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If $\gamma$ is the ratio of specific heats and $R$ is the universal gas constant, then the molar specific heat at constant volume
$\mathrm{C}_{\mathrm{V}}$ is given by
Options:
$\mathrm{C}_{\mathrm{V}}$ is given by
Solution:
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Verified Answer
The correct answer is:
$\frac{\mathrm{R}}{\gamma-1}$
From the Mayer's formula
$$
C_{p}-C_{V}=R
$$
and
$$
\begin{gathered}
\gamma=\frac{C_{p}}{C_{V}} \\
C_{V}=C_{p}
\end{gathered}
$$
$$
\Rightarrow \quad \gamma \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{p}}
$$
Substituting Eq. (ii) in Eq. (i), we get
$$
\begin{aligned}
\gamma C_{V}-C_{V} &=R \\
C_{V}(\gamma-1) &=R \\
C_{V} &=\frac{R}{\gamma-1}
\end{aligned}
$$
$$
C_{p}-C_{V}=R
$$
and
$$
\begin{gathered}
\gamma=\frac{C_{p}}{C_{V}} \\
C_{V}=C_{p}
\end{gathered}
$$
$$
\Rightarrow \quad \gamma \mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{p}}
$$
Substituting Eq. (ii) in Eq. (i), we get
$$
\begin{aligned}
\gamma C_{V}-C_{V} &=R \\
C_{V}(\gamma-1) &=R \\
C_{V} &=\frac{R}{\gamma-1}
\end{aligned}
$$
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