To solve this problem, we will use the first law of thermodynamics, which states:$\\ \Delta U = Q - W \\$Where:$\\$• $\Delta U$ is the change in internal energy.$\\$• $Q$ is the heat added to the system.$\\$• $W$ is the work done by the system.$\\$Since the initial and final states (A and C) are the same for both paths,$\\ \Delta U_{AC} = \Delta U_{ABC} \\$For path ABC:$\\ Q_{ABC} = Q_{AB} + Q_{BC} = 400$ J$+ 100$ J$= 500$ J$\\$The work done in path ABC is the area under the curve, which is the area of the rectangle:$\\ W_{ABC} = (P_B - P_A) \times (V_C - V_A) \\ W_{ABC} = (6 \times 10^4 \,$Pa$- 2 \times 10^4 \,$Pa$) \times (4 \times 10^{-3} \,$m$^3 - 2 \times 10^{-3} \,$m$^3) \\ W_{ABC} = 4 \times 10^4 \,$Pa$\times 2 \times 10^{-3} \,$m$^3 \\ W_{ABC} = 80$ J$\\$Using the first law of thermodynamics for path ABC:$\\ \Delta U_{ABC} = Q_{ABC} - W_{ABC} \\ \Delta U_{ABC} = 500$ J$- 80$ J$= 420$ J$\\$For path AC, the work done is:$\\ W_{AC} = (P_C - P_A) \times (V_C - V_A) \\ W_{AC} = (6 \times 10^4 \,$Pa$- 2 \times 10^4 \,$Pa$) \times (4 \times 10^{-3} \,$m$^3 - 2 \times 10^{-3} \,$m$^3) \\ W_{AC} = 80$ J$\\$Using the first law of thermodynamics for path AC:$\\ \Delta U_{AC} = Q_{AC} - W_{AC} \\ 420$ J$= Q_{AC} - 80$ J$\\$Solving for $Q_{AC}: \\ Q_{AC} = 420$ J$+ 80$ J$= 500$ J$\\$Therefore, the heat absorbed by the system in the process AC is $500$ J.$\\$This corresponds to Option 1.