Question

Consider the methanol(1) + ethanol(2) system at 101.325 kPa. Using Raoult's Law, predict the system temperature and vapor-phase composition for this mixture.

Answer 1

Apply Raoults:

x1*P°1 = y1*PT

x2*P°2 = y2*PT

and

y2 = 1-y1

x2 = 1-x2

so:

x1*P°1 = y1*PT

(1-x1)*P°2 = (1-y1)*PT

substitute Pt:

x1*P°1 = y1*101.3

(1-x1)*P°2 = (1-y1)*101.3

Assume a 50:50 composiiton in liquid

so:

PT = 101.325 kPa = 1.013 bar

0.5*P°1 = y1*1.013

0.5*P°2 = (1-y1)*1.013

Apply vapor pressure for antoine equations:

log10(P) = A − (B / (T + C))

P = vapor pressure (bar) T = temperature (K)

Methanol (A,B,C) = 5.15853 1569.613 -34.846

Ethanol (A,B,C) = 4.92531 1432.526 -61.819

methanol log10(P) = 5.15853 − (1569.613 / (T + -34.846))

ethanol log10(P) = 4.92531 − (1432.526 / (T + -61.819))

substitute:

0.5* (5.15853  − (1569.613  / (T + -34.846))) = y1*1.013

0.5*(4.92531  − (1432.526  / (T + -61.819)) ) = (1-y1)*1.013

solve for T,y1:

y1 = 0.5/1.013* (5.15853  − (1569.613  / (T + -34.846)))

y1 = 0.4935(5.15853  − (1569.613  / (T + -34.846)))

substitute in 2

0.4935*(4.92531  − (1432.526  / (T + -61.819)) ) = (1-0.4935(5.15853  − (1569.613  / (T + -34.846))))

solve for T

0.4935*4.92531 - 0.4935*1432.526 / (T + -61.819) = 1- 0.4935*5.15853 + 0.4935*1569.613 /(T + -34.846)

(0.4935*4.92531 - 1 + 0.4935*5.15853) = 0.4935*1569.613 /(T + -34.846)+0.4935*1432.526 / (T + -61.819)

3.976 = 774.60/(T + -34.846) + 706.9515 / (T + -61.819)

T = 49°C approx