Here is the solution for question 1.2: 1.2 The work done during the constant pressure process The constant pressure process is from state 3 to state 4. From the P-V diagram: Pressure at state 3 and 4, $P_3 = P_4 = 6200 \text{ kPa}$ Volume at state 3, $V_3 = 0.0054 \text{ m}^3$ Volume at state 4, $V_4 = 0.0125 \text{ m}^3$ The formula for work done during a constant pressure process is: $$W = P \times (V_{final} - V_{initial})$$ In this case, $W_{3-4} = P_3 \times (V_4 - V_3)$. Step 1: Substitute the values into the formula. $$W_{3-4} = 6200 \text{ kPa} \times (0.0125 \text{ m}^3 - 0.0054 \text{ m}^3)$$ Step 2: Calculate the change in volume. $$W_{3-4} = 6200 \text{ kPa} \times (0.0071 \text{ m}^3)$$ Step 3: Perform the multiplication. $$W_{3-4} = 44.02 \text{ kJ}$$ The work done during the constant pressure process (3-4) is $\boxed{\text{44.02 kJ}}$.