Under an adiabatic process, the volume of an ideal gas gets doubled.
Question:

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:

  1. (1) (2) ${ }^{\frac{1+\gamma}{2}}$

  2. (2) $\frac{1}{2}$

  3. (3) $\left(\frac{1}{2}\right)^{\gamma}$

  4. (4) $\left(\frac{1}{2}\right)^{\frac{\gamma+1}{2}}$


Correct Option: 1

Solution:

(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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