A cycle followed by an engine (made of one mole of perfect gas in a cyclinder with a piston) is shown in (figure)
A to B: volume constant B to C: adiabatic
C to D: volume constant D to A: adiabatic
`V_(c)=V_(D)= 2V_(A)= 2V_(B)`
(a) In which part of the cycle heat is supplied to the engine fron outside?
(b) In which part of the cycle heat is being given to the surrounding by the engine?
(c ) What is the work done by the engine in one cycle? Write your answer in term of `P_(A), P_(B), V_(A)`.
(d) What is the efficiency of the engine? `(gamma= 5//3 "for the gas")`, `(C_(v)=3/2R "for one mole")`

Explanation

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)]`