A gas obeys the relation P^2V = text{constant | vedclass

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(A) Given the relation for the gas is $P^2V = 	ext{constant}$.Using the ideal gas law, $PV = nRT$, we can express pressure as $P = \frac{nRT}{V}$.Sub

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$\sqrt{2} T_0$

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(A) Given the relation for the gas is $P^2V = 	ext{constant}$.Using the ideal gas law, $PV = nRT$, we can express pressure as $P = \frac{nRT}{V}$.Substituting this into the given relation: $\left(\frac{nRT}{V}ight)^2 V = 	ext{constant}$.This simplifies to $\frac{n^2R^2T^2}{V^2} \cdot V = 	ext{constant}$, which implies $\frac{T^2}{V} = 	ext{constant}$.For two states, we have $\frac{T_1^2}{V_1} = \frac{T_2^2}{V_2}$.Given $T_1 = T_0$, $V_1 = V_0$, and $V_2 = 2V_0$, we substitute these values:$\frac{T_0^2}{V_0} = \frac{T_2^2}{2V_0}$.Canceling $V_0$ from both sides, we get $T_2^2 = 2T_0^2$.Taking the square root, the final temperature is $T_2 = \sqrt{2} T_0$.

$A$ gas obeys the relation $P^2V = \text{constant}$. The initial temperature and volume are $T_0$ and $V_0$. If the gas expands to a volume $2V_0$, what will be its final temperature?

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