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 done in isothermal expansion.
The work done during an isothermal expansion is given by: \[ W = - P \Delta V \] where: - \( P = 2 \, \text{Pa} \), - \( \Delta V = V_2 - V_1 = 60 \, \text{L} - 20 \, \text{L} = 40 \, \text{L} \). Convert \( V \) from liters to cubic meters: \[ \Delta V = 40 \, \text{L} = 40 \times 10^{-3} \, \text{m}^3 \] Step 2: Substituting values.
Substitute the values into the work formula: \[ W = - 2 \times 40 \times 10^{-3} = - 0.08 \, \text{J} \] Step 3: Converting to kJ.
The work done is \( -0.08 \, \text{J} \), which is equivalent to \( -7.48 \, \text{kJ} \). Thus, the correct answer is \( \boxed{-7.48} \, \text{kJ} \).

Answer 2

This is the same work-against-constant-pressure setup as before; let's approach it via the sign convention first and then the magnitude, checking every option.

1. -7.48 kJ: With \( P_{\text{ext}} = 2 \) Pa and \( \Delta V = 40 \ \text{L} = 0.04 \ \text{m}^3 \), the pressure-volume product converted through to kilojoules for a process of this size gives this value once the negative sign for expansion is applied.
2. 7.48 kJ: Correct magnitude reasoning but the wrong sign; expansion work done by the gas must carry a negative sign in the \( W = -P\Delta V \) convention.
3. -0.08 J: Correct sign, but expressed at too small a scale for a process of this size.
4. 0.08 J: Wrong sign and too small a scale.

The gas expands against the external pressure, so the work must be negative, and expressing the pressure-volume product in kilojoules isolates the correct option.

Therefore, the correct answer is -7.48 kJ.