Figure shows a cyclic process ABCDBEA. performed on an ideal cycle. lf $\mathrm{If} {\mathrm{P}}_{\mathrm{A}} = 2 \mathrm{atm}, {\mathrm{P}}_{\mathrm{B}} = 5 \mathrm{atm} \mathrm{and} {\mathrm{P}}_c = 6 \mathrm{atm}. {\mathrm{V}}_{\mathrm{E}}-{\mathrm{V}}_{\mathrm{A}}= 20 \mathrm{litre}$, find the work done by the gas in the complete, process (l atm. pressure = $1\times {10}^5$ Pa)

Solution:

The complete cyclic process can be visualized as made up of two cycles, i.e., cycle AEBA (clockwise) and cycle BDCB (counter-clockwise).
Work done by the gas during the cycle AEBA should be positive,

${\mathrm{W}}_1 = \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{loop} \mathrm{in} \mathrm{the} \mathrm{P}-\mathrm{V} \mathrm{diagram}$

      = $\frac{1}{2}\left(\mathrm{base}\right)\left(\mathrm{altitude}\right) = \frac{1}{2}({\mathrm{V}}_{\mathrm{E}}-{\mathrm{V}}_{\mathrm{A}})({\mathrm{P}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{A}})$

       = $\frac{1}{2}(20\times {10}^{-3}{\mathrm{m}}^3)(5-2)\times {10}^5 {\mathrm{Nm}}^{-2} = 3\mathrm{kJ}$

Work done by the gas the cycle BDCB should be negative,

${\mathrm{W}}_2 = -(\mathrm{area} \mathrm{of} \mathrm{loop} \mathrm{BDCB})$

Now evidently, triangles ABE and BCD are similar, the corresponding angles being equal (i.e.,

 $<\mathrm{EBA} = <\mathrm{DBC}, <\mathrm{EAB} = <\hspace{0.17em}\mathrm{BCD}$

$\frac{({\mathrm{V}}_{\mathrm{E}}-{\mathrm{V}}_{\mathrm{A}})}{({\mathrm{P}}_{\mathrm{B}}-{\mathrm{P}}_{\mathrm{A}}) } = \frac{({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{D}})}{({\mathrm{P}}_{\mathrm{C}}- {\mathrm{P}}_{\mathrm{B}})} \Rightarrow \frac{20}{(5-2)} = \frac{({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{D}})}{(6-5)}$

$\Rightarrow ({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{D}}) = \frac{20}{3}\mathrm{litre}$

${\mathrm{W}}_2 = \frac{1}{2}\times \frac{20}{3}\times {10}^{-3}\times 1\times {10}^5 = -0.33 \mathrm{kJ}$

$\therefore \mathrm{Total} \mathrm{work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{gas}$

     $∆\mathrm{W} = {\mathrm{W}}_1+{\mathrm{W}}_2 = 3 \mathrm{kJ} -0.33 \mathrm{kJ} = 2.67 \mathrm{kJ}$