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, (γ_1)/(γ_2) is :

Let γ_1 be the ratio of molar specific heat at constant… — JEE Main 2023 Physics PYQ

JEE Main 2023 · 24 Jan · 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,

\dfrac{\gamma_1}{\gamma_2}

is :

A.
$\dfrac{21}{25}$
B.
$\dfrac{35}{27}$
C.
$\dfrac{27}{35}$
D.
$\dfrac{25}{21}$

Show answer and solution

Correct option: D

Solution

For a monoatomic gas: degrees of freedom = 3, so Cv = (3/2)R, Cp = (5/2)R, giving γ₁ = 5/3.

For a diatomic rigid rotator: degrees of freedom = 5, so Cv = (5/2)R, Cp = (7/2)R, giving γ₂ = 7/5.

Therefore γ₁/γ₂ = (5/3)/(7/5) = (5/3) × (5/7) = 25/21.

[JEE Main 2023 · 24 Jan · Shift 2]

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