One mole of an ideal gas undergoes a process T = 300 + 2V. Then amount of work done by gas when volume increases from `2m^3` to ` 4m^3`:

Options

A. `"300R In 2 + 4 R`
B. `"2 + 2 In 2"`
C. `"300 R In 2"`
D. None of these

Explanation

To solve the problem of finding the work done by one mole of an ideal gas as it undergoes a specific process, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- The process equation: \( T = 300 + 2V \)
- The initial volume \( V_1 = 2 \, m^3 \)
- The final volume \( V_2 = 4 \, m^3 \)
- The number of moles \( n = 1 \)

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
For one mole of gas (\( n = 1 \)), this simplifies to:
\[ PV = RT \]
Substituting the expression for temperature \( T \):
\[ P = \frac{RT}{V} = \frac{R(300 + 2V)}{V} = \frac{300R}{V} + 2R \]

### Step 3: Set Up the Work Done Integral
The work done \( W \) by the gas during expansion is given by:
\[ W = \int_{V_1}^{V_2} P \, dV \]
Substituting the expression for \( P \):
\[ W = \int_{2}^{4} \left( \frac{300R}{V} + 2R \right) dV \]

### Step 4: Break Down the Integral
The integral can be split into two parts:
\[ W = \int_{2}^{4} \frac{300R}{V} \, dV + \int_{2}^{4} 2R \, dV \]

### Step 5: Solve the First Integral
The first integral:
\[ \int \frac{300R}{V} \, dV = 300R \ln V \]
Evaluating from \( 2 \) to \( 4 \):
\[ 300R \left( \ln 4 - \ln 2 \right) = 300R \ln \left( \frac{4}{2} \right) = 300R \ln 2 \]

### Step 6: Solve the Second Integral
The second integral:
\[ \int 2R \, dV = 2R V \]
Evaluating from \( 2 \) to \( 4 \):
\[ 2R (4 - 2) = 2R \cdot 2 = 4R \]

### Step 7: Combine the Results
Combining both parts of the work done:
\[ W = 300R \ln 2 + 4R \]

### Final Expression for Work Done
Thus, the total work done by the gas when the volume increases from \( 2 \, m^3 \) to \( 4 \, m^3 \) is:
\[ W = 300R \ln 2 + 4R \]