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

Options

A. `(3)/(2)R`
B. `(7)/(2)R`
C. `(9)/(2)R`
D. `(5)/(2)R`

Answer

To solve the problem, we need to determine the molar specific heat at constant volume (\(C_v\)) for a diatomic gas that has one additional vibrational mode, using the law of equipartition of energy.

### Step-by-step Solution:

1. **Understanding Degrees of Freedom**:
- A diatomic gas has translational and rotational degrees of freedom.
- For a diatomic molecule, the degrees of freedom are:
- Translational: 3 (movement in x, y, z directions)
- Rotational: 2 (rotation about two axes, as rotation about the axis of the bond does not contribute significantly)
- Therefore, the total degrees of freedom for a diatomic gas without vibrational modes is:
\[ F_{\text{diatomic}} = 3 \text{ (translational)} + 2 \text{ (rotational)} = 5 \]

2. **Adding Vibrational Degrees of Freedom**:
- When we include one additional vibrational mode, we add 2 degrees of freedom (one for kinetic and one for potential energy associated with the vibrational motion).
- Thus, the total degrees of freedom becomes:
\[ F_{\text{total}} = 5 + 2 = 7 \]

3. **Applying the Equipartition Theorem**:
- According to the law of equipartition of energy, each degree of freedom contributes \(\frac{1}{2}R\) to the molar specific heat at constant volume (\(C_v\)).
- Therefore, we can calculate \(C_v\) as follows:
\[ C_v = \frac{F_{\text{total}}}{2} R = \frac{7}{2} R \]

4. **Final Result**:
- The molar specific heat of the diatomic gas at constant volume with one additional vibrational mode is:
\[ C_v = \frac{7}{2} R \]

### Conclusion:
Thus, the answer to the question is that the molar specific heat of a diatomic gas at constant volume, where the molecule has one additional vibrational mode, is \(\frac{7}{2} R\).