Unformatted text preview: iverse we have Suniverse = Swater + S surr = + 1.45 kcal/K + ( 1.45 kcal/K) = 0. This must be so for a reversible process. (d) If the process were irreversible, Tsurr > 100C, so Ssurr would be less negative. Thus Suniverse > 0. 20.37
The total rate of the entropy change is Stotal/t = S source/t + Swater/t = ( Q/t)/Tsource + (+ Q/t)/Twater = ( 7.50 cal/s)/(513 K) + (+ 7.50 cal/s)/(300 K) = + 0.0104 cal/K s. 20.39
The heat flow to freeze the water is Q = m waterL = (2.5 kg)(3.33 105 J/kg ) = 8.33 105 J. We find the final temperature of the ice from Q = m icecice Tice ; 8.33 105 J = (450 kg)(2100 J/kg C)[T ( 15C)], which gives T = 14.12C. The heating of the ice does not occur at constant temperature. Because the temperature change is small, to estimate the entropy change, we will use the average temperature: Tice,av = !(Tice + T) = ![( 15C) + ( 14.12C)] = 14.56C; The total entropy change is S = Swater + Sice = ( Q/Twater) + (+ Q/Tice,av) = [( 8.33 105 J)/(273.2 K)] + [(+ 8.33 105 J)/(258.6 K)] = 172 J/K = + 1.7 102 J/K. 20.45
(a) For the isothermal process we find the ratio of pressures from the ideal gas equation: (P2iV2/P1V1) = T2/T1= 1, so P2i/P1 = V1 /V2 = 2. For the adiabatic process, P1V1 = P2aV2, so P2a/P1 = (V1/V2) = 2. Because > 1, P2a/P1 > 2, so the final pressure is greater for the adiabatic process. (b) For the isothermal process we have U = 0; Qi = Wi. Thus Si = Wi/T = [nRT ln(V2/V1)]/T = nR ln(1/2) = nR ln 2. For the adiabatic process we have Qa = 0. Thus Sa = 0. (c) Because each process is reversible, the energy change of the universe is zero. For the isothermal process we have Ssurr,i = Si = nR ln 2. For the adiabatic process we have Ssurr,a = Sa = 0. 20.46
(a) Ideal gases do not interact, so each gas expands to twice the volume at constant temperature. Thus we have SN = SA = nR ln(V2/V1) = nR ln(2) = nR ln 2. The total change of the system is Ssys = SN + SA = 2nR ln 2 = 2(1 mol)(8.315 J/mol K) ln 2 = 11.5 J/K. (b) Because there is no interaction with the environment, Senv = 0. (c...
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