An ideal gas expands in such a way that PV^2 | vedclass

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(D) From the ideal gas equation, $PV = nRT$, we have $P = \frac{nRT}{V}$.Substituting this into the given process equation $PV^2 = K$ (where $K$ is a

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Such an expansion is possible only with cooling.

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(D) From the ideal gas equation, $PV = nRT$, we have $P = \frac{nRT}{V}$.Substituting this into the given process equation $PV^2 = K$ (where $K$ is a constant):$(\frac{nRT}{V})V^2 = K$$nRTV = K$$TV = \frac{K}{nR} = 	ext{constant}$.Since $T \propto \frac{1}{V}$, as the gas expands, the volume $V$ increases.Because $V$ increases, the temperature $T$ must decrease to keep the product $TV$ constant.Therefore, the gas undergoes cooling during this expansion.Thus, the correct option is $D$.

An ideal gas expands in such a way that $PV^2 = \text{constant}$ throughout the process.

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