Question 845: Thermodynamics & KTG

Question

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 $$\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}$$ then $$\mathrm{C}_{\mathrm{v}}=$$

Accepted Answer

$$\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$$
Dividing both the sides by $$\mathrm{C}_{\mathrm{v}}$$,
$$\begin{array}{lll}
	herefore & \gamma-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}} & \ldots .\left(\because \frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gammaight) \
	herefore & \mathrm{C}_{\mathrm{v}}=\frac{\mathrm{R}}{\gamma-1}
\end{array}$$

Answer

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

Answer

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

Answer

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

Answer

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

Thermodynamics & KTG

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{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}} then \mathrm{C}_{\mathrm{v}}=

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