Q139 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Physics
An ideal gas at pressure $\mathrm{P}_0$ undergoes an isothermal expansion until its volume is 8.0 times its initial volume. The gas is slowly and adiabatically compressed back to its original volume. If the adiabatic constant of the gas is $\gamma=4 / 3$, then the ratio of the average kinetic energy per molecule in this final state to that in the initial state is
- A1.44
- B1.68
- Correct answerC2.0
- D1.2
Solution

For BC $$ \begin{aligned} & \mathrm{T}_0\left(8 \mathrm{~V}_0\right)^{\gamma-1}=\mathrm{T}^{\prime \prime}\left(\mathrm{V}_0\right)^{\gamma-1} \\ & \Rightarrow \quad \mathrm{T}_0\left(8 \mathrm{~V}_0\right)^{1 / 3}=\mathrm{T}^{\prime \prime} \mathrm{V}_0^{1 / 3} \\ & \Rightarrow \quad \mathrm{T}_0(8)^{1 / 3}=\mathrm{T}^{\prime \prime} \\ & \Rightarrow \quad \mathrm{T}^{\prime \prime}=2 \mathrm{~T}_0 \end{aligned} $$ Now, $\mathrm{U}_{\mathrm{av}} \propto \mathrm{T}$ $$ \frac{\left(\mathrm{U}_{\mathrm{av}}\right)_{\mathrm{f}}}{\left(\mathrm{U}_{\mathrm{av}}\right)_{\mathrm{i}}}=\frac{\mathrm{T}^{\prime \prime}}{\mathrm{T}_0}=\frac{2 \mathrm{~T}_0}{\mathrm{~T}_0}=2 $$
Hence, the correct answer is option: 'C'.