The Antoine Equation gives the saturated vapour pressure (P) as a function of te
ID: 545073 • Letter: T
Question
The Antoine Equation gives the saturated vapour pressure (P) as a function of temperature (T): log10[P] = A-BIT + C)] where P. is in mmHg and T is in "C a. Determine the normal boiling point of benzene and the normal boiling point of toluene. The Antoine constants for benzene and toluene are as follows: Benzene: A = 6.91, B = 1211, C = 221 Toluene: A = 6.95, B = 1344, C = 219 20 marks] b. Assuming ideal behaviour, construct a T-xy diagram for the benzene toluene system at 760 mmHg based on benzene. (NB The temperature extremes of the diagram are the boiling points of the two components.) [40 marks] C. A mixture containing 40 mol% benzene and 60 mol% toluene is fed to a flash tank operating at 760 mmHg. If two thirds (by moles) of the feed is vaporised, what is the operating temperature of the tank and the 40 marks] composition of the two phases formed?Explanation / Answer
For Benzene, Antoine equation is
Log P1sat= 6.91-1211/(t+221)
At 760mm Hg, the boiling point is normal boiling point
Hence log (760)= 6.91-1211/(t+221), t= 79.55 deg.c boilng point of benzene
For toluene, similarly,
Log (760)= 6.95-1344/(t+219), t= boiling point of toluene= 111.3 deg.c
Let 1 represents benzene and 2 represents toluene. P1sat, P2sat are saturation pressures of Benzene and toluene, x1,x2 are mole fractions of benzene and toluene in the liquid phase, y1, y2 are mole fractions of benzene and toluene in the vapor phass. P is total pressure
From Raoult’s law for ideal solutions
,y1P= x1P1sat (1) and y2P= x2p2sat (2)
Addition of Eq.1 and 2 gives
P= x1P1sat+x2P2sat ( y1+y2=1)
P= x1P1sat+(1-x1)P2sat
Hence x1= (P-P2sat)/(P1sat-P2sat) (3)
Since the limits of temperature are from 79.55 deg.c to 111.3 deg,c
Assume some temperature between these two extremes, calculate P1sat, P2sat , use Eq.3 to calculate x1, since x2= 1-x1, use eq.1 to calculate y1 and hence y2=1-y1
The calculations of P1sat and P2sat are done using Antoine equation. The calculations are done in excel.
3. let the feed= 100 mole/hr, let L= liquid and V= vapor= (2/3)*100=66.67, F= L+V, L= 100-66.67=33.33
writing compnent benzene balance ( let y1= mole fraction of benzene in the vapor phase and x1= mole fraction toluene in the liquid phase)
100*0.4= 66.67*y+33.33*x ( x and y are in equilibrium)
40= 66.67x+33.33y (4)
the calculations involve trial and error. Assume one temperature, Calculate P1sat, P2sat, and calculate x1=(P-P2sat)/ (P1sat-P2sat) and y1 = x1P1sat/P and substitute these value in Eq.1. This calculations are to be done till eq.4 is satisfied.
the highlighted data in the table above shows the calculations.
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