According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is:-

Solution :

According to law of equipartition of energy, each degree of freedom contributes :

$$\frac{1}{2}R$$

to molar specific heat at constant volume.

For a diatomic gas :

• Translational degrees of freedom = 3

Contribution :

$$\frac{3}{2}R$$

• Rotational degrees of freedom = 2

Contribution :

$$R$$

• One vibrational mode contributes two degrees of freedom (kinetic + potential)

Contribution :

$$R$$

Therefore,

Total molar specific heat at constant volume :

$$C_V = \frac{3}{2}R + R + R$$

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

Final Answer :

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