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. `3 R`
B. `3.5 R`
C. `4 R`
D. None of these

Explanation

To solve the problem, we need to find the molar heat capacity of an ideal gas undergoing a specific process where the internal energy \( U \) is related to 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 temperature**:
The internal energy \( U \) of an ideal gas can be expressed as:
\[ U = \frac{F}{2} nRT \]
where \( F \) is the degrees of freedom, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature. Given that \( U = V^2 \), we can infer that:
\[ V^2 \propto T \]
This means that temperature \( T \) is proportional to \( V^2 \).

2. **Using the Ideal Gas Law**:
The ideal gas law is given by:
\[ PV = nRT \]
From this, we can express temperature \( T \) as:
\[ T = \frac{PV}{nR} \]
Substituting this into our earlier relationship gives:
\[ V^2 \propto \frac{PV}{nR} \]
Rearranging, we get:
\[ PV \propto V^3 \quad \Rightarrow \quad P \propto V^2 \]

3. **Finding the relationship between pressure and volume**:
From the proportionality \( PV^{-1} = \text{constant} \), we can deduce that:
\[ PV^{-1} = k \quad \text{(some constant)} \]

4. **Relating to the adiabatic process**:
The general form of an adiabatic process is given by:
\[ PV^x = \text{constant} \]
Comparing this with our derived equation \( PV^{-1} = \text{constant} \), we can identify:
\[ x = -1 \]

5. **Finding degrees of freedom \( F \)**:
The relationship between \( \gamma \) and degrees of freedom \( F \) is:
\[ \gamma = 1 + \frac{2}{F} \]
Given \( \gamma = \frac{4}{3} \):
\[ \frac{4}{3} = 1 + \frac{2}{F} \]
Solving for \( F \):
\[ \frac{4}{3} - 1 = \frac{2}{F} \quad \Rightarrow \quad \frac{1}{3} = \frac{2}{F} \quad \Rightarrow \quad F = 6 \]

6. **Calculating the molar heat capacity \( C \)**:
The molar heat capacity \( C \) can be expressed as:
\[ C = C_v + \frac{R}{1 - x} \]
where \( C_v = \frac{F}{2} R \) and \( x = -1 \):
\[ C_v = \frac{6}{2} R = 3R \]
Substituting \( C_v \) and \( x \) into the equation for \( C \):
\[ C = 3R + \frac{R}{1 - (-1)} = 3R + \frac{R}{2} = 3R + \frac{R}{2} = 3R + 0.5R = 3.5R \]

7. **Final Result**:
Thus, the molar heat capacity of the gas for the process is:
\[ C = \frac{7R}{2} \quad \text{or} \quad 3.5R \]