Question

When the number of molecules per unit volume of an ideal gas is \( 0.8 \times 10^{24} \), the mean free path length for its molecules is \( 2.2 \times 10^{-5} \, {m} \). If the number of molecules per unit volume is \( 1.0 \times 10^{24} \), then the mean free path is

• A. \( 17.6 \times 10^{-5} \, {m} \)
• B. \( 1.76 \times 10^{-5} \, {m} \)
• C. \( 3.52 \times 10^{-5} \, {m} \)
• D. \( 35.2 \times 10^{-5} \, {m} \)
• E. \( 8.8 \times 10^{-5} \, {m} \)

Hint:

The mean free path is inversely proportional to the number of molecules per unit volume. If the number of molecules increases, the mean free path decreases.

Answer 1

The Correct Option is B

The mean free path \( \lambda \) is inversely proportional to the number of molecules per unit volume \( n \). 
Hence, we have the relation: \[ \lambda \propto \frac{1}{n} \] Let the initial number of molecules per unit volume be \( n_1 = 0.8 \times 10^{24} \) and the initial mean free path be \( \lambda_1 = 2.2 \times 10^{-5} \, {m} \). 
When the number of molecules per unit volume is increased to \( n_2 = 1.0 \times 10^{24} \), the mean free path \( \lambda_2 \) becomes: \[ \frac{\lambda_2}{\lambda_1} = \frac{n_1}{n_2} \] Substituting the values: \[ \frac{\lambda_2}{2.2 \times 10^{-5}} = \frac{0.8 \times 10^{24}}{1.0 \times 10^{24}} = 0.8 \] Thus: \[ \lambda_2 = 0.8 \times 2.2 \times 10^{-5} = 1.76 \times 10^{-5} \, {m} \] Hence, the correct answer is (B).