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An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6 × 10⁻⁸ s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to
- 0.5 × 10⁻⁸ s
- 4 × 10⁻⁸ s
- 3 × 10⁻⁸ s
- 2 × 10⁻⁷ s
Correct answer: 4 × 10⁻⁸ s
Solution
Mean free time tau = lambda/v ~ (T/P)/sqrt(T) = sqrt(T)/P. tau2 = tau1 * sqrt(T2/T1) * (P1/P2) = 6e-8 * sqrt(500/300) * (1/2) = 3.87e-8 s, closest to 4e-8 s.
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