Question

If molecular diameter of gas $A$ is double of gas $B$, then mean free path of gas $A$ for same molecular density is

• A. $\frac{1}{4}$ times of gas $B$
• B. $\frac{1}{2}$ times of gas $B$
• C. $4$ times of gas $B$
• D. $2$ times of gas $B$

Answer 1

The correct option is A $\frac{1}{4}$ times of gas $B$

Given,
Diameter of gas $A=2\times$ diameter of gas B
$d_A=2d_B---\left(1\right)$
We know, mean free path $\left(\lambda \right)$
$\lambda =\frac{1}{\sqrt{2}n\pi d^2}$
$n\to$ molecular density
For constant value of $n$
$\lambda \propto \frac{1}{d^2}$
For gas B, ${\lambda }_B\propto \frac{1}{d_B^2}----\left(2\right)$
For gas A, ${\lambda }_A\propto \frac{1}{4d_B^2}$ [From (1)]
$\Rightarrow {\lambda }_A=\frac{{\lambda }_B}{4}$ [From (2)]
Mean free path of gas $A$ is $\frac{1}{4}$ times of gas $B$.