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, where the temperature and the pressure are as specified. We've seen that before, 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? Well, 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 going to draw this little container out like this. The mean free path, the distance between, you know, 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 \(r\) 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 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 a conceptual understanding of this problem. What we're really looking for is that radius, the radius of these nitrogen particles. So I'm going to call this \(r_n^2\). So let's go ahead and start off with our lambda equation. We know that \(\lambda\) for nitrogen gas here is going to be \(8 \times 10^{-8}\), and this is going to be equal to when we have \(V_{average} \times T_{average}\), but I'm going to skip this because we don't have any information about the average speed or average time between collisions. And I'm just going to go ahead and write this in terms of, the other ideal gas variables. So we have \(V / \sqrt{2 \times 4 \pi r_n^2 }\). Right? So the radius of the nitrogen particles squared times \(N\), the number of particles. So just don't get confused, this is a little confusing because this \(N\) doesn't stand for nitrogen, it stands for the number of particles, but this \(N\) over here kind of stands for the nitrogen gas. Alright? 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 going to do here is I'm actually going to rewrite this equation to be a little bit more, sort of understandable and legible. So we've got \(1 / \sqrt{2 \times 4 \pi r^2} \times V / N\), and this equals \(\lambda_n^2\). So if I want to figure out what this \(r_n^2\) is, I want to figure out everything else. So what I actually have is I actually have what this \(\lambda\) is. I'm just giving that number outright. It's \(8 \times 10^{-8}\). But what about these other two variables \(V\) and \(N\)? If you'll notice here, we don't have the volume or the number of particles. Instead, what we have is STP, 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 2 of the ideal gas variables, but we're given another of those 2. So what we're going to have to do here is we're going to 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 going to go over here to \(PV = N k_b T\). You may have seen this from a previous video. And so we're going to do here is we're going to divide this \(N\) over and we're going to divide the \(P\) over, and what I end up with is \(V / N\) is equal to \(k_b T / P\). So now what we can do here is in our lambda equation, we can take this \(V / N\) and instead just write it as, variables \(k_b T / P\). Alright? So what I'm going to do here is I'm going to do \(1 / \sqrt{2 \times 4 \pi}\), radius squared times, and this is going to be \(k_b T / P\). This is going to be what my \(\lambda\) is equal to. So now what I'm going to do is I'm going to start just plugging in some numbers over here. Okay? So what I've got here is, so so I've got \(8 \times 10^{-8}\) is equal to \(1 / \sqrt{2 \times 4 \pi}\) radius squared times, and this is going to be, let's see, \(1.38 \times 10^{-23}\). 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 \(1.01 \times 10^5\). Alright. So, again, we're still looking just for this \(r\) over here. What I'm going to do is I'm going to actually because we have one over this whole entire thing here, what I'm going to do is I'm going to move this whole thing up to the other side, and then I'm going to move this \(8 \times 10^{-8}\) down to the denominator. Okay? So what I end up with hereceries, this, and then I'm just this works out to just a number here. When you multiply all this stuff together and you divide the \(8 \times 10^{-8}\) over to the other side, you're going to get \(4.66 \times 10^{-19}\). Okay? So now all I have to do is I'm just going to move over the \(\sqrt{2 \times 4 \pi}\). So when you divide this stuff over to the other side, which you end up with is that \(r_n^2\) is equal to, this is going to be \(2.62 \times 10^{-20}\). 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 \(2.66 \times 10^{-20}\), and what you're going to end up with here is \(1.6 \times 10^{-10}\). Alright? So that is the number, \(1.6 \times 10^{-10}\). 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 nitrogen is going to be a diatomic molecule because we have 2 of the nitrogens. Right? And, so this should kind of make some sense that we have \(1.6 \times 10^{-10}\) because, remember, what we said is that whenever you don't have the radius of monoatomic versus diatomic, you can assume that diatomic is roughly about \(1 \times 10^{-10}\). So this kind of agrees with our expectations. Alright, so that's it for this one, guys. Let me know if you have any questions.