One mole of an ideal gas whose adiabatic exponent is `gamma = 4/3` undergoes a process `P = 200 + 1/V` . Then change in internal energy of gas when volume changes from `2 m^2` to `4 m^3` is:

Options

A. `400 J`
B. `800 J`
C. `1200 J`
D. None of these

Explanation

To solve the problem, we need to calculate the change in internal energy of the gas as it undergoes a change in volume from \(2 \, m^3\) to \(4 \, m^3\). The process is defined by the equation \(P = 200 + \frac{1}{V}\), and we know the adiabatic exponent \(\gamma = \frac{4}{3}\).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles, \(n = 1\)
- Adiabatic exponent, \(\gamma = \frac{4}{3}\)
- Initial volume, \(V_i = 2 \, m^3\)
- Final volume, \(V_f = 4 \, m^3\)

2. **Calculate Initial Pressure \(P_i\):**
\[ P_i = 200 + \frac{1}{V_i} = 200 + \frac{1}{2} = 200.5 \, \text{Pa} \]

3. **Calculate Final Pressure \(P_f\):**
\[ P_f = 200 + \frac{1}{V_f} = 200 + \frac{1}{4} = 200.25 \, \text{Pa} \]

4. **Calculate Initial Temperature \(T_i\) using the Ideal Gas Law:**
\[ PV = nRT \implies T_i = \frac{P_i V_i}{nR} \]
Here, we need to express \(R\) in terms of Joules. For one mole of an ideal gas, \(R \approx 8.314 \, \text{J/(mol K)}\):
\[ T_i = \frac{200.5 \times 2}{1 \times 8.314} \approx \frac{401}{8.314} \approx 48.2 \, K \]

5. **Calculate Final Temperature \(T_f\):**
\[ T_f = \frac{P_f V_f}{nR} \]
\[ T_f = \frac{200.25 \times 4}{1 \times 8.314} \approx \frac{801}{8.314} \approx 96.3 \, K \]

6. **Calculate Change in Temperature \(\Delta T\):**
\[ \Delta T = T_f - T_i = 96.3 - 48.2 = 48.1 \, K \]

7. **Calculate \(C_v\) (Specific Heat at Constant Volume):**
\[ C_v = \frac{R}{\gamma - 1} = \frac{8.314}{\frac{4}{3} - 1} = \frac{8.314}{\frac{1}{3}} = 24.942 \, J/(mol \cdot K) \]

8. **Calculate Change in Internal Energy \(\Delta U\):**
\[ \Delta U = n C_v \Delta T = 1 \times 24.942 \times 48.1 \approx 1200 \, J \]

### Final Answer:
The change in internal energy of the gas when the volume changes from \(2 \, m^3\) to \(4 \, m^3\) is approximately \(1200 \, J\).