MT1-cheatsheet

2046 a ideal gases do not interact so each gas expands

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Unformatted text preview: 6 J/gal)(1 gal)/(7.2 104 J/s) = 1.81 103 s = 30 min. 20.6 (a) Work is positive for an expansion. Because the work done is represented by the area under the PV curve, Wbc > Wac. The net work is the sum of the works done in each leg. For a positive net work, Wbc > 0, so we have an expansion from b to c, and the path must be clockwise. P b Tb = Tc (b) We find the ratio of volumes from the ideal gas equation: a c (Pc/Pa)/(Vc/Va ) = Tc/Ta ; V (1)(Vc/Va ) = Tc/Ta , so Vc/Va = Vc/Vb = Tc/Ta = Tb /Ta . 0 The work done during the isothermal expansion is Wbc = nRTb ln(Vc/Vb) = nRTb ln(Tb/Ta) = (1.0 mol)(8.315 J/mol K)(423 K) ln(423 K/273 K) = 1.54 103 J. The work done during the constant pressure compression is Wca = Pa(Va Vc) = PaVa[1 (Vc/Va)] = nRTa [1 (Vc/Va)] = (1.0 mol)(8.315 J/mol K)(273 K)[1 (423 K/273 K)] = 1.25 103 J. For the isothermal expansion Ubc = 0, so Qbc = Wbc = 1.54 103 J. For the constant volume compression Wab = 0, so Qab = Uab = nCV(Tb Ta) = (1.0 mol)8(8.315 J/mol K)(423 K 273 K) = 1.87 103 J. The efficiency is e = Wnet/Qadded = (Wbc + Wca)/(Qbc + Qab) = (1.54 103 J 1.25 103 J)/(1.54 103 J + 1.87 103 J) = 0.085 = 8.5%. 20.18 For the efficiencies of the engines, we have TL2 QL2 Engine W2 QL1 = QH2 TH2 TL1 Engine W1 QH1 TH1 = 0.65eCarnot = 0.60[1 (TL1/TH1)] = 0.65{1 [(703 K)/(953 K)]} = 0.171; e2 = 0.65eCarnot = 0.60[1 (TL2/TH2)] = 0.65{1 [(553 K)/(688 K)]} = 0.128. Because coal is burned to produce the input heat to the first engine, we need to find Q H1. We relate W1 to QH1 from the efficiency: e1 = W1/QH1 , or W1 = 0.171QH1 . Because the exhaust heat from the first engine is the input heat to the second engine, we have e2 = W2/QH2 = W2/QL1 , or W2 = 0.128QL1 . For the first engine we know that QL1 = QH1 W1 , so we get W2 = 0.128QL1 = 0.128(QH1 0.171QH1) = 0.106QH1 . For the total work, we have W = W1 + W2 = 0.171QH1 + 0.106QH1 = 0.277QH1 . When we use the rate at which this work is done, we get 900 106 W = 0.277(QH1/t), which gives QH1/t = 3.25 109...
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This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at Berkeley.

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