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Unformatted text preview: Solutions to Chemical and Engineering Thermodynamics, 3e 5.13 (Also available as a Mathcad worksheet. The Mathcad solution includes graphs.) (a) Use the ClausiusClapeyron equation H vap ln P2 P1 = R 1 T1  1 T2 1 and graphically taking slope, I find H vap ~ 42700 J mol . T (b) The vapor pressure is low enough that the ideal gas approximation should be validthus Plotting ln P vs. d ln Pvap ln P vap H vap = = dT T RT 2 either graphically or analytically, we find a f H vap ~ 313,600 J mol
5.14 (a) Start with Eqn. (5.46) f = P exp R 1 FV  RT I dPU ln f = F PV  1I dP S RT z H P K V P z H RT K T W
P P 0 0 but 1 d ( PV ) dV dZ dV dP =  =  so P PV V Z V f dZ PV dV Z P 1 = ( Z  1)  1 = ( Z  1)  ln   dV P Z =1 Z V = RT V 1 V = RT V ln z
Z V z F H I K V z FGH IJ K or f 1 = ( Z  1)  ln Z + P RT
V ln V = z FGH RT  P dV V IJ K (Eqn. 5.48) (b) Z = 1 + B(T ) RT B and P = 1+ V V V FG H IJ K
V V = ln f B B 1 =  ln 1 + + P V V RT
V FG H IJ K z LMNFGH RT RT RT B   V V V V IJ OP dV KQ = B 1 ZB  ln Z + B dV =  ln Z 2 V V V = V z Solutions to Chemical and Engineering Thermodynamics, 3e (c) vdW e.o.s. P=
V RT a PV V a  ; Z vdW = =  V b V2 RT V  b V 2 RT
V V = z FGH RT RT RT a V a  P dV =  + 2 dV = RT ln  V V V b V V  b V V = V = = RT ln V a  V b V IJ K z FGH IJ K V so ln f
vdW P V a  + ( Z  1)  ln Z V  b RTV Pb a RT P = ln Z  ln Z   + ( Z  1)  ln Z RT RT PV RT = ln F H I K = ( Z  1)  aP A  ln( Z  B) Z Pb . RT ( RT ) (d) PengRobinson equation of state. Start with where A =
2 and B = V V = z a FG RT  PIJ dV = z LM RT  RT + OPdV HV K N V V  b V (V + b) + b(V  b)Q L V + d1 + 2 ib OP V a = RT ln  lnM V  b 2 2 b N V + d1  2 ib Q M P LV + d1 + 2 ib OP Z a = RT ln  lnM Z  B 2 2 b NV + d1  2 ib Q M P
V V = [See solution to Problem 4.2 for integral]. Therefore f
PR ln P = (Z  1)  ln Z + ln V + 1+ 2 b Z a  ln Z  B 2 2 bRT V + 1 2 b a 2 2bRT ln = (Z  1)  ln( Z  B )  d i V + d1  2 ib LM d MN d i OP i PQ
f P will be gotten from V + 1+ 2 b 5.15 (a) liq vap fH 2 S = f H 2 S ; f vap = P f P , where the fugacity coefficient, a f corresponding states. 20 = 0.2237 89.42 255 + 273.15 . TC , H 2 S = 373.2 K Tr = = 0.8002 373.2 ZC , H 2 S = 0.284, which is reasonably close to 0.27 P , H 2 S = 89.42 bar Pr = C Solutions to Chemical and Engineering Thermodynamics, 3e f = 0.765 , fH 2 S = 20 0765 = 153 bar . . . P (b) For a liquid, from Eqn. (5.418) Fro m Fig. 5.41, f = P vap FfI H PK exp
sat LM NM z P P vap V dP RT OP QP
a f Pf
sat Since Pvap = 6.455 103 Pa at the temperature of interest, we will assume that Also, we will consider the liquid to be incompressible. Thus ~ 1. P vap z P V V dP = RT RT P vap z P dP = V P  P vap RT c h and fH 2 S = P vap exp so that Pressure, Pa fH 2 S, Pa 6,921 9,146 12,960 Reported 6,925 9,175 12,967 LMV c P  P MN RT
7 vap h OP = 6455 expL 0.018( P  6455) O Pa MN 8.314 10 310.6PQ PQ
3 P = 10 10 . 50 107 . 10 108 . 5.16 (also available as a Mathcad worksheet) (a) There are (at least) two ways to solve this problem. One way is to start from f = P exp or RT ln f = P R 1 FV  RT I dP U S RT z H P K V T W
P 0 z FH
P 0 V RT dP P I K RT 8.314 106 MPa m3 mol K ( 27315 + 400 )K 0.310748 . = = m3 kg P P(MPa ) 18.01 g mol 103 kg g P Solutions to Chemical and Engineering Thermodynamics, 3e From Steam Tables T = 400 C P MPa 0.01 0.05 0.10 0.2 0.3 0.4 0.5 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 $ V m3 kg
31.063 6.029 3.103 1.5493 1.0315 0.7726 0.6173 0.5137 0.3843 0.3066 0.2548 0.2178 0.19005 0.16847 0.15120 $ V  RT P
0.0118 0.00596 0.00448 0.00444 0.00433 0.00427 0.00420 0.00421 0.00413 0.00415 0.00416 0.00416 0.00417 0.00417 0.00417 By numerical integration of this data we find that f ~ 0.0084 MPa m3 kg P f 0.0084 MPa m3 kg ln = = 0.027032 P 0.310748 MPa m3 kg RT ln so f P = 0.97333 and f = 1947 MPa . A second way to use the steam tables is to assume that . steam at 400C and 0.01 MPa is an ideal gas. From the steam tables, at these conditions, we have $ $ H = 3279.6 kJ kg ; S = 9.6077 kJ kg K $ $ $ G = H  TS = 3279.6  673.15 9.6077 = 3187.8 kJ kg = 57412.7 kJ kmol = G ( 400 C, 0.01 MPa) = G IG( 400 C, 0.01 MPa) Also G IG( T = 400 C, 2 MPa)  G IG( T = 400 C, 0.01 MPa ) = G IG( T = 400 C, 2 MPa) = 574127 kJ kmol + . = 57412.7 + 8.314 ln 200 = 277603 kJ kmol . Also, from steam tables G (T = 400 C, 2 MPa) = (3247.6  673.15 7 .1271) 18.01 =  2791563 kJ kmol . f G  G IG 2791563 + 277603 . . = exp = exp P RT 8.314 67315 . = 0.9726 f = 0.9726 2 MPa = 1.945 MPa
2 MPa 2 MPa 0.01 z P = 0.01 MPa z V IGdP RT dP P FG H IJ K F H I K Solutions to Chemical and Engineering Thermodynamics, 3e (b) Corresponding states TC = 647.3 K, PC = 22.048 MPa, w = 0.344 Tr = 400 + 273.15 2 = 1.04 ; P = = 0.0907 r 647 .3 22.048 a f From corresponding states chart (actually from Table in Hougen, Watson and Rogatz, Vol. II, p. 601) we have f = 0.983 f = 1.966 MPa P (c) Using the program PR1 we find f = 19.40 bar = 1940 MPa .
Comment: The steam table results are probably the most accurate, and the corresponding states results the least accurate. Note that with the availability of the computer program PR1, the PR e.o.s. is the easiest to use. The results would be even more accurate if the PRSV equation was used. ...
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This homework help was uploaded on 02/06/2008 for the course CBE 103 taught by Professor Rethwisch during the Spring '07 term at University of Iowa.
 Spring '07
 Rethwisch

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