Question

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(40) 3. The molar volume (cm/mol) of a binary liquid mixture at a specific T and P is given by: V = 200x1 + 300x2 + X1 X2 (20x1 + 10x2) (a) Find expressions for the partial molar volumes of species 1 and species 2 at specific T and P (b) Show that the expressions for partial molar volumes follow the summability relation (c) Show that these expressions satisfy the Gibbs/Duhem equation at constant T and P (d) Find the values for the followings – Vi; V2; VO;

Answer 1

* Solution - a * Griven: The molar volume f o con binary liquid mixterre -spaecific (T) and (P) -given as - at ap V = 2003, +300X5+ 36,586 ( 203, +10%) for Bin Binary مل since mixture 2, +2, ) ng = 1-K, * Refalace g = (1-x, ) in above equation – Ñ Ê V = 2004, + 300x(1-3, ) + *(1-x,) * (202,+1001-2,1] V = 2004, + 300 – 3002, +(34,20* [ 20% +10-102, ve 100%, +300 +(2,-2,7) x (:00,+10) Sincer 2 - 300%, +2002, 2-1004, y = -100%, +300 + 1 024*+102,-1023 - 10% since V = -1004+104 -- 1003 – 99% +300 1 =-goue, * differencia de equation - cuith respect tolbei) – vو da, (-1093–902, +300 da,

بل ( a few+bgen ) codfrut b. d you) an an ulure 4,65 = constant Apply above equation + d ( 300 ) Since- dx =) in du Since- da doen =nozeno) NY = -10x d (913) – gox d(X) +ond 1300 ) daly day (7) + kandido da, d constant) zo da , 50 d (300) =0 da S dr =-10 x 3x22 - goxi to = > dy dan - 30 x - go case ① *-- for Binary mixterre proporting og specipe -0 ond species- calcelented os– partial molor mi and =mtr. dm da, Ma = max, don dx m, and properties -0 and species- ulure - ma perstiel ef species

m = total to fend property *0 for specie - mi mov so N, V+Pig, dy da, Sincea V=-1033 – 90%, +300 - 304,3–90 dy day Hg = 1-3y but these velus in above equation- >> ū, >> =) = (-1090,3 – 9ox, +300) +(1-x) * (-307, – 90) V, = (-1013 – 90%, +300) + (-307,2–90+3023+902) -1043 – 90%, +300 30x22 – 90+30 +307$+90% 3021–104,3 – 30x} - -99,49gée, Vi = 202,3 – 302, +210 5) RI + 300-90 =) 3

for species-- Ma m = 50 V, M-, dv diley but the value go (V) on 1 from equation -0 ond equation => 21 = (-10233–90%, +300) – 2, X (-300,– 40) ก Vo = -10x? – 903, +300 + 302? + + 9open =) el 3 203; +300 4 scute Case-(2 Summability Relation given casa m = for < x; mi inary mixture B m xm, tegiñ དུ།

y here - m = total property mi = partial proporty of species- ño = purtial loroperty in Species- 7,1 g mole fraction ay species-06 - so put- may m = 77 img = ago above equation write as ? ? ? ? ? ? =) V = 2,5; +23. Sincer for Binary mixture - H, ha -1 ng = 1-1 => V = r. V, + (1-2,). =) Դ Դ Դ Դ Դ Դ Դ () 11 N=%,. , + v, -u. To > 2,. V; -2,. V, + van put the value of u and to from equation - and equation - 6 (3 => > v = 2, x ( 20303-304,2+210) – 2, x 1 2017}+300) +2012+36 Joget=30x* +2100, - 20hit 3009, +2011+ +300 - 3000* + 2003 + 2100, – 3002, + 300 V = =) + 3.

Ô V=-100? – 904, +300 the summabili so it is sutis atisty nability obfeine fotel probody from individered. Relation. Because wed * Case Cage-3 - - * - Chibbs) Dukem equation on do fined des consternt (T) and (P) - at ulure E xidmi a mpaperetical foroporty of species-i for Binary H4 = mole fraction og Binary mixterre Spoeira- a.dñ trodna co Dividing Both side log (de) » X,X dm dre thax dm, malen co dney г. since Hg =(1-2) مت > Replace rg (1-21) in above equation

ñ k, x , My +(1-4) xd ng dy 0 Replace- m, = T and ma = To =) (e. di +(1-0) x dva d2 co ☺ D from equation 3 V = +2023–3042 +210 Differenciate with respect top) Both sider >> di E 20 x daly del d24 -30% da + dal Š x (210) สัช, sincer d loonstant) da 20) dren da non =) dvi dra 20x327 – 30 XD!, to = > dve da, 661, 26096 6

t from equation ④ 4 こ 2030,3 + 300 Diffosen crude with respect to (42. Both side- =) dr. 20x dai + d ( 300) 2x) བྱ?) da) Since & 293 =3 =) da, j 2(300) dva dal, = 20x38 +0 are cos =) di da, = 60% - put the yolue and do dx, and equution - † in equation - I from equation 6 L.H.S ñ y equation (5 s 2, xr 600;?- 60,1 + (1-2,7 x 600,2 € so 60263 – 696124,2 +6921,2–6043 = 0 herea 2.H.S = Rohis so L.H.S = left hand side Right hand side it is satisfy the gilbe/puhem equation, Ri Hos. = -30

* Case Case - 6 7 O from equation 3 V, = 20243-3032 +210 Wulume put (24 = 1) in y pure specige - obtene yourlion above equation - put n, = 1 in above equation - > = V = 20x1113_ 30x1192+ 380 v; = V = 90- 30+ 210 V, = 200 cmpm moe Anguer valume at infinite Dilution calculated as. but 27 co in above equation V = 7, " = 2081033_ 30x1052+210 . "O = 210 cm3 Imol Answeer

e yola molar volume ö pure spocie-6 from equation 14 Vg = 207,9+300 f put 1,30 und Pig=1 in above seperation for yelume pure species- =) Tg = Va = Va = 20810337 300 => Va = 300 cm3) moe Answer molur volume of-specseem at infinite dilution calculated as- for species a at infinite - dilution [ig sol so put [al 21 = 1 i equation- V ܐܐܢܐ (2018 (0237 300 11 = 320 cm3/mol Anguer