(a) Define two specific heats of a gas. Why is `C_(p) gt C_(v)`?
(b) Shown that for an ideal gas,
`C_(p) = C_(v) +(R )/(J)`

Answer

### Step-by-Step Solution

#### Part (a): Definition of Specific Heats and Explanation of \( C_p > C_v \)

1. **Definition of Specific Heats**:
- **Specific Heat at Constant Volume (\( C_v \))**: It is defined as the amount of heat required to raise the temperature of a unit mass of a gas by 1 degree Celsius (or 1 Kelvin) while keeping the volume constant. Mathematically, it can be expressed as:
\[ C_v = \frac{Q}{m \Delta T} \]
where \( Q \) is the heat added, \( m \) is the mass of the gas, and \( \Delta T \) is the change in temperature.

- **Specific Heat at Constant Pressure (\( C_p \))**: It is defined as the amount of heat required to raise the temperature of a unit mass of a gas by 1 degree Celsius (or 1 Kelvin) while keeping the pressure constant. It can be expressed as:
\[ C_p = \frac{Q}{m \Delta T} \]

2. **Why is \( C_p > C_v \)**:
- When heat is added to a gas at constant volume, all the heat contributes to increasing the internal energy and thus the temperature of the gas. However, when heat is added at constant pressure, some of the heat is used to do work against the external pressure as the gas expands. Therefore, more heat is required to achieve the same temperature increase at constant pressure compared to constant volume.
- This can be summarized as:
\[ C_p = C_v + \text{(work done during expansion)} \]
- Hence, \( C_p > C_v \).

#### Part (b): Derivation of the Relation \( C_p = C_v + \frac{R}{J} \)

1. **Starting with the First Law of Thermodynamics**:
- The first law states:
\[ \Delta Q = \Delta U + \Delta W \]
- Where \( \Delta Q \) is the heat added, \( \Delta U \) is the change in internal energy, and \( \Delta W \) is the work done.

2. **For Constant Volume**:
- At constant volume, no work is done (\( \Delta W = 0 \)), so:
\[ \Delta Q = \Delta U \]
- Thus, we can write:
\[ \Delta Q = m C_v \Delta T \]

3. **For Constant Pressure**:
- At constant pressure, the work done is given by \( \Delta W = P \Delta V \). Therefore:
\[ \Delta Q = \Delta U + P \Delta V \]
- We can express this in terms of specific heats:
\[ \Delta Q = m C_p \Delta T \]

4. **Using the Ideal Gas Law**:
- For an ideal gas, we know that:
\[ PV = nRT \]
- Differentiating gives us:
\[ P \Delta V + V \Delta P = R \Delta T \]
- At constant pressure, \( \Delta P = 0 \), so:
\[ P \Delta V = R \Delta T \]

5. **Substituting Back**:
- Now substituting \( P \Delta V \) in the equation for \( \Delta Q \) at constant pressure:
\[ m C_p \Delta T = m C_v \Delta T + P \Delta V \]
- Replacing \( P \Delta V \) with \( R \Delta T \):
\[ m C_p \Delta T = m C_v \Delta T + m R \]

6. **Dividing by \( m \Delta T \)**:
- Dividing through by \( m \Delta T \) (assuming \( \Delta T \neq 0 \)):
\[ C_p = C_v + \frac{R}{J} \]

### Conclusion
Thus, we have shown that for an ideal gas:
\[ C_p = C_v + \frac{R}{J} \]