An ideal gas has a molar heat capacity `C_v` at constant volume. Find the molar heat capacity of this gas as a function of its volume `V`, if the gas undergoes the following process :
(a) `T = T_0 e^(alpha v)` ,
(b) `p = p_0 e^(alpha v)`,
where `T_0, p_0`, and `alpha` are constants.

Answer

To find the molar heat capacity of an ideal gas as a function of its volume \( V \) for the given processes, we will analyze each process step by step.

### Given:
1. Process (a): \( T = T_0 e^{\alpha V} \)
2. Process (b): \( P = P_0 e^{\alpha V} \)

### Step 1: Analyze Process (a)

1. **Differentiate the Temperature Equation**:
\[ T = T_0 e^{\alpha V} \]
Differentiate with respect to \( V \):
\[ dT = T_0 \alpha e^{\alpha V} dV \]
This can be rewritten as:
\[ \frac{dT}{dV} = \alpha T \]
(Let this be Equation 1)

2. **Apply the First Law of Thermodynamics**:
The first law states:
\[ dQ = dU + dW \]
For an ideal gas, we have:
\[ dQ = C dT, \quad dU = C_v dT, \quad dW = P dV \]
Therefore, substituting these into the first law gives:
\[ C dT = C_v dT + P dV \]
Dividing through by \( dT \):
\[ C = C_v + P \frac{dV}{dT} \]
(Let this be Equation 2)

3. **Substitute \( \frac{dV}{dT} \) from Equation 1**:
From Equation 1, we have:
\[ \frac{dV}{dT} = \frac{1}{\alpha T} \]
Substitute this into Equation 2:
\[ C = C_v + P \cdot \frac{1}{\alpha T} \]

4. **Use the Ideal Gas Law**:
From the ideal gas law \( PV = nRT \) for one mole of gas:
\[ P = \frac{RT}{V} \]
Substitute this into the equation for \( C \):
\[ C = C_v + \frac{RT}{V \alpha T} \]
Simplifying gives:
\[ C = C_v + \frac{R}{\alpha V} \]

### Final Result for Process (a):
\[ C = C_v + \frac{R}{\alpha V} \]

---

### Step 2: Analyze Process (b)

1. **Differentiate the Pressure Equation**:
\[ P = P_0 e^{\alpha V} \]
Differentiate with respect to \( V \):
\[ dP = P_0 \alpha e^{\alpha V} dV \]
This can be rewritten as:
\[ \frac{dP}{dV} = \alpha P \]
(Let this be Equation 3)

2. **Apply the First Law of Thermodynamics**:
Using the same form as before:
\[ C = C_v + P \frac{dV}{dT} \]
We need to find \( \frac{dV}{dT} \).

3. **Use the Ideal Gas Law**:
From the ideal gas law \( PV = nRT \):
\[ T = \frac{PV}{R} \]
Differentiate this with respect to \( V \):
\[ dT = \frac{P}{R} dV + \frac{V}{R} dP \]

4. **Substituting \( dP \)**:
Substitute \( dP \) from Equation 3:
\[ dT = \frac{P}{R} dV + \frac{V}{R} P_0 \alpha e^{\alpha V} dV \]
Factor out \( dV \):
\[ dT = \left( \frac{P}{R} + \frac{V P_0 \alpha e^{\alpha V}}{R} \right) dV \]
Thus:
\[ \frac{dV}{dT} = \frac{R}{P + V P_0 \alpha e^{\alpha V}} \]

5. **Substituting into the First Law**:
Substitute \( \frac{dV}{dT} \) into the first law equation:
\[ C = C_v + P \cdot \frac{R}{P + V P_0 \alpha e^{\alpha V}} \]
Simplifying gives:
\[ C = C_v + \frac{R P}{P + V P_0 \alpha e^{\alpha V}} \]

### Final Result for Process (b):
\[ C = C_v + \frac{R}{1 + \alpha V} \]

### Summary of Results:
- For Process (a):
\[ C = C_v + \frac{R}{\alpha V} \]
- For Process (b):
\[ C = C_v + \frac{R}{1 + \alpha V} \]