Following figure shows two processes A and B for a gas. If ${\mathrm{\Delta Q}}_{\mathrm{A}}$ and ${\mathrm{\Delta Q}}_{\mathrm{B}}$ are the amount of heat absorbed by the system in two cases and [${\mathrm{\Delta U}}_{\mathrm{A}}$ and ${\mathrm{\Delta U}}_{\mathrm{B}}$] are changes in internal energies respectively then

According to the first law of thermodynamics,
Heat supplied $\left(\mathrm{\Delta Q}\right)$ = Work done (W)+change in internal energy of the system$\left(\mathrm{\Delta U}\right)$ ${\mathrm{\Delta Q}}_{\mathrm{A}}={\mathrm{\Delta U}}_{\mathrm{A}}+{\mathrm{W}}_{\mathrm{A}}$
Similarly, for process B, ${\mathrm{\Delta Q}}_{\mathrm{B}}={\mathrm{\Delta U}}_{\mathrm{B}}+{\mathrm{W}}_{\mathrm{B}}$
Now, we know that, work done for a process = area under it’s p-V curve

Thus, it is clear from the above graphs, W_A > W_B Also, since the initial and final state are same in both process, so ${\mathrm{\Delta U}}_{\mathrm{A}}={\mathrm{\Delta U}}_{\mathrm{B}}$ So from Eqs (i) and (ii), We can conclude that ${\mathrm{\Delta Q}}_{\mathrm{A}}>{\mathrm{\Delta Q}}_{\mathrm{B}}$