The molar heat capacity of an ideal gas in a process varies as `C=C_(V)+alphaT^(2)` (where `C_(V)` is mola heat capacity at constant volume and `alpha` is a constant). Then the equation of the process is

Options

A. `Ve^(-((alphaT^(2))/(2R)))=` Constant
B. `Ve^(-((alphaT^(2))/(R)))=` constant
C. `Ve^(-((2alphaT^(2))/(R)))=` constant
D. `Ve^(-((3alphaT^(2))/(2R)))=` constant

Explanation

To solve the problem, we start with the given molar heat capacity of an ideal gas in a process:

\[ C = C_V + \alpha T^2 \]

where \( C_V \) is the molar heat capacity at constant volume, \( \alpha \) is a constant, and \( T \) is the temperature.

### Step 1: Use the First Law of Thermodynamics
According to the first law of thermodynamics, we have:

\[ dQ = dU + dW \]

where:
- \( dQ \) is the heat added to the system,
- \( dU \) is the change in internal energy,
- \( dW \) is the work done by the system.

For an ideal gas, the change in internal energy can be expressed as:

\[ dU = n C_V dT \]

### Step 2: Relate Heat Capacity to Heat Transfer
From the definition of heat capacity, we can write:

\[ dQ = n C dT = n (C_V + \alpha T^2) dT \]

### Step 3: Express Work Done
The work done by the gas during an infinitesimal expansion can be expressed as:

\[ dW = P dV \]

Using the ideal gas law, we know that:

\[ PV = nRT \]

Thus, we can express pressure \( P \) as:

\[ P = \frac{nRT}{V} \]

Substituting this into the expression for work done gives:

\[ dW = \frac{nRT}{V} dV \]

### Step 4: Substitute into the First Law
Now substituting \( dU \) and \( dW \) into the first law gives:

\[ n (C_V + \alpha T^2) dT = n C_V dT + \frac{nRT}{V} dV \]

### Step 5: Simplify the Equation
Cancelling \( n C_V dT \) from both sides results in:

\[ \alpha T^2 dT = \frac{RT}{V} dV \]

### Step 6: Rearranging the Equation
Rearranging gives:

\[ \frac{dV}{V} = \frac{\alpha}{R} T dT \]

### Step 7: Integrate Both Sides
Now we integrate both sides:

\[ \int \frac{dV}{V} = \int \frac{\alpha}{R} T dT \]

The left side integrates to:

\[ \ln V = \frac{\alpha}{R} \cdot \frac{T^2}{2} + \ln k \]

where \( k \) is a constant of integration.

### Step 8: Exponentiate to Solve for V
Exponentiating both sides gives:

\[ V = k e^{\frac{\alpha T^2}{2R}} \]

### Step 9: Rearranging to Find the Final Form
Rearranging this equation leads to:

\[ V e^{-\frac{\alpha T^2}{2R}} = k \]

This shows that the product \( V e^{-\frac{\alpha T^2}{2R}} \) is constant.

### Conclusion
Thus, the equation of the process is:

\[ V e^{-\frac{\alpha T^2}{2R}} = \text{constant} \]