Figure below shows two paths that may be taken by a gas to go from a state A to state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be 460 J, 300 J, 380 J, or 500 J?

In process A to B change in internal energy is ΔUAB=ΔQAB−ΔWAB
process AB is Isochoric, therefore heat added is equal to change in internal energy.
ΔUAB=ΔQAB=400J
process BC change in internal energy is ΔUBC=ΔQBC−ΔWBC
ΔUBC=100−PΔV
ΔUBC=100−(4×10⁴)(4×10⁻⁶−2×10⁻⁶)
ΔUBC=100−(4×10⁴)(2×10⁻⁶)
ΔUBC=100−80J=20J
total change in internal energy will be :
ΔUAC=400−20=380J
heat added in process AC will be:
ΔQAC=ΔUAC+ΔWAC
ΔQAC=380+area of P−V diagram
ΔQAC=380+1/2(6×10⁴+2×10⁴)2×10⁻⁶
ΔQAC=380+80=460J