Question

Relevant Data: R=8.314 J/mol-K, 1 atm=101325 Pa=1.01325 bar = 760 mmHg 1. The nonideal binary solution of n-pentanol (1) and n-hexane (2) at 30 °C has been fitted into a Wilson equation with the parameter values equal to 2,2 =1718 cal/mol and az, = 166.6 cal/mol. The two species at 30 °C have the following molar volumes and vapor pressures, JŮ = 108 cm /mol ſv = 130 cm /mol pel = 3.23 mmHg P"= 187.1 mmHg (a) (10 pts) What is that is relevant to the Henry's law constant of n-hexane dissolved in n-pentanol. (b) (20 pts) Determine the composition of the vapor in equilibrium with the liquid mixture containing 20 mol% n-pentanol.

Answer 1

Given - a 12 = 1718 coulmol. coulmol. 921 166.6 <> 108 cm3/mol دایہ = 130 cm3/001. sat Pi = 3.23 mm 49 sat Pa 187.1 MMH9 (i) Assuming n-pentanol & as component (t.) n - hexase as component (2.). NOW, From Wilson - Equation -; en (7) - ^21 en ( x1 + ^12.X2) + x2 LX +42X2 X2 +2* ^12 Jimilarly, en (92) = - en (x2+421*1) + X, [X1+42X2 112 121 where, Xzt Azit 112 = V2 Exp [- 912 & RT ^21 - N.P Exp [- 420 ] V2 لا RT

2 Now, Temp. (T) = 30 °C From Eq? Eq? (3.) é 13. f 14.) Il- 303 K. 112 - 130 cm3/mol 108 cmomo Exp 1718 Collmoi w [ mo) 1.9412 caliero ; 3036] 112 = 0.07 {R= R= 1.9872 col K-moi ^21 108 cm/ 130 cm3/mol 30 F166.6 Coul mol Еҳр 1.9872 Calk-mo; 303K 121 = 0.63 NOW, ! At infinite dilution of component.(2.), X2 -70 or: X, →1. So, from Eg? (2.), en (82) =, -en lot en (0+ 0.63(1)). 0.6361)) -- [ en (V) = 0.462- 2-[-0.93] 0.07 0.63 I to SO, oto.63. en (12) = 1.392 Yze 4.022

(2.) 93 3 97 X = 0.2, Then X2 = 0.8, From Eq? lil, en 17.) - Len (0.2 +0.07(0.8)) +0.p [ 0.370.0760.0, opet 0.63 0.63(62 en (1) = 1.8951 en (Y) = 6.653 Similarly, en (12) 112 = 21 XtMizxz -es [x2 + 221x3] x] -xs [ Fes[0.8+ 8 +0.63(0:2) X24*21* Ener 0.07 0.2 0.2 +0.07(0.P) 0.63 1 0.8+ 0.63(0:2) en (82) = 0.2) or Y2 = 1.234 NOW, p = sot Pi sot P2 •Y2. X2 1, X,. + P = 3.23 ( 6.653) (0.2) +187.1(1.234) (0.8) PE 203.97 mmng

4 Now, p.y, sat 7, xj (203.97) y, = 3.23 (6.653) (0.2) y, = 0.02 Yi+ y2 = 1 Y2 = 1-0.02 Y2 = 0.978 -