Multiphase Systems

Separation and purification - separate components of mixtures from

one another

Vapor-Liquid Liquid-liquid

water
vapor Extractor
Ethanol
Benzene Decanter
Ethanol liquid 1
Benzene Distillation
liquid 2
Solid-Liquid

liquid Ethanol
Amino acid

Solid
Amino acid
Crystallizer

liquid
Ethanol
 Multiphase Systems

Material transferred from one phase (gas, solid, liquid) to another

• Absorption
Gas molecules are absorbed by liquid
(gas to liquid)
• Distillation
More volatile liquid component moves to
vapor phase (liquid to gas)
Desorption
Liquid molecules are removed by gas (liquid to gas)
• Liquid-liquid extraction
Two liquid phases; molecule moves to
phase where it is most soluble
 Single-Component Phase Diagram

Single species phase-change - freezing, melting, evaporation and

condensation
Single Component Phase Equilibrium
Phase Diagram for H2O

In the one-phase regions,

one can vary either the
temperature, or the
pressure, or both (within
limits) without crossing a
phase line.

DF= 2 + c – 𝜋
=2+1–1
= 2 degrees of freedom.
Single Component Phase Equilibrium
Phase Diagram for H2O

Along a phase line (two

phases in equilibrium
with each other), number
of phases is 2.
Can't change the
temperature and pressure
independently.

DF= 2 + c – 𝜋
=2+1–2
= 1 degrees of freedom.
Single Component Phase Equilibrium
Phase Diagram for H2O

At triple point (three

phases in equilibrium)
but there is only one
point on the diagram
DF= 2 + c – 𝜋
=2+1–3
= 0 degrees of freedom.
 Phase Diagram of CO2

When the concentration of all species in each phase no longer change with time, the phase

are said to be in phase equilibrium.

 Phase Equilibrium

Air Air
Air

Water
T
(liquid) Later Eventually
Initial Equilibrium or Saturated
[Not saturated]
 Vapor Pressure (p*)

the equilibrium pressure of the vapor above its liquid.

a measure of its volatility.

• Highly volatile species tends to be more likely in vapor phase

• The higher the vapor pressure for a given temperature the greater the
volatility of species.

• Vapor pressure is related to boiling - liquids with high vapor pressures

(volatile liquids) will boil at lower temperatures.

• The vapor pressure and hence the boiling point of a liquid mixture depends
on the relative amounts of the components in the mixture.

• Vapor pressure is dependent of temperature

• Distillation - differences in the vapor pressure (volatility) of the

components in the liquid mixture
 Vapor Pressure (p*) Estimation

Ø Four methods can be used to estimate the component vapor pressures:

1. Vapor pressure tables in references eg. Water vapor pressure

2. Clausius - Clapeyron equation (best for ideal gases)

3. Cox charts.

4. Antoine equation (most common)

Water Vapor Pressure (p*)
 Clausius Clapeyron Equation

ΔH v
ln p* =− +B
RT
where
p ∗ = vapor pressure of pure substance
T = absolute temperature
R = gas constant

ΔHˆv = latent heat of vaporization

 Clausius-Clapeyron Equation
ΔH v
ln p* =− +B
RT
ΔHˆ
Plot of ln p* vs 1/T – straight line with a slope − v
and intercept B
R

€
 Example

Estimate the heat of vaporization of a pure component from the

following data:
(Ans: 37 kJ/mol) T(K) p* (kPa)
260 3.3
300 32.5
 Example 6.1-1

The vapor pressure of benzene is measured at two temperatures,

with the following results:
T 1 = 7.6 o C p1* = 40 mmHg
o
T 2 = 15.4 C p2* = 60 mmHg

Calculate the latent heat of vaporization and the parameter B in the

Clausius-Clapeyron equation and then estimate p * at 42.2oC using
this equation.
 Example
The vapor pressure of an organic solvent is 50 mm Hg at 25oC and
200 mm Hg at 45oC. The solvent is the only species in a closed flask
at 35oC and is present in both liquid and vapor states. The volume of
gas above the liquid is 150 mL. Estimate the amount of the solvent
(mol) contained in the gas phase.
Solution:
Vapor Pressure (p*) Estimation
Cox Chart
 Antoine Equation

B
log p* = A −
10
T +C

€
 Example
Calculate the vapor pressure of benzene at 50°C using the Antoine
Equation. Also estimate the normal boiling point of benzene (the vapor
pressure at 1 atm)
Solution:

T = 50oC …. P* = 271.2 mm Hg
P* = 760 mm Hg …… T = 80.1oC
 The Gibbs Phase Rule

Ø Variables can be divided into:

1.Extensive variables: depend on the system size. Examples:

moles, mass and volume.

2.Intensive variables: don’t depend on the system size. Example:

temperature, pressure, concentration, density, mole fractions,
mass fractions, specific volume.
 The Gibbs Phase Rule

The number of intensive variables which should be specified for a

given system at equilibrium (DF) is calculated for nonreactive
system by:
DF = 2 + c − Π
where:
DF :=Degree
2 + cof−freedom
Π
DF = 2 + c −: Number
Π of chemical species
P : Number of phases in a system at equilibrium

If R independent reactions occur among the system components and

the reactions proceed to equilibrium, then

DF = 2 + c − Π − R
 Example

Sulfur dioxide – water – air system

Air + SO2 + water
vapor

Dissolved Air +
dissolved SO2 +
water

DF = 2 + c − Π

Π = 2; c = 3 à DF = 2 + 3 - 2 = 3

Specify 3 intensive variables, temperature, pressure and mole fraction air or SO2 in
the gas or liquid phase
 Example

Ethanol – water system

Ethanol vapor +
water vapor

Ethanol + water

DF = 2 + c − Π

Π = 2; c = 2 à DF = 2 + 2 - 2 = 2

Specify 2 intensive variables, temperature and pressure , all concentrations are fixed
 Example

DF = 2 + c − Π
•Air-water-ammonia

Π = 2; c = 3 à DF = 2 + 3 - 2 = 3

Specify 3 intensive variables, temperature, pressure and any one of the concentration
terms are fixed, all other concentration terms are fixed

•Ethanol-air
Π = 2, c = 2 à DF = 2 + 2 - 2 = 2
Specify 2 intensive variables, temperature and pressure, all the concentration terms
are fixed
 Example 6.2-1
Determine the degree of freedom for each of the following systems at
equilibrium. Specify a feasible set of independent variables for each system.
1. Pure liquid water.
2. A mixture of liquid, solid, and vapor water.
3. A vapor-liquid mixture of acetone and methyl ethyl ketone.

Solution:
1. Π = 1; c = 1 à DF = 2 + 1 - 1 = 2
(Specify 2 intensive variables, T, P)

2. Π = 3, c = 1 à DF = 2 + 1 - 3 = 0

3. Π = 2, c = 2 à DF = 2 + 2 - 2 = 2
(Specify either T and P, or a gas phase mass or mole fraction and either T
or P)
 Gas-Liquid Systems: One Condensable
Component
Systems containing several components, of which only one is capable of existing
as a liquid at the process conditions
Ø Evaporation:
Water vapor
Liquid water
Evaporator Liquid water

Ø Drying:
Transfer
Water vapor
Wet solids of Liquid
Dryer Dried solids into Gas
phase
Ø Humidification:

Dry air
Humidified air
Liquid water Humidifier
 Gas-Liquid Systems: One Condensable
Component

Ø condensation and dehumidification:

Transfer of
component
Air of less water
Wet air from Gas to
Condenser or Liquid
Liquid water
Dehumidifier phase
 Saturation

At saturation and when saturated vapor = the only condensable species,

Partial pressure of the vapor in gas = pure component vapor pressure

Raoult’s Law, single condensable species : pi = yiP = pi*(T)

 Raoult’s Law, Single Condensable Species

Raoult’s Law,
single condensable species :

pi = yiP = pi*(T)

where
pi = partial pressure of the vapor in the gas
yi = mol vapor/ mol total gas
P = total pressure
pi*(T) = pure-component vapor pressure at the system
temperature
 Example

Methane and water are contained at equilibrium in a closed

container at 50°C and 760 mm Hg. Perform the gibbs phase rule
and calculate the molar composition of the gas phase.
Solution:
Π = 2, c = 2 à DF = 2 + 2 - 2 = 2
(Specify T and P , mole fractions can be determined
Raoult’s Law, single condensable species :
pH2O = yH2OP = pH2O*(T)
yH2O = pH2O*(T)/P
= 92.51/760
= 0.121 mol H2O/mol

yCH4 = 1- 0.0121 = 0.879 mol dry air/mol

 Example 6.3-1

Air and liquid water are contained at equilibrium in a closed

chamber at 75oC and 760 mm Hg, calculate the molar
composition of the gas phase.

Compound Range (°C) A B C T p*

Water 60 to 150 7.96681 1668.21 228 75 289.2

Solution:
Raoult’s Law, single condensable species :
pH2O = yH2OP = pH2O*(T)
yH2O = pH2O*(T)/P
= 289.1/760
= 0.381 mol H2O/mol

yDA = 0.609 mol dry air/mol

 Example

The vapor pressure of acetone at 22.7 oC is 200 mm Hg. Liquid

acetone is kept in a sealed flask at 22.7 oC and the gas above the
liquid contains air and acetone vapor at equilibrium and at a
pressure of 960 mm Hg. What is the partial pressure of acetone in
the gas in mm Hg? What is the partial pressure of N2 in the gas in
mm Hg? What is the mole fraction of acetone in the gas?

Solution:
pi = p*i(T) at equilibrium (Raoult’s Law)
Therefore, pi = 200 mm Hg

Since the sum of all partial pressures must equal the total pressure,
and the partial pressure of acetone is 200 mm Hg, the partial
pressure of air must be 760 mm Hg, 79% (600 mm Hg) is from
nitrogen.
yi = p*i(T)/P = 200 mm Hg/960 mm Hg= 0.21
 Saturation, Superheated Vapor

A gas in equilibrium with a liquid must be saturated with the volatile

components of that liquid
w w A w
A A A w A w
A
w w w
A A A
A w A w w A A
A A w w
A A A
w w w w A w

w ww w w w w
w w w w ww w w
w w w w w w w
w w w w w
w
w w w w w
w w w w w w w w w w w
w
w w w w w w w w
w w w w w w w w w w w w w w w w

Beginning of contact Air with Air saturated with vapor

between the Phases superheated vapor
pH 2O = 0.0 mm Hg pH 2 O = y H 2 O P < p *
H 2O (T )
*
pH 2 O = y H 2 O P = pH 2O ( )
Tdp

The partial pressure of a vapor at equilibrium in a gas mixture containing

a single condensable component cannot exceed the vapor pressure of the
pure component at the system temperature.
 Superheated Vapor

w w A w
A w A w A
w w w
A A
w A w w A A
A w w
A A
w w w w A w

w w
w w w w ww w w
w w w w w w w w w w
w w w w
w w w w w w
w w w w w w
w w w w w w w w w w

Air with Air saturated with vapor

superheated vapor
pH 2 O = y H 2 O P < p *
H 2O (T )
*
pH 2 O = y H 2 O P = pH ( )
2O
Tdp

• If pi = p*i, the vapor is saturated; any attempt to increase pi (by adding

more vapor or increasing pressure) will lead to condensation

• If pi =yiP < p*i(T), then the vapor is a superheated vapor (gas mixture
can hold more vapor)
 Dew Point
w w A w
A w A w
A
w w w
A A
w A w w A A
A w w
A A
w w w w A w

w w
w w w w ww w w
w w w w w w w w w w
w w w w
w w w w w w
w w w w w w
w w w w w w w w w w

Air with Air saturated with vapor

superheated vapor
pH 2 O = y H 2 O P < p *
H 2O (T )
*
pH 2 O = y H 2 O P = pH 2O ( )
Tdp

• If a superheated gas containing a single condensable species is cooled at

constant pressure, the temperature at which the vapor becomes saturated is
known as the dew point of the gas [Raoult’s law, pi = yiP = p*i(Tdp)]
• The difference between the temperature, Tg and Tdp is known as degrees
of superheat
For a given pressure, the temperature at which p* = P is the boiling point
 Example
A steam of air exiting a condenser at a temperature of 87oC and a pressure of 1125

mm Hg contains 25 mol % of water. Determine the dew point and the degree of

superheat of the air.

Ans: 74.7oC, 12.3oC

 Example 6.3-2

A stream of air at 100oC and 5260 mm Hg contains 10.0% water by volume.

1. Calculate the dew point and degrees of superheat of the air.

2. Calculate the percentage of the vapor that condenses and the final composition

of the gas phase if the air is cooled to 80oC at constant pressure.

3. Calculate the percentage condensation and the final gas-phase composition if,

Instead of being cooled, the air is compressed isothermally to 8500 mm Hg.

 Example 6.3-2

Table B.3 in Appendix B: Vapor Pressure of Water

Example 6.3-2
Example 6.3-2
Example 6.3-2
 Humidification (gas-liquid systems for single
condensable component)
Relative Saturation (Relative Humidity):
partial pressure of the vapor divided by the vapor pressure
of the vapor at the temperature of the gas
pi
sr (hr ) = ´ 100%
pi (T )
*

Example: a relative humidity of 85% means that the partial pressure of water vapor
equals 85% of the vapor pressure of water at the system temperature.

0% relative saturation - no vapor in the gas

100% relative saturation - the partial pressure of the vapor in the gas is
the same as the vapor pressure
 Humidification

Molal Saturation (Molal Humidity):

express vapor concentration in a gas by using the ratio of

the moles of vapor to the moles of vapor-free gas
pi moles of vapor
sm (hm ) = =
P - pi moles of vapor - free ( dry ) gas

where

pi = partial pressure of water vapor in the air.

P = the total pressure
 Humidification

Absolute Saturation (Absolute Humidity):

refers to the mass of water vapor per mass of bone-dry air:

pi M i mass of vapor
sa (ha ) = =
(P - pi )M dry mass of dry gas

Percentage Saturation (Percentage Humidity):

100 times the actual molal humidity of the air divided by
the saturated molal humidity at the same temperature and
pressure
sm pi
s p ( h p ) = * ×100% = *
( P − p ) ×100%
i

sm pi ( P − pi* )
 Example

Air at 50% relative humidity is cooled isobarically at 1 atm absolute

from 90°C to 25°C. Estimate the dew point and degrees of
superheat of the air at 90°C.
Solution:
Table B.3 in Appendix B: Vapor Pressure of Water
 Example

Solution:
 Example

Air at 50% relative humidity is cooled isobarically at 1 atm

absolute from 90°C to 25°C. How much water condenses (mol)
per 100 mol of feed gas?
Solution:

Basis : 100 mol feed gas

100 mol Q2 mol

y1 mol H2O/mol y2 ( mol H2O/mol)
(1-y1) mol BDA/mol (1-y2) (mol BDA/mol)
T = 90oC, P =760 mm Hg T = 25oC, P =760 mm Hg
Saturated with H2O
Q1 mol H2O (liquid)
 Example
100 mol Q2 mol
y1 mol H2O/mol y2 ( mol H2O/mol)
(1-y1) mol BDA/mol (1-y2) (mol BDA/mol)
T = 90oC, P =760 mm Hg T = 25oC, P =760 mm Hg
Saturated with H2O
Q1 mol H2O (liquid)

y1P = pH2O*(T);
y1 = pH2O*(90)/P
=262.9 mmHg/760mmHg = 0.346 mol H2O/mol

y2 = pH2O*(25)/P
=23.76 mmHg/760mmHg = 0.031 mol H2O/mol

Balance on Dry Air 100 (1-0.346) = (1 – 0.031) Q2 mol

Q2 = 67.49 mol

Total Mole balance 100 mol = Q1 + Q2; (Q2 = 67.49mol)

Q1 = 32.51 mol H2O condensed
 Example
The humidity of air at 30°C and a total pressure of 750 mm Hg absolute (100 kPa) is
0.0055. Calculate (a) the percent relative humidity, (b) the molal humidity, and (c) the
partial pressure of the water vapor in the air. What is the dew point of the air-vapor
mixture?
 Example 6.3-3

Humid air at 75oC, 1.1 bar (825.07 mm Hg), and 30% relative
humidity is fed into a process until at a rate of 1000 m3/h.
Determine:

1.The molar flowrate of water, dry air, and oxygen entering the
process unit.

2.The molal humidity, absolute humidity, and percentage

humidity of the air.

3.The dew point.

 Example 6.3-3

Humid air at 75oC, 1.1 bar (825.07 mm Hg), and 30% relative humidity
is fed into a process until at a rate of 1000 m3/h.
Assumptions : Humid air behaves as an ideal gas.

1000 m3/h
y mol H2O/mol
(1-y) mol BDA/mol
T = 75oC, P =825.07 mm Hg
hr = 30%
Example 6.3-3
 Example

Air at 90oC and 1 atm absolute contains 10 mole % water. A

continuous stream of this air enters a compressor-condenser in
which the temperature is lowered to 15.6oC and the pressure is
raised to 3.00 atm. The air leaving the condenser is then heated
isobarically to 100oC. Calculate the fraction of water that is
condensed from the air, the relative humidity of the air at 100oC
and the ratio of the outlet volume (m3) at 100oC/the volume (m3)
feed air at 90oC.
Solution:
Basis : 1 m3 feed air

15.6oC 100oC
90oC 3 atm 3 atm
1 atm absolute compressor condenser heater
0.1 mol water / mol nc mol hr = ?
0.9 mol dry air / mol yw mol water /mol nair,out mol air
(1- yw) mol dry air/mol
nw mol water
 Example
Solution:
Basis : 1 m3 feed air

15.6oC 100oC
90oC 3 atm 3 atm
1 atm absolute compressor condenser heater
0.1 mol water / mol nc mol hr = ?
0.9 mol dry air / mol yw mol water /mol nair,out mol air
(1- yw) mol dry air/mol
nw mol water

Moles of feed air

nfeed air =
PV
=
(1atm )(1000L )
RT æ L.atm ö
ç 0.08206
mol.K ø÷ ( 363K )
è
= 33.6 mol
nfeed water = 0.1(33.6) = 3.36 mol
nfeed dry air = 0.9(33.6) = 30.24 mol
 Example
Solution:
Basis : 1 m3 feed air

15.6oC 100oC
90oC 3 atm 3 atm
1 atm absolute compressor condenser heater
0.1 mol water / mol nc mol hr = ?
0.9 mol dry air / mol yw mol water /mol nair,out mol air
(1- yw) mol dry air/mol
nw mol water

Mol fraction of water in the air leaving the condenser

pw
yw =
P
from table B.5 , p w = f(T =15.6 o C) = 0.01773 bar
p 0.01773 bar
yw = w = = 5.833x10-3
P 3(1.01325)bar
 Example
Solution: Basis : 1 m3 feed air

15.6oC 100oC
90oC 3 atm 3 atm
1 atm absolute compressor condenser heater
0.1 mol water / mol nc mol hr = ?
0.9 mol dry air / mol yw mol water /mol nair,out mol air
(1- yw) mol dry air/mol
nw mol water
Mass balance around compressor/condenser

nfeed dry air = (1- y w ) nc

30.24 = (1- 5.833x10-3 )nc
nc = 30.41mol

nfeed air = nw + nc
nw = 33.6 - 30.41= 3.19 mol

\ Mass fraction of water that is condensed from the air

3.19 mol water condensed
= = 0.949
3.36 mol water in feed air
 Example
Solution:
Basis : 1 m3 feed air

15.6oC 100oC
90oC 3 atm 3 atm
1 atm absolute compressor condenser heater
0.1 mol water / mol nc mol hr = ?
0.9 mol dry air / mol yw mol water /mol nair,out mol air
(1- yw) mol dry air/mol
nw mol water

The relative humidity of the air at 100oC

pw
sr (hr ) = * x100%
p w (T)
Partial pressure of water in the air leaving the condenser , p w = 0.01773bar
Vapor pressure of water at 100 o C, p *w = f (T =100 o C) =1.01325 bar

0.01773
\ hr = x100 =1.75%
1.01325
 Example
Solution:
Basis : 1 m3 feed air

15.6oC 100oC
90oC 3 atm 3 atm
1 atm absolute compressor condenser heater
0.1 mol water / mol nc mol hr = ?
0.9 mol dry air / mol yw mol water /mol nair,out mol air
(1- yw) mol dry air/mol
nw mol water

The ratio m3 outlet at 100oC/m3 feed air at 90oC.

nair,outRT
Vair,out =
P
nair,out = nc = 30.41mol nair,outRT
Vair,out =
(30.4 mol) æç 0.08206 mol.K
L.atm ö
÷ ( 373K ) m 3
nair,out
P
= nc = 30.41mol
Vair,out = è ø x
(3atm ) 1000L Vair,100o C 0.31m3outlet air
=
= 0.31m3 Vair,90o C m3 feed air
 Example
Methane is burned with atmospheric air in a combustor. The analysis of the
combustion products on a dry basis is as follows;
CO2 10.00 %
O2 2.37 %
CO 0.53 %
N2 87.10 %
Calculate
a) the molar composition of combustion products on a wet basis.
b) the percentage of excess air required.
c) the fractional conversion of methane to carbon monoxide.
d) the dew point of the stack gas, assuming that the stack gas pressure is 760
mm Hg and water is the only condensable species
Basis : 100 mol of dry flue gases
nf mol CH4 10.0 mol CO2
0.53 mol CO
no mol Air 2.37 mol O2
0.21 mol O2 / mol 87.1 mol N2
0.79 mol N2 / mol nw mol H2O
X % excess CH4 + 2O2 à CO2 + 2H2O
CH4 + 1.5O2 à CO + 2H2O
 Example

Basis : 100 mol/h of methane

Nitrogen balance (Input = output)
æ 1mol Air ö
no = ( 87.1 mol N2 ) ç ÷ = 110.25 mol Air
è 0.79 mol N2 ø
Atomic Carbon balance (Input = Output)
æ 1mol C ö æ 1mol C ö æ 1mol C ö
(nf mol CH4 ) ç1mol CH ÷ = (10 mol CO2 ) ç1mol CO ÷ + ( 0.53 CO ) ç1mol CO ÷
è 4 ø è 2 ø è ø
\ nf =10.53 mol CH4

Atomic Hydrogen balance (Output = Input)

æ 2 mol H ö æ 4 mol H ö
(w
n mol H2 )ç
O
1 mol H O
÷ (
= 10.53 mol CH 4)ç
1 mol CH
÷
è 2 ø è 4 ø

\ nw = 21.06 mol H2O

 Example
Theoretical Air
æ 2 mol O2 ö
( )
no2
theo
= (10.53 mol CH4 ) ç
1 mol CH
÷ = 21.06 mol O2
è 4 ø

æ 1mol Air ö
(nair )theo = ( 21.06 mol O2 ) ç 0.21mol O ÷ =100.3 mol Air
è 2 ø

The molar composition of combustion products on a wet basis

Total moles of flue gases (wet)

nT =10 + 0.53 + 2.37 + 87.1+ 21.06 =121.06 mol wet flue gas
ni
The molar composition of flue gases on a wet basis, i.e. y i =
nT
y CO2 = 0.0826 y CO = 0.0044 y H2O = 0.174
y O2 = 0.0196 y N2 = 0.720

Percentage of excess air :- 110.25 - 100.3

% excess air = x 100% » 10%
100.3
 Example

Fractional conversion of methane to carbon monoxide :-

nCO 0.53
= = » 0.05
nCH4 10.53

The dew point of the stack gas, assuming that the stack gas pressure is 760 mm
Hg and water is the only condensable species

pH2O
y H2O = = 0.174 , \ pH2O =132.24 mm Hg
P

Table B-6, Tdp= f (pH2O=132.24 mm Hg (~ 0.18 bar)) = 57.8 oC

 Multicomponent Gas-Liquid Systems

At equilibrium,

w w A w w
w A w C A w
w w w A
A w w C
w w A A A A
w w w w A
w C
w A w w A w C A w w A w
w w w w w C w w w w
w w w w w C w w C w
w w w w
w w w w w C w w
w w w w w
w w w w C w w w C

(Pure Liquid) (Liquid Mixtures)

pi = yi P = pi* (T ) pi = yi P = xi pi* (T )

Raoult’s Law, multiple condensable species : pi = yi P = xi pi* (T )

pi = partial pressure of i
yi = mole fraction of i in vapor phase
P = Total Pressure
xi = mole fraction of i in liquid phase
pi*(T)= vapor pressure of i at temperature T
 Vapor-liquid Equilibria for
Benzene-Toluene (ideal solution/mixtures)

(a) pure toluene; (b) a mixture of equal (c) pure benzene

amounts of toluene and
benzene

p b = ½ P b* and pt = ½ P t*

Note also that although the initial amounts of benzene and toluene in the
solution were equal, more benzene than toluene escapes to the gas phase
because of benzene’s higher vapor pressure.
 Raoult’s Law

For ideal-gas mixture and ideal liquid solution:

ü General Raoult’s law for multicomponent mixtures is:

pi = yi P = xi pi* (T )
ü If the liquid mixture is a diluted one, i.e. xi ≈ 0.0 , Raoult’s law can be
reduced to:

Henry’s Law: pi = yi P = xi H i (T )

ü If the liquid phase is a pure one, Raoult’s law can be reduced into:
pi = yi P = pi* (T )

valid for the entire range of compositions for mixtures of similar substances
Where xi = mole fraction of component i dissolved in the liquid phase
Hi (T)= Henry’s law constant at the system temperature
 Example 6.4-2
Use either Raoults law or Henry ’ s law to solve the following
problems:

1.A gas containing 1.0 mol% ethane is in contact with water at 20oC
and 20.0 atm. Estimate the mole fraction of the dissolved ethane.

Solution:
 Example 6.4-2
Use either Raoult’s law or Henry’s law to solve the following problems:

An equimolar liquid mixture of benzene (B) and toluene (T) is in equilibrium with
its vapor at 30oC. What is the system pressure and the composition of the vapor?

Solution:
Txy and Pxy diagrams
 Vapor-liquid Equilibrium
(Ideal Solutions)

• In a vaporization process of liquid mixture

containing several components at constant Constant P
pressure, the vapor composition will have a
composition different from that of the liquid. Vapor
T phase
• As vaporization proceeds the composition of the Dew point
Liquid & curve
remaining liquid continuously changes and hence vapor
does so its vaporization temperature. phase

• A similar phenomenon occurs if a mixture of Liquid

vapors is subjected to condensation at constant phase
pressure – the composition of vapor and the
Bubble point
condensation temperature both change.
curve

xi,yi
 Vapor-liquid Equilibrium
(Ideal Solutions)

Ideal solutions – Raoult’s and Henry’s law are

obeyed & the gas vapor can be considered
Constant P
as ideal gas
Vapor D
phase
Constant P heating path on a T-x-y diagram T
C
C’ Liquid &
A – subcooled liquid vapor Dew point
curve
B - saturated liquid (Bubble Point Temp.) B phase B’
C - saturated vapor (Dew Point Temp.)
D - superheated vapor
B’ - vapor composition in equilibrium with Liquid
phase
liquid at bubble point. temp.
A Bubble point
C’ - liquid composition in equilibrium with curve
vapor at dew point. temp.
xi,yi

Constant P cooling path on a T-x diagram Constant P, heating path on a T-x-y diagram
 Vapor-liquid Equilibrium (VLE)
It is very important to know conditions at which condensation or evaporation
occurs.

Gas çè Liquid : Distillation

Gas è Liquid : Absorption
Liquid è Gas : Stripping/Desorption
 Vapor-liquid Equilibrium
(Ideal Solutions)

Bubble-point temperature (Tbp) is the temperature at which the first bubble

is formed at constant pressure. Liquid phase composition is given.
å i i (Tbp )
x p *

p i = y i P = x i p*i (Tbp ) hence, å yi =1 = i

(trial & error)
P
Bubble-point pressure (Pbp) of the liquid - the pressure at which the first
bubble vapor forms when the liquid is heated at given temperature
p i = y i Pbp = x i p*i (T )

å i i (T
x p *
)
å yi =1 = i
hence, Pbp = å x i p*i (T )
Pbp i
 Vapor-liquid Equilibrium
(Ideal Solutions)

Dew-point temperature (Tdp) is the temperature at which the first droplet of liquid is
formed at a given pressure. Gas phase composition is given.

= y P = x p (T )
yi P
pi *
hence, å x i = 1 = * (trial & error)
p i (Tdp )
i i i dp

Dew-point pressure (Pdp) is the pressure at which condensation occurs at constant

temperature

yi 1
å x i = 1 = Pdp å p*i (T ) hence, Pdp =
yi
i i
åi p* (T )
i
 Bubble Point Calculation
(Ideal Solutions)

Bubble Pressure – Given all xi’s and T, calculate P at which the first bubble
forms, and yi’s of the bubble

Bubble Temperature – Given all xi’s and P, calculate T at which the first
bubble forms, and yi’s of the bubble
n

∑x i pi* (Tbp ) = P
i=1

Vapor mole fractions in equilibrium with the liquid is

 Bubble Point Calculation
(Ideal Solutions)

Given T,x Calculate P,y Given P,x Calculate T,y

n

∑x i pi* (Tbp ) = P
i=1
 Dew Point Calculation
(Ideal Solutions)

Dew Pressure – Given all yi’s and T, calculate P at which the first droplet
condenses, and xi’s of the droplet

Dew Temperature – Given all yi’s and P, calculate T at which the first droplet
condenses, and xi’s of the droplet
n
yi P
∑p *
=1
i=1 i ( Tdp )

Liquid mole fraction in equilibrium with the vapor can be determined from
 Dew Point Calculation
(Ideal Solutions)

Given T,y Calculate P,x Given P,y Calculate T,x

Txy and Pxy diagrams for benzene-toluene
 Example 6.4-3
1. Calculate the temperature and composition of a vapor in equilibrium with a liquid

that is 40.0 mole% benzene-60 mole% toluene at 1 atm. Is the calculated

temperature a bubble-point or dew-point temperature?

Solution:
Since the liquid composition,x is given, this is a

bubble-point calculation. By trial-&-error, assume

T, calculate p*i(T) using Antoine equations until it

n
satisfy ∑x i pi* (Tbp ) = P
i=1
 Example 6.4-3
2. Calculate the temperature and composition of a liquid in equilibrium with a

gas mixture containing 10.0 mole% benzene., 10.0 mole% toluene, and the

balance nitrogen (which may be considered noncondensable) at 1.0 atm. Is the

calculated temperature a bubble-point or dew-point temperature?

Solution:
Since the gas composition, y is given, this is a dew-point calculation. By trial-&-
n
yi P
error, assume T, calculate p*i(T) using Antoine equations until it satisfy ∑
i=1
*
p (Tdp )
i
=1

The composition of the liquid in equilibrium is calculated using

 Example 6.4-3
3. A gas mixture consisting of 15.0 mole% benzene, 10.0 mole% toluene, and

75.0 mole % nitrogen is compressed isothermally at 80oC until condensation

occurs. At what pressure will condensation begin? What will be the

composition of the initial condensate?

Solution:
At 80oC, the p*i(T) is calculated using Antoine equation.

The composition of the liquid in equilibrium is calculated using

 Example 6.4-4
1. Using the Txy diagram, estimate the bubble-point temperature and the

equilibrium vapor composition associated with a 40 mol % benzene-60 mol %

toluene liquid mixture at 1 atm. If the mixture is steadily vaporized until the

remaining liquid contains 25% benzene, what is the final temperature?

Solution:
 Example 6.4-4
2. Using the Txy diagram, estimate the dew-point temperature and the

equilibrium liquid composition associated with a vapor mixture of benzene

and toluene containing 40 mol % benzene at 1 atm. If condensation proceeds

until the remaining vapor contains 60 % benzene, what is the final

temperature?

Solution:
 Example 6.4-5
A mixture that is 70 mol % benzene and 30 mol % toluene is to be distilled in a

batch distillation column. The column startup procedure calls for charging the

reboiler at the base of the column and slowly adding heat until boiling begins.

Estimate the temperature at which boiling begins and the final composition of the

vapor generated, assuming the system pressure is 760 mm Hg.

Solution:
 Example
A vapor that is 65 mole% styrene and 35 mole% toluene is in equilibrium with
a liquid mixture of the same two species. The system temperature is 86oC. Use
Raoult’s law to estimate the mole fractions of the liquid and the system
pressure.
Solution:

Since the vapor composition is given, 86oC is the dew point temperature and the
system pressure is a dew point pressure
1 1
Pdp = =
yi
åi p* (T ) éê 0.65 + *
0.35 ù
ú
i dp p *
(
êë s dpT ) p T (T )
dp ú
û
pi* can be estimated using Antoine equation.
 Example

pi* can be estimated using Antoine Eq. B

log10 ( p* ) = A −
T +C

ps*(86oC) = 114.3 mm Hg pT*(86oC) = 356.8 mm Hg

1 1
Pdp = =
yi é 0.65 ù
åi p* (T ) ê * + * ú 0.35 Pdp = 150 mm Hg
i dp
êë p s (Tdp ) p T (Tdp )úû

The molar composition of liquid phase

p i = y i P = x i p*i (Tdp )
yi P
xi =
p*i (Tdp )

xs = 0.853 mol styrene/mol

xT = 0.147 mol toluene/mol
 Example
A benzene-toluene mixture containing 30 mole% benzene vapor is placed a
closed container at 1 atm and 115°C. Is the vapor mixture in equilibrium
with its liquid mixture?
Solution:
Assuming ideal solution, so

p i = y i P = x i p*i (T)
( )
y B P = x B p*B (T) ..... 0.3(760) = x B p*B 115o C
y T P = x Tp*T (T) ..... 0.7(760) = x p (115 C)
T
*
T
o

PB*= 1990 mm Hg
PT*= 859 mm Hg

xB = 0.11 and xT = 0.62

xB + xT = 0.73 ….. the mixture is not in equilibrium
 Example

A vapor mixture containing 30 mole% benzene and 70% toluene at 1

atm is cooled isobarically a closed container from an initial temperature
of 115°C.

a. At what temperature does the first drop of condensate form?

What is its composition?

b. At one point during the process the system temperature is

100°C. Determine the mole fraction of benzene in the vapor and
liquid phases.

c. Calculate the ratio of total moles in vapor to total moles liquid at

the system of 100°C

a. At what temperature does the last bubble of vapor condense?

What is its composition?
 Example
A vapor mixture containing 30 mole% benzene and 70% toluene at 1 atm is
cooled isobarically a closed container from an initial temperature of 115°C.

a. At what temperature does the first drop of condensate form? What is its
composition?
Tdp = ? P = 1 atm (760 mm Hg)

Assuming ideal solution, so

p i = y i P = x i p*i (Tdp )
yiP 0.3(760) 0.7(760)
å xi = 1 = = + *
p i (Tdp ) p B (Tdp ) p T (Tdp )
* *

Trial and error until ∑xi=1

Tdp ~ 105°C xB = 0.12 and xT = 0.88
 Example

A vapor mixture containing 30 mole% benzene and 70% toluene at 1 atm is

cooled isobarically a closed container from an initial temperature of 115°C.
b. At one point during the process the system temperature is 100°C.
Determine the mole fraction of benzene and toluene in the vapor and liquid
phases *
p i = y i P = x i p i (T)
(
y B P = x B p*B (T) ..... y B P = x B p*B 100o C )
(
y T P = x Tp*T (T) ..... (1 - y B )P = (1 - x B )p*T 100o C )
(
..... y B (760) = x B p*B 100o C )
.... (1 - y B )(760) = (1 - x B )p (100 C)
*
T
o

PB*= 1350 mm Hg
PT*= 556 mm Hg

xB = 0.24 and yB = 0.45

xT = 0.76 and yT = 0.55
 Example
A vapor mixture containing 30 mole% benzene and 70% toluene at 1 atm is
cooled isobarically a closed container from an initial temperature of 115°C.
c. The ratio (total moles in vapor/total moles liquid) at the system
temperature is 100°C
Basis : F = 1 mol feed
V mol Overall balance :- 1 = L + V .. (1)
yB ~ 0.45
yT ~ 0.55 Benzene Balance:- 0.3(1) = 0.46V + 0.26L .. (2)
F = 1mol T=100oC Solve eq. (1) & (2) simultaneously
xB ~ 0.3
xT ~ 0.7
L = 0.714 mol
L mol V = 0.286 mol
xB ~ 0.24
xT ~ 0.76 V/L = 0.4 mol vapor/mol liquid
 Example

A vapor mixture containing 30 mole% benzene and 70% toluene at 1 atm is

cooled isobarically a closed container from an initial temperature of 115°C.
d. At what temperature does the last bubble of vapor condense? What is its
composition?

last bubble of vapor condenses = first liquid drop vaporises (i.e. at bubble point
temperature (Tbp ))
p i = y i P = x i p*i (Tbp )
P = å Pi = å x i p*i (Tbp ) = x Bp*B (Tbp ) + x Tp*B (Tbp )
i

f (Tbp) = 0.3pB*(Tbp) + 0.7pT*(Tbp) – 760 = 0

Trial and error until f (Tbp) = 0

Tdp ~ 98.5°C

yB = 0.51 and yT = 0.49

 Deviations from Raoult’s law

(a) When Raoult's law is obeyed, a plot of vapor pressure against mole fraction
yields a straight line.

(b) A mixture of acetone and chloroform shows negative deviations from

Raoult's law at 35°C, indicating that the two different molecules prefer each
other's company to their own.

(c) The opposite behavior is shown at 55°C by a benzene-methanol mixture

where the polar and nonpolar molecules prefer the company of their own kind