1. Use equipartition theorem for ideal gases

For an ideal gas, $C_V = \frac{f}{2}R$ where $f$ is the number of active degrees of freedom per molecule.

2. Gas A: rigid diatomic molecule

A rigid diatomic molecule has:

• $3$ translational degrees of freedom
• $2$ rotational degrees of freedom
• no vibrational contribution (since it is rigid)

So, $f_A = 3+2 = 5$ Hence, $C_V^A = \frac{5}{2}R$

3. Gas B: diatomic molecule with one additional vibrational mode

For a diatomic molecule with one vibrational mode active:

• $3$ translational degrees of freedom
• $2$ rotational degrees of freedom
• $1$ vibrational mode contributes $2$ degrees of freedom

So, $f_B = 3+2+2 = 7$ Hence, $C_V^B = \frac{7}{2}R$

4. Take the ratio

$C_V^A : C_V^B = \frac{5}{2}R : \frac{7}{2}R = 5:7$

5. About the given masses

The masses $m$ and $\frac{m}{4}$ do not affect the molar specific heat of ideal gases. Specific heat depends only on the active degrees of freedom.

6. Option check

• A: $7:9$ ✗
• B: $5:7$ ✓
• C: $3:5$ ✗
• D: $5:9$ ✗

Therefore, the correct answer is Option B.