《化工热力学》课程授课教案（讲义）Chapter 12 Solution Thermodynamics：Application

Chapter 12 Solution Thermodynamics:Application 12.1 Liquid-phase Properties from VLE data 12.1.1 Fugacity For species in the vapor mixture Vapor =yoP TPy Accordingto the criterionofequilibrium =(=1,2.,N) =yP Liquid When P≤105a*1=方=yP TP For example Methylethylketone(1)/toluene(2)systemat 50C See Table 12.1 f~x diagram 30 元=P 20 =xh =10 =2P 0=x5 1

1 Chapter 12 Solution Thermodynamics: Application 12.1 Liquid-phase Properties from VLE data 12.1.1 Fugacity For species in the vapor mixture ˆ ˆ v v i i i f y P =  According to the criterion of equilibrium ( ) ˆ ˆ 1,2, , l v i i f f i N = = ˆ ˆ l v i i i f y P =  When P≤105 ˆ 1 v i  ˆl v ˆ i i i f f y P = = For example Methyl ethyl ketone (1) /toluene (2) system at 50℃ See Table 12.1 f ～x diagram Vapor T P {yi} Liquid T P {xi} 1 1 sat f P = 1 20 30 10 ˆ / i f kPa 1 ˆ i f y P = 2 2 ˆ f y P = 1 1 1 ˆid f x f = 2 2 2 ˆid f x f = 2 2 sat f P =

12.1.2 Activity Coefficient 方=PP Easy calculation of activity coefficient is allowed from experimental low-pressure Review Definitionofactivity coefficient Lewis/Randall rule=xx Henry'slaw xi-→0 户回典文 H, Constant T,P Henry'slaw applies to a species as it approaches infinite dilution ina Henry's law binary solution,and the G-D equation insures validity of the Lewis/Randall rule for the other species as it approaches purity Lewis/Randall rule G-D方程可以证明：在一定T、P下，若二元溶液的一个组分逸度符合 Lewis-Randall规则，那么另一个组分逸度必定符合Henry规则 将遵守Lewis/Randal规则的溶液称为理想溶液、或Lewis/Randall规则意 义上的理想溶液

2 i f ˆ i f Hi Constant T, P Henry’s law 12.1.2 Activity Coefficient Easy calculation of activity coefficient is allowed from experimental low-pressure Review Definition of activity coefficient Lewis/Randall rule ˆid i i i f x f = xi→1 Henry’s law xi→0 ˆ lim id i i i x f x H → = 0 ˆ lim i i i x i f H → x  Henry’s law applies to a species as it approaches infinite dilution in a binary solution, and the G-D equation insures validity of the Lewis/Randall rule for the other species as it approaches purity G-D 方程可以证明：在一定 T、P 下，若二元溶液的一个组分逸度符合 Lewis-Randall 规则， 那么另一个组分逸度必定符合 Henry 规则 将遵守 Lewis/Randall 规则的溶液称为理想溶液、或 Lewis/Randall 规则意 义上的理想溶液。 ˆ ˆ ˆ i i i id i i i f f f x f  = = i i i sat i i i i y P y P x f x P  = = ˆ ˆ i i id i f f  = Lewis/Randall rule i f xi

将遵守Henry定律的溶液称为理想稀溶液或Henry定律意义上的理想溶 液。 真实稀溶液的溶剂和溶质分别符合Lewis/Randall规则和Henry定律 12.1.3活度系数的对称归一化 对于理想溶液或Lew心Randal灿l规则意义上的理想溶液方一 For speciesi血eali四，=1活度系数的对称归一化 或对称归一化活度系数 12.1.4活度系数的不对称归一化 对于理想稀溶液或Henry规则意义上的理想溶液方= x,H For species in ideal solution =1 For species in solution >I positive deviation solution 7,<I negative deviation solution 活度系数的不对称归一化或不对称归一化活度系数 12.1.5两种活度系数的关系 对于液体混合物中的ⅰ组分，其逸度不会因采用不同的活度系数而变化 即=x=Hx 对于二元溶液，仅与T,P有关，与浓度无关，可以通过取 x一0时的极限获得 因为lim方=1 H lim =Yi 3

3 将遵守 Henry 定律的溶液称为理想稀溶液或 Henry 定律意义上的理想溶 液。 真实稀溶液的溶剂和溶质分别符合 Lewis/Randall 规则和 Henry 定律。 12.1.3 活度系数的对称归一化 对于理想溶液或 Lewis/Randall 规则意义上的理想溶液 ˆ i i i i f x f  = For species in real solution 1 lim 1 i i x  → = 活度系数的对称归一化 或对称归一化活度系数 12.1.4 活度系数的不对称归一化 对于理想稀溶液或 Henry 规则意义上的理想溶液 * ˆ i i i i f x H  = For species in ideal solution * 1 i  = For species in solution * 1 i   positive deviation solution 1 i   negative deviation solution For species in real dilution solution * 0 lim 1 i i x  → = 活度系数的不对称归一化或不对称归一化活度系数 12.1.5 两种活度系数的关系 对于液体混合物中的 i 组分，其逸度不会因采用不同的活度系数而变化 即 ˆl l i i i i i i i f f x H x    = = 或 * i i l i i H f   = 对于二元溶液，仅与 T,P 有关，与浓度无关，可以通过取 xi→0 时的极限获得 因为 0 lim 1 i i x   → = 0 lim i i i x    → = 0 * 0 0 lim lim lim i i i i x i i i l x i i i x H f      →  →  → = = =

Iny =InIn 即为两种活度系数之间 的关系 类似地推出%区→ 例4-639C°、2MPa下二元溶液中的组分1的逸度为 f=6x-9x+4x MPa 确定在该温度、压力状态下 ()纯组分1的逸度与逸度系数： 2)组分1的亨利系数k1: (3M1与x1的关系式（若组分1的标准状态是以Lewis-Randall定则为基础）。 解 (1)x1=1 有691NA=吾-05 回A=4=m-g+=6 (3)若组分1的标准状态是以Lewis-Randal定则为基础 =立.6-6-9%+4城 x×1 12.2 Models For the Excess Gibbs The definition of Excess property ME =M M id The actualextensive The extensiveproperty ofan property ofasolution ideal solution at the same T,P and x

4 * ln ln ln i i i     = − 即为两种活度系数之间 的关系 类似地推出 * * ( 1) i i i i x    = → 例 4-6 39C°、2MPa 下二元溶液中的组分 1 的逸度为 2 3 1 1 1 1 ˆ f x x x MPa = − + 6 9 4 确定在该温度、压力状态下 (1) 纯组分 1 的逸度与逸度系数； (2) 组分 1 的亨利系数 k1； (3)γ1 与 x1 的关系式（若组分 1 的标准状态是以 Lewis-Randall 定则为基础）。 解 : (1) x1 = 1 f1 =6-9+4=1MPa 1 1 1 0.5 2 f P  = = = (2) 1 1 1 0 1 ˆ lim x f H → x = 1 2 3 1 1 1 1 0 1 6 9 4 lim 6 x x x x H MPa → x − + = = (3) 若组分 1 的标准状态是以 Lewis-Randall 定则为基础 1 1 1 1 ˆ f x f  = 2 3 1 1 1 1 2 1 1 1 1 1 1 ˆ 6 9 4 6 9 4 1 f x x x x x x f x  − + = = = − +  12.2 Models For the Excess Gibbs The definition of Excess property M E ≡ M－ M id at the same T, P and x The actual extensive property of a solution The extensive property of an ideal solution

Ingeneral G R7=8T,P, For liquid at low to moderate Pand constantT 7=8.，) 12.2.1 The Margules equation For binary system GE A2A RT :) 2a x =0 Iny=4 n为=1+5 x2=0ln5= 4 12.2.2 Local-Composition Model The concept of Local-composition Example: for binary solution x=0.5 and x2=0.5 ●o Around 1 molecules Around 2 molecukes 6=0.4andx3=0.6 =0.6and3=0.4 (1)The Wilson equation For binary system G =-n6+N6)-名ln(3+Ax)

5 In general ( , , ) E G g T P x RT = For liquid at low to moderate P and constant T 1 2 ( , , , ) E N G g x x x RT = 12.2.1 The Margules equation For binary system 12 21 1 2 12 1 21 2 E G A A x x RT A x A x   =   + 2 2 12 1 1 12 21 2 2 21 2 2 21 12 1 4 ln 1 2 ln 1 A x b b ac A A x a A x A A x   − −    −  − = +          = +       1 1 12 2 2 21 0 ln 0 ln x A x A     = =  = =  12.2.2 Local-Composition Model The concept of Local-composition ： Example： for binary solution x1 =0.5 and x2 =0.5 （1）The Wilson equation For binary system 1 1 12 2 2 2 21 1 ln( ) ln( ) E G x x x x x x RT = − +  − +  Around 1 molecules x1=0.4 and x2=0.6 Around 2 molecules x1 =0.6 and x2=0.4

ΛaA n%=-l+Ae)++八ex+A Λ21 42 nh=-ln+)+3+入x+As (2)NRTLequation For binary system GE GT G RT= )2 lny1=x号2 G22 +G21) (x2+xG2) (G2 12 G21t21 (x+xG)2 (3)UNIQUACand UNIFAC equation For multi-component system Only the parameters ofbinary system required 12.3 Property Changes of Mixing The Excess properties ofreal solutions s=S-∑xS,+R∑xlnx G=G-∑xG,-RT∑xlnx vE=V-∑xy H=H-∑xH Property changes of mixing AM-2闪 A molar(or unit-mass)solution A molar (or unit-mass)pure- property species property a t the same T,P and x Another ME=△M-△M 6

6 12 21 1 1 12 2 2 1 12 2 2 21 1 21 12 2 2 21 1 1 2 21 1 1 12 2 ln ln( ) ( ) ln ln( ) ( ) x x x x x x x x x x x x x x     = − +  + − +  +    = − +  + − +  +  （2）NRTL equation For binary system 21 21 12 12 1 2 1 2 21 2 1 12 E G G G x x RT x x G x x G   = + + + ( ) ( ) 2 2 21 12 12 1 2 21 2 1 2 21 2 1 12 2 2 12 21 21 2 1 12 2 2 1 12 1 2 21 ln ln G G x x x G x x G G G x x x G x x G           = +      +    +       = +      +    +   （3）UNIQUAC and UNIFAC equation For multi-component system Only the parameters of binary system required 12.3 Property Changes of Mixing The Excess properties of real solutions ln E i i i i i i S S x S R x x = − +   ln E i i i i i i G G x G RT x x = − −   E E i i i i i i V V x V H H x H = − = −   Property changes of mixing i i i   − M M x M  a t the same T, P and x Another E id M M M =  −  A molar (or unit-mass) solution property A molar (or unit-mass) pure￾species property

AM=∑xAM, Property changes of mixing For binary system A=Aw+-金） AM,=AM-4成 (d△M】 Gibbs-Duhem equation (r学-含0 Property changes of mixing S=AS+R∑xlnx,G=AG-RT∑xlnx VE=AV H=△H Example 12.2 The excess enthalpy (heat of mixing)for a liquid mixture of species 1 and 2at fixed T and Pis represented by the equation: HE=xx2(40x+20x) where HE is in Jmol.Determine expressions for AFAξas functionsofx] Solution 12.2 那=+-政所=- dx d H5=xx(40x+20x)> H5=20x-20x Whence dH=20-60 i5=20-60x2+40.xi5=40.x

7  =  M x M  i i Property changes of mixing For binary system 1 1 ( ) 1 2 1 1 1 d M M M x dx d M M M x dx     =  + −         =  −     Gibbs-Duhem equation , ,     1 0 N i i P x T x i M M dT dP x d M T P =           + −  =        Property changes of mixing ln E i i i S S R x x =  +  ln E i i i G G RT x x =  −  E E V V H H =  =  Example 12.2 The excess enthalpy (heat of mixing) for a liquid mixture of species 1 and 2 at fixed T and P is represented by the equation : 1 2 1 2 (40 20 ) E H x x x x = + where H E is in J mol-1 . Determine expressions for 1 2 E E H H as functions of x1 Solution 12.2 1 1 2 1 1 1 (1 ) E E E E E E dH dH H H x H H x dx dx = + − = − 1 2 1 2 (40 20 ) E H x x x x = + 3 1 1 20 20 E H x x = − Whence 2 1 1 20 60 E dH x dx = − 2 3 3 1 1 1 2 1 20 60 40 40 E E H x x H x = − + =