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The molar specific heats of an ideal gas at constant pressure and volume are denoted by $C_P$ and $C_V$ respectively. If $\gamma=\frac{C_P}{C_V}$ and $R$ is the universal gas constant, then $C_V$ is equal to
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Verified Answer
The correct answer is:
$\frac{R}{(\gamma-1)}$
Using,
$C_P-C_V=R$
$\Rightarrow \quad C_V\left(\frac{C_P}{C_V}-1\right)=R$
$(\gamma-1)=\frac{R}{C_V}\left(\because \quad \frac{C_P}{C_V}=\gamma\right)$
or $\quad C_V=\frac{R}{(\gamma-1)}$
$C_P-C_V=R$
$\Rightarrow \quad C_V\left(\frac{C_P}{C_V}-1\right)=R$
$(\gamma-1)=\frac{R}{C_V}\left(\because \quad \frac{C_P}{C_V}=\gamma\right)$
or $\quad C_V=\frac{R}{(\gamma-1)}$
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