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 $B$. If number densities of the gases $A$ and $B$ are $n_{A}$ and $n_{B}$, respectively, the correct option is:
Options
Solution
Sol. [Key Concept: Mean free path λ ∝ 1/(√2 nπd²)]
$\lambda_{A} = \frac{1}{2}\lambda_{B}$
$$\begin{matrix} & \lambda = \frac{1}{\sqrt{2}\pi d^{2}n} \\ & d_{A} = 2d_{B} \\ & \lambda_{A} = \frac{1}{\sqrt{2}\pi d_{A}^{2}n_{A}} \\ & \lambda_{B} = \frac{1}{\sqrt{2}\pi d_{B}^{2}n_{B}} \\ & \frac{\lambda_{A}}{\lambda_{B}} = \frac{1}{2} = \frac{d_{B}^{2}}{d_{A}^{2}}\frac{n_{B}}{n_{A}} \\ & \frac{n_{A}}{n_{B}} = \frac{2 \cdot d_{B}^{2}}{d_{A}^{2}} \\ & \frac{n_{A}}{n_{B}} = \frac{2 \cdot d_{B}^{2}}{4d_{B}^{2}} = \frac{1}{2} \end{matrix}$$
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