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(D) From the ideal gas equation,$PV = \mu RT$,so $P = \frac{\mu RT}{V}$.Given the process law $VP^2 = K$ (where $K$ is a constant).Substituting $P$ in

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$\sqrt{2} T$

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(D) From the ideal gas equation,$PV = \mu RT$,so $P = \frac{\mu RT}{V}$.Given the process law $VP^2 = K$ (where $K$ is a constant).Substituting $P$ in the law: $V \left( \frac{\mu RT}{V} ight)^2 = K$.$V \cdot \frac{\mu^2 R^2 T^2}{V^2} = K$.$\frac{T^2}{V} = \frac{K}{\mu^2 R^2} = 	ext{constant}$.This implies $T^2 \propto V$,or $T \propto \sqrt{V}$.Therefore,$\frac{T_2}{T_1} = \sqrt{\frac{V_2}{V_1}}$.Given $V_1 = V$,$V_2 = 2V$,and $T_1 = T$.$\frac{T_2}{T} = \sqrt{\frac{2V}{V}} = \sqrt{2}$.$T_2 = \sqrt{2} T$.

During an experiment,an ideal gas is found to obey an additional law $VP^2 = \text{constant}$. The gas is initially at temperature $T$ and volume $V$. When the gas expands to a volume $2V$,its temperature will be:

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