Question

Assume Raoult's Law applies to a binary mixture of water and methanol at 70 C.

a) Which species should be considered species (a)? Why?

b) Prepare a Pxy diagram for this mixture

c) What is the composition of the vapor in equilibrium with liquid at x_i for methanol of 0.3?

d) What is the composition of the liquid in equilibrium with vapor at y_i for water of 0.3?

For parts (c) and (d), determine these composition values analytically. Then confirm that they are correct on the Pxy diagram you created in part (b).

Answer 1

the compound having lower boiling point is cosidered as (a). So in this methanol is having lower boiling point than water. Methanol boils at 64.5 deg.c while water boils at 100 deg.c at 1 atm pressure.

Antione constants :

Methanol =               ln Psat(Kpa)= 16.5785- (3638.27/(T-33.650), T in K

water                        ln Psat(Kpa)= 16.3872- 3885.70/(T-42.980) T in K

From Raoults law,   y1P= x1psat and y2P= x2p2sat and hence y1P+y2P = x1p1sat+x2p2sat

P= x1p1sat + x2p2sat,

x1= 1-x2

by varying x1 , we can get different values of P.

1 Calculate P1sat and P2sat at 70 deg.c, from Antoiine equations given

2. Assume x1 and Calcualte P. Calculate y1= x1p1sat/P.

Repeat the procedure to get different values of P.

the plot of P, x, y is drawn and shown below

at x1=0.3,   P= 0.3*P1sat+0.7*p2sat= 0.3*124.3+0.7*31.255=59.1685 Kpa

y1= x1p1sat/P= 0.3*124.3/59.1685=0.630, for y2= 0.3 means y1=0.7,

x1p1sat= y1P and y2P= x2psat

x1= y1P/P1sat and x2= y2P/P2sat

y1P/P1sat +(1-y1)P/P2sat= 1

0.7*P/124.36 +0.3P/31.255 =1, P= 65.67 Kpa

x1= y1P/P1sat= 0.7*65.67/124.36= 0.37 and x2= 1-0.37=0.63

the procedure is shown on the graph.