Question

A container of volume 𝑉 has helium gas in it with 𝑁 number of He atoms. The mean free path of these atoms is 𝜆_He. Another container has argon gas with the same number of Ar atoms in volume 2𝑉 with their mean free path being 𝜆_Ar. Taking the radius of Ar atoms to be 1.5 times the radius of He atoms, the ratio 𝜆_Ar⁄𝜆_He is__________ (rounded off to two decimal places).

Answer 1

Correct Answer: 0.85 - 0.95

We are given two containers:

• Container 1 has helium gas with volume \( V \) and \( N \) number of helium atoms.
• Container 2 has argon gas with volume \( 2V \) and the same number of argon atoms \( N \).

The mean free path of these atoms are denoted as \( \lambda_{\text{He}} \) for helium and \( \lambda_{\text{Ar}} \) for argon.

The goal is to find the ratio \( \frac{\lambda_{\text{Ar}}}{\lambda_{\text{He}}} \), given the following conditions:

• The radius of an argon atom is 1.5 times the radius of a helium atom, i.e., \( r_{\text{Ar}} = 1.5 \cdot r_{\text{He}} \).

From kinetic theory, the mean free path \( \lambda \) is given by the formula:

\[ \lambda = \frac{1}{\sqrt{2} \, n \sigma} \] where \( n \) is the number density of particles, and \( \sigma \) is the collision cross-section. The collision cross-section \( \sigma \) is proportional to the square of the radius of the gas atoms, i.e., \( \sigma \propto r^2 \). Thus, the ratio of the mean free paths for argon and helium can be written as: \[ \frac{\lambda_{\text{Ar}}}{\lambda_{\text{He}}} = \frac{\frac{1}{n_{\text{Ar}} \sigma_{\text{Ar}}}}{\frac{1}{n_{\text{He}} \sigma_{\text{He}}}} = \frac{n_{\text{He}} \sigma_{\text{He}}}{n_{\text{Ar}} \sigma_{\text{Ar}}} \] Since the number of atoms is the same for both gases, the number density \( n \) is inversely proportional to the volume, i.e., \[ n_{\text{Ar}} = \frac{N}{2V}, \quad n_{\text{He}} = \frac{N}{V} \] Thus, the ratio of number densities is: \[ \frac{n_{\text{He}}}{n_{\text{Ar}}} = 2 \] The collision cross-section is proportional to the square of the radius, so: \[ \frac{\sigma_{\text{Ar}}}{\sigma_{\text{He}}} = \left( \frac{r_{\text{Ar}}}{r_{\text{He}}} \right)^2 = (1.5)^2 = 2.25 \] Now, the ratio of the mean free paths is: \[ \frac{\lambda_{\text{Ar}}}{\lambda_{\text{He}}} = \frac{2}{2.25} = \frac{8}{9} \approx 0.89 \] Thus, the ratio of the mean free paths is approximately: \[ \boxed{0.85 \, \text{to} \, 0.95} \]