Let gamma_1 be the ratio of molar specific he | vedclass

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(C) For a monoatomic gas,the degrees of freedom $f = 3$. The ratio of specific heats is $\gamma_1 = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.F

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$\frac{21}{25}$

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(C) For a monoatomic gas,the degrees of freedom $f = 3$. The ratio of specific heats is $\gamma_1 = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.For a rigid diatomic gas,the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). The ratio of specific heats is $\gamma_2 = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}$.Therefore,the ratio $\frac{\gamma_2}{\gamma_1} = \frac{7/5}{5/3} = \frac{7}{5} 	imes \frac{3}{5} = \frac{21}{25}$.

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 a diatomic gas. Considering the diatomic gas molecule as a rigid rotator,the ratio $\frac{\gamma_2}{\gamma_1}$ is

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