An ideal gas is expanding such that PT^2 = te | vedclass

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(C) For an ideal gas, the equation of state is $PV = \mu RT$, which implies $P = \frac{\mu RT}{V}$.Given the process equation $PT^2 = k$ (where $k$ is

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$3$

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(C) For an ideal gas, the equation of state is $PV = \mu RT$, which implies $P = \frac{\mu RT}{V}$.Given the process equation $PT^2 = k$ (where $k$ is a constant).Substituting $P$ in the process equation: $\left(\frac{\mu RT}{V}\right)T^2 = k$.This simplifies to $\mu RT^3 = kV$.Differentiating both sides with respect to $T$: $\mu R(3T^2 dT) = k dV$.Dividing the differential equation by the original equation $\mu RT^3 = kV$:$\frac{3\mu RT^2 dT}{\mu RT^3} = \frac{kdV}{kV}$.This yields $\frac{3dT}{T} = \frac{dV}{V}$.The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.From the derived relation, $\frac{1}{V} \frac{dV}{dT} = \frac{3}{T}$.Therefore, $\gamma = 3/T$.

An ideal gas is expanding such that $PT^2 = \text{constant}$. The coefficient of volume expansion of the gas is: (in $/T$)

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