Question

Consider \( N_i \) number of ideal gas particles enclosed in a volume \( V_1 \). If the volume is changed to \( V_2 \) and the number of particles is reduced by half, the mean free path becomes four times of its initial value. The ratio \( \frac{V_1}{V_2} \) is ........... (Round off to one decimal place).

Hint:

The mean free path depends on the number of particles and the volume; changing these parameters can significantly affect the path length.

Answer 1

Correct Answer: 0.5

The mean free path \( \lambda \) is inversely proportional to the number of particles and the volume. The mean free path for the initial state is \( \lambda_1 \), and after the change, it becomes \( 4\lambda_1 \). Thus, we have: \[ \lambda_1 \propto \frac{V_1}{N_i}. \] After halving the number of particles, \[ \lambda_2 \propto \frac{V_2}{\frac{N_i}{2}} = \frac{2V_2}{N_i}. \] Using the given condition that \( \lambda_2 = 4 \lambda_1 \), we get: \[ \frac{2V_2}{N_i} = 4 \times \frac{V_1}{N_i} \Rightarrow 2V_2 = 4V_1 \Rightarrow V_2 = 2V_1. \] Final Answer: The ratio \( \frac{V_1}{V_2} \) is \( 0.5 \).