流体相平衡的分子热力学--- Prausnitz书的答案solutions.pdf

Advanced Thermodynamics KP8108 Solutions to Selected Problems a companion to J M Prausnitz R N Lichtenthaler and E G de Azevedo Molecular Thermodynamics 2nd ed Prof Bj rn Hafskjold Dep of Chemistry NTNU Assoc Prof Tore Haug Warberg Dep of Chem Eng NTNU May 16 2007 ii Contents Problem 2 2 1 Problem 2 5 2 Problem 2 6 3 Problem 2 7 4 Detailed derivation of Eqs 3 53 and 3 54 5 Problem 3 2 7 Problem 3 3 8 Problem 3 7 10 Problem 3 9 11 Problem 5 2 13 Iteration 1 y1 0 01 14 Iteration 2 y1 3 4835 10 3 15 Problem 5 3 16 Problem 5 16 20 Problem 5 17 24 Iteration 1 y3 0 25 Iteration 2 y3 1 04 10 4 26 Problem 6 2 27 Problem 6 16 30 Problem 7 1 35 Problem 7 3 38 Problem 10 3 41 Method 1 41 Method 2 44 Problem 10 6 44 Problem 10 9 45 Question a 45 Question b 47 Question c method I 50 Question c method II 51 Problem 11 1 53 Problem 11 6 54 Problem 11 9 55 Problem 12 3 57 Problem 12 4 58 Problem 12 7 60 iii iv Chapter 2 Problem 2 2 Forthegivenequationofstate P V n b RT andbyusingtheMaxwellrelation 2A T V 2 A V T we fi nd S V T P T V R V n b P T For an isothermal change S Z S V T dV Z R V n b dV nRln V 2 nb V1 nb nRln P1 P2 Alternatively from 2G T P 2 G P T we fi nd S P T V T P nR P This formula integrates to the same expression for S as above For U we know the diff erential dU TdS pdV and hence U V T T S V T P Using the fi rst of the Maxwell rules from above we get U V T T P T V P 0 Consequently U 0 Alternatively we can write U P T T S P T P V P T which gives U P T T S P T P V P T T V T P P V P T RT P RT P 0 Again we fi nd that U 0 For the next derivative we shall use H U PV and H P T U P T V P V P T 0 nRT P nb PnRT P2 nb 1 For an isothermal change H Z H P T dP Z nbdP nb P2 P1 The changes in Gibbs energy G and Helmholtz energy A are found from the defi nitions of G and A and the results above G H T S nb P2 P1 nRT ln P1 P2 A U T S nRT ln P1 P2 Problem 2 5 Because entropy is constant it would be natural to express the result as an explicit function of S However this is not feasible because the equation of state is expressed in terms of T and V We shall therefore use implicit diff erentiation and fi rst write entropy as a function of volume and temperature diff erentiate use some of the Maxwell relations and then put the diff erential of S V T equal to zero dS S V T dV S T V dT P T V dV CV T dT 0 Thermodynamically the exhaust temperature does not depend on the gas fl ow rate through the turbine and for convenience we choose one mole of gas as the basis for the calculation With the van der Waal equation we fi nd for one mole of gas P RT v b a v2 P T V R v b Inserted into the expression for dS 0 this gives the following diff erential equation be tween v and T at constant entropy dv v b cV R dT T This is the adiabatic equation of state for a van der Waal gas Integration at constant cv gives ln v 2 b v1 b cV R ln T 2 T1 where T2and v2 are unknown It is giventhat thefi nal pressure is atmospheric and by using the equation of state we could express v2by P2 This is however a little cumbersome but since the exhaust pressure is low we try fi rst by approximating the exhaust gas is an ideal gas This gives ln T 2 T1 R cV ln vig 2 b v1 b R cV ln RT2 P2 v1 b 2 where we have approximated vig 2 b vig 2 RT2 P2 Rearrangement gives ln T 2 T1 R cV ln T 1 T2 R cV ln RT1 P2 v1 b R cV 1 R cV ln RT1 P2 v1 b R cP ln RT1 P2 v1 b where we have made use of cig P cig V R With the given numbers we fi nd ln T2 350 273 K 8 314 J K 1 mol 1 33 5J K 1mol 1 ln 82 06cm3 atm mol 1K 1 350 273 K 1 atm 600 45 cm3mol 1 1 123 T2 623exp 1 123 203K Checking the ideal gas assumption Vig 2 RT2 P2 82 06 cm3atm mol 1K 1203 K 1 atm 16700 cm3mol 1 Vvdw 2 16400 cm3mol 1 Note The van der Waal result was found by an iterative solution of the cubic equation in V at 203 K and 1 atm If we want to improve the ideal gas assumption we may insert Vvdw 2 16400 cm3mol 1into the equation above ln T 2 T1 R cV ln vvdw 2 b v1 b 8 314 J K 1mol 1 25 2 J K 1mol 1 ln 16400 45 cm3 mol 1 600 45 cm3mol 1 1 117 T2 623exp 1 117 204K Problem 2 6 The virial expansion of the compression factor is z Pv RT 1 B v C v2 The compression factor for the van der Waal gas is zvdw v RT RT v b a v2 1 1 b v a RT 1 v We recognize the fi rst term on the right hand side as the sum of a geometric series 1 1 b v 1 b v b v 2 3 Ordering the van der Waal compression fcator in increasing powers of 1 v gives zvdw 1 b a RT 1 v b v 2 and we conclude that B b a RT Problem 2 7 We start by recovering one result from Problem 2 2 U P T T V T P P V P T From the given equation of state the volume is z Pv RT 1 BP RT v RT P 1 BP RT RT P a b T2 From this we fi nd V T P R P 2b T3 V P T RT P2 U P T T R P 2b T3 P RT P2 2b T2 Integration at the constant temperature then gives U Z 0 U P T dP 2b 2 4 Chapter 3 Detailed derivation of Eqs 3 53 and 3 54 Start with this defi nition of the chemical potential i A ni V Applied to Eq 3 50 the defi nition leads to i Z V P ni T V nj RT V dV RT ni X j njln P V njRT u i Ts i 1 We note that j 1 is the only term that survives the diff erentiation of the sum This term is ni niln P V niRT ln P V niRT 1 2 Furthermore the standard state of component i is pure ideal gas at temperature T and pressure P 1 bar The chemical potential of i in this state is i g i u i RT Ts i 3 Eqs 2 and 3 into 1 gives i Z V P ni T V nj RT V dV RT ln P V niRT 1 i RT i i Z V P ni T V nj RT V dV RT ln P V niRT 4 The relation between the chemical potential and fugacity is i i RT ln fi f i 5 The fugacity of the standard state is P 1 bar The fugacity is further expressed by the fugacity coeffi cient as fi iPi iyiP i fi yiP 6 5 Combination of Eqs 4 5 and 6 yields RT ln iyiP P Z V P ni T V nj RT V dV RT ln P V niRT RT ln i Z V P ni T V nj RT V dV RT ln P V niRT RT ln y iP P Z V P ni T V nj RT V dV RT ln PV nRT Z V P ni T V nj RT V dV RT lnz 3 53 For a pure component the chemical potential is i G ni i RT ln fi f i i RT ln fi P RT ln P f i RT ln fi P pure i i i RT ln P f i G ni G ni RT ln P P 7 The Gibbs energy is taken from Eq 3 51 in the book G ni Z V P ni RT V dV RT ln P V niRT PV ni u i Ts i 8 Note that G ni h i Ts i u i RT Ts i 9 and PV ni RTz 10 6 Eqs 8 9 and 10 inserted into 7 combine to RT ln fi P pure i G ni G ni RT ln P P Z V P ni RT V dV RT ln P V niRT PV ni u i Ts i u i RT Ts i RT ln P P Z V P ni RT V dV RT ln PV niRT PV ni RT Z V P ni RT V dV RT lnz RTz RT Z V P ni RT V dV RT lnz RT z 1 3 54 Note There is probably an easier way to understand this equation It is illogical to bring in expressions for both A and G and pretend they are independent functions The elegant solution would be to bring in the homogeneous properties of P P T V N and develop Eq 3 54 directly from Eq 3 53 Suggestions are welcome Problem 3 2 The gas mixture is illustrated in Figure 1 If we consider the gas to be a pseudo pure A B xA 0 25 xB 0 75 P 50 bar T 373 K A 0 65 B 0 90 Figure 1 Properties of the gas mixture substance the relation between the chemical potential and the fugacity would be M M RT ln fM f M where M and f M are the chemical potential and the fugacity respectively in the standard state This would correspond to a molar Gibbs energy for the mixture GM M On the 7 other hand the molar Gibbs energy for the mixture is given by Gm X i ixi where the chemical potential of each component is i i RT ln fi f i i RT ln iPi f i Here i is the fugacity coeffi cient of component i Equating GMand Gmgives M RT ln fM f M X i xi i RT X i xiln iPi f i 11 which we can use to express fMin terms of the component properties We should however fi rst make a remark on the relation between the standard states The M is not the same as i even if they refer to the same pressure because M contains the properties of an ideal mixture M X i xi i RT X i xiln xi 12 The standard fugacities which are usually taken to be ideal gas at 1 bar pressure are the same f M f i Equations 11 and 12 give X i xi i RT X i xiln xi RT ln fM f M X i xi i RT X i xiln iPi f i ln fM 1 bar X i xiln iPi 1 bar X i xiln xi With the given data we fi nd ln fM 1 bar 0 25 ln 0 65 0 25 50 0 75 ln 0 90 0 75 50 0 25 ln0 25 0 75 ln0 75 3 7253 fM 1 exp 3 7253 41 5 bar Problem 3 3 The two situations are shown schematically in Figure 2 Equilibrium between gas and liquid at 1 bar may be expressed as gas C2H6 1bar liq C2H6 xC2H6 0 33 10 4 The same equilibrium at 35 bar may be expressed as gas C2H6 35bar liq C2H6 xC2H6 x 8 gasgas H2O C2H6H2O C2H6 P 1 barP 35 bar PH2O 0 0316 barPH2O 0 0316 bar PC2H6 1 barPC2H6 35 bar T 298 KT 298 K liquidliquid H2O C2H6H2O C2H6 xH2O 1 0 xH2O 1 0 xC2H6 0 33 10 4xC2H6 Figure 2 The two equilibrium situations where xC2H6 is the unknown denoted by x In the gas phase the eff ect of a change in the chemical potential at constant temperature is d i vidP where viis the partial molar volume of component i Since there is so little water in the gas phase we assume it to be pure C2H6 in which case we can neglect the eff ect of a change in composition and viis the molar volume of C2H6 i e vi vC2H6 The molar volume is given by the equation of state v RT P 1 aP bP2 where a 7 63 10 3bar 1and b 7 22 10 5bar 2 Integration of the chemical potential at constant T gives gas C2H6 35bar gas C2H6 1bar 35 bar Z 1 bar C2H6 dP T dP RT 35 bar Z 1 bar 1 P a bP dP RT ln 35 bar 1 bar a 34 bar 1 2b 352 12 bar2 In the liquid phase the change in chemical potential is the sum of two contributions one duetothepressurechange andoneduetothecompositionchange Theeff ect ofthechange in pressure is d C2H6 vC2H6dP where vC2H6is the partial molar volume of ethane in the liquid Since the liquid is practically incompressible it may be assumed to be constant over the pressure change in question here and in any case the contribution is negligibly small for the liquid The eff ect of the change in composition is d C2H6 RTdlnaC2H6 where aC2H6 fC2H6 f C2H6 is the activity of ethane We have to assume that the activity 9 coeffi cient of ethane defi ned by C2H6 aC2H6 xC2H6 is constant over the concentration range considered here which leads to liq C2H6 x liq C2H6 x 0 33 10 4 35 bar Z 1 bar vC2H6dP RT xC2H6 x Z xC2H6 0 33 10 4 dln C2H6 RT xC2H6 x Z xC2H6 0 33 10 4 dln xC2H6 RT xC2H6 x Z xC2H6 0 33 10 4 dln xC2H6 RT ln x 0 33 10 4 Equating the changes in the chemical potential of the gas and liquid phases give RT ln x 0 33 10 4 RT ln 35 7 63 10 3 34 1 2 7 22 10 5 352 12 ln x 0 33 10 4 ln 35 7 63 10 3 34 1 2 7 22 10 5 352 12 3 2517 x 0 33 10 4exp 3 2517 8 53 10 4 Problem 3 7 Throttling is an isenthalpic process and we shall therefore consider the conservation of enthalpy in this problem Since the equation of state is given explicitly in the volume as function of P and T we fi nd the enthalpy from the following relation H P Z 0 V T V T P nT dP X i nih0 i where h0 i is the molar enthalpy of pure i at the actual temperature The given volumetric data for the mixture leads to V T V T P nT a b T where a 5 10 5m3mol 1and b 0 2 m3K mol 1 Integration leads to the molar enthalpy h a b T P X i xih0 i where xiis the mole fraction of component i The condition hin houtgives a b Tin Pin X i xih0 i in a b Tout Pout X i xih0 i out a b Tin Pin a b Tout Pout X i xi h0 i out h 0 i in 10 We have h0 i out h0 i in c0 P i Tout Tin and c0 P P i xic0 P i 33 5 J mol 1K 1 Solving for Pin gives Pin a b Tout Pout c0 P Tout Tin a b Tin 5 10 5m3mol 1 0 2m3Kmol 1 200K 105Pa 33 5Jmol 1K 1 200K 300K 5 10 5m3mol 1 0 2m3Kmol 1 300K 55 9 105Pa 55 9 bar Problem 3 9 The internal energy of mixing is using one mole as basis mixu uM x1u1 x2u2 Since the van der Waal equation is explicit in P P RT v b a v2 it is convenient to fi nd the internal energy from u Z v P T P T V nT dv X i xiu0 i where P T P T V nT RT v b a v2 RT v b a v2 Notethat themixingmustbe assumedto occur at constant pressure not at constantvolume which means that we must use diff erent volumes for the lower integration limit for the mixture and the two pure components The molar internal energy for the mixture and each of the pure components are uM aM Z vM dv v2 X i xiu0 i aM vM X i xiu0 i u1 a1 v1 u0 1 u2 a2 v2 u0 2 The change in internal energy for the mixture is mixu aM vM X i xiu0 i x1 a 1 v1 u0 1 x2 a 2 v2 u0 2 aM vM x1a1 v1 x2a2 v2 11 With the given data we fi nd aM X i X j xixj a iaj 1 kij 0 52 106 h 1 04 2 1 04 4 17 1 0 0 1 4 17i bar cm6mol 2 2 24 106bar cm6mol 2 vM X i xivi 0 5 32 2 48 5 cm3mol 1 40 35 cm3mol 1 mixu 2 24 40 35 0 5 1 04 32 2 0 5 4 17 48 5 106 bar cm6mol 2 cm3mol 1 363J mol 1 12 Chapter 5 Problem 5 2 The question here is whether the fugacity of CO2in the compressed gas is higher than the fugacity at saturation conditions The actual temperature is lower than the triple point temperature and the condensed phase in equilibrium with the vapour if it exists is solid To fi nd the saturation pressure we consider how it changes from the vapour pressure of pure CO2 component 1 to its new value after injection of H2 component 2 until the total pressure is 60 bar We therefore consider a charging process as illustrated in Figure 3 The H2is inert in this case in the sense that it does not dissolve in the solid phase The change CO2 g CO2 g H2 g P CO2 0 1392 bar g PCO2 PH2 60 bar T 173 K T 173 K CO2 s s CO2 s T 173 K T 173 K Figure 3 Available data for the state changes in chemical potential of CO2in the gas phase is 1 g RT c Z dln f1 RT ln f 1 c 13 where f1is the fugacity of CO2 The fugacity of pure CO2vapour can be assumed to be equal to its vapour pressure the saturation pressure of 0 1392 bar is quite low but please validate this assumption by numerical calculations The fugacity of CO2in the compressed state must be calculated somehow and we shall be using f1 1y1P where 1and y1 are the the fugacity coeffi cient and the mole fraction of CO2 respectively and P is the total pressure equal to 60 bar The fugacity coeffi cient is estimated from van 13 der Waal s equation z Pv RT v v b a RTv 14 The mixing rules are a X i X j yiyj a iaj b X i yibi Thefugacity coeffi cient of component1 in themixtureis calculated from Eq 3 70 inPraus nitz et al ln 1 ln v v b b1 v b 2 a 1 P jyj a j RTv lnz 15 Numerical values for the van der Waals parameters may be found in several textbooks e g in P W Atkins Physical Chemistry 6th edition Oxford 1998 a atm cm6mol 2b cm3mol 1 CO2 1 3 640 10642 67 H2 2 0 2476 10626 61 Alternativelywemayusethecorrespondingstateprincipleandcalculatea 27 RTc 2 64Pc and b RTc 8Pcfrom critical data Do this on your own and compare the results with the table Because the mole fraction of CO2is low at most 0 01 we make a small error if we choose the van der Waal parameters to be those of pure H2 but for the time being we determine the mixture parameters on the basis of 1 mole CO2 Iteration 1 y1 0 01 This gives a 0 012 3 640 106 2 0 01 0 99 3 640 106 0 2476 106 0 992 0 2476 106 atm cm6mol 2 2 6183 105atm cm6mol 2 b 0 01 42 67 0 99 26 61 cm3mol 1 26 77 cm3mol 1 At 60 bar the molar volume in Eq 14 note that we have to solve a cubic equation in v is v 250 69 cm3mol 1and z 1 048 Inserted into Eq 15 ln 1 ln 250 69 250 69 26 77 42 67 250 69 26 77 2 3 640 106 0 01 3 640 106 0 99 0 2476 106 250 69 82 0567 173 ln1 048 0 292 1 exp 0 292 0 7468 14 Inserting the last result into Eq 13 gives the change in chemical potential of CO2in the gas phase 1 g 83 1439 cm3bar mol 1K 1 173 K ln 0 7468 y 1 60 bar 0 1392 bar 14428 cm3bar mol 1 ln 321 9y1 The change in chemical potential for CO2in the solid phase is 1 s c Z vsdP vs 60 bar Z 0 1392 bar dP 27 6 cm3mol 1 60 0 1392 bar 1652 16 cm3bar mol 1 Here we have assumed that the solid is incompressible and that H2does not dissolves in the solid At equilibrium in the compressed state we have 1 g 1 s which gives 14428 cm3bar mol 1 ln 321 9y1 1652 16 cm3bar mol 1 y1 1 321 9 exp 1652 16 14428 3 4835 10 3 At this point we conclude that Iteration 1 is slightly off and try another calculation Iteration 2 y1 3 4835 10 3 This gives a 0 00352 3 640 106 2 0 0035 0 9965 3 640 106 0 2476 106 0 99652 0 2476 106 atm cm6mol 2 2 5254 105atm cm6mol 2 b 0 0035 42 67 0 9965 26 61 cm3mol 1 26 67 cm3mol 1 and v 251 17 cm3mol 1 and z 1 048 Inserting these numbers into Eq 15 gives ln 1 ln 251 17 251 17 26 77 42 67 251 17 26 77 2 3 640 106 0 0035 3 640 106 0 9965 0 2476 106 251 17 82 0567 173 ln1 048 0 282 1 exp 0 282 0 7543 A small change in 1from 0 7468 to 0 7543 is observed The improved value of the change in chemical potential for the gas phase is calculated once more from Eq 13 1 g 83 1439 cm3bar mol 1K 1 173 K ln 0 7543 y 1 60 bar 0 1392 bar 14428 cm3bar mol 1 ln 325 1y1 15 The solid phase chemical potential is the same as above and we fi nd y1 1 325 1 exp 1652 16 14428 3 4492 10 3 We consider Iteration 2 to be good and accept this as the fi nal result To conclude the calculation we need a material balance using one mole material in total as basis nCO2 g nCO2 s nH2 g 1 mol 16 nH2 g 0 99 mol 17 The mole fraction of CO2in the gas phase is yCO2 nCO2 g nCO2 g nH2 g 0 00345 18 Solution of Eqs 16 18 gives the amount of precipitated CO2 nCO2 s 0 0066mol Problem 5 3 The situation is shown in Figure 4 The question is whether some of the CO2will condense into liquid or solid or not This depends on the pressure and temperature in the gas during the isenthalpic expansion In the following we will denote CH4by 1 and CO2by 2 The 30 mol CO2 g 70 mol CH4 g 70 bar 313 K CO2 s or CO2 l CH4 g 1 bar T Figure 4 The given process critical point of CO2is at Pc 73 8 bar and Tc 304 2 K The triple point is at PT 5 18 bar and TT 216 8 K If the gas cools to a temperature in the range between the critical point and the triple point some CO2may possibly condense to a liquid if it cools to T 216 8 K it may condense to a solid Because the equation of state is given as a virial expansion in molar density it is most convenient to express the enthalpy of the gas as H Z V P T P T V nT dV PV X i niu0 i where niis the number of moles of component i and u0 i is the molar internal energy of pure i in the ideal gas state at the actual temperature The strategy is to fi nd how H depends 16 on V and T