Question

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters $1\AA$ and $2\AA$. The gases may be considered under identical conditions of temperature, pressure, and volume.

Answer 1

Given:

Molecular diameter of first gas $=1\AA$

Molecular diameter of second gas $=2\AA$

Since, Mean free path
$\lambda =\frac{K_{B}T}{\sqrt{2}\pi d^{2}P}$

Where, $d=$ diameter of molecules

The condition for both gases is identical, so $T$ and $P$ will be same for both.

As per given condition,

$\lambda \propto \frac{1}{d^2}$

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{d_2^2}{d_1^2}=\frac{2^2}{1^2}$

${\lambda }_1:{\lambda }_2=4:1$

$\text{Final Answer}:4:1$.