Answer:
Operating Pressure P = 793.716 mmHg
Y_Benzene y1 = 0.541
Explanation:
Given that;
liquid phase leaving the evaporator = 32.5 mole%
Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K
Now let 1 and 2 represent Benzene and Toluene respectively.
Antoine's Constant for these components are;
COMPONENETS A B C
Benzene 1 4.72583 1660.652 -1.461
Toluene 2 4.07827 1343.943 -53.773
Antoine's equation is expressed as;
Ps = 10^(A - (B/(T+C)))
Ps is in Bar and T is in Kelvin
so
P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar
P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar
now here, liquid leaving and vapor are both in equilibrium
composition of liquid leaving are;
X1 = 32.5% = 0.325
X2 = 1 - X1 = 1 - 0.325 = 0.675
Now
Raoult's Law is expressed as;
p × y1=x1 × pis for all components
So for Benzene ; p × y1=x1 × p1s ------let this be equation 1
for Toluene ; p × y2=x2 × p2s ------let this be equation 2
lets add equ 1 and 2
p × y1=x1 × p1s + p × y2=x2 × p2s
p(y1 + y2) = x1 × p1s + x2 × p2s
buy y1 + y2 = 1
therefore we substitute
p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar
we know that 1 Bar = 750.062 mmHg
so p = 1.0582 × 750.062
p = 793.716 mmHg
Also from equation 1
p × y1=x1 × p1s
y1 = (x1 × p1s) / p
y1 = (0.325 × 1.7617) / 1.0582
y1 = 0.541
Therefore;
Operating Pressure P = 793.716 mmHg
Y_Benzene y1 = 0.541
