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(C) From the ideal gas law, $PV = nRT$, we have $P = \frac{nRT}{V}$.Given the process law $VP^2 = 	ext{constant}$.Substituting $P$ in the given law:

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$\sqrt{2} T$

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(C) From the ideal gas law, $PV = nRT$, we have $P = \frac{nRT}{V}$.Given the process law $VP^2 = 	ext{constant}$.Substituting $P$ in the given law: $V \left(\frac{nRT}{V}ight)^2 = 	ext{constant}$.This simplifies to $V \cdot \frac{n^2 R^2 T^2}{V^2} = 	ext{constant}$, which means $\frac{T^2}{V} = 	ext{constant}$.Therefore, $\frac{T_1^2}{V_1} = \frac{T_2^2}{V_2}$.Given $T_1 = T$, $V_1 = V$, and $V_2 = 2V$.Substituting these values: $\frac{T^2}{V} = \frac{T_2^2}{2V}$.$T_2^2 = 2T^2$.$T_2 = \sqrt{2} T$.

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 $V$. What will be the temperature of the gas when it expands to a volume $2V$?

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