Question

The molecular density of a gas is n and diameter of its molecule is d. The mean free path of molecule is:

• A. \(\frac{\pi}{nd^2}\)
• B. \(\frac{1}{\pi n d}\)
• C. \(\frac{1}{\sqrt{2}\pi n d^2}\)
• D. \(\frac{\pi}{3\sqrt{2}\pi n d^2}\)

Hint:

The mean free path is inversely proportional to the density of the gas (\(n\)) and the collision cross-section (\(\pi d^2\)). A denser gas or larger molecules will lead to a shorter mean free path. The \(\sqrt{2}\) factor is a crucial correction that arises from considering the relative speeds of the colliding molecules.

Answer 1

The Correct Option is C

Step 1: Define mean free path. The mean free path (\(\lambda\)) is the average distance a particle travels between collisions with other particles.
Step 2: Model with stationary target particles. Assuming one particle moves through stationary particles, it collides with any particle within a cylinder of radius \(d\) (cross-sectional area \(\sigma = \pi d^2\)). The collision rate is \(n\sigma v\). The mean free path is \(\lambda = \frac{1}{n\sigma} = \frac{1}{\pi n d^2}\).
Step 3: Refine model with relative motion. The stationary target model is inaccurate. Accounting for the relative motion of all particles using the Maxwell-Boltzmann distribution introduces a \(\sqrt{2}\) factor. The corrected formula is \(\lambda = \frac{1}{\sqrt{2} n \sigma} = \frac{1}{\sqrt{2}\pi n d^2}\), where \(n\) is the number density and \(d\) is the molecular diameter.