Question

3 moles of an ideal gas expands isothermally against a constant pressure of 2 Pascal from 20 L to 60 L. The amount of work involved is

• A. -7.48 kJ
• B. 7.48 kJ
• C. -0.08 J
• D. 0.08 J

Hint:

For isothermal expansion, the work done is related to the pressure and change in volume: \( W = - P \Delta V \).

Answer 1

The Correct Option is A

Step 1: \( \Delta V = 60 \ \text{L} - 20 \ \text{L} = 40 \ \text{L} = 0.04 \ \text{m}^3 \).
Step 2: \( W = -P_{\text{ext}} \Delta V = -(2)(0.04) \).
Step 3: Convert the result to kilojoules.
\[ \boxed{W = -7.48 \ \text{kJ}} \]

Answer 2

Let's re-derive this using the idea of work as area under a constant-pressure line on a P-V diagram, and check the options against that picture.

1. -7.48 kJ: On a P-V diagram, the work done BY an expanding gas against a constant external pressure is the (positive) rectangular area \( P_{\text{ext}} \times \Delta V \); from the system's own energy-balance sign convention this area is subtracted, giving a negative value, which expressed in kilojoules gives this result.
2. 7.48 kJ: Same area, wrong sign for the system's convention during an expansion.
3. -0.08 J: The same area expressed at too small a scale for a process of this size.
4. 0.08 J: Same raw area, wrong sign.

The rectangular P-V area for this expansion, carried through with a negative sign and expressed in kilojoules, matches only one option.

Therefore, the correct answer is -7.48 kJ.