For an ideal gas the equation of a process for which the heat capacity of the gas varies with temperatue as `C=(alpha//T(alpha)` is a constant) is given by

Options

A. `VlnT=constant`
B. `VT^(1//(gamma-1))_(e)^(alpha//RT)`=constant
C. `V^(1)/(gamma-1)T^(alpha//RT)`=constant
D. `V^(gamma-1)T`=constant

Explanation

To solve the problem, we need to derive the equation of the process for an ideal gas where the heat capacity \( C \) varies with temperature as \( C = \frac{\alpha}{T} \), where \( \alpha \) is a constant.

### Step-by-Step Solution:

1. **Understanding the First Law of Thermodynamics**:
The first law of thermodynamics states:
\[ dQ = dU + dW \]
where \( dQ \) is the heat added to the system, \( dU \) is the change in internal energy, and \( dW \) is the work done by the system.

2. **Expressing \( dQ \), \( dU \), and \( dW \)**:
For an ideal gas, we can express:
\[ dQ = nC dT, \quad dU = nC_V dT, \quad dW = P dV \]
where \( n \) is the number of moles, \( C \) is the heat capacity, \( C_V \) is the molar heat capacity at constant volume, and \( P \) is the pressure.

3. **Substituting the given heat capacity**:
Given \( C = \frac{\alpha}{T} \), we substitute this into the equation:
\[ n \frac{\alpha}{T} dT = nC_V dT + P dV \]

4. **Using the Ideal Gas Law**:
From the ideal gas law, we know:
\[ PV = nRT \quad \Rightarrow \quad P = \frac{nRT}{V} \]
Substitute this into the equation:
\[ n \frac{\alpha}{T} dT = nC_V dT + \frac{nRT}{V} dV \]

5. **Canceling \( n \)**:
We can cancel \( n \) from all terms:
\[ \frac{\alpha}{T} dT = C_V dT + \frac{RT}{V} dV \]

6. **Rearranging the equation**:
Rearranging gives us:
\[ \frac{\alpha}{T} dT - C_V dT = \frac{RT}{V} dV \]
Factor out \( dT \):
\[ \left(\frac{\alpha}{T} - C_V\right) dT = \frac{RT}{V} dV \]

7. **Dividing by \( T \)**:
Dividing the entire equation by \( T \) gives:
\[ \frac{\alpha}{RT} dT = \frac{1}{\gamma - 1} dT + \frac{1}{V} dV \]

8. **Integrating**:
Now we can integrate both sides. The left side will involve integrating with respect to \( T \) and the right side with respect to \( V \):
\[ \int \frac{\alpha}{RT} dT = \int \left(\frac{1}{\gamma - 1} dT + \frac{1}{V} dV\right) \]

9. **Applying integration rules**:
Using the integration rules:
\[ \frac{\alpha}{R} \ln T = \frac{1}{\gamma - 1} \ln T + \ln V + C \]
where \( C \) is the constant of integration.

10. **Exponentiating both sides**:
Exponentiating both sides leads to:
\[ e^{\frac{\alpha}{R}} = T^{\frac{1}{\gamma - 1}} V \]

11. **Final equation**:
Rearranging gives us the final equation of the process:
\[ V T^{\frac{1}{\gamma - 1}} = e^{\frac{\alpha}{R}} \text{ (constant)} \]

### Conclusion:
The equation of the process for which the heat capacity of the gas varies with temperature as \( C = \frac{\alpha}{T} \) is given by:
\[ V T^{\frac{1}{\gamma - 1}} = \text{constant} \]