《化工热力学》课程授课教案（讲义）Chapter 11 Solution thermodynamics - Theory

Chapter 11 Solution thermodynamics:Theory Purpose: In the chemical,petroleum,and pharmaceutical industries multi-component gases or liquids commonly undergo composition changes as the result of mixing and separation processes,the transfer of species from one phase to another,or chemical reaction. To develop the theoretical foundation for applications of thermodynamics to gas mixtures and liquid solutions 且的 1、了解溶液热力学的基本概念 2、学习溶液热力学的基本原理 3、为相平衡的学习打下基础 content 11.1 Fundamental Property Relation 11.2 The chemical Potential and Phase Equilibrium 11.3 Partial Properties 11.4 Ideal-Gas Mixture 11.5 Fugacity and Fugacity Coefficient:Pure Species 11.6 Fugacity and Fugacity Coefficient:Species in Solution 11.7 Generalized Correlations for the Fugacity Coefficient 11.8 The Ideal Solution 11.9 Excess Properties

Chapter 11 Solution thermodynamics: Theory Purpose: In the chemical, petroleum, and pharmaceutical industries multi-component gases or liquids commonly undergo composition changes as the result of mixing and separation processes, the transfer of species from one phase to another, or chemical reaction. To develop the theoretical foundation for applications of thermodynamics to gas mixtures and liquid solutions 目的 1、了解溶液热力学的基本概念 2、学习溶液热力学的基本原理 3、为相平衡的学习打下基础 content 11.1 Fundamental Property Relation 11.2 The chemical Potential and Phase Equilibrium 11.3 Partial Properties 11.4 Ideal-Gas Mixture 11.5 Fugacity and Fugacity Coefficient: Pure Species 11.6 Fugacity and Fugacity Coefficient: Species in Solution 11.7 Generalized Correlations for the Fugacity Coefficient 11.8 The Ideal Solution 11.9 Excess Properties

11.1 Fundamental Property Relation REVIEW Fundamental Property Relations of Thermodynamics for Homogeneous Phase System For one mol For n mol dU=Tds-Pdv d(nU)=Td(nS)-Pd(nV) dH TdS+Vdp d(nH)=Td(nS)+(nV)d(np) dA=-PdV-SdT d(nA)=-Pd(nV)-(nS)d(nT) dG=VdP-SdT d(nG)=(nV)dp-(nS)dT 1.=-n G 使用这些方程时一定要注意一下几点： 1.恒组分，恒质量体系，也就是封闭体系： 2.均相体系（单相）； 3.平衡态间的变化；只存在体积功： 4.常用于1摩尔的时的性质。 For the more general case of a single-phase,open system nG=g(pT) d(nG)=+Gdp+Gdm+G) on. d0-g2n+1e.p+2ew Chemicalpotential d(nG)=(nV)dP-(nS)dT+udn, In the same way d(nU)=Td(nS)-Pd(nv)+udn

11.1 Fundamental Property Relation REVIEW Fundamental Property Relations of Thermodynamics for Homogeneous Phase System For one mol For n mol dU TdS PdV dH TdS VdP dA PdV SdT dG VdP SdT = − = + = − − = − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) d nU Td nS Pd nV d nH Td nS nV d nP d nA Pd nV nS d nT d nG nV dP nS dT = − = + = − − = − , , ( ) [ ] ( ) [ ] T n P n nG nV P nG nS T  =   = −  使用这些方程时一定要注意一下几点： 1. 恒组分，恒质量体系，也就是封闭体系； 2. 均相体系（单相）； 3. 平衡态间的变化；只存在体积功； 4. 常用于 1 摩尔的时的性质。 For the more general case of a single-phase, open system 1 2 ( , , , ,., ,.) i nG g p T n n n = , ( ) ( ) [ ]P n nG d nG dT T  = +  , ( ) [ ]T n nG dP P  +  , , 1 1 1 ( ) [ ]T P n j nG dn n   +  , , 2 2 2 ( ) [ ]T P n j nG dn n   +  , , , , ( ) ( ) ( ) ( ) [ ] [ ] [ ] P n T n T P n i j i i i nG nG nG d nG dT dP dn T P n     = + +     Chemical potential , , ( ) [ ] j i i T P n i nG n    =  ( ) ( ) ( ) i i i d nG nV dP nS dT dn = − + In the same way ( ) ( ) ( ) i i i d nU Td nS Pd nV dn = − +

d(nH)=Td (nS)+nvdp+>udn, d(n4)）=-nsdT-Pd(nr)+∑4，dm, Fundamental Property Relation A -n on on For the case ofn mole of solution d(nG)=(nV)dp-(nS)dT+udn, For the case of one mole of solution n=1 ni-xi dG=aP-T+∑4k→(0n=p(p=-S 11.2 The chemical Potential and Phase Equilibrium For a closed system consisting of two phases in equilibrium Each individual phase is an open system a phase (nG)°(nV)(nS B phase (nG°(n/P(nS)

( ) ( ) i i i d nH Td nS nVdP dn = + + ( ) ( ) i i i d nA nSdT Pd nV dn = − − + Fundamental Property Relation , , , , , , ( ) [ ] ( ) [ ] ( ) [ ] j i j i j i i nS nV n i nS P n i T nV n i nU n nH n nA n      =   =   =  For the case of n mole of solution ( ) ( ) ( ) i i i d nG nV dP nS dT dn = − + For the case of one mole of solution n=1 ni =xi i i i dG VdP SdT dx = − + → , , ( ) ( ) T n P n G G V S P T   = = −   11.2 The chemical Potential and Phase Equilibrium For a closed system consisting of two phases in equilibrium Each individual phase is an open system αphase β phase ( ) ( ) ( ) . i i nG nV nS n       ( ) ( ) ( ) . i i nG nV nS n      

d(nGy"=(nV)"dp-(nS)"dT+u"dn" d(nG)=(nV)dp-(nS)dr+u dn The total system property is expressed by (nM)=(nM)"+(nM)" dnG)=(nV)dP-(nS)dT+∑dm+∑4'dm Since the two-phase system is closed d(nG)=(nV)dp-(nS)dT Σ城+∑4城=0+3W=0=12 dn"-dn" For a closed system consisting of x phases in equilibrium 4=4=.=40=12N) Multiple phases at the same T and P are in equilibrium when the chemical potential of each species is the same in all phases Attention Key points in this section 11.3 Partial Properties 11.3.1 Definition of the partial molar property the partial molar property of species i in solution - 在恒温、恒压下，物系的容量性质随某种组分摩尔数的变化率叫 做该组分的偏摩尔性质

( ) ( ) ( ) i i i d nG nV dP nS dT dn      = − + ( ) ( ) ( ) i i i d nG nV dP nS dT dn      = − + The total system property is expressed by ( ) ( ) ( ) nM nM nM   = + ( ) ( ) ( ) i i i i i i d nG nV dP nS dT dn dn     = − + +     Since the two-phase system is closed d nG nV dP nS dT ( ) ( ) ( ) = − 0 i i i i i i dn dn         + = i i i i n n n dn dn     + = = − ( 1,2,. ) i i i N     = = For a closed system consisting ofπphases in equilibrium . ( 1,2,. ) i i i i N       = = = = Multiple phases at the same T and P are in equilibrium when the chemical potential of each species is the same in all phases Attention Key points in this section 11.3 Partial Properties 11.3.1 Definition of the partial molar property the partial molar property of species i in solution ( ) , , j i i T P n nM M n    =      在恒温、恒压下，物系的容量性质随某种组分摩尔数的变化率叫 做该组分的偏摩尔性质

Attention Solution properties M H,S,A.U.v,. Pure-species properties M,H.S.4.U.V. Partial properties M:.S.4.0.v. =G偏摩尔自由焓是一种化学位 ATTENTION 偏摩尔性质的三个重要要素：①恒温、恒压；②容量性质；③随 某组分摩尔数的变化率。. 物理意义 在恒温、恒压下，物系中某组分摩尔数的变化所引起物系的一系 列热力学性质的变化。偏摩尔性质的物理意义可通过实验来理解。如： 在一个无限大的、颈部有刻度的容量瓶中，盛入大量的乙醇水溶液， 在乙醇水溶液的温度、压力、浓度都保持不变的情况下，加入1mol 乙醇，充分混合后，量取瓶上的溶液体积的变化，这个变化值即为乙 醇在这个温度、压力和浓度下的偏摩尔体积。 ATTENTION 偏摩尔性质的三个重要要素：①恒温、恒压；②容量性质；③随 某组分摩尔数的变化率。· 溶液性质M:H,S,A,U,V, 纯组分性质M,:H,S,A,U

Attention ： Solution properties M H S A U V : , , , , , . Pure-species properties : , , , , , . M H S A U V i i i i i i Partial properties : , , , , , . M H S A U V i i i i i i ( ) , , j i i i T P n nG G n     = =      偏摩尔自由焓是一种化学位 ATTENTION ： 偏摩尔性质的三个重要要素：①恒温、恒压；②容量性质；③随 某组分摩尔数的变化率。. 物理意义 在恒温、恒压下，物系中某组分摩尔数的变化所引起物系的一系 列热力学性质的变化。偏摩尔性质的物理意义可通过实验来理解。如： 在一个无限大的、颈部有刻度的容量瓶中，盛入大量的乙醇水溶液， 在乙醇水溶液的温度、压力、浓度都保持不变的情况下，加入 1mol 乙醇，充分混合后，量取瓶上的溶液体积的变化，这个变化值即为乙 醇在这个温度、压力和浓度下的偏摩尔体积。 ATTENTION ： 偏摩尔性质的三个重要要素：①恒温、恒压；②容量性质；③随 某组分摩尔数的变化率。. 溶液性质 M H S A U V : , , , , , . 纯组分性质 : , , , , , . M H S A U V i i i i i i

偏摩尔性质M,:万5，A,0,。 偏摩尔自由焓定义为化学位是偏摩尔性质的一个特例，而化学位 的连等式，只是在数值上相等，物理意义完全不同 11.3 Partial Properties 11.3.1 Definition of the partial molar property Thinking(11-1) 0=u)】 on,)aspnm omlh4=ag)） -G on, Attention Chemical potential 10010 on, on, 11.3.2 Equations Relation Molar and Partial Molar Properties For a system (T,P,x1,X2.) any thermodynamic property M(H,U,G,S,etc.)is 1)Summability Relation M=∑xM,nM=∑n,M, Can be used to calculate mixture properties from partial properties? 2)Gibbs/Duhem equation

偏摩尔性质 : , , , , , . M H S A U V i i i i i i 偏摩尔自由焓定义为化学位是偏摩尔性质的一个特例，而化学位 的连等式，只是在数值上相等，物理意义完全不同 11.3 Partial Properties 11.3.1 Definition of the partial molar property Thinking(11-1) ( ) j i i i nS, nV, n nU U n     =      , , ( ) j i i i nS P n nH H n     =      ? only ( ) , , j i i i T P n nG G n     = =      Attention ： Chemical potential , , , , , , , , ( ) ( ) ( ) ( ) [ ] [ ] [ ] [ ] j i j i j i j i i T P n nS nV n nS P n T nV n i i i i nG nU nH nA n n n n          = = = =     11.3.2 Equations Relation Molar and Partial Molar Properties For a system (T, P, x1, x2. ) any thermodynamic property M (H, U, G, S, etc.) is 1) Summability Relation i i i i i i M x M nM n M = =   Can be used to calculate mixture properties from partial properties？ 2) Gibbs/Duhem equation

〔)r+(.dp-∑a-0Σa-0oaRn 作用： (1)检验实验测得的混合物热力学性质数据的正确性； (2)从一个组元的偏摩尔量推算另一组元的偏摩尔量。 3)Generic relation - 11.3.3 Partial Properties in Binary Solutions Generic relation =M- aM →，=M+x dM,=M- dM Summability Relation M=∑xM,→M=xM,+x,M Gibbs/Duhem equation ∑(xd,）=0+xdm,+xdm,=0con.T const.p,T Example 11.3 Solution 11.3 w-()) M-∑xM,(cons.P,T) The molar volume of the binary antifreeze solution

, , 0 i i P x T x M M dT dP x dM T P           + − =        ( ) , 0 ( . , ) i i T P  x dM const P T = 作用： (1)检验实验测得的混合物热力学性质数据的正确性； (2)从一个组元的偏摩尔量推算另一组元的偏摩尔量。 3) Generic relation , , , j i k i k k i k T P x M M M x x        = −             11.3.3 Partial Properties in Binary Solutions Generic relation , , , j i k i k k i k T P x M M M x x        = −             → 1 2 1 dM M M x dx = + 2 1 1 dM M M x dx = − Summability Relation M x M = i i→ M x M x M = + 1 1 2 2 Gibbs/Duhem equation ( ) , 0 . , i i T P x dM const p T  = → 1 1 2 2 x dM x dM const p T + = 0 . , Example 11.3 Solution 11.3 , , ( . , ) i i P x T x i i M M dM dT dP M dx T P M x M const P T       = + +           =   The molar volume of the binary antifreeze solution

V=∑识=g+g =0.3×38.632+0.7×17.765=24.025cm31mol The total number of moles required is 号器-26m The volume of each pure species is ',=xng=0.3x83.246×40.727=1017cm3 '3=x,n'2=0.7×83.246×18.068=1053cm Example 11.4 Solution 11.4 （a)H=400x1+600x2+xx,4(10x+20x）J1mol x2=1-x1 月=400x+601-x)+x1-x)[40x+200-4】,dl-180-60x =600-180x,-20x3J1mol (4) dx 月=1+盟=H+0-)盟 i1=600-180x-20x+(1-x)(-180-60x） =420-60x+40xJ1m0l (B) dH五2=600-180x,-20x-x(-180-60x) ，=H- =600+40xJ1mol (C) Another method m-(ea）,a-[ea on, on, H=600-180x-20x

1 1 2 2 2 3 0.3 38.632 0.7 17.765 24.025 / V xV x V x V i i cm mol = = + =  +  =  The total number of moles required is 2000 83.246 24.025 t V n mol V = = = The volume of each pure species is 3 1 1 1 0.3 83.246 40.727 1017 V x nV cm t = =   = 3 2 2 2 0.7 83.246 18.068 1053 V x nV cm t = =   = Example 11.4 , , , j i k i k k i k T P x M M M x x        = −             Solution 11.4 （a） ( ) 1 2 1 2 1 2 H x x x x x x J mol = + + + 400 600 4 10 20 / x2 = 1-x1 ( ) ( ) ( ) 1 1 1 1 1 1 3 1 1 400 600 1 1 40 20 1 600 180 20 / ( ) H x x x x x x x x J mol A = + − + − + −     = − − → 2 1 1 180 60 dH x dx = − − ( ) 1 2 1 1 1 1 dH dH H H x H x dx dx = + = + − → ( )( ) 3 2 1 1 1 1 1 2 3 1 1 600 180 20 1 180 60 420 60 40 / ( ) H x x x x x x J mol B = − − + − − − = − + 2 1 1 dH H H x dx = − → ( ) 3 2 2 1 1 1 1 3 1 600 180 20 180 60 600 40 / ( ) H x x x x x J mol C = − − − − − = + Another method ( ) , , j i i T P n nM M n    =      , ( ) , , j i i T P n nH H n    =      3 1 1 H x x = − − 600 180 20

M=60a-150n-20g n=60a-180n-20答 A=厂oa7 =600-10-20x36-20子 om Jrpm n=%+场品1品1 五，=420-20×3x+40x a- 6m会-异 :H2=600+40x H=600-180x-20x (b)H,=600-180×1-20×1°=400J1mol H2=600-180×0-20×03=600J1mo (c)=lim=420J /mol =lim H:=lim H =600+40=640J mol 例题4-1 在100℃和0.1013MP下，丙烯腈(1)-乙醛(2)二元混合气 体的摩尔体积为v=RT/P+[a+b+2Gy],a,b,c是常数，其单位 与V的单位一致。试推导偏摩尔体积与组成的关系，并讨论纯组分 (1)的偏摩尔性质和组分(1)在无限稀时的偏摩尔性质。 解：从公式推导偏摩尔性质 民-+0-w瑞

3 3 1 1 2 600 180 20 n nH n x n x n = − − 3 1 1 2 600 180 20 n nH n n n = − − ( ) 2 2 3 1 1 1 2 3 1 1 1 , , 1 2 600 180 20 3 20 T P n nH n n H n n n n n n n     −  = = − −  −        1 2 n n n = + 1 2 1 1 n n n n   = =   2 3 1 1 1 H x x = −  + 420 20 3 40 ( ) 1 3 2 1 3 2 2 2 , , 2 600 20 T P n nH n n H n n n n n     −  = = −        3 2 1 H x = + 600 40 3 1 1 H x x = − − 600 180 20 (b) 3 1 H J mol = −  −  = 600 180 1 20 1 400 / 3 2 H J mol = −  −  = 600 180 0 20 0 600 / (c) 1 1 1 0 lim 420 / x H H J mol  → = = 2 1 2 2 2 0 1 lim lim 600 40 640 / x x H H H J mol  → → = = = + = 例题 4-1 在 100 ℃和 0.1013MPa 下，丙烯腈（1）-乙醛（2）二元混合气 体的摩尔体积为 2 2 1 2 1 2 V RT P ay by cy y = + + +   2  ，a b c , , 是常数，其单位 与 V 的单位一致。试推导偏摩尔体积与组成的关系，并讨论纯组分 （1）的偏摩尔性质和组分（1）在无限稀时的偏摩尔性质。 解：从公式推导偏摩尔性质 ( ) 1 1 1 1 dV V V y dy = + −

Geaeie ratom-假n 2-2海-2+20 7=RTP+a(+2y2)+(2c-b) 对于纯组分(1)片=1=0(y=)=RT/P+a 对于无限稀组分(1)1→0，为→1m=RT/P+2c-b 定义M=mM组分1的无限稀偏摩尔性质 注意：(0必=)≠（以=）≠ 例题4-2 在25℃和0.1MPa时，测得甲醇(1)中水（2)的摩尔体积近似为 了2=18.1-32x2cm3mo1,及纯甲醇的摩尔体积为y=40.7cm3mo'。 试求该条件下的甲醇的偏摩尔体积和混合物的摩尔体积。 解：在保持T、P不变化的情况下，由Gibbs-Duhem方程 x,d+x,d2=0d=-点d2=-正(-6.4x,dk)=-6.4x,d 了d-了-64x,g=40.7-32x(cm3mor V=x+x乃 =x1(40.7-3.2x）+x3(8.1-32x) =40.7x+18.1x2-3.2x 11.3.4 Relations among Partial Properties Every equation that provides a linear relation among

Generic relation , , , j i k i k k i k T P x M M M x x        = −             1 2 1 2 1 2 2 2 2 dV ay by cy cy dy = − − + ( ) ( ) 2 2 1 1 1 2 2 V RT P a y y y c b y = + + + − 2 2 对于纯组分（1） 1 2 y y = = 1, 0 ( ) 1 1 V y RT P a = = + 1 对于无限稀组分（1） 1 2 y y → → 0 , 1 1 1 1 0 lim 2 y V RT P c b V  → = + − = 定义 定义 0 lim i i i y M M  → = 组分 i 的无限稀偏摩尔性质 注意： ( ) 2 2 1 V y V 1  =  ( ) 1 1 2 V y V 1  =  例题 4-2 在 25℃和 0.1MPa 时，测得甲醇（1）中水（2）的摩尔体积近似为 2 2 1 V x = − 18.1 3.2 cm3 mol-1，及纯甲醇的摩尔体积为 1 V = 40.7 cm3 mol-1。 试求该条件下的甲醇的偏摩尔体积和混合物的摩尔体积。 解：在保持 T、P 不变化的情况下，由 Gibbs-Duhem 方程 1 1 2 2 x dV x dV + = 0 ( ) 2 2 1 2 1 1 2 2 1 1 6.4 6.4 x x dV dV x dx x dx x x = − = − − = − 1 2 1 1 2 2 0 6.4 V x V dV x dx = −   2 1 2 V x = − 40.7 3.2 (cm3 mol-1) ( ) ( ) 1 1 2 2 2 2 1 2 2 2 2 1 2 2 40.7 3.2 18.1 3.2 40.7 18.1 3.2 V x V x V x x x x x x x = + = − + − = + − 11.3.4 Relations among Partial Properties Every equation that provides a linear relation among