Question

The vapor pressures of benzene and toluene at 95◦C are, respectively, 155.7 kPa and 63.3 kPa. A gaseous mixture consisting of 50 moles of benzene and 50 moles of toluene was cooled to 95◦C. As a result of the cooling, some of the benzene and toluene condensed. If the total pressure above the condensate was 101.3 kPa, calculate:

(a) the mole fraction of benzene in the condensate (liquid).
(b) the mole fraction of benzene in the gas phase after cooling. (c) the number of moles of benzene in the condensate.

Answer 1

(a) According to Raoult's law

P_A = P^o_A x X_A

P_A = partial pressure of component A

P^o_A = vapour pressure of pure A

X_A = mole fraction A in liquid

say benzene mole fraction in liquid = X_B

P^o_B = 155.7 kPa

P_B = 155.7 kPa x X_B -----------(1)

similarly for toluene

P_T = 63.3 kPa x X_T -----------------(2)

P = total pressure = P_T + P_B = 101.3 kPa given

add 1 and 2

P_B + P_T =101.3 kPa =   63.3 kPa x X_T + 155.7 kPa x X_B -----(3)

also we know that sum of all mole fractions = 1

X_B + X_T = 1

X_T = 1 - X_B

replace in 3

101.3 kPa =   63.3 kPa x (1 - X_B) + 155.7 kPa x X_B

X_B = 0.411

Ans (a) = 0.411

(b)

In gas phase P_B = P x Y_B

P_B = partial pressure of benzene

P = total pressure

Y_B = mole fraction in vapour phase

from equation (1) P_B = 155.7 kPa x 0.411 = 64 kPa

64 kPa = 101.3 kPa x Y_B

Y_B = 0.63

(c) we had total 50 moles benzene and 50 moles toluene

assuming that it behaves as an ideal gas

X_B = n_B/(n_B + n_T)

n_B/X_B = (n_B + n_T) ---(a)

n_T/X_T = (n_B + n_T) ----(b)

n_B/n_T = X_B/X_T = 0.411 / (1 - 0.411) = 0.698

n_B = 0.698n_T

similarly n'_B/n'_T = Y_B/Y_T = 0.63 / (1-0.63) = 1.703

n'_B = 1.703n'_T

also n_T + n'_T = 50 mol

n_B + n'_B = 50 mol

1.703n'_T + 0.698n_T = 50 mol

replace n'_T = 50 - n_T

1.703(50 - n_T)  + 0.698n_T = 50 mol

85.13 - 1.703 n_T + 0.698n_T = 50 mol

35.13 = 1.005nT

n_T = 35 mol

n_T' = 15 mol

n'_B/n'_T  = 1.703

n'_B = 25.545 mol

n_B = 50 - 25.545 mol = 24.455 mol

Ans = 24.455 mol