Figure below shows two paths that may be taken by a gas to go from a state A to a state C.In process A B, 400 J of heat is added to the system and  process B C, 100 J of heat is added to the system. The heat absorbed by the system in the process A C will be

As initial and final points are same soΔUABC=ΔUAC AB is isochoric process.  ΔWAB=0ΔQAB=ΔUAB=400 J BC is isobaric process.  ΔQBC=ΔUBC+ΔWBC100=ΔUBC+6×1044×10-3-2×10-3100=ΔUBC+12×10ΔUBC=100-120=-20 J As, ΔUABC=ΔUAC ΔUAB+ΔUBC=ΔQAC-ΔWAC400-20=ΔQAC-2×104×2×10-3+12×2×10-3×4×104380=ΔQAC-(40+40),ΔQAC=380+80=460 J