Let $\gamma_1$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant…

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 $\frac{\gamma_2}{\gamma_1}$ is
  1. $\frac{37}{21}$
  2. $\frac{27}{35}$
  3. $\frac{21}{25}$
  4. $\frac{35}{27}$

Solution

For monoatomic gas, $\gamma_1=\frac{5}{3}$ For rigid diatomic gas, $\begin{aligned} \gamma_2 & =\frac{7}{5} \\ \therefore \quad \frac{\gamma_2}{\gamma_1} & =\frac{7}{5} \times \frac{3}{5}=\frac{21}{25} \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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