Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between
the gas molecule changes from $\tau_{1}$ to $\tau_{2}$. If $\frac{C_{p}}{C_{v}}=\gamma$ for this
gas then a good estimate for $\frac{\tau_{2}}{\tau_{1}}$ is given by:
Correct Option: 1
(1) We know that Relaxation time,
$T \propto \frac{V}{\sqrt{T}}$
Equation of adiabatic process is $T V^{\gamma-1}=$ constant
$\Rightarrow \quad T \propto \frac{1}{V^{\gamma-1}}$
$\Rightarrow T \propto V^{1+\frac{\gamma-1}{2}}$ using (i)
$\Rightarrow T \propto V^{\frac{1+\gamma}{2}}$
$\Rightarrow \frac{T_{f}}{T}=\left(\frac{2 V}{V}\right)^{\frac{1+\gamma}{2}}=(2)^{\frac{1+\gamma}{2}}$
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