If $\mathrm{C}_{\mathrm{p}}$ and $\mathrm{C}_{\mathrm{v}}$ are molar specific heats of an ideal gas at constant pressure and volume respectively and ' $\gamma$ ' is $\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}$ then $\mathrm{C}_{\mathrm{p}}=$ ( $\mathrm{R}=$ universal gas constant)

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Aicharya · Accepted Answer

Given, $\gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}$. $\begin{aligned} 	herefore \quad & \frac{C_v}{C_p}=\frac{1}{\gamma} \ & \frac{C_v-C_P}{C_p}=\frac{1-\gamma}{\gamma} \ & \frac{-\left(C_P-C_vight)}{C_P}=\frac{1-\gamma}{\gamma} \ & \frac{-R}{C_p}=\frac{1-\gamma}{\gamma} \ 	herefore \quad & \frac{R}{C_p}=\frac{\gamma-1}{\gamma} \ 	herefore \quad & C_P=\frac{\gamma R}{\gamma-1} \end{aligned}$ $\ldots\left(\because C_p-C_v=Right)$

If $\mathrm{C}_{\mathrm{p}}$ and $\mathrm{C}_{\mathrm{v}}$ are molar specific heats of an ideal gas at…

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