Molar heat capacity of gas whose molar heat capacity at constant volume is `C_V`, for process `P = 2e^(2V)`is :

Options

A. `C_V + R/(1+2V)`
B. `C_V +R/(2V)`
C. `C_V + R/V`
D. None of these

Explanation

To find the molar heat capacity of a gas whose molar heat capacity at constant volume is \( C_V \) for the process defined by \( P = 2e^{2V} \), we can follow these steps:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states that:
\[ dQ = dU + dW \]
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.

### Step 2: Express \( dU \) and \( dW \)
For an ideal gas, the change in internal energy can be expressed as:
\[ dU = nC_V dT \]
The work done by the system during a process is given by:
\[ dW = P dV \]

### Step 3: Substitute into the First Law
Substituting the expressions for \( dU \) and \( dW \) into the first law gives:
\[ dQ = nC_V dT + P dV \]

### Step 4: Rearrange for \( dQ \)
Rearranging the equation, we have:
\[ dQ = nC_V dT + P dV \]

### Step 5: Use the Ideal Gas Law
The ideal gas law states:
\[ PV = nRT \]
Given \( P = 2e^{2V} \), we can substitute this into the ideal gas law:
\[ 2e^{2V}V = nRT \]

### Step 6: Differentiate the Ideal Gas Law
Differentiating both sides with respect to \( T \):
\[ \frac{d}{dT}(2e^{2V}V) = \frac{d}{dT}(nRT) \]
Using the product rule on the left side:
\[ 2e^{2V} \frac{dV}{dT} + 4Ve^{2V} \frac{dV}{dT} = nR \]
Factoring out \( \frac{dV}{dT} \):
\[ \left(2e^{2V} + 4Ve^{2V}\right) \frac{dV}{dT} = nR \]

### Step 7: Solve for \( \frac{dV}{dT} \)
Thus, we have:
\[ \frac{dV}{dT} = \frac{nR}{2e^{2V} + 4Ve^{2V}} \]

### Step 8: Substitute \( \frac{dV}{dT} \) into \( dQ \)
Now, substituting \( \frac{dV}{dT} \) back into the expression for \( dQ \):
\[ dQ = nC_V dT + P \left(\frac{nR}{2e^{2V} + 4Ve^{2V}}\right) dT \]
This simplifies to:
\[ dQ = n \left(C_V + \frac{P}{2e^{2V} + 4Ve^{2V}} R\right) dT \]

### Step 9: Identify Molar Heat Capacity \( C \)
From the above equation, we can identify the molar heat capacity \( C \):
\[ C = C_V + \frac{P}{2e^{2V} + 4Ve^{2V}} R \]

### Step 10: Substitute \( P \)
Substituting \( P = 2e^{2V} \):
\[ C = C_V + \frac{2e^{2V}}{2e^{2V} + 4Ve^{2V}} R \]
This simplifies to:
\[ C = C_V + \frac{2R}{2 + 4V} \]

### Final Result
Thus, the molar heat capacity of the gas for the given process is:
\[ C = C_V + \frac{R}{1 + 2V} \]