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One mole of a monoatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points $\left(V_{0}, T_{0}\right)$ and $\left(2 V_{0}, 3 T_{0}\right)$ in a ${V} \cdot T$ diagram. What is the value of the heat capacity of the gas at the point $\left(V_{0}, T_{0}\right) ?$
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Verified Answer
The correct answer is:
$2 R$
Heat capacity of an ideal gas in a thermodynamic process,
$\because$ Ideal gas is monoatomic,
$$
\begin{aligned}
C_{V} &=\frac{f R}{2}=\frac{3 R}{2} \\
C_{\text {process }} &=C_{V}+\frac{p}{n} \cdot \frac{d V}{d T} \\
&=\frac{3}{2} R+\frac{p}{n} \cdot \frac{V_{0}}{2 T} \\
\therefore C_{\text {process }}=\frac{3}{2} R+\frac{n R T}{n 2 T} &[\because p V=n R T] \\
=& \frac{3}{2} R+\frac{R}{2}=2 R
\end{aligned}
$$
$\because$ Ideal gas is monoatomic,
$$
\begin{aligned}
C_{V} &=\frac{f R}{2}=\frac{3 R}{2} \\
C_{\text {process }} &=C_{V}+\frac{p}{n} \cdot \frac{d V}{d T} \\
&=\frac{3}{2} R+\frac{p}{n} \cdot \frac{V_{0}}{2 T} \\
\therefore C_{\text {process }}=\frac{3}{2} R+\frac{n R T}{n 2 T} &[\because p V=n R T] \\
=& \frac{3}{2} R+\frac{R}{2}=2 R
\end{aligned}
$$
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