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

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Heat & Thermodynamics

Let

γ_{1}

be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and

γ_{2}

be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio,

\frac{γ _{1}}{γ _{2}}

(A) $\frac{27}{35}$
(B) $\frac{35}{27}$
(C) $\frac{25}{21}$ ✓ Correct
(D) $\frac{21}{25}$

Answer: (C)

Solution

For a monoatomic gas the degrees of freedom are

f = 3

, giving

γ_{1} = 1 + \frac{2}{f} = \frac{5}{3}

. Treating the diatomic molecule as a rigid rotator gives

f = 5

, three translational and two rotational, so

γ_{2} = \frac{7}{5}

. The required ratio is

\frac{γ _{1}}{γ _{2}} = \frac{5/3}{7/5} = \frac{25}{21}

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option

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