Question 306: Thermodynamics & KTG

Question

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 :

Accepted Answer

For monoatomic gas $\gamma_1=\frac{5}{3}$
For diatomic gas at low temperatures
$$
\begin{aligned}
& \gamma_2=\frac{7}{5} \\
& 	herefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}
\end{aligned}
$$

Answer

$$\frac{35}{27}$$

Answer

$$\frac{25}{21}$$

Answer

$$\frac{21}{25}$$

Answer

$$\frac{27}{35}$$

Thermodynamics & KTG

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 :

Choose the correct answer: