Figure below shows two paths that may be taken by a gas to go from a state $$A$$ to a 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

Solution :
Since, initial and final points are same
So, $$\Delta {U_{A \to B \to C}} = \Delta {U_{A \to C}}\,......\left( {\text{i}} \right)$$
Also $$A \to B$$  is isochoric process
So $$d{W_{A \to B}} = 0$$
and $$dQ = dU + dW$$
So, $$d{Q_{A \to B}} = d{U_{A \to B}} = 400\,J$$

Next $$B \to C$$  is isobaric process
$$\eqalign{ & {\text{So,}}\,\,\,d{Q_{B \to C}} = d{U_{B \to C}} + d{W_{B \to C}} \cr & = d{U_{B \to C}} + p\Delta {V_{B \to C}} \cr & \Rightarrow 100 = d{U_{B \to C}} + 6 \times {10^4}\left( {2 \times {{10}^{ - 3}}} \right) \cr & \Rightarrow d{U_{B \to C}} = 100 - 120 = - 20\,J \cr} $$
From Eq. (i),
$$\eqalign{ & \because \Delta {U_{A \to B \to C}} = \Delta {U_{A \to C}} \cr & \Rightarrow \Delta {U_{A \to B}} + \Delta {U_{B \to C}} = d{Q_{A \to C}} - d{W_{A \to C}} \cr & \Rightarrow 400 + ( - 20) = d{Q_{A \to C}} - \left( {p\Delta {V_A} + } \right.{\text{Area}}\,{\text{of}}\left. {\Delta ABC} \right) \cr & \Rightarrow d{Q_{A \to C}} = 380 + \left( {2 \times {{10}^4} \times 2 \times {{10}^{ - 3}} + \frac{1}{2} \times 2 \times {{10}^{ - 3}} \times 4 \times {{10}^4}} \right) \cr & = 380 + \left( {40 + 40} \right) \cr & d{Q_{A \to C}} = 460\,J \cr} $$