A gas follows VT^2 = text{constant}. The coef | vedclass

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(B) Given the relation for the gas: $VT^2 = C$, where $C$ is a constant.Differentiating both sides with respect to $T$:$\frac{d}{dT}(VT^2) = \frac{d}{

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$-\frac{2}{T}$

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(B) Given the relation for the gas: $VT^2 = C$, where $C$ is a constant.Differentiating both sides with respect to $T$:$\frac{d}{dT}(VT^2) = \frac{d}{dT}(C)$$V(2T) + T^2 \left(\frac{dV}{dT}\right) = 0$Rearranging the terms to find $\frac{dV}{dT}$:$T^2 \left(\frac{dV}{dT}\right) = -2VT$$\frac{dV}{dT} = -\frac{2VT}{T^2} = -\frac{2V}{T}$The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \left(\frac{dV}{dT}\right)$.Substituting the value of $\frac{dV}{dT}$:$\gamma = \frac{1}{V} \left(-\frac{2V}{T}\right)$$\gamma = -\frac{2}{T}$

$A$ gas follows $VT^2 = \text{constant}$. The coefficient of volume expansion of the gas is

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