2 mol of an ideal gas at `27^(@)`C expand isothermally and reversibly
from a volume of 2L to 20L. The work done (in KJ) by the gas is :-

Options

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

Explanation

To solve the problem of calculating the work done by 2 moles of an ideal gas expanding isothermally and reversibly from a volume of 2L to 20L, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of moles (n) = 2 mol
- Initial volume (V1) = 2 L
- Final volume (V2) = 20 L
- Temperature (T) = 27°C

2. **Convert Temperature to Kelvin:**
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore,
\[ T = 27 + 273 = 300 \, K \]

3. **Use the Formula for Work Done in Isothermal Expansion:**
- The work done (W) during isothermal and reversible expansion is given by:
\[ W = -nRT \ln\left(\frac{V2}{V1}\right) \]
- Here, R is the universal gas constant, which is approximately 8.314 J/(mol·K).

4. **Calculate the Ratio of Volumes:**
- Calculate \(\frac{V2}{V1}\):
\[ \frac{V2}{V1} = \frac{20}{2} = 10 \]

5. **Substitute the Values into the Work Formula:**
- Now substitute n, R, T, and the volume ratio into the work formula:
\[ W = -2 \times 8.314 \, \text{J/(mol·K)} \times 300 \, K \times \ln(10) \]

6. **Calculate \(\ln(10)\):**
- The natural logarithm of 10 is approximately 2.303.
- Therefore,
\[ W = -2 \times 8.314 \times 300 \times 2.303 \]

7. **Perform the Calculation:**
- First, calculate \(2 \times 8.314 \times 300 \times 2.303\):
\[ = 2 \times 8.314 \times 300 \times 2.303 \approx 11488.28 \, J \]
- Since the work done is negative (as it is work done by the gas), we have:
\[ W \approx -11488.28 \, J \]

8. **Convert Joules to Kilojoules:**
- To convert from Joules to Kilojoules, divide by 1000:
\[ W \approx -11.488 \, kJ \]

### Final Answer:
The work done by the gas is approximately **-11.488 kJ**.