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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 ) |
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Answer» Solution :(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)]` |
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