• # question_answer A $25\times {{10}^{-3}}{{m}^{3}}$volume cylinder is filled with 1 mol of ${{O}_{2}}$gas at room temperature (300K). The molecular diameter of ${{O}_{2}}$, and its root mean square speed, are found to be 0.3 nm, and 200 m/s, respectively. What is the average collision rate (per second) for an ${{O}_{2}}$ molecule? [JEE Main 10-4-2019 Morning] A) $\tilde{\ }{{10}^{11}}$                 B) $\tilde{\ }{{10}^{13}}$C) $\tilde{\ }{{10}^{10}}$                 D) $\tilde{\ }{{10}^{12}}$

Correct Answer: D

Solution :

$\text{v}=\frac{{{V}_{a\text{v}}}}{\lambda }$                 $\lambda =\frac{RT}{\sqrt{2}\pi {{\sigma }^{2}}{{N}_{A}}P}$ $P=\frac{RT}{V}$ $\Rightarrow \lambda =\frac{V}{\sqrt{2}\pi {{\sigma }^{2}}{{N}_{A}}}$ ${{V}_{a\text{v}}}=\sqrt{\frac{8}{3\pi }}\times {{V}_{rms}}$ $\therefore \text{v}=\frac{200\times \sqrt{2}\pi \times {{\sigma }^{2}}{{N}_{A}}}{25\times {{10}^{-3}}}\times \sqrt{\frac{8}{3\pi }}$ $=17.68\times {{10}^{8}}/\sec .$ $=.1768\times {{10}^{10}}/\sec \tilde{\ }{{10}^{10}}$   This answer does not match with JEE-Answer key

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