1. Use the ideal-gas relation

For any ideal gas, $C_P - C_V = R$ where $R \approx 8.3\ \text{J mol}^{-1}\text{K}^{-1}.$

Check both gases:

• For gas $A$: $29 - 22 = 7$
• For gas $B$: $30 - 21 = 9$

These are close to $R$ (difference due to rounding), so we proceed treating both as ideal gases.

2. Find degrees of freedom from $C_V$

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

For a rigid diatomic molecule (no vibration active):

• 3 translational + 2 rotational degrees of freedom
• so, $f=5$ Therefore, $C_V = \frac{5}{2}R \approx \frac{5}{2}(8.3)=20.75\approx 21$

This matches gas $B$ very well since $C_V=21$. So gas $B$ has no vibrational mode active.

3. Effect of one vibrational mode in a diatomic gas

A vibrational mode contributes 2 degrees of freedom (one kinetic + one potential). Thus for a diatomic gas with one active vibrational mode: $f=5+2=7$ Hence, $C_V = \frac{7}{2}R \approx \frac{7}{2}(8.3)=29.05$

But this is not equal to $22$ for gas $A$.

So we should be careful: in many standard chemistry/thermodynamics conventions for a diatomic gas, the increase in molar heat capacities due to vibrational excitation is:

• translational + rotational only: $C_V = \frac{5}{2}R \approx 20.8$ $C_P = \frac{7}{2}R \approx 29.1$
• if vibration becomes active, heat capacity increases above these values.

Now compare with gas $A$: $C_P = 29, \quad C_V = 22$ These are slightly above the rigid-diatomic values:

• rigid diatomic: $C_V \approx 20.8$, $C_P \approx 29.1$
• given for $A$: $C_V=22$, $C_P=29$

Thus gas $A$ shows a small increase in $C_V$, indicating vibrational contribution has started.

For gas $B$: $C_V=21, \quad C_P=30$ This is essentially the rigid diatomic case within rounding.

Hence,

• $A$ has a vibrational mode active,
• $B$ has none.

4. Check options

• A: $A$ is rigid but $B$ has a vibrational mode
  False.
• B: $A$ has a vibrational mode but $B$ has none
  True.
• C: $A$ has one vibrational mode and $B$ has two
  False.
• D: Both $A$ and $B$ have a vibrational mode each
  False.

5. Final answer

The correct option is $\boxed{\text{B}}$