If `C_(p) and C_(v)` denote the specific heats (per unit mass of an ideal gas of molecular weight `M`), then
where `R` is the molar gas constant.

Options

A. `C_(p)-C_(v)=R`
B. `C_(p)-C_(v)=R//M`
C. `C_(p)-Cv=MR`
D. `C_(p)-C_(v)=R//M^(2)`

Answer

To solve the problem, we need to establish the relationship between the specific heats at constant pressure (\(C_p\)) and constant volume (\(C_v\)) for an ideal gas and the molar gas constant (\(R\)).

### Step-by-Step Solution:

1. **Understand the Definitions**:
- \(C_p\) is the specific heat at constant pressure.
- \(C_v\) is the specific heat at constant volume.
- \(R\) is the molar gas constant.

2. **Use the Known Relation**:
- For an ideal gas, there is a well-known relationship:
\[ C_p - C_v = R \]
This equation relates the difference between the specific heats at constant pressure and volume to the molar gas constant.

3. **Convert to Specific Heats per Unit Mass**:
- The problem states that \(C_p\) and \(C_v\) are given as specific heats per unit mass of an ideal gas with molecular weight \(M\).
- To convert from molar specific heats to specific heats per unit mass, we divide by the molecular weight \(M\):
\[ c_p = \frac{C_p}{M} \quad \text{and} \quad c_v = \frac{C_v}{M} \]

4. **Substituting into the Relation**:
- We substitute \(C_p\) and \(C_v\) in the known relation:
\[ \frac{C_p}{M} - \frac{C_v}{M} = R \]

5. **Simplifying the Equation**:
- Factor out \(1/M\):
\[ \frac{C_p - C_v}{M} = R \]

6. **Rearranging the Equation**:
- Multiply both sides by \(M\):
\[ C_p - C_v = R \cdot M \]

7. **Final Result**:
- Therefore, the relationship between the specific heats per unit mass and the molar gas constant is:
\[ c_p - c_v = \frac{R}{M} \]

### Final Answer:
\[ c_p - c_v = \frac{R}{M} \]