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An ideal gas has molar heat capacity $C_V$ at constant volume. The gas undergoes a process where in the temperature changes as $T=T_0\left(1+\alpha V^2\right)$, where, $T$ and $V$ are temperature and volume respectively, $T_0$ and $\alpha$ are positive constants. The molar heat capacity $C$ of the gas is given as $C=C_V+R f(V)$, where, $f(V)$ is a function of volume. The expression for $f(V)$ is
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Verified Answer
The correct answer is:
$\frac{1+\alpha V^2}{2 \alpha V^2}$
Given, $T=T_0\left(1+\alpha V^2\right)$
$$
\begin{aligned}
& \Rightarrow \quad \frac{d T}{d V}=\mathrm{T}_0 \cdot 2 \alpha V \\
& \Rightarrow \quad d V=\frac{d \mathrm{~T}}{\mathrm{~T}_0 \cdot 2 \alpha V}
\end{aligned}
$$
From first law of thermodynamics,
$$
\begin{array}{cc}
& d Q=d U+d W \\
\Rightarrow & n C \cdot d T=n C_V \cdot d T+p d V \\
\Rightarrow & n C \cdot d T+n C_V \cdot d T+p\left(\frac{d T}{T_0 \cdot 2 \alpha V}\right) \\
\Rightarrow & C=C_V+\left(\frac{p}{n}\right)\left(\frac{1}{T_0 2 \alpha V}\right)
\end{array}
$$
Now as gas is ideal,
or
$$
\begin{aligned}
& p V=n R T \\
& \frac{p}{n}=\frac{R T}{V}=\frac{R\left[T_0\left(1+\alpha V^2\right)\right]}{V}
\end{aligned}
$$
substituting in Eq. (ii), we get,
$$
C=C_V+R \cdot\left(\frac{1+\alpha V^2}{\partial \alpha V^2}\right)
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{d T}{d V}=\mathrm{T}_0 \cdot 2 \alpha V \\
& \Rightarrow \quad d V=\frac{d \mathrm{~T}}{\mathrm{~T}_0 \cdot 2 \alpha V}
\end{aligned}
$$
From first law of thermodynamics,
$$
\begin{array}{cc}
& d Q=d U+d W \\
\Rightarrow & n C \cdot d T=n C_V \cdot d T+p d V \\
\Rightarrow & n C \cdot d T+n C_V \cdot d T+p\left(\frac{d T}{T_0 \cdot 2 \alpha V}\right) \\
\Rightarrow & C=C_V+\left(\frac{p}{n}\right)\left(\frac{1}{T_0 2 \alpha V}\right)
\end{array}
$$
Now as gas is ideal,
or
$$
\begin{aligned}
& p V=n R T \\
& \frac{p}{n}=\frac{R T}{V}=\frac{R\left[T_0\left(1+\alpha V^2\right)\right]}{V}
\end{aligned}
$$
substituting in Eq. (ii), we get,
$$
C=C_V+R \cdot\left(\frac{1+\alpha V^2}{\partial \alpha V^2}\right)
$$
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