2 mole of an ideal gas at `27^(@)C` expands isothermally and reversibly from a volume of 4 litre to 40 litre. The work done (in kJ) by the gas is :

Options

A. `w = - 28.72kJ`
B. `w = - 11.488 kJ`
C. `w = - 5.736 kJ`
D. `w = - 4.988 kJ`

Explanation

To solve the problem of calculating the work done by an ideal gas during isothermal and reversible expansion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Number of moles (n) = 2 moles
- Initial volume (V1) = 4 liters
- Final volume (V2) = 40 liters
- Temperature (T) = 27°C = 27 + 273 = 300 K
- Ideal gas constant (R) = 8.314 J/(mol·K)

2. **Formula for Work Done:**
The work done (W) during an isothermal reversible expansion is given by the formula:
\[ W = -nRT \ln\left(\frac{V_2}{V_1}\right) \]
Since we will use logarithm base 10, we can convert it using:
\[ \ln(x) = 2.303 \log_{10}(x) \]
Therefore, the work done can also be expressed as:
\[ W = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right) \]

3. **Calculate the Volume Ratio:**
\[ \frac{V_2}{V_1} = \frac{40 \text{ L}}{4 \text{ L}} = 10 \]

4. **Calculate the Logarithm:**
\[ \log_{10}(10) = 1 \]

5. **Substitute Values into the Work Done Formula:**
Now substitute the values into the work done formula:
\[ W = -2.303 \times 2 \text{ moles} \times 8.314 \frac{\text{J}}{\text{mol·K}} \times 300 \text{ K} \times 1 \]

6. **Perform the Calculation:**
\[ W = -2.303 \times 2 \times 8.314 \times 300 \]
\[ W = -2.303 \times 2 \times 2494.2 \text{ J} \]
\[ W = -11488.28 \text{ J} \]

7. **Convert to Kilojoules:**
\[ W = -\frac{11488.28}{1000} \text{ kJ} = -11.488 \text{ kJ} \]

8. **Final Answer:**
The work done by the gas during the isothermal expansion is:
\[ W = -11.488 \text{ kJ} \]