For isobaric expansion, the work done (\( W \)) is given by the equation:
\( W = P \Delta V \)
Where \( P = 2 \, \text{bar} = 2 \times 10^5 \, \text{Pa} \), and the volume change \( \Delta V = V_2 - V_1 = 0.6 \, \text{m}^3 - 0.5 \, \text{m}^3 = 0.1 \, \text{m}^3 \).
So,
\( W = 2 \times 10^5 \, \text{Pa} \times 0.1 \, \text{m}^3 = 20 \, \text{kJ}. \)
Thus, the work transfer is \( +20 \, \text{kJ} \).