A gas is found to obey the law P^2V = constan | vedclass

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(D) Given the law for the gas is $P^2V = C$ (constant).From the ideal gas equation,we have $PV = nRT$,which implies $P = \frac{nRT}{V}$.Substituting $

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$\sqrt{3} T_0$

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(D) Given the law for the gas is $P^2V = C$ (constant).From the ideal gas equation,we have $PV = nRT$,which implies $P = \frac{nRT}{V}$.Substituting $P$ into the given law: $(\frac{nRT}{V})^2 V = C$.This simplifies to $\frac{n^2 R^2 T^2}{V^2} \cdot V = C$,or $\frac{T^2}{V} = 	ext{constant}$.Therefore,$T^2 \propto V$.For the initial state,we have $T_0^2 \propto V_0$.For the final state,let the temperature be $T_f$ and volume be $3V_0$,so $T_f^2 \propto 3V_0$.Dividing the two relations: $\frac{T_f^2}{T_0^2} = \frac{3V_0}{V_0} = 3$.Taking the square root on both sides,we get $T_f = \sqrt{3} T_0$.

$A$ gas is found to obey the law $P^2V =$ constant. The initial temperature and volume are $T_0$ and $V_0$. If the gas expands to a volume $3V_0$,its final temperature becomes

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