Find the efficiency of the reversible engine, whose cycle is shown in the above T - S plot :

Formula for efficiency $\mathrm{\eta }=\frac{{\mathrm{Q}}_{\mathrm{s}}-{\mathrm{Q}}_{\mathrm{R}}}{{\mathrm{Q}}_{\mathrm{s}}}\dots ....\left(1\right)$

Where Q_S is heat supplied and Q_R is heat rejected.

$\because \mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}\text{ and }\mathrm{G}$ is a state function. Therefore, change in enthalpy which is equal to heat supplied under constant pressure will be equal to area under curve.
From the diagram, Q_S is area under the sloping line

$\begin{array}{l}=\left(\frac{1}{2}\times \left(2{\mathrm{S}}_{\mathrm{o}}-{\mathrm{S}}_{\mathrm{o}}\right)\times \left(2{\mathrm{T}}_{\mathrm{o}}-{\mathrm{T}}_{\mathrm{o}}\right)\right)\\ +\left(2{\mathrm{S}}_{\mathrm{o}}-{\mathrm{S}}_{\mathrm{o}}\right)\times {\mathrm{T}}_{\mathrm{o}}\\ =\frac{3}{2}{\mathrm{S}}_{\mathrm{o}}{\mathrm{T}}_{\mathrm{o}}\end{array}$

and Q_R is area under the horizontal line

$\begin{array}{l}=\left(2{\mathrm{S}}_{\mathrm{o}}-{\mathrm{S}}_{\mathrm{o}}\right)\times {\mathrm{T}}_{\mathrm{o}}\\ ={\mathrm{S}}_{\mathrm{o}}{\mathrm{T}}_{\mathrm{o}}\end{array}$

Now substituting the value of Q_S and Q_R in (1) we get,

$\mathrm{\eta }=\frac{\frac{3}{2}{\mathrm{S}}_{\mathrm{o}}{\mathrm{T}}_0-{\mathrm{S}}_0{\mathrm{T}}_{\mathrm{o}}}{\frac{3}{2}{\mathrm{S}}_0{\mathrm{T}}_0}$