Let's solve the problem step by step. ## **Step 1: Understand the process on the PV diagram** - **Process 1–2:** From (P1, v1) = (500 kPa, .5 m³/kg) to (P2, v2) = (100 kPa, 1 m³/kg) — a straight line (linear process on PV diagram). - **Process 2–3:** From (P2, v2) = (100 kPa, 1 m³/kg) to (P3, v3) = (400 kPa, 1 m³/kg) — vertical line (isochoric process, volume is constant). Given: - Mass, \( m = 4.000 \) kg --- ## **Step 2: Work done in each process** ### **Process 1–2 (Linear on PV diagram):** The work done (\( W_{12} \)) is the area under the line between 1 and 2: \[ W_{12} = m \times \text{Area under 1-2 line} \] For a straight line: \[ \text{Area} = \text{Area}_{\text{rectangle}} + \text{Area}_{\text{triangle}} \] Where: - Rectangle: Height = lower pressure (100 kPa), Base = (1 - .5) m³/kg = .5 m³/kg - Triangle: Height = (500 - 100) kPa = 400 kPa, Base = .5 m³/kg So, \[ \text{Area}_{\text{rect}} = 100 \times .5 = 50 \ \text{kPa·m}^3/\text{kg} \] \[ \text{Area}_{\text{tri}} = \frac{1}{2} \times .5 \times (500 - 100) = .25 \times 400 = 100 \ \text{kPa·m}^3/\text{kg} \] \[ \text{Total area} = 50 + 100 = 150 \ \text{kPa·m}^3/\text{kg} \] \[ W_{12} = m \times \text{Total area} \] \[ W_{12} = 4.000 \times 150 = 600 \ \text{kPa·m}^3 \] Since 1 kPa·m³ = 1 kJ, \[ W_{12} = 600 \ \text{kJ} \] --- ### **Process 2–3 (Isochoric, constant volume):** No work is done during a constant volume process: \[ W_{23} = \] --- ### **Step 3: Total Work** \[ W_{1-3} = W_{12} + W_{23} = 600 + = 600 \ \text{kJ} \] --- ## **Final Answer** \[ \boxed{600 \ \text{kJ}} \] --- ### **Summary Table** | Process | Work (kJ) | |-----------|-----------| | 1 → 2 | 600 | | 2 → 3 | | | **Total** | **600** | **The total work for process 1–3 is \(\boxed{600}\) kJ.**