The specific heats, `C_(P)` and `C_(V)` of a gas of diatomic molecules, A are given (in units of `Jmol^(-1) K^(-1)`) by 29 and 22 , respectively. Another gas of diatomic molecules , B has the corresponding values 30 and 21. if they are treated as ideal gases, then :

Options

A. A is rigid but B has a vibrational mode.
B. A has vibrational mode but B has none.
C. A has one vibrational mode and B has two.
D. Both A and B have a vibrational mode each.

Answer

To solve the problem, we need to analyze the specific heats of the two diatomic gases, A and B, and determine their degrees of freedom, particularly focusing on the vibrational degrees of freedom.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- For gas A:
- \( C_P = 29 \, J \, mol^{-1} \, K^{-1} \)
- \( C_V = 22 \, J \, mol^{-1} \, K^{-1} \)
- For gas B:
- \( C_P = 30 \, J \, mol^{-1} \, K^{-1} \)
- \( C_V = 21 \, J \, mol^{-1} \, K^{-1} \)

2. **Use Mayer's Relation**:
- Mayer's relation states that:
\[ C_P - C_V = R \]
- For gas A:
\[ R_A = C_P - C_V = 29 - 22 = 7 \, J \, mol^{-1} \, K^{-1} \]
- For gas B:
\[ R_B = C_P - C_V = 30 - 21 = 9 \, J \, mol^{-1} \, K^{-1} \]

3. **Relate \( C_V \) to Degrees of Freedom**:
- The relation between \( C_V \) and the degrees of freedom \( f \) is given by:
\[ C_V = \frac{f}{2} R \]
- For gas A:
\[ 22 = \frac{f_A}{2} \cdot 7 \implies f_A = \frac{22 \cdot 2}{7} = \frac{44}{7} \approx 6.28 \implies f_A \approx 6 \]
- For gas B:
\[ 21 = \frac{f_B}{2} \cdot 9 \implies f_B = \frac{21 \cdot 2}{9} = \frac{42}{9} \approx 4.67 \implies f_B \approx 5 \]

4. **Determine the Degrees of Freedom**:
- For diatomic gases, the degrees of freedom can be broken down as follows:
- Translational: 3
- Rotational: 2
- Vibrational: \( f - 5 \) (if \( f > 5 \))
- For gas A:
- Total degrees of freedom \( f_A = 6 \)
- Therefore, vibrational degrees of freedom = \( 6 - 5 = 1 \)
- For gas B:
- Total degrees of freedom \( f_B = 5 \)
- Therefore, vibrational degrees of freedom = \( 5 - 5 = 0 \)

5. **Conclusion**:
- Gas A has 1 vibrational degree of freedom.
- Gas B has 0 vibrational degrees of freedom.
- Thus, the correct option is that gas A has one vibrational mode and gas B has none.