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

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(D) Given the process equation: $PV^2 = 	ext{constant}$.
Using the ideal gas law $PV = nRT$,we can write $P = \frac{nRT}{V}$.
Substituting $P$ into

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Such an expansion is possible only with cooling.

Vedclass · Answer

(D) Given the process equation: $PV^2 = 	ext{constant}$.
Using the ideal gas law $PV = nRT$,we can write $P = \frac{nRT}{V}$.
Substituting $P$ into the process equation: $(\frac{nRT}{V})V^2 = 	ext{constant}$.
This simplifies to $nRTV = 	ext{constant}$,which means $TV = 	ext{constant}$ (since $nR$ is constant).
Therefore,$T \propto \frac{1}{V}$.
This represents a rectangular hyperbola in the $T-V$ plane,not a parabola or a straight line.
Since the gas expands,$V$ increases,which implies $T$ must decrease. Therefore,such an expansion is possible only with cooling.

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

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