JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • 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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