A gas can expand from `100 mL` to `250 mL` under a constant pressure of `2` atm. The work done by the gas is

Options

A. 30.83 J
B. 25 J
C. 5 kJ
D. 16 J

Explanation

To find the work done by the gas during its expansion, we can follow these steps:

### Step 1: Identify the given values
- Initial volume (V1) = 100 mL
- Final volume (V2) = 250 mL
- External pressure (P_external) = 2 atm

### Step 2: Calculate the change in volume (ΔV)
\[ \Delta V = V2 - V1 = 250 \, \text{mL} - 100 \, \text{mL} = 150 \, \text{mL} \]

### Step 3: Use the formula for work done (W)
The work done by the gas during expansion at constant pressure is given by:
\[ W = -P_{\text{external}} \Delta V \]
Substituting the values:
\[ W = -2 \, \text{atm} \times 150 \, \text{mL} \]

### Step 4: Convert mL to liters
Since 1 L = 1000 mL, we convert 150 mL to liters:
\[ 150 \, \text{mL} = \frac{150}{1000} \, \text{L} = 0.15 \, \text{L} \]

### Step 5: Substitute the volume in liters into the work formula
Now, substituting the volume in liters into the work formula:
\[ W = -2 \, \text{atm} \times 0.15 \, \text{L} \]
Calculating this gives:
\[ W = -0.3 \, \text{atm} \cdot \text{L} \]

### Step 6: Convert work from atm·L to joules
Using the conversion factor \(1 \, \text{atm} \cdot \text{L} = 101.325 \, \text{J}\):
\[ W = -0.3 \, \text{atm} \cdot \text{L} \times 101.325 \, \text{J/atm} \cdot \text{L} \]
Calculating this gives:
\[ W = -30.3975 \, \text{J} \approx -30.39 \, \text{J} \]

### Conclusion
The work done by the gas during its expansion is approximately:
\[ W \approx -30.39 \, \text{J} \]