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: Understand the concept of mean free path. The mean free path (\(\lambda\)) is the average distance a particle (like a molecule) travels between successive collisions with other particles.
Step 2: Consider a simplified model. If we imagine one molecule moving through a gas of stationary molecules, it will collide with any molecule whose center is within a cylinder of radius \(d\) (or cross-sectional area \(\sigma = \pi d^2\)). The volume of this collision cylinder is \(Area \times length = \sigma v t\). The number of collisions is the number of particles in this volume, which is \(n(\sigma v t)\). The rate of collision is \(n\sigma v\). The mean free path would be the average distance traveled per collision, \(\lambda = \frac{v}{n\sigma v} = \frac{1}{n\sigma} = \frac{1}{\pi n d^2}\).
Step 3: Apply the correct model accounting for the motion of all particles. The simplified model is incorrect because it assumes the target molecules are stationary. When the relative motion of all molecules is properly accounted for using the Maxwell-Boltzmann distribution of speeds, an additional factor of \(\sqrt{2}\) appears in the denominator. The correct formula for the mean free path is: \[ \lambda = \frac{1}{\sqrt{2} n \sigma} = \frac{1}{\sqrt{2}\pi n d^2} \] where \(n\) is the number density of molecules and \(d\) is the molecular diameter.