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 calculating the work done by one mole of an ideal gas during the process where the temperature varies with volume as \( T = 300 + 2V \), we can follow these steps:

### Step 1: Understand the Work Done by the Gas
The work done \( W \) by the gas during a volume change is given by the integral of pressure with respect to volume:
\[ W = \int_{V_1}^{V_2} P \, dV \]
where \( V_1 = 2 \, m^3 \) and \( V_2 = 4 \, m^3 \).

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

### Step 3: Set Up the Integral for Work Done
Now, substituting the expression for pressure into the work integral:
\[ W = \int_{2}^{4} \left( \frac{300R}{V} + 2R \right) dV \]

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

### Step 5: Calculate Each Integral
1. **First Integral**:
\[ \int_{2}^{4} \frac{300R}{V} \, dV = 300R \left[ \ln V \right]_{2}^{4} = 300R (\ln 4 - \ln 2) = 300R \ln \left( \frac{4}{2} \right) = 300R \ln 2 \]

2. **Second Integral**:
\[ \int_{2}^{4} 2R \, dV = 2R \left[ V \right]_{2}^{4} = 2R (4 - 2) = 4R \]

### Step 6: Combine the Results
Now, combining both parts of the integral:
\[ W = 300R \ln 2 + 4R \]

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