Question

For the given concentration, if the ratio of the diameters of the molecules of two gases is 1 : 2, then the ratio of their mean free paths is

• A. \( 4 : 1 \)
• B. \( 2 : 1 \)
• C. \( 1 : 1 \)
• D. \( 1 : 2 \)

Hint:

The mean free path is inversely proportional to the square of the diameter of the gas molecules and the number density (concentration). When the concentration is the same for two gases, the ratio of their mean free paths is inversely proportional to the square of the ratio of their diameters.

Answer 1

The Correct Option is A

The mean free path \( \lambda \) of a gas molecule is the average distance a molecule travels between successive collisions.
It is given by the formula: $$ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} $$ where \( d \) is the diameter of the gas molecule and \( n \) is the number density of the gas molecules (number of molecules per unit volume), which is proportional to the concentration.
For two gases at the same concentration, \( n_1 = n_2 = n \).
Let the diameters of the molecules of the two gases be \( d_1 \) and \( d_2 \), and their mean free paths be \( \lambda_1 \) and \( \lambda_2 \), respectively.
The ratio of their diameters is given as \( d_1 : d_2 = 1 : 2 \), so \( \frac{d_1}{d_2} = \frac{1}{2} \).
The mean free paths are: $$ \lambda_1 = \frac{1}{\sqrt{2} \pi d_1^2 n} $$ $$ \lambda_2 = \frac{1}{\sqrt{2} \pi d_2^2 n} $$ The ratio of their mean free paths is: $$ \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{\sqrt{2} \pi d_1^2 n}}{\frac{1}{\sqrt{2} \pi d_2^2 n}} = \frac{\sqrt{2} \pi d_2^2 n}{\sqrt{2} \pi d_1^2 n} = \frac{d_2^2}{d_1^2} = \left( \frac{d_2}{d_1} \right)^2 $$ Since \( \frac{d_1}{d_2} = \frac{1}{2} \), we have \( \frac{d_2}{d_1} = 2 \).
$$ \frac{\lambda_1}{\lambda_2} = (2)^2 = 4 $$ Therefore, the ratio of their mean free paths is \( \lambda_1 : \lambda_2 = 4 : 1 \).