Verified Solution • 2025–26 Session moderate

QJEEMP_2025_11: JEE Main 2025 Physics - Class 11 JEEMP_2025_11 Solution: Consider two different diatomic gases. For the first gas, the…

Consider two different diatomic gases. For the first gas, the molecules are rigid and the ratio of specific heats is given by γ1\gamma_1. For the second gas, the molecules possess additional vibrational modes and the ratio of specific heats is given by γ2\gamma_2.

If CpC_p and CvC_v are the specific heats at constant pressure and constant volume respectively, which of the following options correctly describes the relationship between γ1\gamma_1 and γ2\gamma_2?

Multiple Choice Options

  • A γ2<γ1\gamma_2 < \gamma_1
  • B γ2>γ1\gamma_2 > \gamma_1
  • C γ2=γ1\gamma_2 = \gamma_1
  • D 2γ2=γ12\gamma_2 = \gamma_1

Step-by-Step Verified Solution

Solution & Explanation

1. The Formula for γ\gamma

The ratio of specific heats γ\gamma depends on the degrees of freedom (ff) of the gas molecule:

γ=CpCv=1+2f\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}

2. Analyzing Degrees of Freedom (ff)

  1. For Rigid Diatomic Molecules (γ1\gamma_1): The molecules have 3 translational and 2 rotational degrees of freedom.
  2. f1=5  ⟹  γ1=1+25=75=1.4f_1 = 5 \implies \gamma_1 = 1 + \frac{2}{5} = \frac{7}{5} = 1.4
  3. For Diatomic Molecules with Vibrational Modes (γ2\gamma_2): In addition to the 5 translational and rotational modes, vibrational modes add 2 more degrees of freedom.
  4. f2=7  ⟹  γ2=1+27=97≈1.28f_2 = 7 \implies \gamma_2 = 1 + \frac{2}{7} = \frac{9}{7} \approx 1.28

3. Comparison

Since the denominator (ff) is larger for the second gas, the fraction 2f\frac{2}{f} becomes smaller, making the overall value of γ\gamma smaller.

1.28 &lt; 1.4 \implies \gamma_2 &lt; \gamma_1

Correct Option: (A)

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