Question

The mean free path of molecules in an ideal gas A is half that of another ideal gas B. The diameter of the spherical molecules of gas A is twice the diameter of the molecules of gas B. If number densities of the gases A and B are \(n_A\) and \(n_B\), respectively, then

• A. \(n_A=\dfrac{1}{2}n_B\)
• B. \(n_A=n_B\)
• C. \(n_A=2n_B\)
• D. \(n_A=\dfrac{1}{4}n_B\)

Hint:

Remember the important relation: \[ \lambda=\frac{1}{\sqrt{2}\pi d^2 n} \] Mean free path is inversely proportional to both the square of molecular diameter and the number density. \[ \lambda \propto \frac{1}{d^2 n} \] Always convert proportionality into a ratio before substituting numerical relations.

Answer 1

The Correct Option is A

Concept:

• The mean free path of gas molecules is given by \[ \lambda=\frac{1}{\sqrt{2}\pi d^2 n} \] where itemize

• \(\lambda\) = mean free path

• \(d\) = diameter of molecule

• \(n\) = number density of molecules
Thus, \[ \lambda \propto \frac{1}{d^2 n} \] itemize

Step 1: Write the expression for both gases.
For gas A, \[ \lambda_A = \frac{1}{\sqrt{2}\pi d_A^2 n_A} \] For gas B, \[ \lambda_B = \frac{1}{\sqrt{2}\pi d_B^2 n_B} \] Taking ratio, \[ \frac{\lambda_A}{\lambda_B} = \frac{d_B^2 n_B}{d_A^2 n_A} \]

Step 2: Substitute the given conditions.
Given, \[ \lambda_A=\frac{1}{2}\lambda_B \] and \[ d_A=2d_B \] Substituting into the ratio, \[ \frac{1}{2} = \frac{d_B^2 n_B} {(2d_B)^2 n_A} \] \[ \frac{1}{2} = \frac{d_B^2 n_B} {4d_B^2 n_A} \] \[ \frac{1}{2} = \frac{n_B}{4n_A} \]

Step 3: Solve for the number density ratio.
Cross-multiplying, \[ 4n_A = 2n_B \] \[ n_A = \frac{n_B}{2} \] Therefore, \[ \boxed{ n_A=\frac{1}{2}n_B } \]

Step 4: Choose the correct option.
Hence, \[ \boxed{\text{Option (A)}} \]