Mean Free Path of Gases - Online Tutor, Practice Problems & Exam Prep

The Mean Free Path (λ) represents the average distance gas particles travel before colliding with another particle. It can be calculated using the equation: $\left(\frac{V}{\sqrt{2}4\pi r^2}\right)n$. The mean free path decreases with increasing particle size (r) and density (n). Understanding this concept is crucial for grasping gas behavior under various conditions, such as temperature and pressure.

Mean Free Path

Hey, guys. So throughout your problems, you may be asked to calculate something called the Mean Free Path. In this video, I want to show you what the mean free path is conceptually, the equations for it, and we'll do a quick example together. Let's check this out. Imagine that I have a canister of gas particles that are sealed inside this little container like this. Now, remember these gas particles aren't stationary. They're all moving all over the place. What happens here is if you look at a random gas particle and its path as it's moving throughout the canister, it eventually travels in a straight line before bumping into another particle. All the collisions are elastic, so these things scatter off in straight lines. But basically, it just keeps on bouncing like this over and over again and these particles just go all over the place. The mean free path is really just the average. That just means the mean distance that the gas particles travel before they collide with another particle. I like to use the analogy of the container gas being like a room full of people. If I'm a person walking through this room and I walk in a straight line, eventually, I'm going to bump into somebody else. If I can only walk in a straight line, I'm just going to keep bumping off people as I walk through the room. So, if the container is full of people, the mean free path is the average distance you walk before bumping into somebody else.

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 this distance here is just equal to velocity times time. Right? This is basically just delta x equals v*t. 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, like the distance that they travel, in terms of the variables inside of the ideal gas law, like pressure, volume, and temperature. The equation that you need to know is actually this one:

λ = V 2 ⋅ 4 π r 2 ⋅ n

Unfortunately, there isn't really a good way to memorize this equation, but if you ever need it, it will be given to you on a formula sheet. Notice how in the equation, an increase in the number of particles or the radius of a particle decreases the mean free path, because if the number of particles increases or if the particles are physically larger, the average distance before a collision decreases.

We'll check out our example here. We have oxygen molecules at STP. Remember that STP is a set of conditions where the temperature is 273 Kelvin and the pressure is 1.01 ⋅ 10 5 Pa . So, for the first part, we want to calculate the mean free path of oxygen molecules λ₁₄ . We'll use the equation for which we've been given the average speed of the particles. We don't know the time, so we'll use the more complicated equation as follows:

λ = V 2 ⋅ 4 π r 2 ⋅ n

We know the temperature and pressure but need the volume and number of particles. The ideal gas law helps relate these with:

PV = nRT

Solving for V/n, we get V n = kBT P . Plugging in values, we find the mean free path for oxygen to be about 93 nanometers.

For the second part, we calculate the average time between collisions using:

T αβεραγε = λ ₂ₒ V αβεραγε

The result is approximately 2.07 times 10^-10 seconds, showing the very rapid collision frequency of oxygen molecules at STP.

That's it for this one. Let me know if you have any questions.

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

Calculating Radius of Nitrogen

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.

Here’s what students ask on this topic:

What is the mean free path in gases?

The mean free path (λ) in gases is the average distance a gas particle travels before colliding with another particle. It is a crucial concept in understanding the behavior of gases, particularly in relation to their macroscopic properties like pressure, volume, and temperature. The mean free path can be calculated using the equation:

$\frac{V}{\sqrt{2}4\pi r^{2}n}$

where V is the volume, r is the radius of the particles, and n is the number of particles. The mean free path decreases with increasing particle size and density.

How do you calculate the mean free path of gas molecules?

To calculate the mean free path (λ) of gas molecules, you can use the equation:

$\frac{V}{\sqrt{2}4\pi r^{2}n}$

where V is the volume of the gas, r is the radius of the gas particles, and n is the number of particles. This equation shows that the mean free path is inversely proportional to the particle size and density. In practical problems, you may need to use the ideal gas law to find the necessary variables.

What factors affect the mean free path of gas particles?

The mean free path (λ) of gas particles is affected by several factors:

• Particle Size (r): Larger particles have a shorter mean free path because they collide more frequently.
• Density (n): Higher density means more particles in a given volume, leading to more frequent collisions and a shorter mean free path.
• Temperature (T): Higher temperatures increase the average speed of particles, which can affect the frequency of collisions.
• Pressure (P): Higher pressure increases the density of particles, reducing the mean free path.

These factors are interrelated and can be understood through the ideal gas law and kinetic molecular theory.

Why is the mean free path important in understanding gas behavior?

The mean free path (λ) is important in understanding gas behavior because it provides insight into how gas particles interact with each other. It helps explain macroscopic properties like pressure, volume, and temperature. For example, a shorter mean free path indicates more frequent collisions, which can affect the pressure and temperature of the gas. Understanding the mean free path is also crucial for applications in fields like thermodynamics, fluid dynamics, and material science, where gas behavior under different conditions is studied.

How does temperature affect the mean free path of gas molecules?

Temperature affects the mean free path (λ) of gas molecules by influencing their average speed. As temperature increases, the kinetic energy of the gas particles also increases, causing them to move faster. This increased speed can lead to more frequent collisions, which may reduce the mean free path. However, the exact effect of temperature on the mean free path also depends on other factors like pressure and density. In general, higher temperatures tend to decrease the mean free path due to the increased collision frequency.