A cycle followed by an engine (made of one mole of perfect gas in a cylinder 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 from 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

(a) For the process A to B, volume is constant
So, dV = 0 `therefore` dW = 0
From first law of thermodynamics in,
dQ = dU + đW, dW = 0
`therefore` dQ=dU
Hence, in this process heat supplied is utilised to increases internal energy of the system.
As pressure is increased at constant volume (From Gay-Lussac.s law) i.e. P`prop` T and internal energy is proportional to temperature and hence internal energy increases with temperature `therefore dU gt 0`
`therefore` From dQ=dU, dQ `gt` 0 , so in process AB heat is supplied to the system.
(b)From dQ = dU, dQ > 0, so in process AB heat is supplied to the system. (b) In CD process, volume remains constant but pressure decreases. Hence, from P`prop` T, temperature also decreases so heat is being given to the surrounding by the engine.
(c) To calculate work done by the engine in one cycle , we should calculate work done in each part separately.
`therefore W_(AB)=int_A^P P dV=0` ...(1)
`W_(CD)=int_(V_C)^(V_D) P dV=0`...(2)
`W_(BC)=int_(V_B)^(V_C) P dV=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)` (adiabatic)
`=(P_C V_C -P_B V_B)/(1-gamma)`...(3)
Similarly , `W_(DA)=(P_AV_A-P_DV_D)/(1-gamma)` (adiabatic) ...(4)
B and C are along adiabatic curve BC,
`therefore P_B V_B^gamma =P_C V_C^gamma`
`therefore P_C=P_B(V_B/V_C)^gamma=P_B(1/2)^gamma =2^(-gamma)P_B`...(5)
Similarly, `P_D=2^(-gamma)P_A` ...(6)
Total work done by the engine in one cycle ABCDA
`W=W_(AB)+W_(BC)+W_(CD)+W_(DA)`
`=0+W_(BC)+0+W_(DA)`
`W=1/(1-gamma)(P_C V_C-P_B V_B)+1/(1-gamma)(P_AV_A-P_DV_D)`
`W=1/(1-gamma)[2^(-gamma)P_B(2V_B)-P_BV_B]+1/(1-gamma)[P_A V_A-2^(-gamma)P_A (2V_A)]`
`W=1/(1-gamma)[P_B V_B 2^(-gamma+1) - P_B V_B +P_A V_A -2^(-gamma+1)P_A V_A]`
`therefore W=1/(1-gamma) [P_B V_B (2^(gamma-1)-1) -P_A V_A (2^(1-gamma)-1)]` but `V_A=V_B`
`therefore W=(2^(1-gamma)-1)/(1-gamma)[P_B V_A-P_A V_A]`
`=(2^(1-gamma)-1)/(1-gamma)V_A(P_B-P_A)`
Putting `gamma=5/3` and simplifying
`W=3/2 [1-(1/2)^(2/3)](P_B-P_A)V_A`
(D) Efficiency `eta="Net work"/"Heat supplied"`
`=(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)]`