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

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