As shown in (figure)
(a) Heat is supplied to the engine when the cycle of changes goes from A to B. Pressure is increasing. So does the temperature.
(b) Heat is being given to the surrounding by the engine in the part C to D. When pressure decreases aand so does the temperature.
(c ) To calculate work done by the engine in one cycle, we calculate separately work done is the four parts of cycle
`W_(AB) = int_(V_(A))^(V_(B))pdV=0 , W_(CD) = int_(V_(C))^(V_(D))pdV=0`
`W_(BC) = int_(V_(B))^(V_(C))pdV=K int_(V_(B))^(V_(C))(dV)/(V^(gamma))=K/(1-gamma)[V^(1)-gamma]_(V_(B))^(V_(C))= 1/(1-gamma)[PV]_(V_(B))^(V_(C))= ((P_(C)V_(C)-P_(B)V_(B)))/(1-gamma)`
Similarly, `W_(DA)=(P_(A)V_(A)-P_(D)V_(D))/(1-gamma)`
As B and C lie on same adiabatic curve, therefore,
`P_(B)V_(B)^(gamma)= P_(C)V_(C)^(gamma) or P_(C)=P_(B)((V_(B))/(V_(C)))^(gamma)= P_(B)(1/2)^(gamma)= 2^(-gamma)P_(B)`
Similarly, `P_(D)=2^(-gamma)P_(A)`
Total work done by the engine in one cycle
`W=W_(AB)+W_(BC)+W_(CD)+W_(DA)= W_(BC)+W_(DA)= ((P_(C)V_(C)-P_(B)V_(B)))/(1-gamma)+(P_(A)V_(A)-P_(D)V_(D))/(1-gamma)`
`W=1/(1-gamma)[2^(-gamma)P_(B)(2V_(B))-P_(B)V_(B)+P_(A)V_(A)-2^(-gamma)P_(A)(2V_(A))]`
`W=1/(1-gamma)[(2^(1-gamma)-1)P_(B)V_(B)+(2^(1-gamma)-1)P_(A)V_(A)]`
`W=((2^(1-gamma)-1))/(1-gamma)(P_(B)-P_(A))V_(A)= ((1-2^(1-5//3)))/((5//3-1))(P_(B)-P_(A))V_(A) (:. V_(B)=V_(A))`
`W=3/2[1-(1/2)^(2//3)](P_(B)-P_(A))V_(A)`
(d) Efficiency, `eta= ("Net work done"//"cycle")/("Heat energy supplied"//"cycle")=W/(Q_(AB)`....(ii)
As `W_(AB)=0, therefore, Q_(AB)= U_(AB)= 3/2nRT(T_(B)-T_(A))= 3/2xx1(P_(B)-P_(A))V_(A)`
From (ii), `eta=(3//2[1-(1/2)^(2//3)](P_B-P_(A))V_(A))/(3/2(P_(B)-P_(A))V_(A))=[1-(1/2)^(2//3)]`