Question

Problem 2: For a mixture of benzene and toluene (XA-0.33) at l 00 °C, determine the system P and assumptions made in solving this problem. List Problem 3:

Answer 1

Antoine equation is given by

log₁₀(Psat[mmHg])=A-B/(T[^oC]+C)

A

B

C

toluene	6.95087	1342.31	219.187
benzene	6.87987	1196.76	219.161

At T = 100 C,

saturation pressure of toluene,

$P_{T}^{sat}= 10^{6.95087-\frac{1342.31}{100+219.187}} = 556.5\ mm\ Hg$

For benzene,

$P_{B}^{sat}= 10^{6.87987-\frac{1196.76}{100+219.161}} = 1349.47\ mm\ Hg$

Assuming that the mixture is Ideal and hence follows Raoult's law.

By Raoult's law,

y_BP = x_BP_B^sat

y_TP = x_TP_T^sat

On adding these equation, we get

y_BP + y_TP = x_BP_B^sat + x_TP_T^sat

.: P =  x_BP_B^sat + x_TP_T^sat ........................................as y_B + y_T = 1

.: P = 0.33*1349.47 + (1-0.33)*556.5

.: P = 818.18 mm Hg.......................................................Answer

Now,

y_BP = x_BP_B^sat

.: y_B = (0.33*1349.47) / (818.18)

.: y_B = 0.544287.......................................................Answer