(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

**(a) Definition of Specific Heats:**

1. **Specific Heat at Constant Pressure (Cₚ):**
- Cₚ is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin) at constant pressure. Mathematically, it can be expressed as:
\[ C_p = \frac{dQ_P}{m \cdot dT} \]
- Here, \(dQ_P\) is the heat added at constant pressure, \(m\) is the mass of the substance, and \(dT\) is the change in temperature.

2. **Specific Heat at Constant Volume (Cᵥ):**
- Cᵥ is defined as the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin) at constant volume. It can be expressed as:
\[ C_v = \frac{dQ_V}{m \cdot dT} \]
- Here, \(dQ_V\) is the heat added at constant volume.

3. **Why is Cₚ > Cᵥ?**
- The reason \(C_p\) is greater than \(C_v\) is due to the work done against the external pressure when the gas expands. At constant pressure, when heat is added, the gas does work to expand, which requires additional energy. At constant volume, no work is done, so less heat is required to achieve the same temperature change. Thus, \(C_p > C_v\).

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

1. **First Law of Thermodynamics:**
- According to the first law of thermodynamics, the heat added to the system is equal to the change in internal energy plus the work done by the system:
\[ dQ = dU + dW \]

2. **At Constant Pressure:**
- The heat added at constant pressure can be expressed as:
\[ dQ_P = dU + P dV \]
- For one mole of gas, we can write:
\[ dQ_P = C_p dT \]

3. **At Constant Volume:**
- The heat added at constant volume can be expressed as:
\[ dQ_V = dU \]
- For one mole of gas, we can write:
\[ dQ_V = C_v dT \]

4. **Equating the Two Expressions:**
- From the two equations, we have:
\[ C_p dT = C_v dT + P dV \]
- Rearranging gives:
\[ C_p - C_v = \frac{P dV}{dT} \]

5. **Using the Ideal Gas Law:**
- The ideal gas law states:
\[ PV = nRT \]
- For one mole of gas, we can differentiate this to find:
\[ P dV + V dP = R dT \]
- At constant pressure, \(dP = 0\), thus:
\[ P dV = R dT \]

6. **Substituting Back:**
- Substituting \(P dV\) into our previous equation gives:
\[ C_p - C_v = R \]

7. **Including the Mechanical Equivalent (J):**
- To include the mechanical equivalent \(J\), we can express the relation as:
\[ C_p - C_v = \frac{R}{J} \]
- Therefore, we conclude:
\[ C_p = C_v + \frac{R}{J} \]

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