JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Heat & Thermodynamics

For a diatomic gas, if

γ_{1} = (\frac{Cp}{Cv})

for rigid molecules and

γ_{2} = (\frac{Cp}{Cv})

for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct?
( Cp and Cv are specific heats of the gas at constant pressure and volume)

(A) $γ_{2} > γ_{1}$
(B) $γ_{2} = γ_{1}$
(C) $2 γ_{2} = γ_{1}$
(D) $γ_{2} < γ_{1}$ ✓ Correct

Answer: (D)

Solution

For an ideal gas,

γ = 1 + \frac{2}{f}

, where f is the number of degrees of freedom per molecule.

Rigid diatomic molecules have 3 translational + 2 rotational degrees of freedom, f = 5:

γ_{1} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4

Diatomic molecules with an additional vibrational mode (which contributes 2 more degrees of freedom — kinetic and potential) have f = 7:

γ_{2} = 1 + \frac{2}{7} = \frac{9}{7} \approx 1.29

Since

γ

decreases as f increases,

γ_{2} < γ_{1}

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Heat & Thermodynamics

Solution

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