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

Step 1: Recall the mean free path formula.
The mean free path is $\lambda = \dfrac{1}{\sqrt{2}\,\pi d^2 n}$, so $\lambda \propto \dfrac{1}{d^2 n}$, where $d$ is molecular diameter and $n$ is number density.
Step 2: Write it for both gases.
$\lambda_A = \dfrac{1}{\sqrt2 \pi d_A^2 n_A}$ and $\lambda_B = \dfrac{1}{\sqrt2 \pi d_B^2 n_B}$.
Step 3: Form the ratio.
Dividing one by the other gives $\dfrac{\lambda_A}{\lambda_B} = \dfrac{d_B^2 n_B}{d_A^2 n_A}$.
Step 4: Insert the given conditions.
We are told $\lambda_A = \tfrac12 \lambda_B$ and $d_A = 2 d_B$. So $\dfrac12 = \dfrac{d_B^2 n_B}{(2d_B)^2 n_A} = \dfrac{n_B}{4 n_A}$.
Step 5: Solve for the densities.
Cross-multiplying: $4 n_A = 2 n_B$, hence $n_A = \dfrac{n_B}{2}$.
Step 6: Pick the option.
This is exactly option A.
\[ \boxed{ n_A = \tfrac12 n_B } \]