[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.10,0:00:02.14,Default,,0000,0000,0000,,Let's do some more problems
Dialogue: 0,0:00:02.14,0:00:04.56,Default,,0000,0000,0000,,that involve the ideal gas equation.
Dialogue: 0,0:00:04.56,0:00:06.68,Default,,0000,0000,0000,,Let's say I have a gas in a container
Dialogue: 0,0:00:06.68,0:00:15.32,Default,,0000,0000,0000,,and the current pressure is 3 atmospheres.
Dialogue: 0,0:00:15.32,0:00:19.76,Default,,0000,0000,0000,,And let's say that the volume of the container
Dialogue: 0,0:00:19.76,0:00:27.41,Default,,0000,0000,0000,,is, I don't know, 9 liters.
Dialogue: 0,0:00:27.41,0:00:30.14,Default,,0000,0000,0000,,Now, what will the pressure become
Dialogue: 0,0:00:30.14,0:00:39.28,Default,,0000,0000,0000,,if my volume goes from 9 liters to 3 liters?
Dialogue: 0,0:00:39.28,0:00:42.18,Default,,0000,0000,0000,,So from the first ideal gas equation video
Dialogue: 0,0:00:42.18,0:00:43.35,Default,,0000,0000,0000,,you can kind of have the intuition
Dialogue: 0,0:00:43.35,0:00:46.94,Default,,0000,0000,0000,,that you have a bunch of-- and we're holding--
Dialogue: 0,0:00:46.94,0:00:47.90,Default,,0000,0000,0000,,and this is important.
Dialogue: 0,0:00:47.90,0:00:50.76,Default,,0000,0000,0000,,We're holding the temperature constant
Dialogue: 0,0:00:50.76,0:00:52.54,Default,,0000,0000,0000,,and that's an important thing to realize.
Dialogue: 0,0:00:52.54,0:00:58.38,Default,,0000,0000,0000,,So in our very original intuition
Dialogue: 0,0:00:58.38,0:01:00.35,Default,,0000,0000,0000,,behind the ideal gas equation we said,
Dialogue: 0,0:01:00.35,0:01:02.99,Default,,0000,0000,0000,,look, if we have a certain number of particles
Dialogue: 0,0:01:02.99,0:01:06.85,Default,,0000,0000,0000,,with a certain amount of kinetic energy,
Dialogue: 0,0:01:06.85,0:01:08.83,Default,,0000,0000,0000,,and they're exerting a certain pressure
Dialogue: 0,0:01:08.83,0:01:09.78,Default,,0000,0000,0000,,on their container,
Dialogue: 0,0:01:09.78,0:01:14.37,Default,,0000,0000,0000,,and if we were to make the container smaller,
Dialogue: 0,0:01:14.37,0:01:16.20,Default,,0000,0000,0000,,we have the same number of particles.
Dialogue: 0,0:01:16.20,0:01:17.43,Default,,0000,0000,0000,,n doesn't change.
Dialogue: 0,0:01:17.43,0:01:19.88,Default,,0000,0000,0000,,The average kinetic energy doesn't change,
Dialogue: 0,0:01:19.88,0:01:21.66,Default,,0000,0000,0000,,so they're just going to bump into the walls more.
Dialogue: 0,0:01:21.66,0:01:24.22,Default,,0000,0000,0000,,So that when we make the volume smaller,
Dialogue: 0,0:01:24.22,0:01:26.73,Default,,0000,0000,0000,,when the volume goes up----
Dialogue: 0,0:01:26.73,0:01:27.76,Default,,0000,0000,0000,,when the volume goes down,
Dialogue: 0,0:01:27.76,0:01:30.07,Default,,0000,0000,0000,,the pressure should go up.
Dialogue: 0,0:01:30.07,0:01:32.62,Default,,0000,0000,0000,,So let's see if we can calculate the exact number.
Dialogue: 0,0:01:32.62,0:01:35.43,Default,,0000,0000,0000,,So we can take our ideal gas equation:
Dialogue: 0,0:01:35.43,0:01:41.87,Default,,0000,0000,0000,,pressure times volume is equal to nRT.
Dialogue: 0,0:01:41.87,0:01:44.32,Default,,0000,0000,0000,,Now, do the number of particles change
Dialogue: 0,0:01:44.32,0:01:47.98,Default,,0000,0000,0000,,when I did this situation when I shrunk the volume?
Dialogue: 0,0:01:47.98,0:01:48.65,Default,,0000,0000,0000,,No!
Dialogue: 0,0:01:48.65,0:01:49.76,Default,,0000,0000,0000,,We have the same number of particles.
Dialogue: 0,0:01:49.76,0:01:50.92,Default,,0000,0000,0000,,I'm just shrinking the container,
Dialogue: 0,0:01:50.92,0:01:55.20,Default,,0000,0000,0000,,so n is n, R doesn't change, that's a constant,
Dialogue: 0,0:01:55.20,0:01:57.22,Default,,0000,0000,0000,,and then the temperature doesn't change.
Dialogue: 0,0:01:57.22,0:02:00.32,Default,,0000,0000,0000,,So my old pressure times volume
Dialogue: 0,0:02:00.32,0:02:02.69,Default,,0000,0000,0000,,is going to be equal to nRT,
Dialogue: 0,0:02:02.69,0:02:04.31,Default,,0000,0000,0000,,and my new pressure times volume--
Dialogue: 0,0:02:04.31,0:02:07.95,Default,,0000,0000,0000,,so let me call this P1 and V1.
Dialogue: 0,0:02:07.95,0:02:11.00,Default,,0000,0000,0000,,and then P2 is this----
Dialogue: 0,0:02:11.00,0:02:15.60,Default,,0000,0000,0000,,sorry, that's V2.
Dialogue: 0,0:02:15.60,0:02:21.70,Default,,0000,0000,0000,,so V2 is this, and we're trying to figure out P2.
Dialogue: 0,0:02:21.70,0:02:23.13,Default,,0000,0000,0000,,P2 is what?
Dialogue: 0,0:02:23.13,0:02:31.35,Default,,0000,0000,0000,,Well, we know that P1 times V1 is equal to nRT,
Dialogue: 0,0:02:31.35,0:02:33.40,Default,,0000,0000,0000,,and we also know that since temperature and
Dialogue: 0,0:02:33.40,0:02:35.98,Default,,0000,0000,0000,,the number of moles of our gas stay constant,
Dialogue: 0,0:02:35.98,0:02:40.79,Default,,0000,0000,0000,,that P2 times V2 is equal to nRT.
Dialogue: 0,0:02:40.79,0:02:43.20,Default,,0000,0000,0000,,And since they both equal the same thing,
Dialogue: 0,0:02:43.20,0:02:45.67,Default,,0000,0000,0000,,we can say that the pressure times the volume,
Dialogue: 0,0:02:45.67,0:02:47.80,Default,,0000,0000,0000,,as long as the temperature is held constant,
Dialogue: 0,0:02:47.80,0:02:49.20,Default,,0000,0000,0000,,will be a constant.
Dialogue: 0,0:02:49.20,0:02:55.76,Default,,0000,0000,0000,,So P1 times V1 is going to equal P2 times V2.
Dialogue: 0,0:02:55.76,0:02:57.94,Default,,0000,0000,0000,,So what was P1?
Dialogue: 0,0:02:57.94,0:03:03.23,Default,,0000,0000,0000,,P1, our initial pressure, was 3 atmospheres.
Dialogue: 0,0:03:06.63,0:03:12.02,Default,,0000,0000,0000,,So 3 atmospheres times 9 liters is equal to
Dialogue: 0,0:03:12.02,0:03:15.98,Default,,0000,0000,0000,,our new pressure times 3 liters.
Dialogue: 0,0:03:15.98,0:03:18.99,Default,,0000,0000,0000,,And if we divide both sides of the equation by 3,
Dialogue: 0,0:03:18.99,0:03:24.70,Default,,0000,0000,0000,,we get 3 liters cancel out,
Dialogue: 0,0:03:24.70,0:03:33.64,Default,,0000,0000,0000,,we're left with 9 atmospheres.
Dialogue: 0,0:03:33.64,0:03:34.80,Default,,0000,0000,0000,,And that should make sense.
Dialogue: 0,0:03:34.80,0:03:39.26,Default,,0000,0000,0000,,When you decrease the volume by 2/3
Dialogue: 0,0:03:39.26,0:03:40.30,Default,,0000,0000,0000,,or when you make the volume
Dialogue: 0,0:03:40.30,0:03:42.94,Default,,0000,0000,0000,,1/3 of your original volume,
Dialogue: 0,0:03:42.94,0:03:46.20,Default,,0000,0000,0000,,then your pressure increases by a factor of three.
Dialogue: 0,0:03:46.20,0:03:51.57,Default,,0000,0000,0000,,So this went by times 3, and this went by times 1/3.
Dialogue: 0,0:03:51.57,0:03:52.90,Default,,0000,0000,0000,,That's a useful thing to know in general.
Dialogue: 0,0:03:52.90,0:03:55.20,Default,,0000,0000,0000,,If temperature is held constant,
Dialogue: 0,0:03:55.20,0:03:57.48,Default,,0000,0000,0000,,then pressure times volume
Dialogue: 0,0:03:57.48,0:03:59.11,Default,,0000,0000,0000,,are going to be a constant.
Dialogue: 0,0:03:59.11,0:04:00.96,Default,,0000,0000,0000,,Now, you can take that even further.
Dialogue: 0,0:04:00.96,0:04:06.88,Default,,0000,0000,0000,,If we look at PV equals nRT,
Dialogue: 0,0:04:06.88,0:04:09.16,Default,,0000,0000,0000,,the two things that we know don't change
Dialogue: 0,0:04:09.16,0:04:11.84,Default,,0000,0000,0000,,in the vast majority of exercises we do
Dialogue: 0,0:04:11.84,0:04:13.54,Default,,0000,0000,0000,,is the number of molecules we're dealing with,
Dialogue: 0,0:04:13.54,0:04:15.53,Default,,0000,0000,0000,,and obviously, R isn't going to change.
Dialogue: 0,0:04:15.53,0:04:18.26,Default,,0000,0000,0000,,So if we divide both sides of this by T,
Dialogue: 0,0:04:18.26,0:04:23.16,Default,,0000,0000,0000,,we get PV over T is equal to nR,
Dialogue: 0,0:04:23.16,0:04:24.92,Default,,0000,0000,0000,,or you could say it's equal to a constant.
Dialogue: 0,0:04:24.92,0:04:27.20,Default,,0000,0000,0000,,This is going to be a constant number for any system
Dialogue: 0,0:04:27.20,0:04:28.63,Default,,0000,0000,0000,,where we're not changing
Dialogue: 0,0:04:28.63,0:04:31.52,Default,,0000,0000,0000,,the number of molecules in the container.
Dialogue: 0,0:04:31.52,0:04:33.37,Default,,0000,0000,0000,,So, if we are changing the pressure----
Dialogue: 0,0:04:33.37,0:04:35.65,Default,,0000,0000,0000,,So if initially we start with
Dialogue: 0,0:04:35.65,0:04:40.00,Default,,0000,0000,0000,,pressure one, volume one, and some temperature one
Dialogue: 0,0:04:40.00,0:04:41.50,Default,,0000,0000,0000,,that's going to be equal to this constant.
Dialogue: 0,0:04:41.50,0:04:44.19,Default,,0000,0000,0000,,And if we change any of them,
Dialogue: 0,0:04:44.19,0:04:44.73,Default,,0000,0000,0000,,we go back to
Dialogue: 0,0:04:44.73,0:04:48.86,Default,,0000,0000,0000,,pressure two, volume two, temperature two,
Dialogue: 0,0:04:48.86,0:04:50.47,Default,,0000,0000,0000,,they should still be equal to this constant,
Dialogue: 0,0:04:50.47,0:04:51.47,Default,,0000,0000,0000,,so they equal each other.
Dialogue: 0,0:04:51.47,0:04:55.35,Default,,0000,0000,0000,,So for example, let's say I start off with a
Dialogue: 0,0:04:55.35,0:05:01.08,Default,,0000,0000,0000,,pressure of 1 atmosphere.
Dialogue: 0,0:05:01.08,0:05:05.07,Default,,0000,0000,0000,,and I have a volume of----
Dialogue: 0,0:05:05.07,0:05:08.61,Default,,0000,0000,0000,,I'll switch units here just to do things differently
Dialogue: 0,0:05:08.61,0:05:10.64,Default,,0000,0000,0000,,----2 meters cubed.
Dialogue: 0,0:05:10.64,0:05:20.21,Default,,0000,0000,0000,,And let's say our temperature is 27 degrees Celsius.
Dialogue: 0,0:05:20.21,0:05:21.74,Default,,0000,0000,0000,,Well, and I just wrote Celsius
Dialogue: 0,0:05:21.74,0:05:22.70,Default,,0000,0000,0000,,because I want you to always remember
Dialogue: 0,0:05:22.70,0:05:23.97,Default,,0000,0000,0000,,you have to convert to Kelvin,
Dialogue: 0,0:05:23.97,0:05:27.83,Default,,0000,0000,0000,,so 27 degrees plus 273 will get us
Dialogue: 0,0:05:27.83,0:05:33.15,Default,,0000,0000,0000,,exactly to 300 Kelvin.
Dialogue: 0,0:05:33.15,0:05:39.53,Default,,0000,0000,0000,,And let's say that our new temperature is
Dialogue: 0,0:05:39.53,0:05:40.63,Default,,0000,0000,0000,,Actually let's figure out what the new temperature
Dialogue: 0,0:05:40.63,0:05:41.42,Default,,0000,0000,0000,,is going to be.
Dialogue: 0,0:05:41.42,0:05:46.27,Default,,0000,0000,0000,,Let's say our new pressure is 2 atmospheres.
Dialogue: 0,0:05:46.27,0:05:47.88,Default,,0000,0000,0000,,The pressure has increased.
Dialogue: 0,0:05:47.88,0:05:50.01,Default,,0000,0000,0000,,Let's say we make the container smaller,
Dialogue: 0,0:05:50.01,0:05:52.49,Default,,0000,0000,0000,,so 1 meter cubed.
Dialogue: 0,0:05:52.49,0:05:55.10,Default,,0000,0000,0000,,So the container has been decreased by half
Dialogue: 0,0:05:55.10,0:05:56.68,Default,,0000,0000,0000,,and the pressure is doubled by half.
Dialogue: 0,0:05:56.68,0:05:57.59,Default,,0000,0000,0000,,So you could guess.
Dialogue: 0,0:05:57.59,0:06:02.15,Default,,0000,0000,0000,,You know, we have made the pressure higher----
Dialogue: 0,0:06:02.15,0:06:08.18,Default,,0000,0000,0000,,Let me make the container even smaller.
Dialogue: 0,0:06:08.18,0:06:08.77,Default,,0000,0000,0000,,Actually, no.
Dialogue: 0,0:06:08.77,0:06:10.71,Default,,0000,0000,0000,,Let me make the pressure even larger.
Dialogue: 0,0:06:10.71,0:06:14.26,Default,,0000,0000,0000,,Let me make the pressure into 5 atmospheres.
Dialogue: 0,0:06:14.26,0:06:16.94,Default,,0000,0000,0000,,Now we want to know what the second temperature is
Dialogue: 0,0:06:16.94,0:06:18.81,Default,,0000,0000,0000,,and we set up our equation.
Dialogue: 0,0:06:18.81,0:06:19.53,Default,,0000,0000,0000,,And so we have
Dialogue: 0,0:06:19.53,0:06:28.10,Default,,0000,0000,0000,,2/300 atmosphere meters cubed per Kelvin
Dialogue: 0,0:06:28.10,0:06:32.69,Default,,0000,0000,0000,,is equal to 5/T2, our new temperature,
Dialogue: 0,0:06:32.69,0:06:40.15,Default,,0000,0000,0000,,and then we have 1,500 is equal to 2T2.
Dialogue: 0,0:06:40.15,0:06:41.37,Default,,0000,0000,0000,,Divide both sides by 2.
Dialogue: 0,0:06:41.37,0:06:46.90,Default,,0000,0000,0000,,You have T2 is equal to 750 degrees Kelvin,
Dialogue: 0,0:06:46.90,0:06:48.31,Default,,0000,0000,0000,,which makes sense, right?
Dialogue: 0,0:06:48.31,0:06:50.54,Default,,0000,0000,0000,,We increased the pressure so much
Dialogue: 0,0:06:50.54,0:06:53.29,Default,,0000,0000,0000,,and we decreased the volume at the same time
Dialogue: 0,0:06:53.29,0:06:55.64,Default,,0000,0000,0000,,that the temperature just had to go up.
Dialogue: 0,0:06:55.64,0:06:56.55,Default,,0000,0000,0000,,Or if you thought of it the other way,
Dialogue: 0,0:06:56.55,0:06:58.18,Default,,0000,0000,0000,,maybe we increased the temperature
Dialogue: 0,0:06:58.18,0:06:59.50,Default,,0000,0000,0000,,and that's what drove the pressure
Dialogue: 0,0:06:59.50,0:07:00.69,Default,,0000,0000,0000,,to be so much higher,
Dialogue: 0,0:07:00.69,0:07:03.87,Default,,0000,0000,0000,,especially since we decreased the volume.
Dialogue: 0,0:07:03.87,0:07:05.40,Default,,0000,0000,0000,,I guess the best way to think about is
Dialogue: 0,0:07:05.40,0:07:08.23,Default,,0000,0000,0000,,this pressure went up so much,
Dialogue: 0,0:07:08.23,0:07:10.20,Default,,0000,0000,0000,,it went up by factor of five,
Dialogue: 0,0:07:10.20,0:07:12.48,Default,,0000,0000,0000,,it went from 1 atmosphere to 5 atmospheres,
Dialogue: 0,0:07:12.48,0:07:14.37,Default,,0000,0000,0000,,because on one level
Dialogue: 0,0:07:14.37,0:07:18.03,Default,,0000,0000,0000,,we shrunk the volume by a factor of 1/2,
Dialogue: 0,0:07:18.03,0:07:19.68,Default,,0000,0000,0000,,so that should have doubled the pressure,
Dialogue: 0,0:07:19.68,0:07:21.90,Default,,0000,0000,0000,,so that should have gotten us to two atmospheres.
Dialogue: 0,0:07:21.90,0:07:23.78,Default,,0000,0000,0000,,And then we made the temperature a lot higher,
Dialogue: 0,0:07:23.78,0:07:25.41,Default,,0000,0000,0000,,so we were also bouncing into the container.
Dialogue: 0,0:07:25.41,0:07:27.90,Default,,0000,0000,0000,,We made the temperature 750 degrees Kelvin,
Dialogue: 0,0:07:27.90,0:07:29.89,Default,,0000,0000,0000,,so more than double the temperature,
Dialogue: 0,0:07:29.89,0:07:33.88,Default,,0000,0000,0000,,and then that's what got us to 5 atmospheres.
Dialogue: 0,0:07:33.88,0:07:37.99,Default,,0000,0000,0000,,Now, one other thing that you'll probably hear about
Dialogue: 0,0:07:37.99,0:07:39.69,Default,,0000,0000,0000,,is the notion of what happens
Dialogue: 0,0:07:39.69,0:07:42.48,Default,,0000,0000,0000,,at standard temperature and pressure.
Dialogue: 0,0:07:42.48,0:07:44.04,Default,,0000,0000,0000,,Let me delete all of the stuff over here.
Dialogue: 0,0:07:44.04,0:07:47.57,Default,,0000,0000,0000,,Standard temperature and pressure.
Dialogue: 0,0:07:47.57,0:07:51.53,Default,,0000,0000,0000,,Let me delete all this stuff that I don't need.
Dialogue: 0,0:07:52.89,0:07:56.81,Default,,0000,0000,0000,,Standard temperature and pressure.
Dialogue: 0,0:07:56.81,0:07:57.47,Default,,0000,0000,0000,,And I'm bringing it up
Dialogue: 0,0:07:57.47,0:07:58.69,Default,,0000,0000,0000,,because even though it's called
Dialogue: 0,0:07:58.69,0:07:59.88,Default,,0000,0000,0000,,standard temperature and pressure,
Dialogue: 0,0:07:59.88,0:08:03.70,Default,,0000,0000,0000,,and sometimes called STP,
Dialogue: 0,0:08:03.70,0:08:05.74,Default,,0000,0000,0000,,unfortunately for the world,
Dialogue: 0,0:08:05.74,0:08:07.84,Default,,0000,0000,0000,,they haven't really standardized
Dialogue: 0,0:08:07.84,0:08:13.74,Default,,0000,0000,0000,,what the standard pressure and temperature are.
Dialogue: 0,0:08:13.74,0:08:15.92,Default,,0000,0000,0000,,I went to Wikipedia and I looked it up.
Dialogue: 0,0:08:15.92,0:08:16.88,Default,,0000,0000,0000,,And the one that you'll probably see
Dialogue: 0,0:08:16.88,0:08:19.99,Default,,0000,0000,0000,,in most physics classes and most standardized tests
Dialogue: 0,0:08:19.99,0:08:23.94,Default,,0000,0000,0000,,is standard temperature is 0 degrees celsius,
Dialogue: 0,0:08:23.94,0:08:26.84,Default,,0000,0000,0000,,which is, of course, 273 degrees Kelvin.
Dialogue: 0,0:08:26.84,0:08:30.30,Default,,0000,0000,0000,,And standard pressure is 1 atmosphere.
Dialogue: 0,0:08:30.30,0:08:31.24,Default,,0000,0000,0000,,And here on Wikipedia,
Dialogue: 0,0:08:31.24,0:08:38.53,Default,,0000,0000,0000,,they wrote it as 101.325 kilopascals,
Dialogue: 0,0:08:38.53,0:08:41.34,Default,,0000,0000,0000,,or a little more than 101,000 pascals.
Dialogue: 0,0:08:41.34,0:08:44.25,Default,,0000,0000,0000,,of course, a pascal is a newton per square meter.
Dialogue: 0,0:08:44.25,0:08:45.97,Default,,0000,0000,0000,,In all of this stuff, the units are really
Dialogue: 0,0:08:45.97,0:08:47.66,Default,,0000,0000,0000,,the hardest part to get a hold of.
Dialogue: 0,0:08:47.66,0:08:49.74,Default,,0000,0000,0000,,But let's say that we assume
Dialogue: 0,0:08:49.74,0:08:50.68,Default,,0000,0000,0000,,these are all different
Dialogue: 0,0:08:50.68,0:08:52.20,Default,,0000,0000,0000,,standard temperatures and pressures
Dialogue: 0,0:08:52.20,0:08:54.88,Default,,0000,0000,0000,,based on different standard-making bodies.
Dialogue: 0,0:08:54.88,0:08:55.78,Default,,0000,0000,0000,,So they can't really agree with each other.
Dialogue: 0,0:08:55.78,0:08:57.26,Default,,0000,0000,0000,,But let's say we took this as
Dialogue: 0,0:08:57.26,0:09:00.89,Default,,0000,0000,0000,,the definition of standard temperature and pressure.
Dialogue: 0,0:09:00.89,0:09:04.60,Default,,0000,0000,0000,,So we're assuming that temperature
Dialogue: 0,0:09:04.60,0:09:07.23,Default,,0000,0000,0000,,is equal to 0 degrees Celsius,
Dialogue: 0,0:09:07.23,0:09:11.20,Default,,0000,0000,0000,,which is equal to 273 degrees Kelvin.
Dialogue: 0,0:09:11.20,0:09:15.26,Default,,0000,0000,0000,,And pressure, we're assuming, is 1 atmosphere,
Dialogue: 0,0:09:15.26,0:09:16.08,Default,,0000,0000,0000,,which could also be written as
Dialogue: 0,0:09:16.08,0:09:22.44,Default,,0000,0000,0000,,101.325 or 3/8 kilopascals.
Dialogue: 0,0:09:22.44,0:09:26.35,Default,,0000,0000,0000,,So my question is if I have an ideal gas
Dialogue: 0,0:09:26.35,0:09:30.02,Default,,0000,0000,0000,,at standard temperature and pressure,
Dialogue: 0,0:09:30.02,0:09:36.45,Default,,0000,0000,0000,,how many moles of that do I have in 1 liter?
Dialogue: 0,0:09:36.45,0:09:37.58,Default,,0000,0000,0000,,No, let me say that the other way.
Dialogue: 0,0:09:37.58,0:09:40.87,Default,,0000,0000,0000,,How many liters will 1 mole take up?
Dialogue: 0,0:09:40.87,0:09:43.78,Default,,0000,0000,0000,,So let me say that a little bit more.
Dialogue: 0,0:09:43.78,0:09:46.38,Default,,0000,0000,0000,,So n is equal to 1 mole.
Dialogue: 0,0:09:46.38,0:09:48.94,Default,,0000,0000,0000,,So I want to figure out what my volume is.
Dialogue: 0,0:09:48.94,0:09:50.66,Default,,0000,0000,0000,,So if I have 1 mole of a gas,
Dialogue: 0,0:09:50.66,0:09:55.56,Default,,0000,0000,0000,,I have 6.02 times 10 to 23 molecules of that gas.
Dialogue: 0,0:09:55.56,0:09:58.46,Default,,0000,0000,0000,,It's at standard pressure, 1 atmosphere,
Dialogue: 0,0:09:58.46,0:10:01.00,Default,,0000,0000,0000,,and at standard temperature, 273 degrees,
Dialogue: 0,0:10:01.00,0:10:03.46,Default,,0000,0000,0000,,what is the volume of that gas?
Dialogue: 0,0:10:03.46,0:10:07.74,Default,,0000,0000,0000,,So let's apply PV is equal to nRT.
Dialogue: 0,0:10:07.74,0:10:10.10,Default,,0000,0000,0000,,Pressure is 1 atmosphere,
Dialogue: 0,0:10:10.10,0:10:11.75,Default,,0000,0000,0000,,but remember we're dealing with atmospheres.
Dialogue: 0,0:10:11.75,0:10:15.36,Default,,0000,0000,0000,,1 atmosphere times volume
Dialogue: 0,0:10:15.36,0:10:16.66,Default,,0000,0000,0000,,that's what we're solving for.
Dialogue: 0,0:10:16.66,0:10:18.04,Default,,0000,0000,0000,,I'll do that in purple
Dialogue: 0,0:10:18.04,0:10:22.01,Default,,0000,0000,0000,,is equal to 1 mole, we have 1 mole of the gas,
Dialogue: 0,0:10:22.01,0:10:29.31,Default,,0000,0000,0000,,times R, times temperature, times 273.
Dialogue: 0,0:10:29.31,0:10:31.79,Default,,0000,0000,0000,,Now this is in Kelvin; this is in moles.
Dialogue: 0,0:10:31.79,0:10:39.51,Default,,0000,0000,0000,,We want our volume in liters.
Dialogue: 0,0:10:39.51,0:10:41.56,Default,,0000,0000,0000,,So which version of R should we use?
Dialogue: 0,0:10:41.56,0:10:44.41,Default,,0000,0000,0000,,Well, we're dealing with atmospheres.
Dialogue: 0,0:10:44.41,0:10:46.61,Default,,0000,0000,0000,,We want our volume in liters,
Dialogue: 0,0:10:46.61,0:10:48.03,Default,,0000,0000,0000,,and of course, we have moles in Kelvin,
Dialogue: 0,0:10:48.03,0:10:50.53,Default,,0000,0000,0000,,so we'll use this version, 0.082.
Dialogue: 0,0:10:50.53,0:10:52.21,Default,,0000,0000,0000,,So this is 1,
Dialogue: 0,0:10:52.21,0:10:54.87,Default,,0000,0000,0000,,so we can ignore the 1 there, the 1 there.
Dialogue: 0,0:10:54.87,0:10:56.39,Default,,0000,0000,0000,,So the volume is equal to
Dialogue: 0,0:10:56.39,0:11:02.20,Default,,0000,0000,0000,,0.082 times 273 degrees Kelvin,
Dialogue: 0,0:11:02.20,0:11:19.23,Default,,0000,0000,0000,,and that is 0.082 times 273 is equal to 22.4 liters.
Dialogue: 0,0:11:19.23,0:11:21.43,Default,,0000,0000,0000,,So if I have any ideal gas,
Dialogue: 0,0:11:21.43,0:11:24.08,Default,,0000,0000,0000,,and all gases don't behave ideally ideal,
Dialogue: 0,0:11:24.08,0:11:25.48,Default,,0000,0000,0000,,but if I have an ideal gas
Dialogue: 0,0:11:25.48,0:11:26.93,Default,,0000,0000,0000,,and it's at standard temperature,
Dialogue: 0,0:11:26.93,0:11:29.10,Default,,0000,0000,0000,,which is at 0 degrees Celsius,
Dialogue: 0,0:11:29.10,0:11:30.42,Default,,0000,0000,0000,,or the freezing point of water,
Dialogue: 0,0:11:30.42,0:11:32.42,Default,,0000,0000,0000,,which is also 273 degrees Kelvin,
Dialogue: 0,0:11:32.42,0:11:33.71,Default,,0000,0000,0000,,and I have a mole of it,
Dialogue: 0,0:11:33.71,0:11:37.56,Default,,0000,0000,0000,,and it's at standard pressure, 1 atmosphere,
Dialogue: 0,0:11:37.56,0:11:42.48,Default,,0000,0000,0000,,that gas should take up exactly 22.4 liters.
Dialogue: 0,0:11:42.48,0:11:44.80,Default,,0000,0000,0000,,And if you wanted to know how many meters cubed
Dialogue: 0,0:11:44.80,0:11:46.38,Default,,0000,0000,0000,,it's going to take up.
Dialogue: 0,0:11:46.38,0:11:50.99,Default,,0000,0000,0000,,well, you could just say 22.4 liters times----
Dialogue: 0,0:11:50.99,0:11:53.24,Default,,0000,0000,0000,,now, how many meters cubed are there----
Dialogue: 0,0:11:53.24,0:11:57.50,Default,,0000,0000,0000,,so for every 1 meter cubed, you have 1,000 liters.
Dialogue: 0,0:11:57.50,0:11:59.63,Default,,0000,0000,0000,,I know that seems like a lot, but it's true.
Dialogue: 0,0:11:59.63,0:12:02.48,Default,,0000,0000,0000,,Just think about how big a meter cubed is.
Dialogue: 0,0:12:02.48,0:12:09.36,Default,,0000,0000,0000,,So this would be equal to 0.0224 meters cubed.
Dialogue: 0,0:12:09.36,0:12:12.45,Default,,0000,0000,0000,,If you have something at 1 atmosphere, a mole of it,
Dialogue: 0,0:12:12.45,0:12:14.75,Default,,0000,0000,0000,,and at 0 degrees Celsius.
Dialogue: 0,0:12:14.75,0:12:16.08,Default,,0000,0000,0000,,Anyway, this is actually
Dialogue: 0,0:12:16.08,0:12:17.71,Default,,0000,0000,0000,,a useful number to know sometimes.
Dialogue: 0,0:12:17.71,0:12:22.25,Default,,0000,0000,0000,,They'll often say you have 2 moles
Dialogue: 0,0:12:22.25,0:12:25.29,Default,,0000,0000,0000,,at standard temperature and pressure.
Dialogue: 0,0:12:25.29,0:12:26.97,Default,,0000,0000,0000,,How many liters is it going to take up?
Dialogue: 0,0:12:26.97,0:12:29.61,Default,,0000,0000,0000,,Well, 1 mole will take up this many,
Dialogue: 0,0:12:29.61,0:12:31.78,Default,,0000,0000,0000,,and so 2 moles at standard temperature and pressure
Dialogue: 0,0:12:31.78,0:12:33.44,Default,,0000,0000,0000,,will take up twice as much,
Dialogue: 0,0:12:33.44,0:12:34.80,Default,,0000,0000,0000,,because you're just taking PV equals nRT
Dialogue: 0,0:12:34.80,0:12:36.27,Default,,0000,0000,0000,,and just doubling.
Dialogue: 0,0:12:36.27,0:12:38.79,Default,,0000,0000,0000,,Everything else is being held constant.
Dialogue: 0,0:12:38.79,0:12:40.99,Default,,0000,0000,0000,,The pressure, everything else is being held constant,
Dialogue: 0,0:12:40.99,0:12:43.04,Default,,0000,0000,0000,,so if you double the number of moles,
Dialogue: 0,0:12:43.04,0:12:44.21,Default,,0000,0000,0000,,you're going to double the volume it takes up.
Dialogue: 0,0:12:44.21,0:12:46.11,Default,,0000,0000,0000,,Or if you half the number of moles,
Dialogue: 0,0:12:46.11,0:12:47.67,Default,,0000,0000,0000,,you're going to half the volume it takes up.
Dialogue: 0,0:12:47.67,0:12:49.66,Default,,0000,0000,0000,,So it's a useful thing to know that in liters
Dialogue: 0,0:12:49.66,0:12:52.02,Default,,0000,0000,0000,,at standard temperature and pressure,
Dialogue: 0,0:12:52.02,0:12:52.91,Default,,0000,0000,0000,,where standard temperature and pressure
Dialogue: 0,0:12:52.91,0:12:56.52,Default,,0000,0000,0000,,is defined as 1 atmosphere and 273 degrees Kelvin,
Dialogue: 0,0:12:56.52,0:13:00.16,Default,,0000,0000,0000,,an idea gas will take up 22.4 liters of volume