Here are the solutions to your questions: --- 1.1 The work done during adiabatic compression The process from point 1 to 2 is adiabatic compression ($PV^\gamma = C$). Given: $P_1 = 100 \text{ kPa}$, $V_1 = 0.082 \text{ m}^3$, $V_2 = 0.0054 \text{ m}^3$, $\gamma = 1.4$. Step 1: Calculate the pressure at point 2 ($P_2$) using the adiabatic relation. $$P_1V_1^\gamma = P_2V_2^\gamma$$ $$P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma$$ $$P_2 = 100 \text{ kPa} \left(\frac{0.082 \text{ m}^3}{0.0054 \text{ m}^3}\right)^{1.4}$$ $$P_2 = 100 \text{ kPa} (15.185)^{1.4}$$ $$P_2 = 100 \text{ kPa} \times 39.810$$ $$P_2 = 3981 \text{ kPa}$$ Step 2: Calculate the work done during the adiabatic compression. The work done for a reversible adiabatic process is given by: $$W_{1-2} = \frac{P_2V_2 - P_1V_1}{1-\gamma}$$ $$W_{1-2} = \frac{(3981 \text{ kPa})(0.0054 \text{ m}^3) - (100 \text{ kPa})(0.082 \text{ m}^3)}{1-1.4}$$ $$W_{1-2} = \frac{21.4974 \text{ kJ} - 8.2 \text{ kJ}}{-0.4}$$ $$W_{1-2} = \frac{13.2974 \text{ kJ}}{-0.4}$$ $$W_{1-2} = -33.2435 \text{ kJ}$$ The negative sign indicates work is done on the gas. $$\boxed{-33.24 \text{ kJ}}$$ 1.2 The work done during the constant pressure process The process from point 3 to 4 is a constant pressure process. Given: $P_3 = 6200 \text{ kPa}$, $V_3 = 0.0054 \text{ m}^3$, $V_4 = 0.0125 \text{ m}^3$. Step 1: Calculate the work done for the constant pressure process. $$W_{3-4} = P_3(V_4 - V_3)$$ $$W_{3-4} = 6200 \text{ kPa} (0.0125 \text{ m}^3 - 0.0054 \text{ m}^3)$$ $$W_{3-4} = 6200 \text{ kPa} (0.0071 \text{ m}^3)$$ $$W_{3-4} = 44.02 \text{ kJ}$$ $$\boxed{44.02 \text{ kJ}}$$ 1.3 The volume at point 3 and point 5 in litres From the diagram: $V_3 = 0.0054 \text{ m}^3$ and $V_5 = 0.082 \text{ m}^3$. To convert from cubic meters to litres, multiply by 1000 ($1 \text{ m}^3 = 1000 \text{ L}$). Step 1: Convert $V_3$ to litres. $$V_3 = 0.0054 \text{ m}^3 \times 1000 \text{ L/m}^3 = 5.4 \text{ L}$$ Step 2: Convert $V_5$ to litres.