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1. Obtain the equilibrium mole fractions (liquid and vapor) of a
mixture of water, methanol and ethanol at 100 kPa and 70°C. You can
assume the mixture is ideal.
2. Create one plot where the equilibrium mole fraction of benzene
in the vapor phase is plotted against the equilibrium mole fraction
of benzene in the liquid phase for three different pressures (0.2
bar, 0.6 bar and 1 bar). Comment on the effect of pressure on the
relative volatility
1. to obtain the equilibrium mole fractions of mixture of water ,methanol & ethanol at 100 kPa & 70oC
assuming ideal mixture
using antoine equation
ln Psat = A-B/(T+C)
component | A | B | C |
water | 16.2620 | 3799.89 | -46.80 |
methanol | 16.5938 | 3644.30 | -33.39 |
ethanol | 16.6758 | 3674.49 | -46.70 |
therefore
by solving equations at 70o(343.15K)
ln Psatwater=16.2620-3799.89/(343.15-46.80)
lnPsatwater=3.43969529
Psatwater=31.1774566 kPa
similarly for methanol
Psatmethanol=125.071kPa
for ethanol
Psatethanol=72.3001454 kPa
now by using Raoults law
yiP=xiPisat
as &
100 = xw*31.1774566 +xM*125.071+xE*72.3001454 we can put xw=1-xM-xE
100 =(1-xM-xE)*31.1774566 +xM*125.071+xE*72.3001454
68.82=93.89*xM+41.12*xE
by trial & error we get xM=0.6 & xE=0.3 so
xw=0.1
yM=1.25*0.6=0.75
yw=0.3117xw=0.031
yE=0.723xE=0.216
checking answer
yw+yM+yE=0.75+0.031+0.216=0.9971
hence solution is appropriate
Q.2
to create plot where equilibrium mole fraction of benzene in the vapor is plotted against the equilibrium mole fraction of benzene in liquid phase .
yP=xPsat
for 3 different P=0.2,0.6,1 bar
for P=0.2 bar
we can plot on graph at different P
& to comment on effect of pressure on relative volatility
volatility of component is defined as partial pressure of component i to its mole fraction in liquid phase.
as the partial pressure is higher volatility increases
but the relative volatility is independant of total pressure. total pressure doesnt affect relative volatility.