Question 933: Thermodynamics & KTG

Question

The molar specific heats of an ideal gas at constant pressure and volume are denoted by '$$\mathrm{C}_{\mathrm{p}}$$' and '$$C_v$$' respectively. If $$\gamma=\frac{C_p}{C_v}$$ and '$$R$$' is universal gas constant, then $$C_v$$ is equal to

Accepted Answer

$$\begin{aligned} & \gamma=\frac{C_p}{C_v}, 	ext { Also } C_p-C_v=R \ & C_p=\gamma C_v \ 	herefore & \gamma C_v-C_v=R \ 	herefore & C_v(\gamma-1)=R \ 	herefore & C_v=\frac{R}{\gamma-1}\end{aligned}$$

Answer

$$\frac{\mathrm{R}}{\gamma-1}$$

Answer

$$\gamma \mathrm{R}$$

Answer

$$\frac{1+\gamma}{1-\gamma}$$

Answer

$$\frac{\gamma-1}{\mathrm{R}}$$

Thermodynamics & KTG

The molar specific heats of an ideal gas at constant pressure and volume are denoted by '$\mathrm{C}_{\mathrm{p}}' and 'C_v' respectively. If \gamma=\frac{C_p}{C_v} and 'R' is universal gas constant, then C_v$ is equal to

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