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NEET 2026 (Re-neet) · Mean Free Path & Real Gases

Mean Free Path & Real Gases question from NEET 2026 (Re-neet)

Q1NEET 2026 · RE-NEETMedium

Mean Free Path & Real Gases

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:

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Answer: (a)

$\lambda \propto \dfrac{1}{n d^2}$ , so $\dfrac{\lambda_A}{\lambda_B} = \dfrac{n_B d_B^2}{n_A d_A^2}$ .
$\dfrac{1}{2} = \dfrac{n_B d_B^2}{n_A (2d_B)^2} = \dfrac{n_B}{4n_A}$
$4n_A = 2n_B \Rightarrow n_A = \dfrac{1}{2}n_B$
So the answer is (a).

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