An ideal gas has molar heat capacity C_V at c | vedclass

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(B) Given the process equation: $T = T_0(1 + \alpha V^2)$.Differentiating with respect to $V$: $\frac{dT}{dV} = T_0(2\alpha V) \Rightarrow dV = \frac{

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$\frac{1+\alpha V^2}{2 \alpha V^2}$

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(B) Given the process equation: $T = T_0(1 + \alpha V^2)$.Differentiating with respect to $V$: $\frac{dT}{dV} = T_0(2\alpha V) \Rightarrow dV = \frac{dT}{2\alpha V T_0}$.From the first law of thermodynamics: $dQ = dU + dW$.For $n$ moles: $nC dT = nC_V dT + P dV$.Dividing by $n dT$: $C = C_V + \frac{P}{n} \frac{dV}{dT}$.Substituting $dV/dT = \frac{1}{2\alpha V T_0}$: $C = C_V + \frac{P}{n} \frac{1}{2\alpha V T_0}$.Using the ideal gas law $PV = nRT$, we have $\frac{P}{n} = \frac{RT}{V}$.Substituting $T = T_0(1 + \alpha V^2)$: $\frac{P}{n} = \frac{R T_0(1 + \alpha V^2)}{V}$.Now, substitute this into the expression for $C$: $C = C_V + \left[ \frac{R T_0(1 + \alpha V^2)}{V} \right] \left[ \frac{1}{2\alpha V T_0} \right]$.Simplifying: $C = C_V + R \left( \frac{1 + \alpha V^2}{2\alpha V^2} \right)$.Comparing with $C = C_V + Rf(V)$, we get $f(V) = \frac{1 + \alpha V^2}{2\alpha V^2}$.

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