Question

An ideal gas enclosed in a cylinder is at pressure $2$ atm and temperature $300\,\text{K}$. The mean time between two successive collisions is $6\times10^{-8}\,\text{s}$. If the pressure is doubled and temperature is increased to $500\,\text{K}$, the mean time between two successive collisions will be close to

• A. $3\times10^{-6}\,\text{s}$
• B. $4\times10^{-8}\,\text{s}$
• C. $2\times10^{-7}\,\text{s}$
• D. $5\times10^{-8}\,\text{s}$

Hint:

Mean collision time varies as $\displaystyle \tau \propto \frac{\sqrt{T}}{P}$. Increasing temperature increases $\tau$, while increasing pressure decreases it.

Answer 1

The Correct Option is B

Step 1: Mean time between collisions $\tau$ is inversely proportional to number density and average speed: \[ \tau \propto \frac{1}{n\bar{v}} \]
Step 2: For an ideal gas, \[ n \propto \frac{P}{T}, \qquad \bar{v} \propto \sqrt{T} \]
Step 3: Hence, \[ \tau \propto \frac{1}{\left(\frac{P}{T}\right)\sqrt{T}} = \frac{\sqrt{T}}{P} \]
Step 4: Therefore, \[ \frac{\tau_2}{\tau_1} = \frac{\sqrt{T_2}/P_2}{\sqrt{T_1}/P_1} \] Given: \[ T_1=300\,\text{K}, \quad P_1=2\,\text{atm} \] \[ T_2=500\,\text{K}, \quad P_2=4\,\text{atm} \]
Step 5: Substitute: \[ \frac{\tau_2}{\tau_1} = \frac{\sqrt{500}/4}{\sqrt{300}/2} = \frac{2\sqrt{500}}{4\sqrt{300}} = \frac{\sqrt{5}}{2\sqrt{3}} = \sqrt{\frac{5}{12}} \approx 0.645 \]
Step 6: Hence, \[ \tau_2 \approx 0.645 \times 6\times10^{-8} \approx 3.9\times10^{-8}\,\text{s} \approx 4\times10^{-8}\,\text{s} \]