Thermodynamics Question 2

Question 2 - 24 January - Shift 2

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

(1) $\frac{27}{35}$

(2) $\frac{35}{27}$

(3) $\frac{25}{21}$

(4) $\frac{21}{25}$

Show Answer

Answer: (3)

Solution:

Formula: Heat Capacity Ratio

For monoatomic gas $\gamma_1=\frac{5}{3}$

For diatomic gas at low temperatures

$\gamma_2=\frac{7}{5}$

$\therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}$