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Question
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
VC = VD = 2VA = 2VB
- In which part of the cycle heat is supplied to the engine from outside?
- In which part of the cycle heat is being given to the surrounding by the engine?
- What is the work done by the engine in one cycle? Write your answer in term of PA, PB, VA.
- What is the efficiency of the engine?
(γ = `5/3` for the gas), (Cv = `3/2` R for one mole)
Solution
a. For process AB (which is the isochoric process), volume is constant.
So, dV = 0
⇒ dW = 0
dQ = dU + dW = dU
⇒ dQ = dU = Change in internal energy
Hence, in this process heat supplied is utilised to increase, the internal energy of the system.
b. For the process CD (which is also an isochoric process), volume is constant but the pressure decreases. Hence, temperature also decreases (because PαT) so heat is given to the surroundings.
c. To calculate work done by the engine in one cycle, we calculate work done in each part separately.
(i) `W_(AB) = int_A^B PdV` = 0
(ii) `W_(CD) = int_C^D PdV` = 0 ....(As V is constant, dV = 0)
(iii) `W_(BC) = int_B^C PdV`
= `K int_B^C (dV)/(V^γ)`
= `K int_(V_B)^(V_C) V^-γ dV`
= `K/(1 - γ) [V^(1 - γ)]_(V_B)^(V_C)`
= `(K[V_C^(1 - γ) - V_B])/(1 - γ)` .....(PVγ = K for an adiabatic change)
= `([(P_CV_C^γ)(V_C^(1 - γ)) - (P_BV_B^γ)(V_B^(1 - γ))])/((1 - γ))`
Similarly, `W_(DA) = int_(V_D)^(V_A) PdV`
= `1/((1 - γ)) (P_AV_A - P_DV_D)` ......[∵ BC is an adiabatic process]
Since B and C lies on adiabatic curve BC,
∴ `P_BV_B^γ = P_CV_C^γ`
`P_C = P_B(V_B/V_C)^γ`
= `P_B(1/2)^γ`
= `2^(-γ)P_B`
Similarly, `P_D = 2^(-γ)P_A`
Total work done by the engine in one cycle ABCDA
`W = W_(AB) + W_(BC) + W_(CD) + W_(DA) = W_(BC) + W_(DA)`
= `((P_CV_C - P_BV_B))/(1 - γ) + ((P_AV_A - P_DV_D))/(1 - γ)`
`W = 1/(1 - γ) [2^(-γ) P_B (2V_B) - P_BV_B + P_AV_A - 2^(-γ) P_B (2V_B)]`
= `1/(1 - γ) [P_AV_B (2^(-γ + 1) - 1) - P_AV_A (2^(-γ + 1) - 1)`
= `1/(1 - γ) (2^(1 - γ) - 1) (P_B - P_A)V_A`
= `3/2 [1 - (1/2)^(2/3)](P_B - P_A)V_A`
d. Heat (Q) is supplied to the engine only from A to B. Thus Q = ΔQAB = ΔUAB = nCVΔT
= `3/2 R(T_B - T_A)` .....[As CV = (3/2)R and ΔT = TB – TA]
= `P_CV_C^γ = P_BV_B^γ`,
`P_C = P_B(V_B/V_C)^γ`
= `P_B(1/2)^γ`
= `P_B2^-γ` .....(As VB/VC = 1/2)
Thus, `P_CV_C^γ = (P_B2^-γ)(2V_B) = 2^(1 - γ)P_BV_B`
= `3/2 (P_BV_B - P_AV_A)`
= `3/2 (P_B - P_A)V_A` .....(PV = RT and VB = VA)
The efficiency of the engine,
η = `W/Q = (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)]`
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