Let $\gamma_1$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $\gamma_2$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, $\frac{\gamma_1}{\gamma_2}$ is :
Solution
For monoatomic gas $\gamma_1=\frac{5}{3}$<br/><br/>
For diatomic gas at low temperatures<br/><br/>
$$
\begin{aligned}
& \gamma_2=\frac{7}{5} \\\\
& \therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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