If for hydrogen `C_(P) - C_(V) = m` and for nitrogen `C_(P) - C_(V) = n`, where `C_(P)` and `C_(V)` refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between `m` and `n` is (molecular weight of hydrogen = 2 and molecular weight or nitrogen = 14)

Options

A. `n = 14 m`
B. `n = 7m`
C. `m = 7n`
D. `m = 14 n`

Explanation

To solve the problem, we need to understand the relationship between the specific heats at constant pressure \( C_P \) and constant volume \( C_V \) for gases, and how they relate to the molecular weights of the gases involved.

### Step-by-Step Solution:

1. **Understanding the Relation Between \( C_P \) and \( C_V \)**:
The difference between the specific heats at constant pressure and constant volume is given by the equation:
\[ C_P - C_V = R \]
where \( R \) is the specific gas constant.

2. **Specific Heats Per Unit Mass**:
For specific heats per unit mass, we can express the specific heats as:
\[ C_P = \frac{R}{M} + C_V \]
where \( M \) is the molecular weight of the gas.

3. **Applying the Relation to Hydrogen**:
For hydrogen (molecular weight \( M_H = 2 \)):
\[ C_{P,H} - C_{V,H} = R_H \]
This can be expressed in terms of \( m \):
\[ m = C_{P,H} - C_{V,H} = \frac{R}{M_H} = \frac{R}{2} \]

4. **Applying the Relation to Nitrogen**:
For nitrogen (molecular weight \( M_N = 14 \)):
\[ C_{P,N} - C_{V,N} = R_N \]
This can be expressed in terms of \( n \):
\[ n = C_{P,N} - C_{V,N} = \frac{R}{M_N} = \frac{R}{14} \]

5. **Finding the Relation Between \( m \) and \( n \)**:
Now, we can find the ratio \( \frac{m}{n} \):
\[ \frac{m}{n} = \frac{\frac{R}{2}}{\frac{R}{14}} = \frac{14}{2} = 7 \]

6. **Conclusion**:
Therefore, the relationship between \( m \) and \( n \) is:
\[ m = 7n \]

### Final Answer:
The relation between \( m \) and \( n \) is \( m = 7n \).

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