Introductory Chemical Engineering Thermodynamics

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Unformatted text preview: Solutions to Chemical and Engineering Thermodynamics, 3e 5.18 (a) Assume the vapor phase is ideal, and that H vap is approximately constant (or an average H vap can be used). P2 H vap 1 1 =- - P R T2 T1 1 vap ln ln F 2.026 I = - H F 1 - 1 I H 1013 K R H 222 .0 + 27315 178.0 + 27315K . . . vap FG H IJ K (1) H R = 3.52 103 K H vap = 2.93 104 J mol (b) H vap(T) = H(sat. vap,T )- H(sat. liq, T) IG IG = H (sat. vap, T ) - H ( T ) - H (sat. liq. , T ) - H (T ) H vap( T ) = TC T 200 + 27315 . (c) Tr = = = 0.851 TC 2831 + 27315 . . LMF H - H I MNGH T JK IG C - sat. vap, T FG H - H IJ H T K IG IG sat. liq., T OP PQ = 44.69 J mol K FG H - H IJ H T K IG C = 5.06 J mol K and sat. vap Tr = 0 .851 FG H - H IJ H T K C sat. liq. Tr = 0 .851 H vap(T) = 556.45 K 44.69 - 506 = 2205 104 J mol . . (d) The reason for the discrepancy is probably not the inaccuracy of corresponding states (since ZC = 0.272 which is close to 0.27) but rather the assumption of an ideal vapor phase in the Clausius-Clapeyron equation. We correct for gas-phase nonideality below. at T = 178 C , Tr = 0.811 , Z = 082 . T = 222 C , Tr = 0.890 , Z = 0.71 The average value of the compressibility is Z= We now replace eqn. 1 with ln P2 - H = P ZR 1 vap 1 (0.82 + 0.71) = 0.765 2 FG 1 - 1 IJ H HT T K 2 1 vap = 0.765 2.93 104 J mol c h = 2.24 104 J mol which is in much better agreement with the result of part (c). A better way to proceed would be to compute the compressibility as a function of temperature, i.e., find Z = Z (T , P ) and then integrate dP H vap P = dT Z (T , P ) RT 2 Solutions to Chemical and Engineering Thermodynamics, 3e 5.21 (also available as a Mathcad worksheet) 5.21 From Eq. 5.7-4 T=P*V*T/H 3 bar 5 10 . Pa Water T 6 m . 1000 . bar . 0.0906 . 10 . 273.1 . K g joule 333.8 . g 6 m . 1000 . bar . 0.01595 . 10 . 289.8 . K g 3 T = 7.412 K Acetic acid T 187 . Tin T joule g 3 T = 2.472 K 6 m . 1000 . bar . 0.00389 . 10 . 505.0 . K g 58.6 . joule g 3 T = 3.352 K Bismuth T 6 m . 1000 . bar . 0.00342 . 10 . 544 . K g joule 52.7 . g T = 3.53 K 5.22 (also available as a Mathcad worksheet) 5.22 i R 8.314 R. c . ( T ) 1234 2 0 .. 10 a 298 a T, K Yi 298 391.6 485.2 578.8 672.4 766 859.6 953.2 1.047 . 10 1.14 . 10 1.234 . 10 3 3 3 H( T , a , b , c ) Ag(s) Yi b .T Tmax Tmin Tmin 14710 b 0.328 c 0 Tmax Tmin . ( i) 10 Hm, J/mol H Yi , a , b , c 1.215 . 10 1.212 . 10 1.21 . 10 1.207 . 10 1.205 . 10 1.202 . 10 1.197 . 10 1.194 . 10 1.192 . 10 5 5 5 5 5 5 5 5 5 5 5 1.2 . 10 1.189 . 10 Solutions to Chemical and Engineering Thermodynamics, 3e Ag(l) Yi Tmax 2485 Tmin Tmin 1234 a T, K Yi 1.234 . 10 1.359 . 10 1.484 . 10 1.609 . 10 1.734 . 10 1.86 . 10 2.11 . 10 2.36 . 10 1.985 . 10 2.235 . 10 2.485 . 10 3 3 3 3 3 3 3 3 3 3 3 14260 b 0.458 c 0 Tmax Tmin . ( i) 10 Hsub, J/mol H Yi , a , b , c 1.139 . 10 1.134 . 10 1.129 . 10 1.124 . 10 1.12 . 10 1.11 . 10 1.115 . 10 1.105 . 10 1.096 . 10 1.091 . 10 5 5 5 5 5 5 5 5 5 5 5 1.1 . 10 BeO(s) Yi Tmax 2800 Tmin Tmin 298 a T, K Yi 298 548.2 798.4 1.049 . 10 1.299 . 10 1.549 . 10 1.799 . 10 2.049 . 10 2.55 . 10 3 3 3 3 3 3 3 3 34230 b 0.869 c 0 Tmax Tmin . ( i) 10 Hm, J/mol H Yi , a , b , c 2.824 . 10 2.806 . 10 2.788 . 10 2.77 . 10 2.752 . 10 2.734 . 10 2.716 . 10 2.698 . 10 2.68 . 10 2.662 . 10 2.644 . 10 5 5 5 5 5 5 5 5 5 5 5 2.3 . 10 2.8 . 10 Solutions to Chemical and Engineering Thermodynamics, 3e Ge(s) Yi Tmax 1210 Tmin Tmin 298 a T, K Yi 298 389.2 480.4 571.6 662.8 754 845.2 936.4 1.028 . 10 1.119 . 10 1.21 . 10 3 3 3 20150 b 0.395 c 0 Tmax Tmin . ( i) 10 Hm, J/mol H Yi , a , b , c 1.665 . 10 1.662 . 10 1.659 . 10 1.656 . 10 1.654 . 10 1.651 . 10 1.648 . 10 1.645 . 10 1.642 . 10 1.639 . 10 1.636 . 10 5 5 5 5 5 5 5 5 5 5 5 Mg(s) Yi Tmax 924 Tmin Tmin 298 a T, K Yi 298 360.6 423.2 485.8 548.4 611 673.6 736.2 798.8 861.4 924 7780 b 0.371 c 0 Tmax Tmin . ( i) 10 DHm, J/mol H Yi , a , b , c 6.376 . 10 6.357 . 10 6.338 . 10 6.318 . 10 6.299 . 10 6.28 . 10 6.261 . 10 6.241 . 10 6.222 . 10 6.203 . 10 6.183 . 10 4 4 4 4 4 4 4 4 4 4 4 Solutions to Chemical and Engineering Thermodynamics, 3e Mg(l) Yi Tmax 1380 Tmin Tmin 924 a T, K Yi 924 969.6 3 1.015 . 10 3 1.061 . 10 7750 b 0.612 c 0 Tmax Tmin . ( i) 10 DHm, J/mol H Yi , a , b , c 5.973 . 10 5.95 . 10 5.927 . 10 5.904 . 10 5.88 . 10 5.857 . 10 5.834 . 10 5.811 . 10 5.788 . 10 5.764 . 10 5.741 . 10 4 4 4 4 4 4 4 4 4 4 4 1.106 . 10 3 3 1.152 . 10 3 1.198 . 10 1.243 . 10 3 3 1.289 . 10 3 1.334 . 10 1.38 . 10 3 NaCl(s) Tmax Yi 1074 Tmin Tmin Yi 298 375.6 453.2 530.8 608.4 686 763.6 841.2 918.8 996.4 1.074 . 10 3 298 a T, K 12440 b 0.391 c 0.46 . 10 3 Tmax Tmin . ( i) 10 DHm, J/mol H Yi , a , b , c 1.021 . 10 1.017 . 10 1.012 . 10 1.006 . 10 9.94 . 10 5 5 5 5 5 4 4 4 4 4 4 1 . 10 9.871 . 10 9.799 . 10 9.721 . 10 9.639 . 10 9.552 . 10 Solutions to Chemical and Engineering Thermodynamics, 3e Si(s) Tmax Yi 1683 Tmin Tmin Yi 1.2 . 10 1.248 . 10 1.297 . 10 1.345 . 10 1.393 . 10 1.442 . 10 1.49 . 10 1.538 . 10 1.586 . 10 1.635 . 10 1.683 . 10 3 3 3 3 3 3 3 3 3 3 3 1200 a T, K 18000 b 0.444 c 0 Tmax Tmin . ( i) 10 DHm, J/mol H Yi , a , b , c 1.452 . 10 1.45 . 10 1.449 . 10 1.447 . 10 1.445 . 10 1.443 . 10 1.442 . 10 1.44 . 10 1.438 . 10 1.436 . 10 1.434 . 10 5 5 5 5 5 5 5 5 5 5 5 5.23 (also available as a Mathcad worksheet) 5.23 The metal tin undergoes a transition from a gray phase to a white phase at 286 K at ambien pressure. Given that the enthalpy change of this transition is 2090 kJ/mole and that the volume change of this transition is -4.35 cm3/mole, compute the temperature at which this transition occurs at 100 bar. 5 10 . Pa From Eq. 5.7-4 T=P*V*T/H 3 bar T 6 m . 286 . K 99 . bar . 4.35 . 10 . mole 2090000 . joule mole T = 5.893 10 3 K 5.24 For the solid-liquid transition FG P IJ H T K FG P IJ H ln T K = eq H P TV ln T f FG H IJ K = eq 127 J g H = = 964 .3 J cc 01317 cc g V . f = 964 .3 J cc = 964 .3 10 6 J m3 = 9643 bar = 9.643 108 Pa P = P1 + 9643 ln 2 T2 T1 eq Solutions to Chemical and Engineering Thermodynamics, 3e T TP = T1 exp R P - P U = 278.7 expR P - 10 Pa U S 9.643 10 Pa V S9.643 10 Pa V T W T W TP TP 5 1 8 8 (1) For the solid-vapor transition, assuming an ideal vapor phase ln P2 H sub 1 1 =- - P R T2 T1 1 H sub R 2 1 - ln FG IJ H K lna P P f = = 1 T2 - 1 T1 ln 26.67 13.33 = -5513 K . 3.696 10-3 - 3822 10-3 a f P2 1 1 = - 5513 - P T2 T1 1 FG H IJ K OP Q -1 and TP T L1 = M - 0.1814 10 NT 1 -3 P TP ln P 1 = 1000 3.696 - 0.1814 ln PTP 26.67 c h (2) Solving Eqns. (1) and (2) simultaneously gives P TP = 0.483 bar = 483 kPa and T TP = 278.7 K . [The melting temperature of benzene ~ triple point temperature = 553 C = 278.7 K , which agrees . exactly with our prediction]. 5.25 First, at 298.15 K, lets relate the Gibbs free energy at any pressure to the value given at 1 bar. kg g 1 mol ( P - 1) bar 3510 3 1000 m kg 12 g G dia (298.15, P ) = G dia ( 298, P = 1 bar ) + bar m3 8.314 10-5 298 .15K mol K kg g 1 mol ( P - 1) bar 2220 3 1000 kg 12 g m G g (29815, P) = G g (298 , P = 1 bar ) + bar m 3 8.314 10 -5 298.15K mol K F GG GH I JJ JK F GG GH I JJ JK Note that V dia = Vg = 1 m3 1 kg g m3 12 = 3.4188 10-6 3510 kg 1000 g mol mol 1 m3 1 kg g m3 12 = 5.4054 10 -6 2220 kg 1000 g mol mol Therefore Solutions to Chemical and Engineering Thermodynamics, 3e G dia (298.15, P ) - G g ( 29815, P ) = G dia ( 298, P = 1 bar ) - G g ( 298, P = 1 bar ) F ( P - 1)bar (3.4188 - 5.4054) 10 m I G mol J +G GH 8.314 10 bar mK 298.15 K JJK mol -6 3 -5 3 at equilibrium at 298.15 K we have 0 = 2900 - 0 - ( P - 1) 8.0143 10-5 P = 1+ 2900 = 36185 106 bar = 36.185 Mbar . 8.0143 10- 5 To find the transition pressure at other temperatures we use the Clapeyron equation J (2.377 - 5740 ) . S Pa m3 P mol K = = 1 3 J T sat V m (3.4188 - 5.4054) 10-6 mol Pa bar . = 16928 106 = 16.928 K K which indicates that for every degree K increase about 298.15 K we need to increase the pressure by 16.928 bar. However, this is a small percentage increase compared to the 36.185 MPa pressure required at 298.15 K. So the transition is essentially (within engineering accuracy) only very weakly dependent on pressure. F I H K Solutions to Chemical and Engineering Thermodynamics, 3e 5.36 FG P IJ H T K (also available as a Mathcad worksheet) = sat H T V >> V V ~ V L V Assume V V = ZRT ' P FG P IJ H T K but = sat H ln P 2 ZRT P T FG H IJ K = sat H ZRT 2 FG ln P IJ H T K Thus = sat 5622.7 43552 - . - 4 .70504 ln T T T F H I K =+ 56227 4.70504 . 1 - = 2 (5622.7 - 4.70504T ) 2 T T T H 1 = (5622.7 - 4 .70504T ) ZRT 2 T 2 or H = ZR (56227 - 4.70504T ) = 31,602 J mol at 75C . Z= but PV B = 1 + = Z = 0.9539 RT V so B = 0.9539 - 1 = -0.04607 ; B = -0.04607V V Then V = 0.9539 RT P . To find P use ln Pvap = 43552 - . Pvap = 0.8736 bar V= and 0.9539 8.314 10-5 (273.15 + 75) = 31606 10 -2 m3 mol . 0.8736 5622.7 - 4 .70504 ln( 27315 + 75) . (273.15 + 75) 31,602 = 0.9539 8.314 (5622.7 - 4.70504 (273.15 + 75)) B = -1456 10-3 m3 mol . ...
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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.

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