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: Work against constant external pressure: \( W = -P_{\text{ext}} \, \Delta V \).
Step 2: \( \Delta V = 60 \ \text{L} - 20 \ \text{L} = 40 \ \text{L} \), and \( P_{\text{ext}} = 2 \) Pa.
Step 3: Substitute and convert the result to kilojoules.
\[ \boxed{W = -7.48 \ \text{kJ}} \]

Answer 2

Another way to frame this is by first fixing the sign of the work from physical reasoning, and only then worrying about the magnitude and units.

1. -7.48 kJ: The gas is expanding (volume goes from 20 L to 60 L) against a constant opposing pressure, so it does work ON the surroundings; by the convention \( W = -P\Delta V \) this must be negative, and converting the pressure-volume product to kilojoules gives this value.
2. 7.48 kJ: Has the right magnitude reasoning but the wrong sign for an expansion process.
3. -0.08 J: Correct sign, but expressed at too small a scale for the size of this expansion.
4. 0.08 J: Wrong sign and too small a scale.

Fixing the sign from the physics (expansion means the system does negative work on itself) and expressing the result in kilojoules leaves only one option standing.

Therefore, the correct answer is -7.48 kJ.