Question

A mixture of benzene and toluene is in a tank whose pressure is maintained at 2.0 atm. Observation through a sight glass on the vessel shows that half of the tank volume is occupied by liquid. Analysis shows that the liquid contains 50.0 wt% benzene and the balance toluene. Raoult's law applies at all composition s, and the density of the liquid is essentially independent of temperature. Estimate (a) the temperature and (b) the fraction of the system mass in the liquid phase

Answer 1

Apply Raoult's law:

x1*Pº1 = y1*PT

x2*Pº2 = y2*PT

where x1 and x2 are liquid fractions (1 for B and 2 for T) and y1 and y2 respectivley for gas phase

PT = Total pressure and Pº is partial pressure of each

recall that

x1+x2 = 1 so x2 = 1-x1

and y1+y2 = 1 so y2 = 1-y1

and PT = 2 atm = 2.027 bar

substituting

x1*Pº1 = y1*2.027

(1-x1)*Pº2 = (1-y1)*2.027

For the pressures:

For partial Pressure, Apply Antoine equation

log10(P) = A ? (B / (T + C))

Pº = 10^(A- B/(T+C))

where P is in bar and T in K

for benzene = 4.72583   1660.652   -1.461

for tolune = 4.23679 1426.448 -45.957

So the equation gets

Pº1 = 10^(4.72583- 1660.652/(T-1.461))

Pº2 = 10^(4.23679- 1426.448/(T-45.957))

substitute

x1*(10^(4.72583- 1660.652/(T-1.461))) = y1*2.027

(1-x1)*(10^(4.23679- 1426.448/(T-45.957))) = (1-y1)*2.027

liquid is x1 = 0.5 and x2 = 0.5 so

0.5*(10^(4.72583- 1660.652/(T-1.461))) = y1*2.027

0.5*(10^(4.23679- 1426.448/(T-45.957))) = (1-y1)*2.027

2 unkown (T and y1)

You could solv enaalytically or iteration

I will use iteration:

0.5*(10^(4.72583- 1660.652/(T-1.461))) = y1*2.027

0.5*(10^(4.23679- 1426.448/(T-45.957))) = (1-y1)*2.027

y1 = 0.2466 * (10^(4.72583- 1660.652/(T-1.461))) (EQN1)

y1 = 1-0.2466*(10^(4.23679- 1426.448/(T-45.957))) (EQN2)

iterate

0.2466 * (10^(4.72583- 1660.652/(T-1.461))) = 1-0.2466*(10^(4.23679- 1426.448/(T-45.957)))

from the iteration:

T = 390 K

T = 92 C

and y1 = 0.69

so benzene is richer

Benzene 0.5 in liquid and 0.69 in gas

Toluene 0.5 in liquid and 0.31 in gas