For a diatomic gas, if \gamma_1=(\fracC pC v) for rigid molecules and \gamma_2=(\fracC pC v) 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)

Heat and Thermodynamics PYQ — JEE Main 2025 · 22 Jan · Shift 2 · Q66 with Solution

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Correct answer: A

Use degrees of freedom to find $C_v$ , $C_p$ , and $\gamma$ .

For an ideal gas with $f$ active degrees of freedom,

C_v = \frac{f}{2}R, \qquad C_p = C_v + R = \frac{f+2}{2}R

So,

\gamma = \frac{C_p}{C_v} = \frac{\frac{f+2}{2}R}{\frac{f}{2}R} = \frac{f+2}{f}

Rigid diatomic molecule

A rigid diatomic molecule has:

$3$ translational degrees of freedom
$2$ rotational degrees of freedom

So total active degrees of freedom:

f_1 = 5

Hence,

\gamma_1 = \frac{f_1+2}{f_1} = \frac{7}{5} = 1.4

Diatomic molecule with vibrational modes active

If vibrational modes are also active, a diatomic molecule has:

$3$ translational
$2$ rotational
$2$ vibrational contributions

Why $2$ for vibration? Because each vibrational mode contributes:

$1$ kinetic
$1$ potential

For a diatomic molecule, there is one vibrational mode, so contribution is $2$ .

Thus,

f_2 = 7

Therefore,

\gamma_2 = \frac{f_2+2}{f_2} = \frac{9}{7} \approx 1.286

Compare $\gamma_1$ and $\gamma_2$

\gamma_1 = \frac{7}{5} = 1.4, \qquad \gamma_2 = \frac{9}{7} \approx 1.286

Thus,

\gamma_2 < \gamma_1

Check options

A: $\gamma_2 < \gamma_1$ ✅
B: $\gamma_2 > \gamma_1$ ❌
C: $\gamma_2 = \gamma_1$ ❌
D: $2\gamma_2 = \gamma_1$ ❌ since $2\cdot \frac{9}{7} = \frac{18}{7} \ne \frac{7}{5}$

Therefore, the correct option is A.

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View written solutionFree

Correct answer: A

Use degrees of freedom to find $C_v$ , $C_p$ , and $\gamma$ .

For an ideal gas with $f$ active degrees of freedom,

C_v = \frac{f}{2}R, \qquad C_p = C_v + R = \frac{f+2}{2}R

So,

\gamma = \frac{C_p}{C_v} = \frac{\frac{f+2}{2}R}{\frac{f}{2}R} = \frac{f+2}{f}

Rigid diatomic molecule

A rigid diatomic molecule has:

$3$ translational degrees of freedom
$2$ rotational degrees of freedom

So total active degrees of freedom:

f_1 = 5

Hence,

\gamma_1 = \frac{f_1+2}{f_1} = \frac{7}{5} = 1.4

Diatomic molecule with vibrational modes active

If vibrational modes are also active, a diatomic molecule has:

$3$ translational
$2$ rotational
$2$ vibrational contributions

Why $2$ for vibration? Because each vibrational mode contributes:

$1$ kinetic
$1$ potential

For a diatomic molecule, there is one vibrational mode, so contribution is $2$ .

Thus,

f_2 = 7

Therefore,

\gamma_2 = \frac{f_2+2}{f_2} = \frac{9}{7} \approx 1.286

Compare $\gamma_1$ and $\gamma_2$

\gamma_1 = \frac{7}{5} = 1.4, \qquad \gamma_2 = \frac{9}{7} \approx 1.286

Thus,

\gamma_2 < \gamma_1

Check options

A: $\gamma_2 < \gamma_1$ ✅
B: $\gamma_2 > \gamma_1$ ❌
C: $\gamma_2 = \gamma_1$ ❌
D: $2\gamma_2 = \gamma_1$ ❌ since $2\cdot \frac{9}{7} = \frac{18}{7} \ne \frac{7}{5}$

Therefore, the correct option is A.

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