1. Use adiabatic relation

For an ideal gas in an adiabatic process, $PV^\gamma=\text{constant}$ Also, using $PV=nRT$, $TV^{\gamma-1}=\text{constant}$

Given the volume doubles: $V_2=2V_1$ So, $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$ $\Rightarrow \frac{T_2}{T_1}=\left(\frac{V_1}{V_2}\right)^{\gamma-1}=\left(\frac{1}{2}\right)^{\gamma-1}$

2. Relate mean collision time to temperature and volume

Mean collision time can be estimated as $\tau \sim \frac{\lambda}{\bar v}$ where $\lambda$ is mean free path and $\bar v$ is average molecular speed.

For an ideal gas,

• Mean free path: $\lambda \propto \frac{1}{n} \propto V$ (since number of molecules is fixed, $n \propto 1/V$)

• Average speed: $\bar v \propto \sqrt{T}$

Hence, $\tau \propto \frac{V}{\sqrt{T}}$

Therefore, $\frac{\tau_2}{\tau_1}=\frac{V_2}{V_1}\sqrt{\frac{T_1}{T_2}}$

3. Substitute the adiabatic temperature change

We have $\frac{V_2}{V_1}=2$ and $\frac{T_1}{T_2}=2^{\gamma-1}$ Thus, $\frac{\tau_2}{\tau_1}=2\sqrt{2^{\gamma-1}}$ $=2\cdot 2^{(\gamma-1)/2}$ $=2^{1+(\gamma-1)/2}$ $=2^{(1+\gamma)/2}$

4. Match with options

$\frac{\tau_2}{\tau_1}=2^{\frac{1+\gamma}{2}}$ This corresponds to Option A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer also gives A, so they agree.