The volume of a gas expands by `0.25 m^(3)` at a constant pressure of `10^(3)Nm^(2)`. The work done is equal to

Options

A. `2.5 erg`
B. `250 J`
C. `250 watt`
D. `250 newton`

Answer

To find the work done when a gas expands at constant pressure, we can use the formula:

\[ W = -P \Delta V \]

Where:
- \( W \) is the work done,
- \( P \) is the pressure,
- \( \Delta V \) is the change in volume.

### Step-by-Step Solution:

1. **Identify the given values**:
- Pressure \( P = 10^3 \, \text{N/m}^2 \) (or 1000 N/m²)
- Change in volume \( \Delta V = 0.25 \, \text{m}^3 \)

2. **Substitute the values into the work formula**:
\[ W = -P \Delta V \]
\[ W = - (10^3 \, \text{N/m}^2) \times (0.25 \, \text{m}^3) \]

3. **Calculate the product**:
\[ W = - (1000) \times (0.25) \]
\[ W = - 250 \, \text{N m} \]

4. **Convert N m to Joules**:
Since \( 1 \, \text{N m} = 1 \, \text{J} \):
\[ W = - 250 \, \text{J} \]

5. **Interpret the result**:
The negative sign indicates that the work is done by the system during expansion.

### Final Answer:
The work done is \( -250 \, \text{J} \).

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