Rankers Physics

Uncategorized: Practice Problem & Solution

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 the universal gas constant, then $C_v$ is equal to: (2013)
$\gamma R$
$\frac{1+\gamma}{1-\gamma}$
$\frac{R}{(\gamma-1)}$
$\frac{(\gamma-1)}{R}$

Solution Explained:

To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:

From Mayer's relation, $C_p - C_v = R$. Given $\gamma = \frac{C_p}{C_v} \implies C_p = \gamma C_v$. Substituting, $\gamma C_v - C_v = R \implies C_v(\gamma - 1) = R \implies C_v = \frac{R}{\gamma - 1}$.

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