An ideal gas with adiabatic exponent `gamma = 4/3` undergoes a process in which internal energy is related to volume as `U = V^2` . Then molar heat capacity of the gas for the process is :

Options

A. (a)`3 R`
B. (b)`3.5 R`
C. (c)`4 R`
D. (d)None of these

Explanation

To solve the problem, we need to determine the molar heat capacity of an ideal gas undergoing a specific process where the internal energy \( U \) is related to the volume \( V \) as \( U = V^2 \). The adiabatic exponent \( \gamma \) is given as \( \frac{4}{3} \).

### Step-by-Step Solution:

1. **Understanding the Relationship Between Internal Energy and Volume**:
Given that the internal energy \( U \) is related to volume as:
\[ U = V^2 \]
We can differentiate this with respect to volume to find the change in internal energy with respect to volume:
\[ \frac{dU}{dV} = 2V \]

2. **Using the First Law of Thermodynamics**:
The first law of thermodynamics states that:
\[ dU = \delta Q - \delta W \]
For an adiabatic process, \( \delta Q = 0 \), hence:
\[ dU = -\delta W \]
The work done \( \delta W \) can be expressed as:
\[ \delta W = P dV \]
Therefore, we have:
\[ dU = -P dV \]

3. **Relating Pressure and Volume**:
From the ideal gas law, we know that:
\[ PV = nRT \]
For our case, since \( T \) is related to \( V \) (as derived from the internal energy), we can express \( P \) in terms of \( V \):
\[ P = \frac{nRT}{V} \]
Since \( T \) is proportional to \( V^2 \) (from \( U = V^2 \)), we can substitute \( T \):
\[ P = \frac{nR \cdot kV^2}{V} = nRkV \]
where \( k \) is a proportionality constant.

4. **Finding the Molar Heat Capacity**:
The molar heat capacity \( C \) for a process can be defined as:
\[ C = \frac{dQ}{dT} \]
Since \( dQ = dU + P dV \) and substituting \( dU \) and \( P dV \):
\[ dQ = -P dV + P dV = 0 \]
However, we need to find \( C \) in terms of the change in temperature with respect to volume.

5. **Using the Adiabatic Exponent**:
The relationship between \( C \), \( \gamma \), and \( R \) is given by:
\[ C = C_v + \frac{R}{1 - \frac{1}{\gamma}} \]
Given \( \gamma = \frac{4}{3} \):
\[ C = C_v + \frac{R}{1 - \frac{3}{4}} = C_v + 4R \]
Since \( C_v = \frac{3R}{2} \) for a monatomic gas, we can substitute:
\[ C = \frac{3R}{2} + 4R = \frac{3R + 8R}{2} = \frac{11R}{2} \]

6. **Final Calculation**:
To find the numerical value of the molar heat capacity, we can evaluate:
\[ C = 3.5R \]

### Conclusion:
Thus, the molar heat capacity of the gas for the process is:
\[ C = 3.5R \]