`C_p`and `C_v` are specific heats at constant pressure and constant volume respectively. It is observed that `C_p - C_v = a` for hydrogen gas
`C_p = C_V = b` for nitrogen gas
The correct relation between a and b is:

Options

A. `a = 28 b`
B. `a = 1/ 14 b`
C. `a = b`
D. `a = 14 b`

Explanation

To solve the problem, we need to find the relationship between \( a \) and \( b \) given the specific heats \( C_p \) and \( C_v \) for hydrogen and nitrogen gases.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- \( C_p \): Specific heat at constant pressure.
- \( C_v \): Specific heat at constant volume.
- The relationship between \( C_p \) and \( C_v \) for any ideal gas is given by:
\[ C_p - C_v = R \]
where \( R \) is the universal gas constant.

2. **Given Information**:
- For hydrogen gas:
\[ C_p - C_v = a \]
- For nitrogen gas:
\[ C_p = C_v = b \]

3. **Expressing \( a \) for Hydrogen**:
- From the relationship \( C_p - C_v = R \), we can express \( a \) for hydrogen:
\[ a = R \]
- The molar mass of hydrogen (\( H_2 \)) is 2 g/mol.

4. **Expressing \( b \) for Nitrogen**:
- For nitrogen gas, since \( C_p = C_v = b \), we can also use the same relationship:
\[ C_p - C_v = R \implies b - b = 0 \]
- However, we need to find \( b \) in terms of \( R \). The molar mass of nitrogen (\( N_2 \)) is 28 g/mol.

5. **Relating \( a \) and \( b \)**:
- For hydrogen, we can express \( a \) in terms of molar mass:
\[ a = \frac{R}{M_H} = \frac{R}{2} \]
- For nitrogen, we express \( b \):
\[ b = \frac{R}{M_N} = \frac{R}{28} \]

6. **Finding the Ratio \( \frac{a}{b} \)**:
- Now, we can find the ratio \( \frac{a}{b} \):
\[ \frac{a}{b} = \frac{\frac{R}{2}}{\frac{R}{28}} = \frac{28}{2} = 14 \]
- Thus, we can express \( a \) in terms of \( b \):
\[ a = 14b \]

### Conclusion:
The correct relation between \( a \) and \( b \) is:
\[ a = 14b \]