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ემზადებით გამოცდისთვის? მოემზადეთ ამ გაკვეთილის დახმარებით შემდეგ თემაზე: Gases and kinetic molecular theory

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# Ideal gas equation example 2

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Let's do some more problems that involve the ideal gas equation. Let's say I have a gas in a container and the current pressure is 3 atmospheres. And let's say that the volume of the container is 9 liters. Now, what will the pressure become if my volume goes from 9 liters to 3 liters? So from the first ideal gas equation video, you can kind of have the intuition, that you have a bunch of -- and we're holding-- and this is important. We're holding the temperature constant, and that's an important thing to realize. So in our very original intuition behind the ideal gas equation we said, look, if we have a certain number of particles with a certain amount of kinetic energy, and they're exerting a certain pressure on their container, and if we were to make the container smaller, we have the same number of particles. n doesn't change. The average kinetic energy doesn't change, so they're just going to bump into the walls more. So that when we make the volume smaller, when the volume goes down, the pressure should go up. So let's see if we can calculate the exact number. So we can take our ideal gas equation: pressure times volume is equal to nRT. Now, do the number of particles change when I did this situation when I shrunk the volume? No! We have the same number of particles. I'm just shrinking the container, so n is n, R doesn't change, that's a constant, and then the temperature doesn't change. So my old pressure times volume is going to be equal to nRT, and my new pressure times volume -- so let me call this P1 and V1. That's V2. V2 is this, and we're trying to figure out P2. P2 is what? Well, we know that P1 times V1 is equal to nRT, and we also know that since temperature and the number of moles of our gas stay constant, that P2 times V2 is equal to nRT. And since they both equal the same thing, we can say that the pressure times the volume, as long as the temperature is held constant, will be a constant. So P1 times V1 is going to equal P2 times V2. So what was P1? P1, our initial pressure, was 3 atmospheres. So 3 atmospheres times 9 liters is equal to our new pressure times 3 liters. And if we divide both sides of the equation by 3, we get 3 liters cancel out, we're left with 9 atmospheres. And that should make sense. When you decrease the volume by 2/3 or when you make the volume 1/3 of your original volume, then your pressure increases by a factor of three. So this went by times 3, and this went by times 1/3. That's a useful thing to know in general. If temperature is held constant, then pressure times volume are going to be a constant. Now, you can take that even further. If we look at PV equals nRT, the two things that we know don't change in the vast majority of exercises we do is the number of molecules we're dealing with, and obviously, R isn't going to change. So if we divide both sides of this by T, we get PV over T is equal to nR, or you could say it's equal to a constant. This is going to be a constant number for any system where we're not changing the number of molecules in the container. So if initially we start with pressure one, volume one, and some temperature one that's going to be equal to this constant. And if we change any of them, if we go back to pressure two, volume two, temperature two, they should still be equal to this constant, so they equal each other. So for example, let's say I start off with a pressure of 1 atmosphere. and I have a volume of-- I'll switch units here just to do things differently-- 2 meters cubed. And let's say our temperature is 27 degrees Celsius. Well, and I just wrote Celsius because I want you to always remember you have to convert to Kelvin, so 27 degrees plus 273 will get us exactly to 300 Kelvin. Let's figure out what the new temperature is going to be. Let's say our new pressure is 2 atmospheres. The pressure has increased. Let's say we make the container smaller, so 1 meter cubed. So the container has been decreased by half and the pressure is doubled by half. Let me make the container even smaller. Actually, no. Let me make the pressure even larger. Let me make the pressure into 5 atmospheres. Now we want to know what the second temperature is, and we set up our equation. And so we have 2/300 atmosphere meters cubed per Kelvin is equal to 5/T2, our new temperature, and then we have 1,500 is equal to 2 T2. Divide both sides by 2. You have T2 is equal to 750 degrees Kelvin, which makes sense, right? We increased the pressure so much and we decreased the volume at the same time that the temperature just had to go up. Or if you thought of it the other way, maybe we increased the temperature and that's what drove the pressure to be so much higher, especially since we decreased the volume. I guess the best way to think about is this pressure went up so much, it went up by factor of five, it went from 1 atmosphere to 5 atmospheres, because on one level we shrunk the volume by a factor of 1/2, so that should have doubled the pressure, so that should have gotten us to two atmospheres. And then we made the temperature a lot higher, so we were also bouncing into the container. We made the temperature 750 degrees Kelvin, so more than double the temperature, and then that's what got us to 5 atmospheres. Now, one other thing that you'll probably hear about is the notion of what happens at standard temperature and pressure. Let me delete all of the stuff over here. Standard temperature and pressure. Let me delete all this stuff that I don't need. Standard temperature and pressure. And I'm bringing it up because even though it's called standard temperature and pressure, and sometimes called STP, unfortunately for the world, they haven't really standardized what the standard pressure and temperature are. I went to Wikipedia and I looked it up. And the one that you'll probably see in most physics classes and most standardized tests is standard temperature is 0 degrees celsius, which is, of course, 273 degrees Kelvin. And standard pressure is 1 atmosphere. And here on Wikipedia, they wrote it as 101.325 kilopascals, or a little more than 101,000 pascals. And of course, a pascal is a newton per square meter. In all of this stuff, the units are really the hardest part to get a hold of. But let's say that we assume that these are all different standard temperatures and pressures based on different standard-making bodies. So they can't really agree with each other. But let's say we took this as the definition of standard temperature and pressure. So we're assuming that temperature is equal to 0 degrees Celsius, which is equal to 273 degrees Kelvin. And pressure, we're assuming, is 1 atmosphere, which could also be written as 101.325 or 3/8 kilopascals. So my question is if I have an ideal gas at standard temperature and pressure, how many moles of that do I have in 1 liter? No, let me say that the other way. How many liters will 1 mole take up? So let me say that a little bit more. So n is equal to 1 mole. So I want to figure out what my volume is. So if I have 1 mole of a gas, I have 6.02 times 10 to 23 molecules of that gas. It's at standard pressure, 1 atmosphere, and at standard temperature, 273 degrees, what is the volume of that gas? So let's apply PV is equal to nRT. Pressure is 1 atmosphere, but remember we're dealing with atmospheres. 1 atmosphere times volume-- that's what we're solving for. I'll do that and purple-- is equal to 1 mole times R times temperature, times 273. Now this is in Kelvin; this is in moles. We want our volume in liters. So which version of R should we use? Well, we're dealing with atmospheres. We want our volume in liters, and of course, we have moles and Kelvin, so we'll use this version, 0.082. So this is 1, so we can ignore the 1 there, the 1 there. So the volume is equal to 0.082 times 273 degrees Kelvin, and that is 0.082 times 273 is equal to 22.4 liters. So if I have any ideal gas, and all gases don't behave ideally ideal, but if I have an ideal gas and it's at standard temperature, which is at 0 degrees Celsius, or the freezing point of water, which is also 273 degrees Kelvin, and I have a mole of it, and it's at standard pressure, 1 atmosphere, that gas should take up exactly 22.4 liters. And if you wanted to know how many meters cubed it's going to take up, well, you could just say 22.4 liters times-- now, how many meters cubed are there -- so for every 1 meter cubed, you have 1,000 liters. I know that seems like a lot, but it's true. Just think about how big a meter cubed is. So this would be equal to 0.0224 meters cubed. If you have something at 1 atmosphere, a mole of it, and at 0 degrees Celsius. Anyway, this is actually a useful number to know sometimes. They'll often say, you have 2 moles at standard temperature and pressure. How many liters is it going to take up? Well, 1 mole will take up this many, and so 2 moles at standard temperature and pressure will take up twice as much, because you're just taking PV equals nRT and just doubling. Everything else is being held constant. The pressure, everything else is being held constant, so if you double the number of moles, you're going to double the volume it takes up. Or if you half the number of moles, you're going to half the volume it takes up. So it's a useful thing to know that in liters at standard temperature and pressure, where standard temperature and pressure is defined as 1 atmosphere and 273 degrees Kelvin, an ideal gas will take up 22.4 liters of volume.