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The molar specific heat of an ideal gas at constant pressure and constant volume is $\mathrm{C}_{\mathrm{p}}$ and $\mathrm{C}_{\mathrm{v}}$ respectively. If $\mathrm{R}$ is universal gas constant and $\gamma=\frac{C_p}{C_y}$ then $C_v=$
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Verified Answer
The correct answer is:
$\frac{\mathrm{R}}{\gamma-1}$
$$
C_p-C_v=R
$$
Dividing both the sides by $\mathrm{C}_{\mathrm{v}}$,
$$
\begin{array}{ll}
\therefore & \gamma-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} \\
\therefore & \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}
\end{array}
$$
C_p-C_v=R
$$
Dividing both the sides by $\mathrm{C}_{\mathrm{v}}$,
$$
\begin{array}{ll}
\therefore & \gamma-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} \\
\therefore & \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}
\end{array}
$$
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