Question

The mean free path for a gas, with molecular diameter $d$ and number density $n$ can be expressed as :

• A. $\frac {1}{\sqrt{2} n \pi d}$
• B. $\frac {1}{\sqrt{2} n \pi d^2}$
• C. $\frac {1}{\sqrt{2} n^2 \pi d^2}$
• D. $\frac {1}{\sqrt{2} n^2 \pi^2 d^2}$

Answer 1

The Correct Option is B

To understand the solution to this question, let's first review the concept of the mean free path and determine how it relates to molecular diameter and number density in a gas.

The mean free path is defined as the average distance a molecule travels between collisions. For a gas, it is dependent on the molecular diameter (\(d\)) and the number density (\(n\)), which is the number of molecules per unit volume.

The formula for the mean free path \((\lambda)\) is given by:

\(\lambda = \frac{1}{\sqrt{2} n \pi d^2}\)

This formula arises from kinetic theory and can be derived considering a molecule moving through the gas, while other molecules are considered stationary. The collision cross-section, which represents the effective area for collision and is determined by the circle formed around two approaching molecules, is given by \(\pi d^2\).

Using the formula above, we can analyze each option:

1. The option \(\frac{1}{\sqrt{2} n \pi d}\) is not correct, as the factor of \(d^2\) is essential for accounting for the area of collision.
2. The option \(\frac{1}{\sqrt{2} n \pi d^2}\) matches the derived formula, making it the correct answer.
3. The option \(\frac{1}{\sqrt{2} n^2 \pi d^2}\) is incorrect because it incorrectly squares the number density, which should only appear linearly, as each molecule contributes individually to the probability of collision.
4. The option \(\frac{1}{\sqrt{2} n^2 \pi^2 d^2}\) is similarly flawed due to incorrect squaring of both terms.

By applying this logic and understanding the derivation, we can conclude that the mean free path for a gas with molecular diameter \(d\) and number density \(n\) is given by \(\frac{1}{\sqrt{2} n \pi d^2}\).