Video: The Mean Free Path and the Effective Radius of a Molecule

The mean free path for methane at a temperature of 273 K and a pressure of 1.22 × 10⁵ Pa is 4.62 × 10⁻⁸ m. Find the effective radius 𝑟 of the methane molecule.

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Video Transcript

The mean free path for methane at a temperature of 273 kelvin and a pressure of 1.22 times 10 to the power of five pascals is 4.62 times 10 to the power of negative eight meters. Find the effective radius 𝑟 of the methane molecule.

Alright, so in this question, we’ve got methane at a temperature of 273 kelvin and a pressure of 1.22 times 10 to the power of five pascals. We know that its mean free path is 4.62 times 10 to the power of negative eight meters. We need to find the effective radius 𝑟 of the methane molecule. To do this, we first need to recall how the mean free path is calculated.

The mean free path 𝑙 is given by 𝑘𝐵𝑇 over square root of two 𝜋 multiplied by 𝑑 squared 𝑝, where as we’ve already mentioned 𝑙 is the mean free path, 𝑇 represents the temperature of the gas, 𝑑 is the effective diameter of a particle of the gas, and 𝑝 is the pressure of the gas. 𝑘𝐵 and root two 𝜋 are just constants. And of course, we’re assuming that the gas we’re dealing with is an ideal gas.

Now, in this question, we’re trying to find the effective radius of the methane molecule. However, in this question, we only deal with the effective diameter. Luckily for us, however, there’s a relationship between the radius of a sphere and the diameter of a sphere. And that relationship is that the diameter is twice as large as the radius. So using the equation that we’ve been given, we can calculate the effective diameter and then just divide this by two to give us the effective radius, which means that we need to rearrange the mean free path equation to solve for the effective diameter.

We can do this by first multiplying both sides of the equation by 𝑑 squared over 𝑙. The 𝑙s cancel on the left-hand side and the 𝑑 squares cancel on the right, leaving us with 𝑑 squared is equal to 𝑘𝐵𝑇 over root two 𝜋𝑝𝑙. At this point, we can take the square root of both sides of the equation, which leaves us with 𝑑 is equal to the square root of 𝑘𝐵𝑇 over root two 𝜋𝑝𝑙.

Now, as we’ve said earlier, 𝑟 is equal to 𝑑 over two. In other words, the radius is equal to the diameter divided by two. So if we divide both sides of the equation by two, then we get the radius. And at this point, we know all of the quantities on the right-hand side of the equation. So we just need to substitute in our values and find out the value of 𝑟. This is what that looks like, a little bit messy. However, once we evaluate this on our calculator, we find that 𝑟 the effective radius is equal to 1.939356... and so on and so forth times 10 to the power of negative 10 meters.

Now, all of the quantities we’ve used from the question have been given to us to three significant figures. Therefore, we need to give our answer to three significant figures as well.

And so our final answer is that the effective radius 𝑟 is 1.94 times 10 to the power of negative 10 meters to three significant figures.

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