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JEE Main 2023 (6 Apr, Shift 1) · Mean Free Path & Real Gases

Mean Free Path & Real Gases question from JEE Main 2023 (6 Apr, Shift 1)

Q12JEE Main 2023 · 6 APR · SHIFT 1Medium

Mean Free Path & Real Gases

The number of air molecules per cm $^{3}$ increased from $3 \times 10^{19}$ to $12 \times 10^{19}$ . The ratio of collision frequency of air molecules before and after the increase in number respectively is:

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Answer: (b)

Collision frequency $f = \sqrt{2}\,n\pi d^2\bar v \propto n$ (same $T$ , same gas).
$\dfrac{f_1}{f_2} = \dfrac{3\times10^{19}}{12\times10^{19}} = 0.25$
So the answer is (b) 0.25.

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