JEE Physics 2023 Question with Solution

Given: A diatomic gas at constant volume has one additional vibrational mode.

Find: The molar specific heat $C_v$ according to the law of equipartition of energy.

Diatomic gas molecules possess two rotational and three translational degrees of freedom. The molecule adds two more degrees of freedom corresponding to one vibrational mode.

Thus, the total degrees of freedom are:

$$f = 3\,(\text{translational}) + 2\,(\text{rotational}) + 2\,(\text{vibrational}) = 7$$

Applying the formula for molar specific heat at constant volume:

$$C_v = \frac{f}{2}R = \frac{7}{2}R$$

Therefore, the molar specific heat is $\frac{7}{2}R$. The solution states the correct option is B, but this conflicts with the listed options; the value $\frac{7}{2}R$ matches option D.

Given: A diatomic molecule with translational, rotational, and one vibrational mode.

Find: The value of $C_v$.

For a diatomic gas:

• Translational degrees of freedom = $3$
• Rotational degrees of freedom = $2$
• One vibrational mode contributes $2$ degrees of freedom in equipartition

So,

$$f = 3 + 2 + 2 = 7$$

Now use

$$C_v = \frac{f}{2}R$$

Hence,

$$C_v = \frac{7}{2}R$$

So the defensible correct option from the given list is D.