If `C_(p) and C_(v)` denoted the specific heats of unit mass of nitrogen at constant pressure and volume respectively, then

Options

A. `C_p - C_V = R//28`
B. `C_p - C_v = = R//14`
C. `C_p - C_v = R`
D. `C_p -C_v = 28R`

Explanation

To solve the problem regarding the specific heats of nitrogen at constant pressure and volume, we follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between the specific heats at constant pressure (\(C_p\)) and at constant volume (\(C_v\)) for an ideal gas is given by the equation:
\[ C_p - C_v = R \]
where \(R\) is the universal gas constant.

2. **Converting Units**:
The problem states that we need to express the specific heats in terms of Joules per kilogram Kelvin (J/kg·K). The gas constant \(R\) is typically given in Joules per mole Kelvin (J/mol·K).

3. **Using Molar Mass**:
To convert \(R\) from J/mol·K to J/kg·K, we need to divide \(R\) by the molar mass of nitrogen. The molar mass of nitrogen (\(N_2\)) is approximately 28 g/mol, which is equivalent to 0.028 kg/mol.

4. **Setting Up the Equation**:
By substituting the molar mass into the equation, we have:
\[ C_p - C_v = \frac{R}{M} \]
where \(M\) is the molar mass of nitrogen in kg. Thus, we can rewrite the equation as:
\[ C_p - C_v = \frac{R}{0.028} \]

5. **Final Expression**:
Therefore, the final expression relating \(C_p\) and \(C_v\) for nitrogen in J/kg·K is:
\[ C_p - C_v = \frac{R}{28} \]

### Conclusion:
The relationship between the specific heats of nitrogen at constant pressure and volume is:
\[ C_p - C_v = \frac{R}{28} \]