Question:
A \(25\times 10^{-3}\,\mathrm{m^3}\) volume cylinder is filled with 1 mol of \(\mathrm{O_2}\) gas at room temperature (300 K). The molecular diameter of \(\mathrm{O_2}\), 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 \(\mathrm{O_2}\) molecule ?
A \(25\times 10^{-3}\,\mathrm{m^3}\) volume cylinder is filled with 1 mol of \(\mathrm{O_2}\) gas at room temperature (300 K). The molecular diameter of \(\mathrm{O_2}\), 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 \(\mathrm{O_2}\) molecule ?
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Collision rate \(= \frac{v}{\lambda}\) with \(\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}\) and \(n = \frac{N_A}{V}\).
Updated On: Nov 5, 2026
- \(\sim 10^{13}\)
- \(\sim 10^{12}\)
- \(\sim 10^{11}\)
- \(\sim 10^{10}\)
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The Correct Option is D
Solution and Explanation
Number density \(n = \frac{6.02\times 10^{23}}{25\times 10^{-3}} = 2.4\times 10^{25}\,\mathrm{m^{-3}}\).
Mean free path \(\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}\), with \(d = 0.3\times 10^{-9}\) m: \(\pi d^2 = 2.83\times 10^{-19}\,\mathrm{m^2}\).
\(\sqrt{2}\,\pi d^2 n = 1.414\times 2.83\times 10^{-19}\times 2.4\times 10^{25} \approx 9.6\times 10^{6}\,\mathrm{m^{-1}}\), so \(\lambda \approx 1.0\times 10^{-7}\) m.
Collision rate \(= \frac{v}{\lambda} = \frac{200}{1.0\times 10^{-7}} \approx 2\times 10^{9}\,\mathrm{s^{-1}}\), which is closest to \(\sim 10^{10}\) per second.
Answer: (4) \(\sim 10^{10}\)
Mean free path \(\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}\), with \(d = 0.3\times 10^{-9}\) m: \(\pi d^2 = 2.83\times 10^{-19}\,\mathrm{m^2}\).
\(\sqrt{2}\,\pi d^2 n = 1.414\times 2.83\times 10^{-19}\times 2.4\times 10^{25} \approx 9.6\times 10^{6}\,\mathrm{m^{-1}}\), so \(\lambda \approx 1.0\times 10^{-7}\) m.
Collision rate \(= \frac{v}{\lambda} = \frac{200}{1.0\times 10^{-7}} \approx 2\times 10^{9}\,\mathrm{s^{-1}}\), which is closest to \(\sim 10^{10}\) per second.
Answer: (4) \(\sim 10^{10}\)
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