In: Other
Given:
Vapor–liquid equilibrium data in mole fractions for the
system
acetone–air–water at 1 atm (101.3 kPa) are as follows:
| y acetone in air | 0.004 | 0.008 | 0.014 | 0.017 | 0.019 | 0.020 | 
| x acetone in water | 0.002 | 0.004 | 0.006 | 0.008 | 0.010 | 0.012 | 
Find:
(a) Plot the data as (1) moles acetone per mole air versus
moles
acetone per mole water, (2) partial pressure of acetone versus g
acetone
per g water, and (3) y versus x. (b) If 20 moles of gas
containing
0.015 mole-fraction acetone is contacted with 15 moles of
water,
what are the stream compositions? Solve graphically. Neglect
water/
air partitioning.
(a) 1. Here data is given between mole fractions so first we convert in the form of mole ratio


Where Y is the mole ratio that is

similarly

| X | Y | 
| 0.002004 | 0.00401 | 
| 0.004016 | 0.00806 | 
| 0.006036 | 0.0142 | 
| 0.008064 | 0.0173 | 
| 0.01010 | 0.0194 | 
| 0.01214 | 0.0204 | 

2. now we have to plot partial pressure of acetone verses g acetone per gram of water
Partial Pressure

Where PT is the total pressure = 101.3 kPa
To calculate the g acetone per g of water we multiply by molecular weight of acetone and divide by molecular weight of water
(g acetone /g water) X'=

| P acetone (kPa) | X' | 
| 0.4052 | 0.00645 | 
| 0.8104 | 0.0129 | 
| 1.4182 | 0.01936 | 
| 1.722 | 0.0258 | 
| 1.9247 | 0.032 | 
| 2.026 | 0.03872 | 

(3) For y verses x we can directly plot from the given data

(b)
20 mole of gas containing 0.015 mole fraction of acetone
Water = 15 moles
Let G = Moles of gas free from acetone = 20 - 20 
0.015
G= 19.7 moles of acetone free gas
Pure water L = 15 moles
Let the counter current contact and
Y1 = moles ratio for gas at inlet = 0.0152
X1 = mole ratio for water outlet
Y2 = Mole ratio for Gas outlet
X2 mole ratio for water inlet = 0 (Water is pure initially)
We apply acetone mole balance
X1 and Y2 are in euillibrium
GY1+ LX2 =GY2 + LX1
G(Y1-Y2)=LX1
We put Y2 =Y
X1=X
Therfore Y=Y1-(L/G)X
When we draw this line in Graph A1 . Where this line intersect with the equillibrium line it gives us stram composition
Y= -(0.76)X + 0.0152

The Red line provides the by the equation of line
The stream composition
X= 0.005 (moles of acetone / mole of air)
Y= 0.012 (mole of acetone/ moles of water)