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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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