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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2019
  • Hard
  • MCQ

A \(25\times10^{-3}\, m^3\) volume cylinder is filled with \(1\, mol\) of \(O_2\) gas at room temperature \((300\, K)\). 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?

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Answer & Solution

Correct Answer

(A) \(\sim 10^{10}\)

Step-by-step Solution

Detailed explanation

\(v=\frac{V_{a v}}{\lambda}\)

\(\lambda=\frac{\mathrm{RT}}{\sqrt{2} \pi \sigma^{2} \mathrm{N}_{\mathrm{A}} \mathrm{P}}\)

\(\sigma=2 \times 0.3 \times 10^{-9}\)

\(P=\frac{R T}{V}\)

\(\Rightarrow \quad=\frac{V}{\sqrt{2} \pi \sigma^{2} N_{A}}\)

\(V_{\mathrm{av}}=\sqrt{\frac{8}{3 \pi}} \times V_{\mathrm{rms}}\)

\(\begin{aligned} \therefore \quad \mathrm{v} =\frac{200 \times \sqrt{2} \pi \times \sigma^{2} \mathrm{N}_{\mathrm{A}}}{25 \times 10^{-3}} \times \sqrt{\frac{8}{3 \pi}} \\ =17.68 \times 10^{8} / \mathrm{sec} \\ =0.1768 \times 10^{10} / \mathrm{sec}-10^{10} \end{aligned}\)

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