Question

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 - Vj; 12; V9;

Answer 1

(10) 1 1 17 Va 2004, + 30042 + 2,42(20x, tok2 ) Put , x ₂ = (1-x, g - 2009, +300(1-,) + m2(1-12) (203, +10 (1-2, 1) = 2000, + 300-3000, +(2, -4,29 ( 20 %, +10-109,) --100M, +300+ (2,-2) (10,+10) 5-1007, +300+10/ +103, -10%,3 -10 842 -904, +300 - 103,3 - 2 2 Now Partial property of component is calculated as reo miamta dm 2 dai m2 man, dm ani ai ū vt 12 dv du da dv dne 2 -go- 3on, ū = (-901, +300 -104,3 ) + 2(-90-303,2) --90x, +300-104,3 – 90x2 - 30%,2 x2 =-goxo-gou2-ion, 3 - 304,272 +300 Ja ven, пр. d221 . (-904, + 300-101, 3) - Mil-90-303,2) -90 , +300-100,3 +964, +30 31 Jy = 2013 $300 (2) 3 3

w (b) Summability Relation is geuen by menim, y v at J, it H. 2 T novi + H2 Uz a nic-gon,-gon2-104,3 - 30x, ²M27.300) + M2 (204; 3+300) = -902,2- 90X, 22-104,4- 307,369 + 300%, + 20%,3 m2 +300kg 5-902,2- 90%(1-21)-102,4-302,3(1-2, ) + 300W,+ 207,3(1-%,) + 300C 1-3, ) s-goni? - goni+g6n, 2 102,4-30 M 37 303,4 +306X, +20x23 - 20/0,4 + 300 – 30$x, - 9 0 * 1 - 10 % 3 4 3 0 0 4 = v mixture is written as Hence, expressions for partial notar nolumes follow the Summability Melation CE ) Gibbs . Quhem equatish for binary niem, +42d ña zo dening both side by da, in aboue aq miler hun) + x + x2 (4m Replace mi = v ma a vi n D 3 du, from ean 2 الاه 3 ū = -GON, ~90X2 -10 W 3- 304,212 +300 3 -900, goli-1,3-1001 3on, Cl-?) +300 AGA =-96 v, - got gów, -104,3 - 301,2+30 +300 ✓ = -90.+200,3 – 30m+30 0 3 3

Arts defferentisteng Wihat (se, I both sides 60 m, 2 - 60m, 091, from equ V2 i 20 20 2,3 +300 differentiating d V₂ Nrit xa 1 2 60 n,? day from equ * G4 du f 12 du 0 dono duo LUS niloor, 2 bon, J + 22 (604,2) - E soon,3-601,2 +601? x2 - 6 x 3 60 %, + 6 o wº - ) 12+601²-603 = 60 43 -60 a Rus ma (laulo) 2 LUS arus so gt satisfy Gibbs/ Duhen equ at constant 14p. Q 1 m, 2 M, CM, 31.0) 9 m2 #ivi ū, laut now rool -30m 12 +300 i -gon, -9042-ion, 5-90+207,3 309,2 +300 Per vi -90 + 20-30+300 Via 200 3 Putn, =lo cm3 Imal

#V₂ = Je (at x2 = 1:0) 7, 20W,3 +300 Put 12 21-0 V2 a 20.+300. V2² 320 cm 3 /mol ū, O Vo ī liens # (u lim ( 2 ) 11,0 71230 (-90.+ 20 on? 3.-304,2+300) ū lim Ngo -Got 300 = 210 cm3 /mol ū À 210 SI - lim lūz) 24-0 a lim (201,3 +300 720 . - 20 + 300 . 320 do Ja 2 320/cm3 Imal