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: Understanding the Question:
We need to derive the expression for molar heat capacity at constant volume ($C_v$) in terms of the gas constant ($R$) and the adiabatic index ($\gamma$).

Step 2: Key Formula or Approach:
1. Mayer's Relation: $C_p - C_v = R$.
2. Definition of $\gamma$: $\gamma = \frac{C_p}{C_v} \implies C_p = \gamma C_v$.

Step 3: Detailed Explanation:
Substitute $C_p = \gamma C_v$ into Mayer's Relation:
$(\gamma C_v) - C_v = R$
Factor out $C_v$:
$C_v(\gamma - 1) = R$
Solve for $C_v$:
$C_v = \frac{R}{\gamma - 1}$.

Step 4: Final Answer:
The value is $\frac{R}{\gamma - 1}$, which corresponds to option (A).