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 find the change in internal energy of the gas when the volume changes from \(2 \, m^3\) to \(4 \, m^3\), we can follow these steps:

### Step 1: Understand the Process
The process described is an adiabatic process, meaning no heat is exchanged with the surroundings. Therefore, we can use the first law of thermodynamics:
\[ \Delta U = Q - W \]
Since \(Q = 0\) for an adiabatic process, we have:
\[ \Delta U = -W \]

### Step 2: Determine the Work Done
The work done by the gas during an adiabatic process can be calculated using the formula:
\[ W = \int P \, dV \]
Given the pressure \(P = 200 + \frac{1}{V}\), we need to integrate this expression from \(V = 2 \, m^3\) to \(V = 4 \, m^3\).

### Step 3: Set Up the Integral
The work done can be expressed as:
\[ W = \int_{2}^{4} \left(200 + \frac{1}{V}\right) dV \]

### Step 4: Calculate the Integral
Now we can calculate the integral:
\[ W = \int_{2}^{4} 200 \, dV + \int_{2}^{4} \frac{1}{V} \, dV \]
Calculating the first part:
\[ \int_{2}^{4} 200 \, dV = 200 \times (4 - 2) = 200 \times 2 = 400 \]
Calculating the second part:
\[ \int_{2}^{4} \frac{1}{V} \, dV = \ln(V) \bigg|_{2}^{4} = \ln(4) - \ln(2) = \ln\left(\frac{4}{2}\right) = \ln(2) \]
Thus, the total work done is:
\[ W = 400 + \ln(2) \]

### Step 5: Calculate the Change in Internal Energy
Since \(\Delta U = -W\):
\[ \Delta U = -\left(400 + \ln(2)\right) = -400 - \ln(2) \]

### Step 6: Substitute Values
For the numerical evaluation, we can approximate \(\ln(2) \approx 0.693\):
\[ \Delta U \approx -400 - 0.693 \approx -400.693 \, J \]

### Final Result
The change in internal energy of the gas when the volume changes from \(2 \, m^3\) to \(4 \, m^3\) is approximately:
\[ \Delta U \approx -400.693 \, J \]