A thermodynamic process is shown in the given figure. The pressures and volumes corresponding to some points in the figure are : P_A = 3 x 10⁴ Pa, P_B = 8 x 10⁴ Pa and V_A = 2 x 10^-3 m³, VD = 5 x 10^-3 m³.

In the process AB, 600 J of heat is added to the system. The change in internal energy of the system in process AC would be : 

560 J

By the graph :

W_AB = 0 and

W_BC = 8 x 10⁴ (5-4) x 10^-3 = 240 J

W_AC = W_AB + W_BC = 0 + 240 = 240 J

Now, \(\Delta Q_{AC} = \Delta Q_{AB} + \Delta Q_{BC} \)

\(\Delta Q_{AC}\) = 600 + 200 = 800 J

By First Law of Thermodynamics : \(\Delta Q_{AC} = \Delta U_{AC} + \Delta W_{AC}\)

800 = \(\Delta U_{AC} + 240 \)

\(\Rightarrow \Delta U_{AC} = 560 J\)