Question

For a liquid mixture of benzene and toluene of equal composition by mass, perform the following:

a) Using Raoult’s law , calculate the total pressure and mole fractions of each substance in a gas phase containing only benzene and toluene vapor in equilibrium with the given liquid mixture at 60^oC.

b) Prove that the bubble point of the liquid mixture at pressure, P = 0.715 atm is 80^OC

c) Briefly describe what would happen if the mixture were heated slowly, eventually reaching a temperature of 80^oC

d) Suppose you had not been given the bubble point in part (b), but had been asked to determine it. How would you have done so

Answer 1

let mass of benzene= 50 gm = mass of toluene ( their composition is same by mass)

molar masses ( gm/mole) : Benzene =78 and toluene= 92

moles= mass/molar mass

moles : Benzene= 50/78= 0.64 and toluene= 50/92= 0.54, total moles= molar of benzene+ moles of toluene= 0.64+0.54=1.18

Mole fraction = moles/total moles

Mole fraction : Benzene (x1) =0.64/1.18= 0.54 and toluene (x2)= 1-0.54=0.46

Antoine equations

Benzene : log P(mm Hg)= 6.87987-1196.76/(t+219.161), t in deg.c

Toluene : log P(mm Hg)= 6.95087-1342.31/(t+219.187), t in deg.c

for a solutions obeying Raoult’s law,

, y1P= x1P1sat (1) and y2P= x2P2satc(2), where P1sat, P2sat are vapor pressures of benzene and toluene respectively, y1, y2 are mole fractions of benzene and toluene in the vapor phase, P is total pressure,x1 and x2 are mole fractions of benzene and toluene in the liquid phase

Addition of eq.1 and 2 gives

P=x1P1sat +x2P2sat, at 60 deg.c, log P1sat(mm Hg)= 2.592881, P1sat= 391.635 mm Hg

For toluene, P2sat= 138.9778

Total pressure= 0.54*391.635+0.46*138.9778 = 275.4127 mm Hg

From Eq.1, y1= 0.54*391.635/275.4127=0.77 and y2=1-0.77=0.23

Given 0.715= (0.54*P1sat+0.46*P2sat)/760 atm

Assume some temperature such that LHS of the above equation matches with RHS

At 80 deg.c, the calculations of P1sat= 757.6738 mm Hg and P2sat= 291.3029 mm Hg

Hence RHS of above equation = (0.54*757.6738+0.46*291.3029)/760 atm =0.715 atm

when the mixture is heated slowly, benzene being more volatile than toluene as seen by the vapor pressure of benzene more than that of toluene, benzene vaporizes and gets enriched in the vapor phase and toluene gets enriched in the liquid phase.

when bubble point pressure is not given

P= x1P1sat+x2P2sat, assume some temperature, calculate P1sat, P2sat and calculate the pressure, the bubble point pressure.