A gas expands from `2m^3` to `2.5m^3` at constant pressure `10^3` pa, the work done is

Options

A. 50 J
B. 25 J
C. 500 J
D. 250 J

Answer

To find the work done by the gas during its expansion at constant pressure, we can use the formula for work done in an isobaric process:

**Step 1: Identify the initial and final volumes.**
- Initial volume \( V_i = 2 \, m^3 \)
- Final volume \( V_f = 2.5 \, m^3 \)

**Step 2: Calculate the change in volume (\( \Delta V \)).**
\[ \Delta V = V_f - V_i = 2.5 \, m^3 - 2 \, m^3 = 0.5 \, m^3 \]

**Step 3: Identify the constant pressure.**
- Pressure \( P = 10^3 \, Pa \)

**Step 4: Use the formula for work done at constant pressure.**
The work done \( W \) is given by:
\[ W = P \Delta V \]

**Step 5: Substitute the values into the formula.**
\[ W = (10^3 \, Pa) \times (0.5 \, m^3) \]

**Step 6: Calculate the work done.**
\[ W = 10^3 \times 0.5 = 500 \, J \]

Thus, the work done by the gas during the expansion is **500 joules**.

---