Q 11-12-149JEE MainJEE Main 2019 (12 Jan, Shift 2)Medium
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 $6\times10^{-8}$ s. If the pressure is doubled and temperature is increased to $500$ K, the mean time between two successive collisions will be close to:
Answer: (D) $4\times10^{-8}$ s
Mean free path $\lambda \propto \dfrac TP$ and mean speed $v \propto \sqrt T$, so the time between collisions
$$\tau = \frac\lambda v \propto \frac{\sqrt T}{P}$$
$$\tau_2 = 6\times10^{-8}\times\sqrt{\frac{500}{300}}\times\frac12 \approx 3.9\times10^{-8} \approx 4\times10^{-8}\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics