Start":"03:39.905 ","End":"03:42.185","Text":"Since our pressure is in atmosphere,"},{"Start":"03:42.185 ","End":"03:50.760","Text":"we take the volume of R as 0.082 atmospheres times liters per mole per K. Of course,"},{"Start":"03:50.760 ","End":"03:56.113","Text":"the volume has to be in liters,"},{"Start":"03:56.113 ","End":"03:59.055","Text":"because R has liters in it."},{"Start":"03:59.055 ","End":"04:01.700","Text":"The volume is 0.1 liters,"},{"Start":"04:01.700 ","End":"04:05.135","Text":"that\u0027s a 100 milliliters, is 0.1 liters."},{"Start":"04:05.135 ","End":"04:09.350","Text":"Now we can calculate the number of moles of oxygen."},{"Start":"04:09.350 ","End":"04:17.635","Text":"Number of moles of oxygen is PV divided by RT because we have the gas law, PV=nRT."},{"Start":"04:17.635 ","End":"04:20.840","Text":"The pressure is 0.963 atmospheres."},{"Start":"04:20.840 ","End":"04:23.180","Text":"The volume is 0.1 liters."},{"Start":"04:23.180 ","End":"04:28.220","Text":"R is 0.082 atmospheres times liters per mole per K,"},{"Start":"04:28.220 ","End":"04:31.345","Text":"and the temperature is 298 Kelvin."},{"Start":"04:31.345 ","End":"04:33.075","Text":"Now we can look at the units,"},{"Start":"04:33.075 ","End":"04:34.740","Text":"atmosphere goes with atmosphere,"},{"Start":"04:34.740 ","End":"04:36.085","Text":"liter with liter,"},{"Start":"04:36.085 ","End":"04:38.990","Text":"K to the power minus 1 times K is 1,"},{"Start":"04:38.990 ","End":"04:44.540","Text":"and we\u0027re left with 1 over moles to the power minus 1,"},{"Start":"04:44.540 ","End":"04:46.415","Text":"which is just moles."},{"Start":"04:46.415 ","End":"04:48.319","Text":"After we do the arithmetic,"},{"Start":"04:48.319 ","End":"04:53.630","Text":"we have 3.94 times 10 power minus 3 moles."},{"Start":"04:53.630 ","End":"04:57.634","Text":"That\u0027s the number of moles of oxygen."},{"Start":"04:57.634 ","End":"05:01.265","Text":"We\u0027re asked to calculate the mass."},{"Start":"05:01.265 ","End":"05:06.410","Text":"We know the relationship between the mass and the number of moles."},{"Start":"05:06.410 ","End":"05:10.075","Text":"The mass is simply the number of moles times the molar mass."},{"Start":"05:10.075 ","End":"05:16.125","Text":"The number of moles is 3.94 times 10 to the power of minus 3 moles,"},{"Start":"05:16.125 ","End":"05:21.920","Text":"and the molar mass of oxygen is 32.0 grams per mole."},{"Start":"05:21.920 ","End":"05:25.635","Text":"Moles times moles to power minus 1 is 1."},{"Start":"05:25.635 ","End":"05:27.330","Text":"We\u0027re left with grams."},{"Start":"05:27.330 ","End":"05:32.865","Text":"When we do the arithmetic we get 0.126 grams."},{"Start":"05:32.865 ","End":"05:36.510","Text":"Now, it\u0027s a small amount, a small mass."},{"Start":"05:36.510 ","End":"05:40.800","Text":"It would be difficult to measure such a mass."},{"Start":"05:40.800 ","End":"05:45.970","Text":"This is a very good way of finding out the mass using the ideal gas law."},{"Start":"05:45.970 ","End":"05:52.980","Text":"So in this video we learned about how to apply Dalton\u0027s law to collect gas over water."}],"ID":18161},{"Watched":false,"Name":"Exercise 1","Duration":"6m 57s","ChapterTopicVideoID":31932,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34207},{"Watched":false,"Name":"Exercise 2","Duration":"5m 32s","ChapterTopicVideoID":31933,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34208},{"Watched":false,"Name":"Exercise 3","Duration":"3m 13s","ChapterTopicVideoID":31934,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34209},{"Watched":false,"Name":"Exercise 4 part a","Duration":"4m 26s","ChapterTopicVideoID":31930,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34210},{"Watched":false,"Name":"Exercise 4 part b","Duration":"3m 29s","ChapterTopicVideoID":31931,"CourseChapterTopicPlaylistID":80096,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34211}],"Thumbnail":null,"ID":80096},{"Name":"Kinetic Theory of Gases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Kinetic Molecular Theory of Gases - Part 1","Duration":"6m 40s","ChapterTopicVideoID":17419,"CourseChapterTopicPlaylistID":80097,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:02.580","Text":"In previous videos,"},{"Start":"00:02.580 ","End":"00:05.010","Text":"we discuss the simple gas laws."},{"Start":"00:05.010 ","End":"00:07.350","Text":"These are experimental laws."},{"Start":"00:07.350 ","End":"00:11.490","Text":"Now we need a theory that explains these laws."},{"Start":"00:11.490 ","End":"00:17.145","Text":"We\u0027re going to talk about the Kinetic Molecular Theory, abbreviated as KMT."},{"Start":"00:17.145 ","End":"00:19.965","Text":"The Kinetic Molecular Theory is a simple theory"},{"Start":"00:19.965 ","End":"00:23.235","Text":"which can be used to obtain the simple gas laws."},{"Start":"00:23.235 ","End":"00:26.250","Text":"Now we\u0027re going to discuss the basic assumptions,"},{"Start":"00:26.250 ","End":"00:30.825","Text":"a better word is postulates of KMT theory."},{"Start":"00:30.825 ","End":"00:34.605","Text":"These are the underlying assumptions of the theory."},{"Start":"00:34.605 ","End":"00:38.964","Text":"The 1st assumption is that the particles,"},{"Start":"00:38.964 ","End":"00:41.000","Text":"they could be atoms or molecules,"},{"Start":"00:41.000 ","End":"00:44.585","Text":"are very small compared to the distance between them."},{"Start":"00:44.585 ","End":"00:47.960","Text":"The volume of each particle can be assumed to be"},{"Start":"00:47.960 ","End":"00:52.089","Text":"negligible compared to the volume of the container."},{"Start":"00:52.089 ","End":"00:58.385","Text":"We have very small particles and large distances between them."},{"Start":"00:58.385 ","End":"01:01.610","Text":"We don\u0027t really notice what the volume is."},{"Start":"01:01.610 ","End":"01:03.485","Text":"The volume is negligible."},{"Start":"01:03.485 ","End":"01:10.040","Text":"The 2nd postulate is that the particles are in constant random straight line motion."},{"Start":"01:10.040 ","End":"01:16.130","Text":"Here they are, they\u0027re moving all different directions but in a straight line."},{"Start":"01:16.130 ","End":"01:18.980","Text":"Now the collisions with the walls of the container"},{"Start":"01:18.980 ","End":"01:22.040","Text":"are the source of the pressure exerted by the gas."},{"Start":"01:22.040 ","End":"01:24.130","Text":"We have our particles moving,"},{"Start":"01:24.130 ","End":"01:29.900","Text":"they hit the container and that exerts a pressure on the walls of the container,"},{"Start":"01:29.900 ","End":"01:31.595","Text":"and that gives us a pressure."},{"Start":"01:31.595 ","End":"01:37.174","Text":"The 3rd postulate is that the infrequent collisions only occasionally do they collide,"},{"Start":"01:37.174 ","End":"01:43.585","Text":"between the particles are hard collisions similar to collisions between billiard balls."},{"Start":"01:43.585 ","End":"01:50.284","Text":"Right here we have they\u0027re approaching each other, the light balls,"},{"Start":"01:50.284 ","End":"01:52.370","Text":"when they hit each other,"},{"Start":"01:52.370 ","End":"01:55.650","Text":"they rebound, they separate,"},{"Start":"01:55.650 ","End":"01:57.780","Text":"and the particles are not deformed."},{"Start":"01:57.780 ","End":"02:02.805","Text":"That\u0027s like hard balls hitting each other and then they separate."},{"Start":"02:02.805 ","End":"02:05.600","Text":"In addition, the particles neither repel"},{"Start":"02:05.600 ","End":"02:08.570","Text":"nor attract each other when they\u0027re not colliding."},{"Start":"02:08.570 ","End":"02:12.530","Text":"Each particle is going its way,"},{"Start":"02:12.530 ","End":"02:16.025","Text":"uninfluenced by the particle near to each other."},{"Start":"02:16.025 ","End":"02:18.890","Text":"They neither attract each other nor repel each other."},{"Start":"02:18.890 ","End":"02:21.230","Text":"Now the 4th postulate is very important."},{"Start":"02:21.230 ","End":"02:25.490","Text":"It tells us that the average kinetic energy of a collection of"},{"Start":"02:25.490 ","End":"02:30.565","Text":"gas particles is proportional to the temperature in Kelvin of the gas."},{"Start":"02:30.565 ","End":"02:33.350","Text":"The higher the kinetic energy,"},{"Start":"02:33.350 ","End":"02:35.390","Text":"the greater the temperature."},{"Start":"02:35.390 ","End":"02:37.940","Text":"This explains the meaning of temperature."},{"Start":"02:37.940 ","End":"02:40.520","Text":"We\u0027re going to use this quite a lot."},{"Start":"02:40.520 ","End":"02:45.005","Text":"We can use these postulates to explain Boyle\u0027s law."},{"Start":"02:45.005 ","End":"02:47.960","Text":"How do you do it? You consider the pressure produced"},{"Start":"02:47.960 ","End":"02:50.750","Text":"by collisions of the particles with the walls of a box."},{"Start":"02:50.750 ","End":"02:53.930","Text":"The form you get out is this."},{"Start":"02:53.930 ","End":"02:55.895","Text":"Now I\u0027m not going to prove it,"},{"Start":"02:55.895 ","End":"03:01.445","Text":"but just to state Boyle\u0027s law in the context of this theory."},{"Start":"03:01.445 ","End":"03:04.985","Text":"The pressure is equal to the number of moles"},{"Start":"03:04.985 ","End":"03:08.825","Text":"times Avogadro\u0027s number times this expression,"},{"Start":"03:08.825 ","End":"03:15.410","Text":"which is the mass of each particle times the average of the square of the speed,"},{"Start":"03:15.410 ","End":"03:17.480","Text":"we\u0027ll explain this soon,"},{"Start":"03:17.480 ","End":"03:20.825","Text":"divided by 3 times the volume 3V."},{"Start":"03:20.825 ","End":"03:27.500","Text":"Now, Mu^2, average or average of Mu^2 is the mean or average,"},{"Start":"03:27.500 ","End":"03:30.890","Text":"is the same thing, square speed of the particles."},{"Start":"03:30.890 ","End":"03:32.930","Text":"We\u0027ll explain this in a minute."},{"Start":"03:32.930 ","End":"03:36.800","Text":"We\u0027re going to explain what\u0027s the mean square speed."},{"Start":"03:36.800 ","End":"03:40.040","Text":"Supposing we have several particles moving at different speeds."},{"Start":"03:40.040 ","End":"03:41.785","Text":"I\u0027m going to call the speed u."},{"Start":"03:41.785 ","End":"03:46.175","Text":"The average or mean speed is u with a bar on the top."},{"Start":"03:46.175 ","End":"03:47.629","Text":"That\u0027s the average."},{"Start":"03:47.629 ","End":"03:53.660","Text":"The average or mean of the squares of the speeds is u^2 with a bar on top."},{"Start":"03:53.660 ","End":"03:59.130","Text":"The square root of the mean u^2 is called U_rms,"},{"Start":"03:59.130 ","End":"04:06.520","Text":"root RMS is root mean square."},{"Start":"04:06.520 ","End":"04:10.535","Text":"That\u0027s the square root of the mean of the square."},{"Start":"04:10.535 ","End":"04:13.900","Text":"We have root mean square."},{"Start":"04:13.900 ","End":"04:17.300","Text":"I\u0027m going to take an example to show what this means."},{"Start":"04:17.300 ","End":"04:21.454","Text":"Consider 5 particles with speed 350, 400,"},{"Start":"04:21.454 ","End":"04:23.143","Text":"450, 500,"},{"Start":"04:23.143 ","End":"04:26.810","Text":"and 550 meters per second."},{"Start":"04:26.810 ","End":"04:34.160","Text":"We\u0027re asked to calculate the mean of the square of the speed and the root mean square."},{"Start":"04:34.160 ","End":"04:36.220","Text":"Let\u0027s first take the mean."},{"Start":"04:36.220 ","End":"04:38.615","Text":"In order to get the mean or the average,"},{"Start":"04:38.615 ","End":"04:43.640","Text":"we add up all the different speeds and divide by the number of particles,"},{"Start":"04:43.640 ","End":"04:51.950","Text":"so 350 plus 400 plus 450 plus 500 plus 550 and we divide that by 5."},{"Start":"04:51.950 ","End":"04:56.915","Text":"Do that we get 450 meters per second."},{"Start":"04:56.915 ","End":"05:00.530","Text":"Now we want the mean of the squares of the speeds."},{"Start":"05:00.530 ","End":"05:04.040","Text":"We need to take each speed and square it."},{"Start":"05:04.040 ","End":"05:07.880","Text":"Then add them up and divide by the number of particles."},{"Start":"05:07.880 ","End":"05:16.885","Text":"It\u0027s 350^2 plus 400^2 plus 450^2 plus 500^2 plus 550^2 divide by 5."},{"Start":"05:16.885 ","End":"05:23.450","Text":"We do that we get 207,500 meters squared per second squared."},{"Start":"05:23.450 ","End":"05:25.520","Text":"Before we go on,"},{"Start":"05:25.520 ","End":"05:33.365","Text":"I should point out that the mean of u^2 is not necessarily equal to the mean squared."},{"Start":"05:33.365 ","End":"05:35.105","Text":"How can I see this?"},{"Start":"05:35.105 ","End":"05:41.465","Text":"If I take the mean of 450 meters per second and square it,"},{"Start":"05:41.465 ","End":"05:49.840","Text":"I\u0027ll get 202,500 meters^2 per second^2."},{"Start":"05:49.840 ","End":"05:54.365","Text":"That\u0027s not equal to 207,500."},{"Start":"05:54.365 ","End":"05:57.710","Text":"Here we have 202,500,"},{"Start":"05:57.710 ","End":"06:00.320","Text":"and here we have 207,500."},{"Start":"06:00.320 ","End":"06:03.050","Text":"These are not equal."},{"Start":"06:03.050 ","End":"06:06.470","Text":"Now, we want to calculate the root mean square."},{"Start":"06:06.470 ","End":"06:11.140","Text":"We take the root of the mean of the square."},{"Start":"06:11.140 ","End":"06:12.645","Text":"That works out."},{"Start":"06:12.645 ","End":"06:21.905","Text":"We\u0027re taking the square root of 207,500 and that\u0027s 456 meters per second."},{"Start":"06:21.905 ","End":"06:26.450","Text":"We can see that it\u0027s not equal to the mean of the speeds."},{"Start":"06:26.450 ","End":"06:31.325","Text":"Mean of the speed was 450 and this is 456."},{"Start":"06:31.325 ","End":"06:36.965","Text":"This is the first part of our description of the kinetic theory,"},{"Start":"06:36.965 ","End":"06:40.500","Text":"and in the next video we\u0027ll continue with it."}],"ID":18166},{"Watched":false,"Name":"Kinetic Molecular Theory of Gases - Part 2","Duration":"7m 17s","ChapterTopicVideoID":17420,"CourseChapterTopicPlaylistID":80097,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.115","Text":"In this video, we\u0027re going to continue our discussion of the kinetic theory."},{"Start":"00:05.115 ","End":"00:09.450","Text":"We\u0027re going to start with discussing the connection between the mean square speed,"},{"Start":"00:09.450 ","End":"00:11.580","Text":"which we discussed in the previous video,"},{"Start":"00:11.580 ","End":"00:13.635","Text":"and the mean kinetic energy."},{"Start":"00:13.635 ","End":"00:15.285","Text":"What\u0027s the kinetic energy?"},{"Start":"00:15.285 ","End":"00:16.890","Text":"That\u0027s the energy of motion."},{"Start":"00:16.890 ","End":"00:18.520","Text":"Kinetic is motion."},{"Start":"00:18.520 ","End":"00:21.885","Text":"It\u0027s the energy of a particle of mass m,"},{"Start":"00:21.885 ","End":"00:24.548","Text":"that\u0027s e_k, e for energy,"},{"Start":"00:24.548 ","End":"00:25.915","Text":"K for kinetic,"},{"Start":"00:25.915 ","End":"00:27.930","Text":"equal to 1/2mu^2,"},{"Start":"00:27.930 ","End":"00:33.675","Text":"where m is the mass and u is the speed."},{"Start":"00:33.675 ","End":"00:35.730","Text":"That\u0027s from simple physics."},{"Start":"00:35.730 ","End":"00:39.510","Text":"If we want the average kinetic energy of a particle,"},{"Start":"00:39.510 ","End":"00:42.345","Text":"we put the average sign above the e_k,"},{"Start":"00:42.345 ","End":"00:45.450","Text":"and then put the average sign above u^2."},{"Start":"00:45.450 ","End":"00:48.605","Text":"We get the average is equal to 1/2m,"},{"Start":"00:48.605 ","End":"00:50.750","Text":"the average of u^2."},{"Start":"00:50.750 ","End":"00:53.300","Text":"Now, I didn\u0027t put the average over the m because I"},{"Start":"00:53.300 ","End":"00:56.435","Text":"assumed that every particle has the same mass."},{"Start":"00:56.435 ","End":"01:00.995","Text":"Now, supposing the average kinetic energy of a mole not just a single particle,"},{"Start":"01:00.995 ","End":"01:03.080","Text":"then that\u0027s the average of E_k."},{"Start":"01:03.080 ","End":"01:05.120","Text":"Now, a E for a mole,"},{"Start":"01:05.120 ","End":"01:08.660","Text":"and then multiply by N_A which is Avogadro\u0027s number."},{"Start":"01:08.660 ","End":"01:15.590","Text":"We get the average is equal to N_A times the average for a single particle."},{"Start":"01:15.590 ","End":"01:17.755","Text":"That\u0027s 1/2N_A."},{"Start":"01:17.755 ","End":"01:21.735","Text":"Then we have mu^2 average from before."},{"Start":"01:21.735 ","End":"01:25.680","Text":"That\u0027s equal to 1/2M, molar mass,"},{"Start":"01:25.680 ","End":"01:27.210","Text":"1/2 the molar mass,"},{"Start":"01:27.210 ","End":"01:30.380","Text":"because N_A times M is the molar mass,"},{"Start":"01:30.380 ","End":"01:32.740","Text":"the average of u^2."},{"Start":"01:32.740 ","End":"01:36.170","Text":"Now, we can take the first expression of this chain,"},{"Start":"01:36.170 ","End":"01:37.835","Text":"and the last one,"},{"Start":"01:37.835 ","End":"01:39.890","Text":"and equate them we get E_k,"},{"Start":"01:39.890 ","End":"01:47.030","Text":"the average kinetic energy is equal to 1/2 the molar mass times the average of u^2."},{"Start":"01:47.030 ","End":"01:49.130","Text":"That\u0027s a very important equation."},{"Start":"01:49.130 ","End":"01:51.695","Text":"I\u0027m going to highlight it."},{"Start":"01:51.695 ","End":"01:55.760","Text":"Now we\u0027re going to derive the ideal gas equation from"},{"Start":"01:55.760 ","End":"02:00.605","Text":"Boyle\u0027s law in the form that we had Boyle\u0027s law in the previous video."},{"Start":"02:00.605 ","End":"02:03.350","Text":"In the previous video we said that Boyle\u0027s law,"},{"Start":"02:03.350 ","End":"02:07.130","Text":"the pressure is equal to n times n,"},{"Start":"02:07.130 ","End":"02:09.755","Text":"the number of moles times Avogadro\u0027s number,"},{"Start":"02:09.755 ","End":"02:12.290","Text":"N_A, times m,"},{"Start":"02:12.290 ","End":"02:18.010","Text":"the mass, the average of u^2 divided by 3 times the volume 3V."},{"Start":"02:18.010 ","End":"02:24.005","Text":"Now, if we multiply each side of this equation by V and divide by n,"},{"Start":"02:24.005 ","End":"02:29.710","Text":"we get PV is equal to N_A times m,"},{"Start":"02:29.710 ","End":"02:32.230","Text":"the average of u^2 divided by 3."},{"Start":"02:32.230 ","End":"02:34.745","Text":"Now I want to isolate the kinetic energy,"},{"Start":"02:34.745 ","End":"02:37.775","Text":"which if you remember was divided by 2."},{"Start":"02:37.775 ","End":"02:40.265","Text":"I\u0027m going to divide this expression by"},{"Start":"02:40.265 ","End":"02:44.620","Text":"2 and also multiply it by 2 to compensate for that."},{"Start":"02:44.620 ","End":"02:48.855","Text":"Now I get N_A times 2/3 m,"},{"Start":"02:48.855 ","End":"02:50.685","Text":"the average of u^2/2."},{"Start":"02:50.685 ","End":"02:57.215","Text":"The 2 and the 2 cancel so you can be sure that this is the same expression as before."},{"Start":"02:57.215 ","End":"03:04.315","Text":"Now, I have N_A times 2/3 times the average kinetic energy of a single particle."},{"Start":"03:04.315 ","End":"03:06.950","Text":"N_A times the average kinetic energy of"},{"Start":"03:06.950 ","End":"03:10.880","Text":"a single particle is the average kinetic energy of a mole."},{"Start":"03:10.880 ","End":"03:15.960","Text":"Now I have 2/3 e_k average."},{"Start":"03:15.960 ","End":"03:19.260","Text":"If I take the first expression,"},{"Start":"03:19.260 ","End":"03:21.845","Text":"PV, and the last one,"},{"Start":"03:21.845 ","End":"03:24.920","Text":"and I multiply both sides of them by n,"},{"Start":"03:24.920 ","End":"03:31.084","Text":"I get PV is equal to n times 2/3 E_k average."},{"Start":"03:31.084 ","End":"03:37.040","Text":"Now I have PV. We\u0027re approaching the ideal gas equation,"},{"Start":"03:37.040 ","End":"03:38.630","Text":"not quite there yet."},{"Start":"03:38.630 ","End":"03:40.100","Text":"Now, if you recall,"},{"Start":"03:40.100 ","End":"03:43.670","Text":"the fourth postulate of the kinetic theory is that"},{"Start":"03:43.670 ","End":"03:48.205","Text":"the average kinetic energy is proportional to the temperature in Kelvin."},{"Start":"03:48.205 ","End":"03:53.105","Text":"The average kinetic energy is proportional to the temperature."},{"Start":"03:53.105 ","End":"03:54.980","Text":"I\u0027ve written proportional to n,"},{"Start":"03:54.980 ","End":"03:57.995","Text":"I\u0027ve kept the n times T,"},{"Start":"03:57.995 ","End":"04:00.625","Text":"which is the temperature in Kelvin."},{"Start":"04:00.625 ","End":"04:04.200","Text":"Now we have PV proportional to nT."},{"Start":"04:04.200 ","End":"04:07.250","Text":"If I take the proportionality constant to be R,"},{"Start":"04:07.250 ","End":"04:09.260","Text":"the gas constant,"},{"Start":"04:09.260 ","End":"04:11.060","Text":"we\u0027ll get the ideal gas equation,"},{"Start":"04:11.060 ","End":"04:17.855","Text":"PV is equal to nRT equation with which you\u0027re all very familiar."},{"Start":"04:17.855 ","End":"04:21.800","Text":"Now we\u0027re going to discuss the connection between the rms speed,"},{"Start":"04:21.800 ","End":"04:25.370","Text":"the root mean squared speed and the average kinetic energy."},{"Start":"04:25.370 ","End":"04:31.670","Text":"Now, we saw before that PV is equal to n times 2/3 the average kinetic energy,"},{"Start":"04:31.670 ","End":"04:34.730","Text":"and it\u0027s also equal to nRT."},{"Start":"04:34.730 ","End":"04:37.980","Text":"We can equate these two expressions."},{"Start":"04:37.980 ","End":"04:39.990","Text":"Cancel the R,"},{"Start":"04:39.990 ","End":"04:42.455","Text":"cancel the n,"},{"Start":"04:42.455 ","End":"04:48.435","Text":"and we get 2/3 E_k average is equal to RT."},{"Start":"04:48.435 ","End":"04:52.305","Text":"Now we want to isolate the E_k average,"},{"Start":"04:52.305 ","End":"04:54.735","Text":"so we can multiply by 3/2."},{"Start":"04:54.735 ","End":"04:57.355","Text":"Multiply each side of the equation by 3/2."},{"Start":"04:57.355 ","End":"05:01.030","Text":"The 2/3 cancels with the 3/2,"},{"Start":"05:01.030 ","End":"05:03.030","Text":"2/3 essentially is just one."},{"Start":"05:03.030 ","End":"05:05.915","Text":"We get E_k is equal to 3_2 RT."},{"Start":"05:05.915 ","End":"05:08.560","Text":"This is another very important equation."},{"Start":"05:08.560 ","End":"05:13.150","Text":"Tells us the average kinetic energy is 3/2 RT."},{"Start":"05:13.150 ","End":"05:17.705","Text":"Now, we know that a molecule moving has 3 degrees of freedom."},{"Start":"05:17.705 ","End":"05:20.530","Text":"It can move in the x direction,"},{"Start":"05:20.530 ","End":"05:22.935","Text":"the y direction, and the z direction."},{"Start":"05:22.935 ","End":"05:30.240","Text":"We say that the average kinetic energy in each degree of freedom is 1/2 RT."},{"Start":"05:30.710 ","End":"05:34.160","Text":"This is very important equation that gives us"},{"Start":"05:34.160 ","End":"05:38.000","Text":"a connection between the temperature and the kinetic energy."},{"Start":"05:38.000 ","End":"05:42.650","Text":"Now we can go on with this equation and write the average kinetic energy"},{"Start":"05:42.650 ","End":"05:47.799","Text":"as 1/2 the molar mass times u^2 average."},{"Start":"05:47.799 ","End":"05:49.800","Text":"That\u0027s on the left side."},{"Start":"05:49.800 ","End":"05:53.180","Text":"The right-hand side, we\u0027ve still got the 3/2 RT."},{"Start":"05:53.180 ","End":"05:58.505","Text":"Now we want to isolate the average of u^2."},{"Start":"05:58.505 ","End":"06:00.260","Text":"How do we do that?"},{"Start":"06:00.260 ","End":"06:02.965","Text":"We can cancel the 2\u0027s,"},{"Start":"06:02.965 ","End":"06:08.000","Text":"and then all we have to do is divide by the molar mass."},{"Start":"06:08.000 ","End":"06:12.125","Text":"We have the average of u^2,"},{"Start":"06:12.125 ","End":"06:14.780","Text":"3RT divided by the molar mass."},{"Start":"06:14.780 ","End":"06:17.390","Text":"We want to take the square root of that,"},{"Start":"06:17.390 ","End":"06:20.105","Text":"so we get u root Mw^2."},{"Start":"06:20.105 ","End":"06:25.310","Text":"The square root of that is 3RT divided by the molar mass."},{"Start":"06:25.310 ","End":"06:27.995","Text":"We have another important equation."},{"Start":"06:27.995 ","End":"06:35.955","Text":"The root Mw^2 speed is equal to the square root of 3RT divided by the molar mass."},{"Start":"06:35.955 ","End":"06:37.760","Text":"This tells us something very important."},{"Start":"06:37.760 ","End":"06:41.465","Text":"It tells us that as the temperature increases,"},{"Start":"06:41.465 ","End":"06:43.730","Text":"the speed will increase."},{"Start":"06:43.730 ","End":"06:45.335","Text":"That\u0027s fairly obvious."},{"Start":"06:45.335 ","End":"06:50.975","Text":"But in addition, as the molar mass increases,"},{"Start":"06:50.975 ","End":"06:53.330","Text":"the speed will decrease."},{"Start":"06:53.330 ","End":"06:58.430","Text":"That\u0027s because we know that heavy bodies move more slowly"},{"Start":"06:58.430 ","End":"07:04.729","Text":"than light bodies if we give them the same amount of energy."},{"Start":"07:04.729 ","End":"07:10.744","Text":"We\u0027re going to use this equation in the next few videos."},{"Start":"07:10.744 ","End":"07:16.710","Text":"In this video, we completed our discussion of the kinetic theory."}],"ID":18167},{"Watched":false,"Name":"Maxwell-Boltzmann Distribution","Duration":"8m 20s","ChapterTopicVideoID":17421,"CourseChapterTopicPlaylistID":80097,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"In the previous 2 videos,"},{"Start":"00:01.950 ","End":"00:05.310","Text":"we discussed the kinetic molecular theory of gases."},{"Start":"00:05.310 ","End":"00:07.080","Text":"In this video, we\u0027re going to discuss"},{"Start":"00:07.080 ","End":"00:10.680","Text":"the Maxwell-Boltzmann distribution of molecular speeds."},{"Start":"00:10.680 ","End":"00:14.250","Text":"This distribution can be derived from the kinetic theory of"},{"Start":"00:14.250 ","End":"00:19.290","Text":"gases and it relates both to molecular or atomic speeds."},{"Start":"00:19.290 ","End":"00:22.220","Text":"Here\u0027s our distribution. On the x-axis,"},{"Start":"00:22.220 ","End":"00:28.900","Text":"we have speed that\u0027s from 0-4,500 meters per second in this particular case."},{"Start":"00:28.900 ","End":"00:30.260","Text":"On the y-axis,"},{"Start":"00:30.260 ","End":"00:34.910","Text":"we have the percentage of molecules that have that particular speed."},{"Start":"00:34.910 ","End":"00:41.300","Text":"In addition, we have indicated here u_modal that\u0027s the maximum of the graph."},{"Start":"00:41.300 ","End":"00:47.395","Text":"That\u0027s the most probable speed: u_modal, u_m is u_modal."},{"Start":"00:47.395 ","End":"00:51.020","Text":"We also have u_average which is the same as u_mean."},{"Start":"00:51.020 ","End":"00:55.235","Text":"We have u_rms root mean squared speed:"},{"Start":"00:55.235 ","End":"01:01.235","Text":"u_rms that we\u0027ve discussed several times and always comes in that order."},{"Start":"01:01.235 ","End":"01:07.355","Text":"u_rms is always larger than u_average which is always larger than u_modal."},{"Start":"01:07.355 ","End":"01:09.295","Text":"In this particular case,"},{"Start":"01:09.295 ","End":"01:12.870","Text":"u_modal is 1,500 meters per second."},{"Start":"01:12.870 ","End":"01:14.030","Text":"We can see it here."},{"Start":"01:14.030 ","End":"01:15.770","Text":"That\u0027s the maximum."},{"Start":"01:15.770 ","End":"01:19.355","Text":"It occurs at 1,500 meters per second."},{"Start":"01:19.355 ","End":"01:21.680","Text":"u_average is to the right of that."},{"Start":"01:21.680 ","End":"01:24.160","Text":"That\u0027s 1,700 meters per second."},{"Start":"01:24.160 ","End":"01:30.215","Text":"Here it is, and u_root mean squared is 1,800 meters per second."},{"Start":"01:30.215 ","End":"01:33.845","Text":"That\u0027s to the right of u_average and u_modal."},{"Start":"01:33.845 ","End":"01:38.330","Text":"If we want to understand the shape of the distribution,"},{"Start":"01:38.330 ","End":"01:40.985","Text":"we need to look at the mathematical expression."},{"Start":"01:40.985 ","End":"01:43.595","Text":"We see that there are 3 terms."},{"Start":"01:43.595 ","End":"01:48.455","Text":"The first term here is called a normalization factor."},{"Start":"01:48.455 ","End":"01:52.880","Text":"It ensures that if we sum up all the fractions,"},{"Start":"01:52.880 ","End":"01:54.440","Text":"we will get 1."},{"Start":"01:54.440 ","End":"01:56.420","Text":"If you want to express it mathematically,"},{"Start":"01:56.420 ","End":"02:03.100","Text":"you can say that the integral of f(u) du is equal to 1."},{"Start":"02:03.100 ","End":"02:06.050","Text":"Now the second term is u squared."},{"Start":"02:06.050 ","End":"02:09.925","Text":"Here we have it. u squared is like a parabola."},{"Start":"02:09.925 ","End":"02:13.590","Text":"The third expression is called a Gaussian."},{"Start":"02:13.590 ","End":"02:19.115","Text":"It\u0027s exponential minus some factors u squared."},{"Start":"02:19.115 ","End":"02:24.225","Text":"It\u0027s symmetrical like a bell-shaped curve."},{"Start":"02:24.225 ","End":"02:28.835","Text":"So that we multiply the parabola times the Gaussian,"},{"Start":"02:28.835 ","End":"02:35.975","Text":"we get that now the Gaussian becomes distorted and it has a long, long tail."},{"Start":"02:35.975 ","End":"02:39.250","Text":"The distribution is asymmetric with a long tail."},{"Start":"02:39.250 ","End":"02:41.600","Text":"This is very important physically."},{"Start":"02:41.600 ","End":"02:44.840","Text":"It has all sorts of important physical ramifications."},{"Start":"02:44.840 ","End":"02:47.315","Text":"For example, if the average speed is low,"},{"Start":"02:47.315 ","End":"02:49.610","Text":"although the average speed may be low,"},{"Start":"02:49.610 ","End":"02:54.725","Text":"there are always molecules in the tail that have a higher speed."},{"Start":"02:54.725 ","End":"02:59.585","Text":"For example, if I put a glass of water on the table at room temperature,"},{"Start":"02:59.585 ","End":"03:04.670","Text":"there will always be some molecules of water that have enough energy to escape,"},{"Start":"03:04.670 ","End":"03:09.490","Text":"that\u0027s why water evaporates even when it\u0027s not at its boiling point."},{"Start":"03:09.490 ","End":"03:14.405","Text":"Now it\u0027s also important to know what happens when we change the temperature,"},{"Start":"03:14.405 ","End":"03:16.010","Text":"or we change the molar mass."},{"Start":"03:16.010 ","End":"03:18.500","Text":"How does the distribution change?"},{"Start":"03:18.500 ","End":"03:20.465","Text":"In order to understand this,"},{"Start":"03:20.465 ","End":"03:26.090","Text":"we need to once again look at the mathematical expression for the distribution,"},{"Start":"03:26.090 ","End":"03:30.115","Text":"and also the mathematical expression for u_modal,"},{"Start":"03:30.115 ","End":"03:32.880","Text":"u_average, and u_rms."},{"Start":"03:32.880 ","End":"03:36.855","Text":"Now we see that u_modal, u_average,"},{"Start":"03:36.855 ","End":"03:43.475","Text":"and u_rms are all proportional to the square root of the temperature,"},{"Start":"03:43.475 ","End":"03:49.384","Text":"and they\u0027re inversely proportional to the square root of the molar mass."},{"Start":"03:49.384 ","End":"03:52.370","Text":"Let\u0027s see how this works out in the curves."},{"Start":"03:52.370 ","End":"03:56.675","Text":"Here we have the Maxwell-Boltzmann distribution for 3 cases."},{"Start":"03:56.675 ","End":"04:00.440","Text":"Here we have oxygen at 273 Kelvin."},{"Start":"04:00.440 ","End":"04:03.515","Text":"Oxygen at a higher temperature of 1,000 Kelvin,"},{"Start":"04:03.515 ","End":"04:07.565","Text":"and we have hydrogen gas at 273 Kelvin."},{"Start":"04:07.565 ","End":"04:11.735","Text":"Let\u0027s first see what happens when we compare these 2."},{"Start":"04:11.735 ","End":"04:14.120","Text":"Oxygen at 273,"},{"Start":"04:14.120 ","End":"04:16.370","Text":"and oxygen at 1,000."},{"Start":"04:16.370 ","End":"04:23.630","Text":"We see as predicted for the expressions for the speeds that at higher temperature,"},{"Start":"04:23.630 ","End":"04:26.330","Text":"the u_rms and u_modal,"},{"Start":"04:26.330 ","End":"04:31.835","Text":"and u_average will be greater than at lower temperatures because"},{"Start":"04:31.835 ","End":"04:37.895","Text":"all the u\u0027s are proportional to the square root of the temperature."},{"Start":"04:37.895 ","End":"04:39.905","Text":"The higher the temperature,"},{"Start":"04:39.905 ","End":"04:42.380","Text":"the higher the speed."},{"Start":"04:42.380 ","End":"04:43.670","Text":"That\u0027s these 2."},{"Start":"04:43.670 ","End":"04:46.550","Text":"We can also look at the mathematical expression,"},{"Start":"04:46.550 ","End":"04:49.910","Text":"and work out that at lower temperature,"},{"Start":"04:49.910 ","End":"04:53.944","Text":"the distribution will be narrower than at higher temperature."},{"Start":"04:53.944 ","End":"04:58.105","Text":"In addition, we can compare a heavier molecule,"},{"Start":"04:58.105 ","End":"05:00.600","Text":"O_2 with a lighter molecule,"},{"Start":"05:00.600 ","End":"05:03.020","Text":"H_2 at the same temperature,"},{"Start":"05:03.020 ","End":"05:05.600","Text":"in this case, 273 Kelvin."},{"Start":"05:05.600 ","End":"05:10.440","Text":"What we need to know here is all the u\u0027s are inversely"},{"Start":"05:10.440 ","End":"05:16.924","Text":"proportional to 1 over the square root of the molar mass."},{"Start":"05:16.924 ","End":"05:19.114","Text":"The higher the molar mass,"},{"Start":"05:19.114 ","End":"05:21.260","Text":"the smaller the speed,"},{"Start":"05:21.260 ","End":"05:25.800","Text":"and that\u0027s why O_2 at 273 Kelvin they are"},{"Start":"05:25.800 ","End":"05:31.400","Text":"smaller lower speed than hydrogen at 273 Kelvin."},{"Start":"05:31.400 ","End":"05:33.330","Text":"Let\u0027s look at an example."},{"Start":"05:33.330 ","End":"05:38.165","Text":"Calculate u_rms for oxygen at 100 degrees Celsius."},{"Start":"05:38.165 ","End":"05:40.205","Text":"Now before we do the calculation,"},{"Start":"05:40.205 ","End":"05:42.395","Text":"we need to know 2 important things."},{"Start":"05:42.395 ","End":"05:47.405","Text":"Whenever we talk about the gas constant in relation to energy,"},{"Start":"05:47.405 ","End":"05:54.925","Text":"we need to use this value: R is equal to 8.3145 joules per mole per Kelvin."},{"Start":"05:54.925 ","End":"05:57.620","Text":"We also need to remember what joule is."},{"Start":"05:57.620 ","End":"06:04.920","Text":"Joule is kilogram times meter squared times second^minus 2."},{"Start":"06:04.920 ","End":"06:06.645","Text":"How can we remember that?"},{"Start":"06:06.645 ","End":"06:13.680","Text":"We have to remember that kinetic energy is equal to half mass times u squared,"},{"Start":"06:13.680 ","End":"06:15.270","Text":"we get the right units,"},{"Start":"06:15.270 ","End":"06:17.465","Text":"thus mass is kilograms,"},{"Start":"06:17.465 ","End":"06:19.800","Text":"and speed is meters per second,"},{"Start":"06:19.800 ","End":"06:23.690","Text":"and we\u0027re squaring meters squared per second squared."},{"Start":"06:23.690 ","End":"06:25.635","Text":"Here\u0027s our calculation."},{"Start":"06:25.635 ","End":"06:33.350","Text":"u_rms is equal to the square root of 3RT divided by the molar mass."},{"Start":"06:33.350 ","End":"06:35.480","Text":"Let\u0027s put in the numbers."},{"Start":"06:35.480 ","End":"06:39.875","Text":"We have 3 times 8.3145 joules"},{"Start":"06:39.875 ","End":"06:44.660","Text":"per Kelvin per mole where the temperature is 100 degrees Celsius."},{"Start":"06:44.660 ","End":"06:47.615","Text":"That\u0027s 373.15 Kelvin,"},{"Start":"06:47.615 ","End":"06:49.970","Text":"and the molar mass is almost 32."},{"Start":"06:49.970 ","End":"06:53.000","Text":"Here I\u0027ve written with rather greater accuracy,"},{"Start":"06:53.000 ","End":"06:56.845","Text":"31.999 grams per mole."},{"Start":"06:56.845 ","End":"07:02.405","Text":"Now, there we can write the joule in terms of kilograms and meters and seconds."},{"Start":"07:02.405 ","End":"07:08.180","Text":"We have joule is kilograms times meters squared per second squared."},{"Start":"07:08.180 ","End":"07:10.970","Text":"Then we note that there something else we need to do,"},{"Start":"07:10.970 ","End":"07:15.500","Text":"because here we had the molar mass in grams."},{"Start":"07:15.500 ","End":"07:17.630","Text":"Chemists like to use grams per mole,"},{"Start":"07:17.630 ","End":"07:20.780","Text":"but here we have kilograms from the joules,"},{"Start":"07:20.780 ","End":"07:25.730","Text":"so we have to convert the grams per mole to kilograms per mole,"},{"Start":"07:25.730 ","End":"07:32.345","Text":"so it\u0027s 31.999 times 10^minus 3 kilograms per mole."},{"Start":"07:32.345 ","End":"07:34.110","Text":"Now we can do the calculation."},{"Start":"07:34.110 ","End":"07:35.885","Text":"First, let\u0027s look at the units."},{"Start":"07:35.885 ","End":"07:38.550","Text":"Kilograms cancels with kilograms,"},{"Start":"07:38.550 ","End":"07:42.360","Text":"mole_minus 1 can cross with mole_minus 1."},{"Start":"07:42.360 ","End":"07:45.810","Text":"Kelvin_minus 1 times Kelvin is just 1,"},{"Start":"07:45.810 ","End":"07:50.210","Text":"so we\u0027re left with meters squared per second squared."},{"Start":"07:50.210 ","End":"07:55.520","Text":"When we work out the arithmetic we get square root of 2.9087"},{"Start":"07:55.520 ","End":"08:01.890","Text":"times 10^5 meters squared times second_minus 2,"},{"Start":"08:01.890 ","End":"08:04.880","Text":"and we take the square root of all that,"},{"Start":"08:04.880 ","End":"08:09.385","Text":"we get 539.3 meters per second."},{"Start":"08:09.385 ","End":"08:13.530","Text":"That\u0027s our root mean square speed."},{"Start":"08:13.530 ","End":"08:20.760","Text":"In this video, we talked about the Maxwell-Boltzmann distribution of molecular speeds."}],"ID":18168},{"Watched":false,"Name":"Graham s Law of Effusion","Duration":"9m 5s","ChapterTopicVideoID":17422,"CourseChapterTopicPlaylistID":80097,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.400","Text":"In previous videos, we found expressions for"},{"Start":"00:02.400 ","End":"00:06.090","Text":"the average kinetic energy and the root-mean-square speed."},{"Start":"00:06.090 ","End":"00:11.220","Text":"We\u0027ll begin this video by reiterating what we learned before."},{"Start":"00:11.220 ","End":"00:14.260","Text":"The average kinetic energy is the first thing."},{"Start":"00:14.260 ","End":"00:19.110","Text":"The average kinetic energy of a mole of molecules is 3/2 RT,"},{"Start":"00:19.110 ","End":"00:20.970","Text":"where R is the gas constant."},{"Start":"00:20.970 ","End":"00:24.855","Text":"The kinetic energy of a gas is proportional to its temperature."},{"Start":"00:24.855 ","End":"00:27.210","Text":"What\u0027s very important is that,"},{"Start":"00:27.210 ","End":"00:31.500","Text":"it\u0027s the same for all ideal gases at the same temperature."},{"Start":"00:31.500 ","End":"00:33.510","Text":"Root-mean-square speed."},{"Start":"00:33.510 ","End":"00:40.775","Text":"The root-mean-square speed is equal to the square root of 3RT divided by Mw."},{"Start":"00:40.775 ","End":"00:44.990","Text":"We see this is proportional to square root of the temperature in Kelvin,"},{"Start":"00:44.990 ","End":"00:48.895","Text":"and inversely proportional to the molar mass."},{"Start":"00:48.895 ","End":"00:51.785","Text":"As the molar mass increases,"},{"Start":"00:51.785 ","End":"00:54.425","Text":"the speed will decrease."},{"Start":"00:54.425 ","End":"00:59.855","Text":"Now, we can use these concepts to study diffusion of gases,"},{"Start":"00:59.855 ","End":"01:03.590","Text":"and another concept called effusion of gases."},{"Start":"01:03.590 ","End":"01:06.830","Text":"First, let\u0027s talk about diffusion of gases."},{"Start":"01:06.830 ","End":"01:09.710","Text":"Diffusion is the random motion of gas molecules from"},{"Start":"01:09.710 ","End":"01:13.495","Text":"a region of high concentration to a region of low concentration."},{"Start":"01:13.495 ","End":"01:17.600","Text":"Eventually, the gas will be evenly distributed throughout the container."},{"Start":"01:17.600 ","End":"01:19.730","Text":"If we have a container,"},{"Start":"01:19.730 ","End":"01:26.210","Text":"and we release the gas in the corner after a certain amount of time,"},{"Start":"01:26.210 ","End":"01:31.070","Text":"the gas molecules will have traveled all over the container,"},{"Start":"01:31.070 ","End":"01:33.155","Text":"and be evenly distributed."},{"Start":"01:33.155 ","End":"01:35.600","Text":"Now supposing we have 2 or more gases,"},{"Start":"01:35.600 ","End":"01:38.945","Text":"eventually we\u0027ll get a homogeneous mixture of gases."},{"Start":"01:38.945 ","End":"01:43.860","Text":"Supposing we have a container with a partition in the middle,"},{"Start":"01:43.860 ","End":"01:47.765","Text":"and on the left-hand side we have a gas,"},{"Start":"01:47.765 ","End":"01:51.215","Text":"which we will indicate by black dots."},{"Start":"01:51.215 ","End":"01:53.240","Text":"On the right-hand side, we have a gas,"},{"Start":"01:53.240 ","End":"01:56.605","Text":"which we will indicate by red dots."},{"Start":"01:56.605 ","End":"02:00.235","Text":"Supposing we remove the partition,"},{"Start":"02:00.235 ","End":"02:03.325","Text":"the red and the black dots will move all"},{"Start":"02:03.325 ","End":"02:07.825","Text":"over the container until we get a homogeneous mixture."},{"Start":"02:07.825 ","End":"02:11.740","Text":"Now we can see from what we\u0027ve learned before that"},{"Start":"02:11.740 ","End":"02:17.285","Text":"the lighter gas will move faster than the heavier gas."},{"Start":"02:17.285 ","End":"02:23.065","Text":"Of course, it will take some time before we get a complete homogeneous mixture."},{"Start":"02:23.065 ","End":"02:28.180","Text":"Unfortunately, this is not enough to get a good equation for diffusion."},{"Start":"02:28.180 ","End":"02:30.760","Text":"Diffusion is quite a complicated process,"},{"Start":"02:30.760 ","End":"02:36.120","Text":"but there\u0027s another process which is more amenable to the simple treatment,"},{"Start":"02:36.120 ","End":"02:39.210","Text":"and that\u0027s called effusion."},{"Start":"02:39.210 ","End":"02:42.850","Text":"What\u0027s effusion? Effusion happens when"},{"Start":"02:42.850 ","End":"02:46.400","Text":"gases escape through a small orifice or hole into a vacuum."},{"Start":"02:46.400 ","End":"02:50.215","Text":"The lighter gas escapes faster than the heavier gas."},{"Start":"02:50.215 ","End":"02:53.300","Text":"Supposing we have a container,"},{"Start":"02:53.300 ","End":"02:59.510","Text":"on one side we have a gas,"},{"Start":"02:59.540 ","End":"03:02.200","Text":"and the other side we have nothing."},{"Start":"03:02.200 ","End":"03:07.195","Text":"That\u0027s a vacuum. A vacuum is when there are no molecules present."},{"Start":"03:07.195 ","End":"03:10.420","Text":"We have a little hole here."},{"Start":"03:10.420 ","End":"03:12.405","Text":"Supposing we open the hole,"},{"Start":"03:12.405 ","End":"03:15.220","Text":"then the molecules from the gas on the left-hand side will"},{"Start":"03:15.220 ","End":"03:19.465","Text":"escape through the hole into the right-hand side."},{"Start":"03:19.465 ","End":"03:22.115","Text":"If we have 2 gases,"},{"Start":"03:22.115 ","End":"03:28.975","Text":"then the lighter gas will escape faster than the heavier gas."},{"Start":"03:28.975 ","End":"03:32.590","Text":"Now, there is a law called Graham\u0027s law of effusion,"},{"Start":"03:32.590 ","End":"03:33.940","Text":"which is a simple law."},{"Start":"03:33.940 ","End":"03:40.390","Text":"It says, the rate of effusion for Gas A compared to the rate of effusion for Gas B"},{"Start":"03:40.390 ","End":"03:48.760","Text":"is equal to the root-mean-square speed of A divided by root mean squared speed of B."},{"Start":"03:48.760 ","End":"03:55.700","Text":"Because obviously, whoever has a higher speed will diffuse faster."},{"Start":"03:55.700 ","End":"04:02.560","Text":"Now (u_rms)A is the square root of 3RT divided by the molar mass of"},{"Start":"04:02.560 ","End":"04:10.300","Text":"molecules A and of B is square root of 3RT divided by the molar mass of B."},{"Start":"04:10.300 ","End":"04:15.585","Text":"3RT cancels top and bottom and then Mw_A"},{"Start":"04:15.585 ","End":"04:21.070","Text":"divided by 1 over Mw_B is just Mw_B over Mw_A,"},{"Start":"04:21.070 ","End":"04:23.350","Text":"and we have the square root of that."},{"Start":"04:23.350 ","End":"04:28.280","Text":"Here we have the rate of effusion of Gas A divided by rate of effusion of"},{"Start":"04:28.280 ","End":"04:33.770","Text":"Gas B is equal to the molar mass of B divided by molar mass of A."},{"Start":"04:33.770 ","End":"04:35.180","Text":"See the order changes."},{"Start":"04:35.180 ","End":"04:37.775","Text":"We have A at the top here and B in the bottom here,"},{"Start":"04:37.775 ","End":"04:42.080","Text":"and here we have the inverse because the rate of effusion"},{"Start":"04:42.080 ","End":"04:46.970","Text":"of a gas is inversely proportion to the square root of its molar mass."},{"Start":"04:46.970 ","End":"04:50.465","Text":"This is called Graham\u0027s law of effusion."},{"Start":"04:50.465 ","End":"04:52.765","Text":"Let\u0027s solve a problem."},{"Start":"04:52.765 ","End":"04:58.175","Text":"If 10 to power minus 4 moles of hydrogen gas effuses through a small hole,"},{"Start":"04:58.175 ","End":"05:03.265","Text":"50 seconds, how much nitrogen gas would effuse in 50 seconds?"},{"Start":"05:03.265 ","End":"05:07.970","Text":"First thing to notice, nitrogen gas is much heavier than hydrogen gas,"},{"Start":"05:07.970 ","End":"05:12.530","Text":"so we expect less to effuse in a particular time."},{"Start":"05:12.530 ","End":"05:14.960","Text":"Here we\u0027ve written 50 seconds,"},{"Start":"05:14.960 ","End":"05:20.825","Text":"but it really doesn\u0027t matter what time as long as they\u0027re both studied for the same time."},{"Start":"05:20.825 ","End":"05:22.685","Text":"This is what we said before."},{"Start":"05:22.685 ","End":"05:24.965","Text":"As nitrogen is heavier than hydrogen,"},{"Start":"05:24.965 ","End":"05:29.090","Text":"we expect less nitrogen to effuse in the same time."},{"Start":"05:29.090 ","End":"05:34.960","Text":"Now, the rate of effusion is the amount that effuses divide by the time of effusion."},{"Start":"05:34.960 ","End":"05:38.220","Text":"The time is the same for both gases,"},{"Start":"05:38.220 ","End":"05:42.830","Text":"we can write that the rate of effusion of nitrogen compared to the rate of effusion of"},{"Start":"05:42.830 ","End":"05:45.590","Text":"hydrogen is equal to the number of"},{"Start":"05:45.590 ","End":"05:48.785","Text":"moles of nitrogen divided by the number of moles of hydrogen."},{"Start":"05:48.785 ","End":"05:51.200","Text":"Because the time cancels,"},{"Start":"05:51.200 ","End":"05:54.620","Text":"time appears in the top and the bottom, and it cancels."},{"Start":"05:54.620 ","End":"05:58.550","Text":"That is equal. We have the expression for the Graham\u0027s law."},{"Start":"05:58.550 ","End":"06:01.100","Text":"This is equal to the square root of"},{"Start":"06:01.100 ","End":"06:04.370","Text":"the molar mass of hydrogen compared to the molar mass of nitrogen,"},{"Start":"06:04.370 ","End":"06:07.715","Text":"remember nitrogen is at the top here and at the bottom here."},{"Start":"06:07.715 ","End":"06:11.870","Text":"The molar mass of hydrogen is 2.016,"},{"Start":"06:11.870 ","End":"06:15.545","Text":"the molar mass of nitrogen is 28.01."},{"Start":"06:15.545 ","End":"06:20.120","Text":"Take the square root of that, we get 0.268."},{"Start":"06:20.120 ","End":"06:22.595","Text":"Now we can solve the problem,"},{"Start":"06:22.595 ","End":"06:28.920","Text":"the moles of nitrogen compared to the moles of hydrogen are equal to 0.268."},{"Start":"06:28.920 ","End":"06:32.765","Text":"We can write that the moles of nitrogen is equal to the moles of hydrogen,"},{"Start":"06:32.765 ","End":"06:38.725","Text":"we are multiplying both sides by the moles of hydrogen times 0.268."},{"Start":"06:38.725 ","End":"06:40.370","Text":"We have 10 to the minus 4,"},{"Start":"06:40.370 ","End":"06:45.110","Text":"that\u0027s the moles of hydrogen that\u0027s given times 0.268."},{"Start":"06:45.110 ","End":"06:47.015","Text":"We multiply these out,"},{"Start":"06:47.015 ","End":"06:51.305","Text":"we get 2.68 times 10 to the minus 5 moles,"},{"Start":"06:51.305 ","End":"06:53.630","Text":"which is, as expected,"},{"Start":"06:53.630 ","End":"06:59.030","Text":"smaller than 10 to the minus 4 moles of hydrogen."},{"Start":"06:59.030 ","End":"07:05.000","Text":"Now, there are other problems that we can solve using Graham\u0027s law."},{"Start":"07:05.000 ","End":"07:11.165","Text":"For example, we might be asked to calculate the distance traveled in a specific time."},{"Start":"07:11.165 ","End":"07:15.185","Text":"We know that the lighter gas will travel further in a particular time."},{"Start":"07:15.185 ","End":"07:18.065","Text":"U is equal to dx by dt."},{"Start":"07:18.065 ","End":"07:20.840","Text":"If you don\u0027t want to use calculus,"},{"Start":"07:20.840 ","End":"07:26.450","Text":"then you know that u is equal to the distance Delta x divided by the time"},{"Start":"07:26.450 ","End":"07:33.380","Text":"Delta t. Distance is proportional to speed."},{"Start":"07:33.380 ","End":"07:35.375","Text":"We can use the same law."},{"Start":"07:35.375 ","End":"07:37.070","Text":"Distance traveled by hydrogen,"},{"Start":"07:37.070 ","End":"07:39.470","Text":"compared to distance traveled by nitrogen is equal to"},{"Start":"07:39.470 ","End":"07:44.210","Text":"square root of the molar mass of nitrogen compared to the molar mass of hydrogen."},{"Start":"07:44.210 ","End":"07:48.085","Text":"Remember, hydrogen is at the top here and the bottom here."},{"Start":"07:48.085 ","End":"07:51.170","Text":"Alternatively, we might be asked to calculate"},{"Start":"07:51.170 ","End":"07:54.200","Text":"the time taken to travel a particular distance."},{"Start":"07:54.200 ","End":"07:59.300","Text":"We can imagine that if we have a heavy molecule and a light molecule,"},{"Start":"07:59.300 ","End":"08:02.360","Text":"then the light molecule will take"},{"Start":"08:02.360 ","End":"08:07.471","Text":"less time to travel a particular distance than the heavy molecule."},{"Start":"08:07.471 ","End":"08:11.315","Text":"The lighter gas will take less time to travel the same distance."},{"Start":"08:11.315 ","End":"08:14.270","Text":"Now, we can do that mathematically."},{"Start":"08:14.270 ","End":"08:17.090","Text":"The speed is equal to the exponent t,"},{"Start":"08:17.090 ","End":"08:20.180","Text":"or if you prefer not to use calculus,"},{"Start":"08:20.180 ","End":"08:24.550","Text":"u is equal to Delta x divided by Delta t,"},{"Start":"08:24.550 ","End":"08:29.870","Text":"so the speed is inversely proportional to the time."},{"Start":"08:29.870 ","End":"08:34.790","Text":"This means that when we compare the 2 times,"},{"Start":"08:34.790 ","End":"08:38.480","Text":"time taken by hydrogen compared to time taken by nitrogen,"},{"Start":"08:38.480 ","End":"08:41.945","Text":"we have to have the lighter one on top."},{"Start":"08:41.945 ","End":"08:44.060","Text":"Because the time taken by hydrogen is going to"},{"Start":"08:44.060 ","End":"08:45.890","Text":"be smaller than the time taken by nitrogen,"},{"Start":"08:45.890 ","End":"08:48.470","Text":"so that\u0027s going to be less than 1."},{"Start":"08:48.470 ","End":"08:53.510","Text":"The mass of hydrogen is smaller than the mass of nitrogen."},{"Start":"08:53.510 ","End":"08:55.640","Text":"This will also be less than one,"},{"Start":"08:55.640 ","End":"08:58.640","Text":"which shows that this is the way we need to do it."},{"Start":"08:58.640 ","End":"09:04.890","Text":"In this video, we learned about Graham\u0027s law of effusion."}],"ID":18169},{"Watched":false,"Name":"Exercise 1","Duration":"5m 24s","ChapterTopicVideoID":31936,"CourseChapterTopicPlaylistID":80097,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34212},{"Watched":false,"Name":"Exercise 2","Duration":"4m 49s","ChapterTopicVideoID":31937,"CourseChapterTopicPlaylistID":80097,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34213},{"Watched":false,"Name":"Exercise 3","Duration":"4m 49s","ChapterTopicVideoID":31935,"CourseChapterTopicPlaylistID":80097,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34214}],"Thumbnail":null,"ID":80097},{"Name":"Nonideal Gases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Nonideal Gases - Theory","Duration":"8m 55s","ChapterTopicVideoID":17423,"CourseChapterTopicPlaylistID":80098,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"In previous videos, we spoke a great deal about ideal gases."},{"Start":"00:04.380 ","End":"00:08.745","Text":"In this video, we\u0027ll discuss deviations from ideal behavior."},{"Start":"00:08.745 ","End":"00:13.125","Text":"Let\u0027s begin by defining the compressibility factor."},{"Start":"00:13.125 ","End":"00:16.980","Text":"The ideal gas law is PV = nRT,"},{"Start":"00:16.980 ","End":"00:19.815","Text":"which you probably know by heart by now."},{"Start":"00:19.815 ","End":"00:23.610","Text":"If we divide both sides of the equation by nRT,"},{"Start":"00:23.610 ","End":"00:27.465","Text":"we get PV divided by nRT = 1."},{"Start":"00:27.465 ","End":"00:32.535","Text":"So PV over nRT is called the compressibility factor."},{"Start":"00:32.535 ","End":"00:36.810","Text":"We can see it\u0027s equal to 1 for ideal gases."},{"Start":"00:36.810 ","End":"00:38.580","Text":"It\u0027s 1 for ideal gases,"},{"Start":"00:38.580 ","End":"00:42.360","Text":"and it can be greater or less than 1 for non-ideal gases."},{"Start":"00:42.360 ","End":"00:47.750","Text":"The compressibility factor is a measure of how ideal a gas is."},{"Start":"00:47.750 ","End":"00:50.900","Text":"Now we should note that a particular gas can exhibit"},{"Start":"00:50.900 ","End":"00:55.340","Text":"the ideal gas behavior for certain values of pressure and temperature,"},{"Start":"00:55.340 ","End":"00:58.925","Text":"and non-ideal behavior at other temperatures and pressures."},{"Start":"00:58.925 ","End":"01:05.285","Text":"There\u0027s no such thing as a particular gas that\u0027s always ideal or always non-ideal."},{"Start":"01:05.285 ","End":"01:08.525","Text":"It all depends on the temperature and pressure."},{"Start":"01:08.525 ","End":"01:15.515","Text":"Now we\u0027re going to mention some experimental facts any good theory should explain."},{"Start":"01:15.515 ","End":"01:19.685","Text":"All gases behave ideally at low pressure and high temperatures."},{"Start":"01:19.685 ","End":"01:25.624","Text":"Conversely, they behave non-ideally at high pressure and low temperatures."},{"Start":"01:25.624 ","End":"01:30.590","Text":"At high pressures, the compressibility of all gases is greater than 1."},{"Start":"01:30.590 ","End":"01:33.020","Text":"If we look at experimental results,"},{"Start":"01:33.020 ","End":"01:36.875","Text":"we see that at 300 Kelvin temperature,"},{"Start":"01:36.875 ","End":"01:39.560","Text":"and below a pressure of 10 bar,"},{"Start":"01:39.560 ","End":"01:42.365","Text":"all these molecules like hydrogen, helium,"},{"Start":"01:42.365 ","End":"01:47.420","Text":"carbon monoxide, nitrogen, oxygen, behave almost ideally."},{"Start":"01:47.420 ","End":"01:51.080","Text":"The compressibility factor is approximately 1."},{"Start":"01:51.080 ","End":"01:54.595","Text":"Two examples of non-ideal behavior."},{"Start":"01:54.595 ","End":"01:58.430","Text":"NH_3, ammonia and sulfur hexafluoride,"},{"Start":"01:58.430 ","End":"02:02.255","Text":"SF_6, which are large in size."},{"Start":"02:02.255 ","End":"02:07.747","Text":"Then for them, PV divided by nRT is approximately 0.88."},{"Start":"02:07.747 ","End":"02:10.490","Text":"So it is a non-ideal behavior."},{"Start":"02:10.490 ","End":"02:18.465","Text":"They exhibit non-ideal behavior under these circumstances at 300 Kelvin and 10 bar."},{"Start":"02:18.465 ","End":"02:22.295","Text":"Well, what are the reasons for deviations from ideal behavior?"},{"Start":"02:22.295 ","End":"02:25.165","Text":"There are two main reasons,"},{"Start":"02:25.165 ","End":"02:26.795","Text":"which we\u0027ll discuss now."},{"Start":"02:26.795 ","End":"02:30.050","Text":"The first one is the volume of the molecules."},{"Start":"02:30.050 ","End":"02:32.855","Text":"One of the postulates of KMT,"},{"Start":"02:32.855 ","End":"02:34.268","Text":"kinetic molecular theory,"},{"Start":"02:34.268 ","End":"02:35.570","Text":"that we learned about,"},{"Start":"02:35.570 ","End":"02:40.055","Text":"is that the volume of the molecules is negligible compared to the total volume."},{"Start":"02:40.055 ","End":"02:42.335","Text":"Now, this isn\u0027t always true."},{"Start":"02:42.335 ","End":"02:45.110","Text":"It\u0027s certainly true at low pressures."},{"Start":"02:45.110 ","End":"02:46.925","Text":"However, at high pressures,"},{"Start":"02:46.925 ","End":"02:48.740","Text":"the volume of the gas decreases"},{"Start":"02:48.740 ","End":"02:53.450","Text":"significantly so that the volume of the molecules cannot be ignored."},{"Start":"02:53.450 ","End":"02:55.400","Text":"We can draw this."},{"Start":"02:55.400 ","End":"02:57.590","Text":"When the pressure is low,"},{"Start":"02:57.590 ","End":"03:00.500","Text":"then the volume might be quite large."},{"Start":"03:00.500 ","End":"03:05.210","Text":"So the volume of the individual molecules isn\u0027t really important."},{"Start":"03:05.210 ","End":"03:08.695","Text":"However, at high pressure,"},{"Start":"03:08.695 ","End":"03:10.405","Text":"the volume is small."},{"Start":"03:10.405 ","End":"03:12.690","Text":"We\u0027re compressing the gas."},{"Start":"03:12.690 ","End":"03:18.730","Text":"Then the volume of the individual molecules is significant compared to the total volume."},{"Start":"03:18.730 ","End":"03:20.740","Text":"So at high pressures,"},{"Start":"03:20.740 ","End":"03:25.390","Text":"we cannot ignore the volume of the actual molecules."},{"Start":"03:25.390 ","End":"03:27.820","Text":"Now, what about intermolecular forces?"},{"Start":"03:27.820 ","End":"03:31.795","Text":"That\u0027s the second reason for deviation from ideal behavior."},{"Start":"03:31.795 ","End":"03:36.700","Text":"Intermolecular forces are the forces between molecules."},{"Start":"03:36.700 ","End":"03:39.100","Text":"Now, if you remember, one of the postulates of"},{"Start":"03:39.100 ","End":"03:44.032","Text":"kinetic molecular theory is that the molecules neither attract or repel each other,"},{"Start":"03:44.032 ","End":"03:45.760","Text":"when they\u0027re not actually colliding."},{"Start":"03:45.760 ","End":"03:47.585","Text":"They just ignore each other."},{"Start":"03:47.585 ","End":"03:50.005","Text":"Now, this is not true."},{"Start":"03:50.005 ","End":"03:54.205","Text":"Molecules usually attract each other as they approach each other."},{"Start":"03:54.205 ","End":"03:55.810","Text":"When they\u0027re far away,"},{"Start":"03:55.810 ","End":"03:56.980","Text":"they attract each other."},{"Start":"03:56.980 ","End":"03:59.515","Text":"It\u0027s called van der Waals forces."},{"Start":"03:59.515 ","End":"04:01.360","Text":"When they get very close,"},{"Start":"04:01.360 ","End":"04:03.280","Text":"they repel each other."},{"Start":"04:03.280 ","End":"04:09.430","Text":"These attractive forces cause the pressure exerted on the walls to decrease."},{"Start":"04:09.430 ","End":"04:11.110","Text":"How does that work out?"},{"Start":"04:11.110 ","End":"04:13.960","Text":"Supposing we have a molecule and it\u0027s going to hit"},{"Start":"04:13.960 ","End":"04:18.865","Text":"the wall of the container to exert pressure on the wall."},{"Start":"04:18.865 ","End":"04:23.320","Text":"Now if there are other molecules that are attracting it,"},{"Start":"04:23.320 ","End":"04:25.300","Text":"that are pulling it towards them,"},{"Start":"04:25.300 ","End":"04:28.300","Text":"then the molecule will take more time,"},{"Start":"04:28.300 ","End":"04:32.435","Text":"will be less likely to exert pressure on the walls."},{"Start":"04:32.435 ","End":"04:36.470","Text":"The pressure exerted will be smaller than"},{"Start":"04:36.470 ","End":"04:41.315","Text":"the pressure in the absence of these additional molecules."},{"Start":"04:41.315 ","End":"04:44.590","Text":"Now these intermolecular forces,"},{"Start":"04:44.590 ","End":"04:48.994","Text":"attractive forces, are more important at low temperature,"},{"Start":"04:48.994 ","End":"04:50.930","Text":"when the molecules are moving slowly."},{"Start":"04:50.930 ","End":"04:52.130","Text":"When they\u0027re moving slowly,"},{"Start":"04:52.130 ","End":"04:53.345","Text":"they feel each other."},{"Start":"04:53.345 ","End":"04:55.430","Text":"When they\u0027re moving very fast,"},{"Start":"04:55.430 ","End":"04:58.160","Text":"then they don\u0027t take any notice of each other."},{"Start":"04:58.160 ","End":"05:01.940","Text":"Now, there are lots of equations for non-ideal gases."},{"Start":"05:01.940 ","End":"05:03.590","Text":"But the easiest equation,"},{"Start":"05:03.590 ","End":"05:04.790","Text":"the most simple equation,"},{"Start":"05:04.790 ","End":"05:06.920","Text":"is the van der Waals equation."},{"Start":"05:06.920 ","End":"05:09.500","Text":"Remember, we had attractive forces,"},{"Start":"05:09.500 ","End":"05:12.335","Text":"also called van der Waals forces."},{"Start":"05:12.335 ","End":"05:16.957","Text":"The simplest equation for non-ideal gas is a van der Waals equation,"},{"Start":"05:16.957 ","End":"05:21.865","Text":"proposed by Johannes van der Waals in the late 19th century."},{"Start":"05:21.865 ","End":"05:23.665","Text":"Here\u0027s his equation."},{"Start":"05:23.665 ","End":"05:26.645","Text":"There\u0027s a factor related to pressure."},{"Start":"05:26.645 ","End":"05:31.060","Text":"Pressure P plus an^2 divided by V^2."},{"Start":"05:31.060 ","End":"05:32.170","Text":"n is the number of moles,"},{"Start":"05:32.170 ","End":"05:33.870","Text":"V is the volume."},{"Start":"05:33.870 ","End":"05:35.876","Text":"P, of course, is a pressure."},{"Start":"05:35.876 ","End":"05:38.910","Text":"There\u0027s a volume factor,"},{"Start":"05:38.910 ","End":"05:41.560","Text":"V minus nb = nRT."},{"Start":"05:42.560 ","End":"05:50.396","Text":"You can see that if an^2 over V^2 is small,"},{"Start":"05:50.396 ","End":"05:54.600","Text":"if an^2 over V^2 is much less than the pressure,"},{"Start":"05:54.600 ","End":"05:59.225","Text":"and if nb is much less than the volume,"},{"Start":"05:59.225 ","End":"06:03.200","Text":"this will just be like the ideal gas equation."},{"Start":"06:03.200 ","End":"06:05.390","Text":"This is ignored and this is ignored."},{"Start":"06:05.390 ","End":"06:08.165","Text":"We just have PV = nRT."},{"Start":"06:08.165 ","End":"06:12.185","Text":"We know when a non-ideal gas will become an ideal gas,"},{"Start":"06:12.185 ","End":"06:17.550","Text":"when an^2 over V^2 is small compared to P,"},{"Start":"06:17.550 ","End":"06:20.515","Text":"and when nb is small compared to V."},{"Start":"06:20.515 ","End":"06:23.750","Text":"Now the parameters a and b are different for different gases."},{"Start":"06:23.750 ","End":"06:27.169","Text":"So this is not a totally general equation,"},{"Start":"06:27.169 ","End":"06:32.525","Text":"it changes from gas to gas depending on the values of a and b."},{"Start":"06:32.525 ","End":"06:36.620","Text":"Now the term an^2 over V^2 accounts for"},{"Start":"06:36.620 ","End":"06:43.025","Text":"the intermolecular forces and the term nb for the volume of the molecules."},{"Start":"06:43.025 ","End":"06:45.665","Text":"If nb is significant,"},{"Start":"06:45.665 ","End":"06:49.625","Text":"then the volume will be reduced, V minus nb."},{"Start":"06:49.625 ","End":"06:52.700","Text":"The actual volume will be reduced."},{"Start":"06:52.700 ","End":"06:57.840","Text":"In a minute we\u0027ll see what the an^2 over V^2 does to the pressure."},{"Start":"06:57.840 ","End":"07:00.890","Text":"It\u0027s sufficient to see this moment that"},{"Start":"07:00.890 ","End":"07:05.555","Text":"the term an^2 over V^2 accounts for the intermolecular forces."},{"Start":"07:05.555 ","End":"07:09.230","Text":"The term nb for the volume of the molecules."},{"Start":"07:09.230 ","End":"07:11.165","Text":"Let\u0027s see what happens to the pressure."},{"Start":"07:11.165 ","End":"07:15.890","Text":"We have a pressure term times a volume term equal to nRT."},{"Start":"07:15.890 ","End":"07:20.560","Text":"Let\u0027s divide both sides of the equation by V minus nb,"},{"Start":"07:20.560 ","End":"07:28.670","Text":"so we get P plus an^2 over V^2 equal to nRT divided by V minus nb."},{"Start":"07:28.670 ","End":"07:30.853","Text":"This canceled here,"},{"Start":"07:30.853 ","End":"07:35.610","Text":"and here it\u0027s V minus nb. That\u0027s our equation."},{"Start":"07:35.610 ","End":"07:39.005","Text":"Now, we can write an equation just for the pressure."},{"Start":"07:39.005 ","End":"07:43.714","Text":"The pressure is equal to nRT over V minus nb."},{"Start":"07:43.714 ","End":"07:47.715","Text":"We\u0027re subtracting an^2 over V^2."},{"Start":"07:47.715 ","End":"07:56.825","Text":"Here we have pressure = nRT over V minus nb minus an^2 over V^2."},{"Start":"07:56.825 ","End":"08:02.360","Text":"Now, a and b are always positive."},{"Start":"08:02.360 ","End":"08:08.135","Text":"That means that an^2 divided by V^2 is also positive."},{"Start":"08:08.135 ","End":"08:13.220","Text":"That means it\u0027s reducing the pressure because it\u0027s negative sign."},{"Start":"08:13.220 ","End":"08:15.460","Text":"So the pressure will be smaller."},{"Start":"08:15.460 ","End":"08:17.640","Text":"We can say, approximately,"},{"Start":"08:17.640 ","End":"08:22.790","Text":"that P real for real gas is approximately P ideal,"},{"Start":"08:22.790 ","End":"08:24.320","Text":"if we ignore the nb,"},{"Start":"08:24.320 ","End":"08:29.380","Text":"it\u0027s nRT over V. That\u0027s the pressure for an ideal gas,"},{"Start":"08:29.380 ","End":"08:32.115","Text":"minus an^2 over V^2,"},{"Start":"08:32.115 ","End":"08:38.870","Text":"so the real pressure will be the ideal pressure minus a positive quantity."},{"Start":"08:38.870 ","End":"08:44.675","Text":"That means that P real will be smaller than P ideal."},{"Start":"08:44.675 ","End":"08:50.375","Text":"So in this video, we learned about van der Waals equation for non-ideal gases."},{"Start":"08:50.375 ","End":"08:55.260","Text":"In the next video, we\u0027ll solve a problem using this equation."}],"ID":18170},{"Watched":false,"Name":"Nonideal Gases - Example","Duration":"4m 47s","ChapterTopicVideoID":17424,"CourseChapterTopicPlaylistID":80098,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.930","Text":"In the previous video,"},{"Start":"00:01.930 ","End":"00:05.350","Text":"we discussed Van der Waals equation for non ideal gases."},{"Start":"00:05.350 ","End":"00:09.850","Text":"Now, we\u0027ll solve a problem based on this equation. Here\u0027s our problem."},{"Start":"00:09.850 ","End":"00:14.230","Text":"Calculate the pressure exerted by 0.5 moles of CO_2,"},{"Start":"00:14.230 ","End":"00:19.855","Text":"carbon dioxide gas in a volume of 2 liters at 273 Kelvin."},{"Start":"00:19.855 ","End":"00:24.190","Text":"First use the ideal gas equation and then Van der Waals equation."},{"Start":"00:24.190 ","End":"00:26.605","Text":"We\u0027re given the following details."},{"Start":"00:26.605 ","End":"00:33.550","Text":"For CO_2, a is equal to 3.658 bars times liter squared per mole squared."},{"Start":"00:33.550 ","End":"00:38.000","Text":"B is equal to 0.0429 liters per mole."},{"Start":"00:38.000 ","End":"00:39.730","Text":"Since we\u0027re working in bars,"},{"Start":"00:39.730 ","End":"00:41.785","Text":"where bars is definition of a,"},{"Start":"00:41.785 ","End":"00:45.885","Text":"then we need R the gas constant also in bars."},{"Start":"00:45.885 ","End":"00:52.855","Text":"R is equal to 0.08315 bars times liter per mole per Kelvin."},{"Start":"00:52.855 ","End":"00:56.120","Text":"Let\u0027s start off with the ideal gas equation."},{"Start":"00:56.120 ","End":"00:58.800","Text":"We know that PV equals nRT,"},{"Start":"00:58.800 ","End":"01:00.520","Text":"so P for an ideal gas,"},{"Start":"01:00.520 ","End":"01:02.125","Text":"I\u0027ve written it as P ideal,"},{"Start":"01:02.125 ","End":"01:06.625","Text":"is equal to nRT divided by V. Let\u0027s substitute the numbers."},{"Start":"01:06.625 ","End":"01:10.235","Text":"For moles, we have 0.5 moles,"},{"Start":"01:10.235 ","End":"01:16.644","Text":"R we have 0.08315 bars times liters per mole per Kelvin."},{"Start":"01:16.644 ","End":"01:19.945","Text":"For the temperature, we have 273 Kelvin,"},{"Start":"01:19.945 ","End":"01:23.490","Text":"and it\u0027s divided by the volume which is 2 liters."},{"Start":"01:23.490 ","End":"01:25.410","Text":"Let\u0027s first look at the units."},{"Start":"01:25.410 ","End":"01:28.965","Text":"We have mole times mole^minus 1 which is 1,"},{"Start":"01:28.965 ","End":"01:32.460","Text":"K^minus 1 times K which is 1."},{"Start":"01:32.460 ","End":"01:34.510","Text":"Liter cancels with liter,"},{"Start":"01:34.510 ","End":"01:36.380","Text":"and all we\u0027re left with is bar."},{"Start":"01:36.380 ","End":"01:37.940","Text":"We work out the numbers."},{"Start":"01:37.940 ","End":"01:41.225","Text":"We get 5.675 bars."},{"Start":"01:41.225 ","End":"01:43.160","Text":"That\u0027s the number we get,"},{"Start":"01:43.160 ","End":"01:46.775","Text":"that sense that we get for the pressure when use the ideal gas equation."},{"Start":"01:46.775 ","End":"01:51.110","Text":"Let\u0027s see now what happens when we use Van der Waals equation."},{"Start":"01:51.110 ","End":"01:57.260","Text":"We saw in the last video that the pressure of a real gas according to Van der Waals"},{"Start":"01:57.260 ","End":"02:03.560","Text":"is nRT divided by V minus nb minus n^2 over v^2."},{"Start":"02:03.560 ","End":"02:05.860","Text":"Let\u0027s substitute the numbers."},{"Start":"02:05.860 ","End":"02:13.160","Text":"Once again, nRT is the same as we had before to ideal gas is 0.5 moles times R which is"},{"Start":"02:13.160 ","End":"02:20.660","Text":"0.08315 bars times liters per mole per Kelvin and it\u0027s divided by 2 liters,"},{"Start":"02:20.660 ","End":"02:30.245","Text":"which is the volume minus n times b. N is 0.5 moles and b is 0.0429 liters per mole."},{"Start":"02:30.245 ","End":"02:32.015","Text":"Then we have the second part,"},{"Start":"02:32.015 ","End":"02:37.960","Text":"minus an^2 over V. N squared is 0.5^2 mole^2,"},{"Start":"02:37.960 ","End":"02:42.410","Text":"a is 3.658 bars times liter^2 per mole^2,"},{"Start":"02:42.410 ","End":"02:47.520","Text":"and the volume^2 is 2^2 times liters^2."},{"Start":"02:47.520 ","End":"02:50.205","Text":"Let\u0027s first look at the unit."},{"Start":"02:50.205 ","End":"02:56.420","Text":"Here we have in the denominator moles times multiply minus 1, that\u0027s 1."},{"Start":"02:56.420 ","End":"02:59.615","Text":"Now we have liter minus liter and that\u0027s good."},{"Start":"02:59.615 ","End":"03:04.755","Text":"The top we have moles times 4^minus 1 is 1."},{"Start":"03:04.755 ","End":"03:08.640","Text":"Kelvin^minus 1 times Kelvin is 1."},{"Start":"03:08.640 ","End":"03:11.570","Text":"We have bar time liter on the top and liter on the bottom,"},{"Start":"03:11.570 ","End":"03:13.775","Text":"so the liter cancels."},{"Start":"03:13.775 ","End":"03:16.580","Text":"All we\u0027re left with is the bar."},{"Start":"03:16.580 ","End":"03:18.875","Text":"Let\u0027s look at the 2nd part."},{"Start":"03:18.875 ","End":"03:22.325","Text":"We have liters^2 at the top and liters^2 in the bottom."},{"Start":"03:22.325 ","End":"03:26.750","Text":"We have moles^2 times mole^minus 2, so that\u0027s 1."},{"Start":"03:26.750 ","End":"03:29.950","Text":"All we\u0027re left once again is bar."},{"Start":"03:29.950 ","End":"03:31.970","Text":"When we calculate the numbers,"},{"Start":"03:31.970 ","End":"03:35.780","Text":"we get 5.735 bars for the first part,"},{"Start":"03:35.780 ","End":"03:40.765","Text":"minus 0.229 bars for the second part,"},{"Start":"03:40.765 ","End":"03:43.080","Text":"and when we subtract these two numbers,"},{"Start":"03:43.080 ","End":"03:46.819","Text":"we get 5.506 bars."},{"Start":"03:46.819 ","End":"03:51.980","Text":"We can already see this isn\u0027t the same as the ideal gas result."},{"Start":"03:51.980 ","End":"03:54.725","Text":"In fact, it\u0027s smaller than we predicted before."},{"Start":"03:54.725 ","End":"03:57.890","Text":"In the previous video, we predicted that P real should be"},{"Start":"03:57.890 ","End":"04:01.315","Text":"smaller than P ideal and it is indeed."},{"Start":"04:01.315 ","End":"04:04.005","Text":"Let\u0027s compare P real and P ideal."},{"Start":"04:04.005 ","End":"04:07.140","Text":"P real is 5.506 bars,"},{"Start":"04:07.140 ","End":"04:11.010","Text":"and P ideal is 5.675 bars and the ratio of the"},{"Start":"04:11.010 ","End":"04:16.210","Text":"two we\u0027re going to multiply it by a 100 to make it into percentages,"},{"Start":"04:16.210 ","End":"04:19.135","Text":"which gives us 97 percent."},{"Start":"04:19.135 ","End":"04:23.375","Text":"We see that P real is 97 percent of P ideal."},{"Start":"04:23.375 ","End":"04:28.415","Text":"We wouldn\u0027t have gone too far wrong in this case if we just use the ideal gas equation."},{"Start":"04:28.415 ","End":"04:32.690","Text":"But a more exact answer is given by using Van der Waals,"},{"Start":"04:32.690 ","End":"04:40.135","Text":"we see that there\u0027s a small error in using the ideal gas equation of about 3 percent."},{"Start":"04:40.135 ","End":"04:42.980","Text":"In this video, we solved a problem that"},{"Start":"04:42.980 ","End":"04:47.190","Text":"compares the ideal gas equation with Van der Waals equation."}],"ID":18171}],"Thumbnail":null,"ID":80098}]

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