JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 7)
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 :
$$\frac{35}{27}$$
$$\frac{25}{21}$$
$$\frac{21}{25}$$
$$\frac{27}{35}$$
Explanation
For monoatomic gas $\gamma_1=\frac{5}{3}$
For diatomic gas at low temperatures
$$ \begin{aligned} & \gamma_2=\frac{7}{5} \\\\ & \therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21} \end{aligned} $$
For diatomic gas at low temperatures
$$ \begin{aligned} & \gamma_2=\frac{7}{5} \\\\ & \therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21} \end{aligned} $$
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