For one mole of an ideal gas` (C_(p) and C_(v)` are molar heat capacities at constant presure and constant volume respectively)

Options

A. `C_(p) - C_(v) = R`
B. `C_(p) - C_(v) = R/2`
C. `C_(p) - C_(v) = -2R`
D. `C_(p) - C_(v) = 0`

Answer

To solve the problem regarding the relationship between the molar heat capacities \( C_p \) and \( C_v \) for one mole of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Start with the definition of enthalpy (H)**:
\[ H = U + PV \]
where \( H \) is enthalpy, \( U \) is internal energy, \( P \) is pressure, and \( V \) is volume.

2. **Differentiate the enthalpy equation**:
\[ dH = dU + d(PV) \]

3. **Use the ideal gas law**:
For one mole of an ideal gas, we can express \( PV \) using the ideal gas equation:
\[ PV = nRT \]
where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

4. **Relate heat capacities to differentials**:
The molar heat capacities at constant pressure and volume are defined as:
\[ nC_p = \frac{dH}{dT} \quad \text{and} \quad nC_v = \frac{dU}{dT} \]

5. **Substituting into the differentiated enthalpy equation**:
Substitute \( dH \) and \( dU \) into the equation:
\[ nC_p dT = nC_v dT + d(PV) \]

6. **Differentiate \( PV \)**:
Since \( PV = nRT \), we have:
\[ d(PV) = d(nRT) = nR dT \]
(For one mole, \( n = 1 \)).

7. **Substituting back into the equation**:
Now substitute \( d(PV) \) back into the equation:
\[ nC_p dT = nC_v dT + nR dT \]

8. **Simplifying the equation**:
Since \( n = 1 \) mole, we can simplify:
\[ C_p dT = C_v dT + R dT \]
Dividing through by \( dT \) (assuming \( dT \neq 0 \)):
\[ C_p = C_v + R \]

9. **Final result**:
Therefore, we conclude that:
\[ C_p - C_v = R \]

### Conclusion:
The relationship for one mole of an ideal gas is:
\[ C_p - C_v = R \]