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 solve the problem of calculating the work done during the expansion of a gas, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The change in volume (ΔV) = 0.25 m³
- The constant pressure (P) = 10³ N/m²

2. **Recall the Formula for Work Done:**
The work done (W) during the expansion of a gas at constant pressure is given by the formula:
\[ W = -P \Delta V \]
Here, the negative sign indicates that work is done by the system during expansion.

3. **Substitute the Known Values into the Formula:**
Substitute the values of pressure and change in volume into the formula:
\[ W = - (10^3 \, \text{N/m}^2) \times (0.25 \, \text{m}^3) \]

4. **Calculate the Work Done:**
- First, calculate the product:
\[ W = - (1000 \, \text{N/m}^2) \times (0.25 \, \text{m}^3) = - 250 \, \text{N m} \]
- Since 1 N m = 1 Joule, we can express this as:
\[ W = - 250 \, \text{J} \]

5. **Interpret the Result:**
The negative sign indicates that the work is done by the gas on the surroundings during the expansion. Therefore, the magnitude of work done is 250 Joules.

### Final Answer:
The work done is **-250 Joules** (or 250 Joules done by the system).