Q167 / 180 · KCET 2008 · Physics
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
- A$\gamma \mathrm{R}$
- B$\frac{(\gamma-1) R}{\gamma}$
- Correct answerC$\frac{\mathrm{R}}{\gamma-1}$
- D$\frac{\gamma \mathrm{R}}{\gamma-1}$
Solution
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
$$
\begin{aligned}
\gamma C_{V}-C_{V} &=R \\
C_{V}(\gamma-1) &=R \\
C_{V} &=\frac{R}{\gamma-1}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.