An ideal gas with $\mathrm{\gamma }$ = 1.5 is initially at state A where its pressure, temperature and volume are, P_A, T_A & V_A. It is taken first through an isobaric expansion to state (P_B, T_B, V_B). Then, it is taken through an adiabatic expansion to state C (P_C, T_C, V_C). The work done during the processes AB and BC are equal i.e., W_AB = W_BC = W. Also, P_C = 8/27 P_A. Then

Work done during isobaric expansion = P_A(V_B - V_A)
Work done during adiabatic expansion $=\frac{{\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{C}}{\mathrm{V}}_{\mathrm{C}}}{\mathrm{\gamma }-1}=2\left({\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{C}}{\mathrm{V}}_{\mathrm{C}}\right)$
Work is equal $\Rightarrow {\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}=2{\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}-2{\mathrm{P}}_{\mathrm{C}}{\mathrm{V}}_{\mathrm{C}} .........\left(\mathrm{i}\right)$
Now, ${\mathrm{P}}_{\mathrm{C}}=\frac{8}{27}{\mathrm{P}}_{\mathrm{A}}=\frac{8}{27}{\mathrm{P}}_{\mathrm{B}} \Rightarrow$using ${\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}^{\mathrm{\gamma }}={\mathrm{P}}_{\mathrm{C}}{\mathrm{V}}_{\mathrm{C}}^{\mathrm{\gamma }},$ we get ${\mathrm{V}}_{\mathrm{C}}=\frac{9}{4}{\mathrm{V}}_{\mathrm{B}}$
Putting this, P_B = P_A and ${\mathrm{P}}_{\mathrm{C}}=\frac{8}{27}{\mathrm{P}}_{\mathrm{A}}\text{ in (i),}$

${\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}=2{\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}-2\left(\frac{8}{27}{\mathrm{P}}_{\mathrm{A}}\right)\left(\frac{9}{4}{\mathrm{V}}_{\mathrm{B}}\right)$

$\Rightarrow {\mathrm{V}}_{\mathrm{B}}=3{\mathrm{V}}_{\mathrm{A}}$

So, ${\mathrm{V}}_{\mathrm{C}}=\frac{9}{4}{\mathrm{V}}_{\mathrm{B}}=\frac{27}{4}{\mathrm{V}}_{\mathrm{A}}$

$\Rightarrow \text{ Work, }\mathrm{W}={\mathrm{P}}_{\mathrm{A}}\left(3{\mathrm{V}}_{\mathrm{A}}\right)-{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}=2{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}$

${\mathrm{\Delta Q}}_{\mathrm{A}\to \mathrm{B}}={\mathrm{nC}}_{\mathrm{p}}\mathrm{\Delta T}=\frac{{\mathrm{C}}_{\mathrm{p}}}{\mathrm{R}}\left({\mathrm{P}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}\right)=6{\mathrm{P}}_{\mathrm{A}}{\mathrm{V}}_{\mathrm{A}}$