Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is : [Given that 1 L bar = 100 J]

Options

A. 30 J
B. `-30 J`
C. `5kJ`
D. `25J`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial volume (V1) = 0.1 L
- Final volume (V2) = 0.25 L
- External pressure (P_ext) = 2 bar

2. **Calculate the Change in Volume (ΔV)**:
\[ \Delta V = V2 - V1 = 0.25 \, \text{L} - 0.1 \, \text{L} = 0.15 \, \text{L} \]

3. **Use the Formula for Work Done (W)**:
For an isothermal irreversible process, the work done by the gas is given by:
\[ W = -P_{\text{ext}} \times \Delta V \]
Here, we need to ensure that the pressure is in the correct units (bar) and the volume is in liters.

4. **Substitute the Values into the Formula**:
\[ W = -2 \, \text{bar} \times 0.15 \, \text{L} = -0.30 \, \text{L bar} \]

5. **Convert Work Done from L bar to Joules**:
We know that:
\[ 1 \, \text{L bar} = 100 \, \text{J} \]
Therefore, to convert the work done into joules:
\[ W = -0.30 \, \text{L bar} \times 100 \, \text{J/L bar} = -30 \, \text{J} \]

6. **Final Answer**:
The work done by the gas is:
\[ W = -30 \, \text{J} \]