60.) A three-step cycle is undergone by 4.5 mol of an ideal diatomic gas: (1) th
ID: 2111609 • Letter: 6
Question
60.) A three-step cycle is undergone by 4.5 mol of an ideal diatomic gas: (1) the temperature of the gas is increased from 150 K to 440 K at constant volume; (2) the gas is then isothermally expanded to its original pressure; (3) the gas is then contracted at constant pressure back to its original volume. Throughout the cycle, the molecules rotate but do not oscillate. What is the efficiency of the cycle?Explanation / Answer
suffixes: 0 = initial, N (1-3) = after phase N given: n = 3.4; T0 = 160; T1 = 400 ==> T1-T0 = 240 Cv, diatomic, no vibration = 5R/2 = 20.7861804962158 J/mol-K Phase 1 (isochoric) Q1 = Cv*n*(T1-T0) = 16961.5232849121 J W1 = 0 P/T = nR/V = constant ==> P1/P0 = T1/T0 = 400/160 = 2.5 Phase 2 (isothermal) given: P2 = P0 T2 = T1 = 400 PV = nRT = constant ==> V2/V1 = P1/P2 = P1/P0 = 2.5 W2 = P1*V1*ln(V2/V1) = nRT*ln(2.5) = 3.4*8.314472*400*ln(2.5) = 10361.1243896218 J Q2 = W2 Phase 3 (isobaric) given: Since it'a a cycle, V3 = V0: P = P0: T3 = T0 W3 = P(V3-V2) = n*R*(T3-T2) = n*R*(T0-T1) = 3.4*8.314472*(-240) = -6784.60931396484 J Q3 = W*Cp/R, where Cp = Cv+R, thus Q = W*(1+Cv/R) = -6784.60931396484*3.5 = -23746.132598877 J Totals Source heat Qin = Q1+Q2 = 27322.6476745339 J Waste heat Qout = -Q3 = 23746.132598877 J W = W1+W2+W3 = Qin-Qout = 3576.51507565696 J eta = W/Qin = 0.1309
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