Initial and final points of both processes A and B are same. Therefore, internal energy in both the processes will be the same because internal energy is a state variable, independent of the path taken.

The area under the P-V curve gives the work done on the system. From the graph, it can be seen the area under the curve for process A is more than the area under the curve for process B., therefore, work done on the system in process A ΔW₁ is more than work done on the system in process B ΔW₂

ΔW₁ > ΔW₂ …..(i)

According to First law of thermodynamics,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied/extracted to/from the system

ΔU=change in internal energy

ΔW=work done by/on the system

For process A

ΔQ₁=ΔU+ΔW₁ …(ii)

For process B

ΔQ₂=ΔU+ΔW₂ …(iii)

From equation (i) ,(ii) and (iii) it is clear that ΔQ₁ > ΔQ_2.