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: What we need.
For an ideal gas we have two molar heat capacities, $C_p$ (at constant pressure) and $C_v$ (at constant volume). We are told $\gamma = \frac{C_p}{C_v}$ and we must write $C_v$ using $R$ and $\gamma$.

Step 2: The two facts we lean on.
Mayer's relation tells us $C_p - C_v = R$. The given ratio tells us $C_p = \gamma C_v$.

Step 3: Swap one into the other.
Put $C_p = \gamma C_v$ into Mayer's relation: $\gamma C_v - C_v = R$.

Step 4: Pull out the common factor.
Both terms on the left have $C_v$, so $C_v(\gamma - 1) = R$.

Step 5: Solve for $C_v$.
Divide both sides by $(\gamma - 1)$ to get $C_v = \dfrac{R}{\gamma - 1}$.

Step 6: Quick sanity check.
For a monatomic gas $\gamma = \frac{5}{3}$, so $C_v = \frac{R}{2/3} = \frac{3R}{2}$, which is the known value. Good, so the answer is option (A).
\[ \boxed{C_v = \frac{R}{\gamma - 1}} \]