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