Question

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 universal gas constant, then $C_v$ is equal to

• A. $\frac{R}{\gamma - 1}$
• B. $\gamma R$
• C. $\frac{1 + \gamma}{1 - \gamma}$
• D. $(\gamma - 1)R$

Hint:

This is a fundamental thermodynamic identity. $C_v$ always relates to the gas constant and the degree of freedom, while $\gamma$ depends on the atomicity of the gas.

Answer 1

The Correct Option is A

Step 1: Bring in Mayer's relation.
For an ideal gas, $C_p - C_v = R$. This connects the two heat capacities to the gas constant.

Step 2: Use the ratio.
Since $\gamma = \tfrac{C_p}{C_v}$, we have $C_p = \gamma C_v$.

Step 3: Substitute.
Put this into Mayer's relation: $\gamma C_v - C_v = R$, so $C_v(\gamma - 1) = R$.

Step 4: Solve for $C_v$.
\[ \boxed{C_v = \frac{R}{\gamma - 1}} \]