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(B) Given the law for the ideal gas process: $VP^2 = K$ (where $K$ is a constant).From the ideal gas equation, we have $PV = nRT$, which implies $P =

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$T' = \sqrt{2} T$

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(B) Given the law for the ideal gas process: $VP^2 = K$ (where $K$ is a constant).From the ideal gas equation, we have $PV = nRT$, which implies $P = \frac{nRT}{V}$.Substituting this into the given law:$V \left( \frac{nRT}{V} \right)^2 = K$$V \left( \frac{n^2 R^2 T^2}{V^2} \right) = K$$\frac{n^2 R^2 T^2}{V} = K$Since $n$ and $R$ are constants, we can write $T^2 \propto V$, or $T \propto \sqrt{V}$.Therefore, $\frac{T_2}{T_1} = \sqrt{\frac{V_2}{V_1}}$.Given $V_1 = V$, $T_1 = T$, and $V_2 = 2V$, we have:$\frac{T'}{T} = \sqrt{\frac{2V}{V}} = \sqrt{2}$.Thus, $T' = \sqrt{2} T$.

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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