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21. Kinetic Theory of Ideal Gases

1

concept

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Hey guys. So throughout your problems you may be asked to calculate something called the mean free path. So in this video, I want to show you what that mean. Free path is conceptually the equations for it and then we'll do a quick example together. Let's check this out. So imagine that I have a canister of gas particles that are sealed inside of this little container like this. Now remember these gas particles aren't stationary, they're all moving all over the place. So what happens here is if you take a if you look at a random gas particle and look at its path as it's moving throughout the canister, what happens is that eventually it travels in a straight line before bumping into another particle. Now all the collisions are elastic. So these things sort of like scatter off in straight lines. But basically what happens is it just keeps on bouncing like this over and over again and these particles just go all over the place. So the mean free path is really just the average, right? That just means mean distance that the gas particles travel before they collide with another particle. And the idea here is that we can actually calculate this by using some of the macroscopic variables. The analogy I like to use here is I like to imagine that the container of gas is like a room full of people. Right? And so if I'm like a person that's walking through this room right here, if I walk in a straight line eventually I'm gonna bump into somebody else and if I can only walk in a straight line, I'm just going to keep bumping off of people as I walk through the room. So the analogy is that if you fill the container full of people, the mean free path is, it's the average distance you walk before bumping into somebody else. All right, So how do we calculate it? Well, this is really just a distance. A lot of equations will use lambda for this. Some textbooks might use l most of them will use lambda and really this distance here is just equal to velocity times time. Right? This is basically just delta X equals V. T. So, there's nothing new here. That's one way to write this. But remember that the goal of the kinetic molecular theory is that we want to express the microscopic properties of the particles, right? The for instance, the distance that they travel in terms of the variables inside of the ideal gas law, like pressure, volume and temperature. So the equation that you need to know is actually this one. It's gonna be the volume divided by the square root of two times four pi r squared times N. Unfortunately, there isn't really a good way to memorize this equation, but if you ever need it, it's going to be given to you on a formula sheet or something like that. So what I want to point out here is two things. It depends on the volume. But notice how the end is in the denominator and what that means is that the the number of particles increases if there are more people in the room, right? If there are more particles and more people in the room then the average distance before you bump into somebody else will go down. And the other thing is that this letter are here actually want to keep this if this little are here is the radius of the particle, is the physical size of these things. If that also increases than the average distance decreases. If the if the if the particles or if the people are bigger than the average distance that you travel before bumping into somebody else also decreases. Alright, so that's really all there is to know. One thing I want to point out here is that in some problems you may not be given with this radius is so you can just assume these values in case you're not. So that's really all there is to it. Let's go ahead and check out our example here. So we have a we have oxygen molecules at S. T. P. So remember that S. T. P. Is just a set of conditions where the temperature is equal to 2 73 kelvin and the pressure is equal to 1.1 times 10 to the fifth pascal's. Alright, so what we have here is in the first part we want to calculate what's the mean free path of oxygen molecules. So we want to calculate lambda. But this is gonna be lambda for 02. So, if we want to use this equation here, we just have to figure out we're gonna use VT. Or we're going to use this more complicated equation. We actually don't. We're only told what the average speed of the particles are. We don't know what the time is. So, we're gonna have to use the more complicated one. So, what happens here is we're gonna write V divided by square root of two times four pi r squared times N. All right. So, here's what happens. So, we do we have the volume? Do we have the number of particles? And the answer is no. We actually don't have V or N. So, we actually can't figure out just yet what they mean? Free path is. So, I'm gonna have to go over here and I'm gonna have to figure out what V and N are. So, I've got V as an unknown and N as an unknown. But notice how we actually have more information about this problem. Notice how we all know we know what the temperature is. It's 2 73. And we also know what the pressure is. So what happens here is that we have P that's known? We have t that's known but that V R N. N are unknown. And notice how these equations are. These variables here are the four variables that are involved in the ideal gas law. So the idea here is that by using the ideal gas law, we're gonna be able to write an expression for V over N. So we want V over N. Is equal to something, some kind of expression. And I want to be able to calculate what that is because then if we get a number for this, we can just plug this back into our main free path equation. Alright, so if I want to start out with the ideal gas law, this is PV equals capital, N K B T. Notice how I'm using the capital end because I have capital and in my mean free path equation. So all I have to do here is everyone V over N. Then I'm gonna have to move the end over to the other side and the P over to the other side. And what you get here is you're gonna get a K B T. Is equal to r C r k B T, divided by pressure is equal to V over N. So now we can do is even though we don't know V over N. We can use the variables that we do know like the temperature and pressure to actually get an expression for V over N. So if you go ahead and plug the sandwich you get is 1.38 times 10 to the -23. Remember that's the bolts, one constant, The temperature is 2 73 and then the pressure is 1.1 times 10 to the fifth pascal's So if you plug this into your calculator, you're going to get 3.73 3. times 10 to the minus 26. And so that is what the over N is. Alright, so now we can do is this V over N. Which is just this number here, we're just gonna plug this back into the mean free path equation. Alright, so, I'm just gonna scoot this down here. Are mean free path for oxygen is going to be I'm going to do this as one over square root of two times four pi And then we have the radius of the oxygen molecules which are actually given to us 1.5 times 10 to minus 10. 1.5 times 10 to minus 10. And then now we have to do is square that. And then over here, on the outside of this parentheses, I'm gonna multiply this. So this is gonna be 3.73 times 10 to the minus 26. All right, so, we have to multiply apply those things. And what you get is the mean free path that's equal to 9. Times 10 to the -8 meters. So, this should make some sense that this is a very small distance. Right? Because if you have oxygen molecules there's a ton of them and you know, the distances between the molecules is going to be very, very small. So this is actually about 93 nanometers. So, it's a very, very small distance here. Alright, so that's the first part. So, let's take a look at the second part now. And the second part. Now that we know that the mean free path is we want to calculate the average time between the collisions. So, in order to figure out that this is gonna be part B, we want to figure out t average the average time between collisions. And to do that. We're just going to use the first half of the equation for the mean free path. We're just going to use this piece right here. All right, so, we're going to use that the lambda 02 is equal to V average times T average. And we actually have what both of these things are now, right? We have the lambda 02, we have the V average. So all I have to do is just divide them. So, Lando two divided by the average is going to be t average. And now I'm just gonna plug stuff in. So this is going to be 9.33 times 10 to the minus eight, divided by the average velocity, which is gonna be 450 m per second here. And if you work this out, you're gonna get uh you're gonna get 2.7 times 10 to the minus 10 seconds. So this is very, very, very short amount of time. These things are colliding billions of times a second. And that makes sense because there's so many particles and they're all flying around at very high speeds and they don't have to travel very, a very far distance. So the time between collisions is going to be very small. Alright guys, so that's it for this one, Let me know if you have any questions.

2

Problem

Laboratory environments can achieve pressures of 3.5×10^{-13} atm and temperatures of 300K. Calculate the mean free path (in km) of air molecules, which you can assume are diatomic.

A

1.04 × 10^{-32} km

B

6.65 km

C

660 km

D

8.22 × 10^{-10} km

3

example

6m

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Alright, everybody. So the mean free path of nitrogen particles at STP remember that means standard temperature and pressure. It's just a set of conditions in which the temperature is this number and the pressure is this number. We've seen that before. All right now, we're told that this mean free path is just this number over here. What we want to calculate in this problem is what's the radius of the nitrogen particles? So which variable is that we remember in our lambda equation, the one for mean free path, if you write it in terms of ideal gas variables, remember the mean free path depends on a couple of things like the volume of whatever container you're in. So for example, I'm gonna draw this little container out like this. The mean free path, the distance between, you know That each particle travels before colliding into another one depends on the overall volume of the container. That's v. It also depends on the number of molecules that you have in that container. Obviously if you have more of these molecules, the distances between collisions get shorter and shorter, but it also depends on this little are over here, which is basically the radius of the particles themselves. So what happens is you can imagine if the particles get way, way bigger, if you start having really, really big particles inside this container, then the average distance between the them colliding into each other is going to be much much smaller. That's kind of what's going on here. It's kind of conceptual understanding of this problem, What we're really looking for is that radius the radius of these nitrogen particles, I'm gonna call this our end too. So let's go ahead and start off with our lambda equation. We know that lambda for nitrogen gas here is going to be eight point or eight times 10 to the negative eight, and this is going to be equal to well, we have the average times t average, but I'm gonna skip this because we don't have any information about the average speed or average time between collisions. And I'm just gonna go ahead and write this in terms of Um the other ideal gas variables. So we have the over square to two times four pi RN two. Right? So the radius of the nitrogen particles squared times n the number of particles. So just don't give you confusing. This is kind of confusing because this end doesn't stand for nitrogen, it stands for the number of particles, but this end over here kind of stands for the nitrogen gas, right? So just don't don't get confused with that. So this is really what we're looking for here. So let's go ahead and start working out all these variables. Alright, so what I'm gonna do here is I'm actually gonna rewrite this equation to be a little bit more um sort of understandable and legible. So we've got one of our four pi sorry, one over squared of two times four pi radius squared times V over N. And this equals lambda and two. So if I want to figure out what this RN too, is I want to figure out everything else. So what I actually have is I actually have what this land is. I'm just given that number out. Right, it's eight times 10 to the -8. But what about these other two variables of V and N. If you'll notice here, we don't have the the volume or the number of particles. Instead, what we have is S. T. P. But that just means that the temperature is this number and the pressure is this other number over here. So, we have a situation where we need to of the ideal gas variables but were given another of those two. So we're gonna have to do here is we're gonna have to use the trick where we go over and use the ideal gas law to represent these variables in terms of other variables. Right? So we're gonna go over here to PV equals and we're gonna use N K B. T. You may have seen this from a previous video and so we're gonna do here is we're gonna divide this end over and we're gonna divide the P over. And what I end up with is V over N. Is equal to K. B. T divided by P. So now what we can do here is in our lambda equation, we can take this V over N. And instead just write it as variables KB T Overpay. Alright. So what I'm gonna do here is I'm going to do one over squared of two times four pi radius squared times. And this is gonna be K. B. T. Overpay. This is gonna be my what my lambda is equal to. So now what I'm gonna do is I'm gonna start just plugging in some numbers over here. Okay so what I've got here is um so I've got eight times 10 to the minus eight is equal to one over square root of two times four pi radius squared times. And this is going to be let's see 1.38 times to the -23. Then we have the temperature which is 273 Kelvin, right? That's what STP means. And then divided by the pressure, which is going to be 1.01 times 10 to the fifth. Alright, so again we're still looking just for this are over here. What I'm gonna do is I'm gonna actually because we have one over this whole entire thing here, what I'm gonna do is I'm gonna move this whole thing up to the other side and then I'm gonna move this eight times 10 to the minus eight down to the denominator. Okay so what I end up with here is basically they're just gonna trade places like this and what I end up with here is squared of two times four pi r into squared equals. And then just this works out to just a number here. When you multiply all this stuff together and you divide the eight times 10 to the minus eight over to the other side, you're gonna get 4.66 times 10 to the minus 19. Okay, so now all I have to do is I'm just gonna move over the square root of two in the four pi So when you divide this stuff over to the other side, what you end up with is that RN two squared is equal to um This is going to be 2.62 times 10 to the minus 20. So now the last thing I have to do is just take the square root of this number here. So what I get is that the radius of nitrogen gas is equal to the square root of 2.66 times 10 to the minus 20. And what you're gonna end up with here is 1.6 times 10 to the -10. Alright, so that is the No. times 10 to the -10. That is the average radius or that that's not the average, that's the radius of nitrogen particles according to this equation here. Now, if you look at this number here, remember that the nitrogen is going to be a di atomic molecules because we have two of the nitrogen is right. And um so this should kind of make some sense that we have 1.6 times 10 to minus 10. Because remember what we said, is that whenever you don't have the radius of mono atomic versus di atomic, you can assume that di atomic is roughly about one times 10 to the minus 10. So this kind of agrees with our expectations. Alright, so that's for this one. Guys, let me know if you have any questions.

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