Question

Assuming the validity of Raoult’s law, do the following calculations for the benzene(1)/toluene(2) system:

(a) Given x1 = 0. 33 and T = 100°C, find y1 and P.

(b) Given y1 = 0. 33 and T = 100°C, find x1 and P.

(c) Given x1 = 0. 33 and P = 120 kPa, find y1 and T.

(d) Given y1 = 0. 33 and P = 120 kPa, find x1 and T.

(e) Given T = 105°C and P = 120 kPa, find x1 and y1.

(f) For part (e), if the overall mole fraction of benzene is z1 = 0. 33, what molar fraction of the two-phase system is vapor?

(g) Why is Raoult’s law likely to be an excellent VLE model for this system at the

stated (or computed) conditions

Answer 1

a) Using Raoult's law, y1 = x1P1sat/P and P = P1sat*x1/y1, where P1sat is the vapor pressure of pure benzene at the given temperature. Assuming ideal behavior, P1sat at 100°C is 101.3 kPa, so y1 = (0.33)(101.3 kPa)/(1 atm) = 33.5 kPa and P = (101.3 kPa)(0.33)/(0.33) = 101.3 kPa.

(b) Using Raoult's law, x1 = y1P/P1sat. Substituting as before, x1 = (33.5 kPa)(101.3 kPa)/(1 atm) = 3430 ppm and P = (101.3 kPa)(0.33)/(0.33) = 101.3 kPa.

(c) Using Raoult's law, y1 = x1P1sat/P and T = T1sat*y1/(x1P1sat), where T1sat is the boiling point of pure benzene at the given pressure. Assuming ideal behavior, T1sat at 120 kPa is 104.4°C, so y1 = (0.33)(101.3 kPa)/(120 kPa) = 0.28 and T = (104.4°C)(0.28)/(0.33) = 89.0°C.

(d) Using Raoult's law, x1 = y1P/P1sat and T = T1sat*y1/(x1P1sat). Substituting as before, x1 = (33.5 kPa)(120 kPa)/(101.3 kPa) = 39.7% and T = (104.4°C)(0.33)/(0.28) = 123°C.

(e) Using Raoult's law, x1 = P1sat1/(P1sat1 + P1sat2) and y1 = P1sat1*x1/P. Substituting as before, x1 = 0.497 and y1 = 0.415.

(f) The vapor mole fraction is given by y1 = z1*x1/(x1 + (1 - x1)*K), where K is the equilibrium constant for the system. Assuming ideal behavior, K = P2sat/P1sat = 1.3,

Step-by-step explanation:

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