WEBVTT

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Let's do some more problems

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that involve the ideal gas equation.

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Let's say I have a gas in a container

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and the current pressure is 3 atmospheres.

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And let's say that the volume of the container

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is, I don't know, 9 liters.

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Now, what will the pressure become

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if my volume goes from 9 liters to 3 liters?

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So from the first ideal gas equation video

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you can kind of have the intuition

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that you have a bunch of-- and we're holding--

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and this is important.

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We're holding the temperature constant

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and that's an important thing to realize.

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So in our very original intuition

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behind the ideal gas equation we said,

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look, if we have a certain number of particles

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with a certain amount of kinetic energy,

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and they're exerting a certain pressure

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on their container,

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and if we were to make the container smaller,

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we have the same number of particles.

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n doesn't change.

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The average kinetic energy doesn't change,

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so they're just going to bump into the walls more.

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So that when we make the volume smaller,

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when the volume goes up----

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when the volume goes down,

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the pressure should go up.

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So let's see if we can calculate the exact number.

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So we can take our ideal gas equation:

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pressure times volume is equal to nRT.

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Now, do the number of particles change

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when I did this situation when I shrunk the volume?

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No!

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We have the same number of particles.

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I'm just shrinking the container,

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so n is n, R doesn't change, that's a constant,

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and then the temperature doesn't change.

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So my old pressure times volume

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is going to be equal to nRT,

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and my new pressure times volume--

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so let me call this P1 and V1.

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and then P2 is this----

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sorry, that's V2.

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so V2 is this, and we're trying to figure out P2.

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P2 is what?

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Well, we know that P1 times V1 is equal to nRT,

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and we also know that since temperature and

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the number of moles of our gas stay constant,

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that P2 times V2 is equal to nRT.

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And since they both equal the same thing,

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we can say that the pressure times the volume,

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as long as the temperature is held constant,

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will be a constant.

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So P1 times V1 is going to equal P2 times V2.

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So what was P1?

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P1, our initial pressure, was 3 atmospheres.

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So 3 atmospheres times 9 liters is equal to

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our new pressure times 3 liters.

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And if we divide both sides of the equation by 3,

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we get 3 liters cancel out,

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we're left with 9 atmospheres.

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And that should make sense.

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When you decrease the volume by 2/3

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or when you make the volume

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1/3 of your original volume,

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then your pressure increases by a factor of three.

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So this went by times 3, and this went by times 1/3.

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That's a useful thing to know in general.

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If temperature is held constant,

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then pressure times volume

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are going to be a constant.

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Now, you can take that even further.

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If we look at PV equals nRT,

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the two things that we know don't change

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in the vast majority of exercises we do

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is the number of molecules we're dealing with,

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and obviously, R isn't going to change.

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So if we divide both sides of this by T,

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we get PV over T is equal to nR,

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or you could say it's equal to a constant.

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This is going to be a constant number for any system

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where we're not changing

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the number of molecules in the container.

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So, if we are changing the pressure----

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So if initially we start with

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pressure one, volume one, and some temperature one

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that's going to be equal to this constant.

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And if we change any of them,

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we go back to

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pressure two, volume two, temperature two,

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they should still be equal to this constant,

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so they equal each other.

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So for example, let's say I start off with a

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pressure of 1 atmosphere.

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and I have a volume of----

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I'll switch units here just to do things differently

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----2 meters cubed.

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And let's say our temperature is 27 degrees Celsius.

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Well, and I just wrote Celsius

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because I want you to always remember

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you have to convert to Kelvin,

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so 27 degrees plus 273 will get us

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exactly to 300 Kelvin.

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And let's say that our new temperature is

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Actually let's figure out what the new temperature

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is going to be.

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Let's say our new pressure is 2 atmospheres.

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The pressure has increased.

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Let's say we make the container smaller,

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so 1 meter cubed.

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So the container has been decreased by half

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and the pressure is doubled by half.

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So you could guess.

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You know, we have made the pressure higher----

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Let me make the container even smaller.

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Actually, no.

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Let me make the pressure even larger.

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Let me make the pressure into 5 atmospheres.

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Now we want to know what the second temperature is

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and we set up our equation.

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And so we have

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2/300 atmosphere meters cubed per Kelvin

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is equal to 5/T2, our new temperature,

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and then we have 1,500 is equal to 2T2.

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Divide both sides by 2.

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You have T2 is equal to 750 degrees Kelvin,

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which makes sense, right?

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We increased the pressure so much

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and we decreased the volume at the same time

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that the temperature just had to go up.

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Or if you thought of it the other way,

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maybe we increased the temperature

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and that's what drove the pressure

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to be so much higher,

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especially since we decreased the volume.

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I guess the best way to think about is

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this pressure went up so much,

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it went up by factor of five,

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it went from 1 atmosphere to 5 atmospheres,

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because on one level

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we shrunk the volume by a factor of 1/2,

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so that should have doubled the pressure,

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so that should have gotten us to two atmospheres.

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And then we made the temperature a lot higher,

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so we were also bouncing into the container.

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We made the temperature 750 degrees Kelvin,

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so more than double the temperature,

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and then that's what got us to 5 atmospheres.

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Now, one other thing that you'll probably hear about

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is the notion of what happens

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at standard temperature and pressure.

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Let me delete all of the stuff over here.

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Standard temperature and pressure.

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Let me delete all this stuff that I don't need.

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Standard temperature and pressure.

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And I'm bringing it up

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because even though it's called

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standard temperature and pressure,

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and sometimes called STP,

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unfortunately for the world,

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they haven't really standardized

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what the standard pressure and temperature are.

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I went to Wikipedia and I looked it up.

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And the one that you'll probably see

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in most physics classes and most standardized tests

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is standard temperature is 0 degrees celsius,

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which is, of course, 273 degrees Kelvin.

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And standard pressure is 1 atmosphere.

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And here on Wikipedia,

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they wrote it as 101.325 kilopascals,

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or a little more than 101,000 pascals.

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of course, a pascal is a newton per square meter.

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In all of this stuff, the units are really

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the hardest part to get a hold of.

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But let's say that we assume

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these are all different

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standard temperatures and pressures

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based on different standard-making bodies.

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So they can't really agree with each other.

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But let's say we took this as

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the definition of standard temperature and pressure.

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So we're assuming that temperature

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is equal to 0 degrees Celsius,

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which is equal to 273 degrees Kelvin.

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And pressure, we're assuming, is 1 atmosphere,

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which could also be written as

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101.325 or 3/8 kilopascals.

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So my question is if I have an ideal gas

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at standard temperature and pressure,

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how many moles of that do I have in 1 liter?

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No, let me say that the other way.

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How many liters will 1 mole take up?

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So let me say that a little bit more.

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So n is equal to 1 mole.

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So I want to figure out what my volume is.

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So if I have 1 mole of a gas,

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I have 6.02 times 10 to 23 molecules of that gas.

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It's at standard pressure, 1 atmosphere,

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and at standard temperature, 273 degrees,

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what is the volume of that gas?

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So let's apply PV is equal to nRT.

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Pressure is 1 atmosphere,

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but remember we're dealing with atmospheres.

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1 atmosphere times volume

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that's what we're solving for.

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I'll do that in purple

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is equal to 1 mole, we have 1 mole of the gas,

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times R, times temperature, times 273.

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Now this is in Kelvin; this is in moles.

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We want our volume in liters.

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So which version of R should we use?

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Well, we're dealing with atmospheres.

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We want our volume in liters,

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and of course, we have moles in Kelvin,

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so we'll use this version, 0.082.

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So this is 1,

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so we can ignore the 1 there, the 1 there.

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So the volume is equal to

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0.082 times 273 degrees Kelvin,

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and that is 0.082 times 273 is equal to 22.4 liters.

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So if I have any ideal gas,

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and all gases don't behave ideally ideal,

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but if I have an ideal gas

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and it's at standard temperature,

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which is at 0 degrees Celsius,

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or the freezing point of water,

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which is also 273 degrees Kelvin,

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and I have a mole of it,

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and it's at standard pressure, 1 atmosphere,

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that gas should take up exactly 22.4 liters.

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And if you wanted to know how many meters cubed

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it's going to take up.

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well, you could just say 22.4 liters times----

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now, how many meters cubed are there----

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so for every 1 meter cubed, you have 1,000 liters.

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I know that seems like a lot, but it's true.

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Just think about how big a meter cubed is.

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So this would be equal to 0.0224 meters cubed.

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If you have something at 1 atmosphere, a mole of it,

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and at 0 degrees Celsius.

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Anyway, this is actually

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a useful number to know sometimes.

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They'll often say you have 2 moles

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at standard temperature and pressure.

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How many liters is it going to take up?

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Well, 1 mole will take up this many,

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and so 2 moles at standard temperature and pressure

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will take up twice as much,

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because you're just taking PV equals nRT

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and just doubling.

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Everything else is being held constant.

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The pressure, everything else is being held constant,

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so if you double the number of moles,

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you're going to double the volume it takes up.

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Or if you half the number of moles,

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you're going to half the volume it takes up.

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So it's a useful thing to know that in liters

00:12:49.656 --> 00:12:52.015
at standard temperature and pressure,

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where standard temperature and pressure

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is defined as 1 atmosphere and 273 degrees Kelvin,

00:12:56.516 --> 00:13:00.159
an idea gas will take up 22.4 liters of volume