Vapor-Liquid Equilibrium
(VLE) at Low Pressures
Chapter 8
Why study VLE?
 Many chemical and environmental
processes involve vapor-liquid
equilibria
 Drying
 Distillation
 Evaporation
Consider the ammonia production
process discussed in Chapter 1
Recycled Product
Ammonia and
unreacted feed
3 moles H2
1 mole N2
N2 + 3 H2 ⇋ Controlled
2NH3
by VLE
Chiller
Condenses
most of the
ammonia
Separator
Bleed
Stream
Reactor
partially
NH
3 vapor+
Nconverts
2 and H2
H2 and N2
to NH3
Liquid
~15%
Ammonia + N2 and H2conversion
Most of what we’ve discussed so far
in the course is VLE
 Raoult’s law is a vapor-liquid
equilibrium estimating equation
 Henry’s law is vapor-liquid equilibrium
estimating equation
 The biggest use of VLE analysis is in
distillation
 Check out the great picture in the text
book – page 160
 Distillation is separation by boiling point
Distillation Columns
Simple VLE Measurement Device
Othmer Still
Some standard conventions used in
VLE
 The lowest-boiling component (most
volatile) is usually called species a.
 The next lowest is species b etc.
 Tables are arranged similarly
Data from Table 8.1
Boiling
Temp
Mole Fraction Acetone in
Liquid
Mole Fraction Acetone in
Vapor
100
0
0
74.8
0.05
0.6381
68.53
0.1
0.7301
65.26
0.15
0.7716
63.59
0.2
0.7916
61.87
0.3
0.8124
60.75
0.4
0.8269
59.95
0.5
0.8387
59.12
0.6
0.8532
58.29
0.7
0.8712
57.49
0.8
0.895
56.68
0.9
0.9335
56.3
0.95
0.9627
56.15
1
1
This data is
used in
examples 8.1
to 8.3, and is
plotted on the
next slide
Acetone-Water Composition
1
Equilibrium
Curve
Acetone fraction in gas
0.9
0.8
0.7If
all we ever dealt with were
0.6binary systems, and if tables like
0.5the one used to create this figure
were available for all combinations
0.4
of species, we wouldn’t need VLE
0.3
correlations
0.2
Reference
Curve
0.1
0
0
Lowest boiling
point component
0.2
0.4
0.6
0.8
Acetone fraction in liquid
1
For Multispecies systems, there is
no simple graph we can make
 The K factor can be
used to help solve
this problem
yi
Ki 
xi
 Relative volatility is
an additional
approach

K More volatilespecies
K Less volatilespecies
ya xb

yb x a
Volatility Measures
Relative volatility and "K" factors
100
Relative Volatility, 
10
K, acetone
1
If
 is greater than 1.5 to 2
over the whole range of
composition, then distillation K,
is water
almost always the cheapest
0.1
0
0.2 technique
0.4
0.6
0.8
separation
Mole Fraction in the Liquid Phase of Acetone
1
Mathematical Treatment of LowPressure VLE
 The tables and graphs we’ve just
looked at are easy to interpret
 It would be nice if we could calculate
values, instead of reading them off
graphs
Our starting point is that the fugacity of
the gas phase equals the fugacity of the
liquid phase
fi
(1)
1
 fi
 yi f i
(1)
i
( 2)
(1) 0

( 2)
i
i
x fi
For ideal gases
the activity
coefficient is one
( 2)0
For gases we usually choose the total
pressure as the standard fugacity
For liquids we usually choose the pure
component partial pressure as the
standard fugacity
P
( 2)
i
Pi
yi P
i 
0
xi Pi
0
The activity coefficient of the liquid
phase

( 2)
i
yi P
i 
0
xi Pi
Partial pressure of
the gas
In example 8.2, we use this equation to
find the activity coeffients for water and for
acetone at a number of concentrations
Example 8.2
yacetonePTotal
 acetone 
0
xacetonePacetone
 water
y water PTotal

0
xwater Pwater
The pure
component
vapor pressures
are a function of
temperature,
and can be
calculated from
the Antoine
equation at the
appropriate
boiling points.
Calculated activity coefficients
activity coefficient
10
acetone
water
1
0
0.2
0.4
0.6
0.8
Mole fraction acetone in the liquid phase
1
Raoult’s Law
yi P
i 
0
xi Pi
yi P   i xi Pi
If we rearrange this
equation, we find…
0
Which is the same as Raoult’s
law, except for the activity
coefficient. Raoult’s law
applies to ideal solutions,
where the activity coefficient
is one.
The acetone water system is not ideal, since the
calculated activity coefficients are not one!!
How much would we be off if we
assumed ideal solution behavior for
acetone?
At 1 atm and
Xacetone=0.05
Experimental
values from
Table 8.1
Calculated
values using
=1
Equilibrium
boiling
temperature
74.8
96.4
Mole fraction
0.6381
0.1656
acetone in the
The calculational details
vapor phase
are in Example 8.3
The Four most common types of
Low Pressure VLE
 Ideal Solution Behavior
 1
 Positive Deviations from Ideal
 1
Solution Behavior
 Negative Deviations from Ideal
 1
Solution Behavior
 Two-Liquid Phase --Heteroazeotropes
Ideal Solution Behavior – Type I behavior
0.1
Activity Coefficient, 
1.0
 Consider a benzenetoluene system
 The two species are
very similar
chemically, and
behave as an ideal
solution
 Benzene has the lower
boiling, and therefore
the higher vapor
pressure
10
The activity coefficient of both species is one at all
concentrations – Figure 8.7b
BenzeneToluene=1
At any P and T
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of benzene
in the liquid phase, xa
Pi  xi Pi
0
800
400
yi P   i xi Pi
0
0
1
Pressure, torr
 Since the activity
coefficients of both
species is one,
they both follow
Raoult’s law
 1
1200
Ideal Solution Behavior
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of benzene
in the liquid phase, xa
Ideal Solution Behavior
Figure 8.7a
PToluene  xToluene 407 torr
PTotal  PBenzene  PToluene
1200
800
PBenzene
400
PBenzene  xBenzene1021 torr
PTotal
PToluene
0
Pi  xi Pi
0
Pressure, torr – at 90 C
yi P   i xi Pi
0
 1
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of benzene
in the liquid phase, xa
 1
0.8
1.0
Not to Scale
0.4
0.6
Equilibrium
curve
0.2
Reference
curve
0
 For an ideal
solution, the
mole fraction of
the most
volatile
component, is
always higher in
the gas than in
the liquid
Mole fraction of benzene
in the liquid phase
Ideal Solution Behavior
Figure 8.7c
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of benzene
in the liquid phase, xa
Ideal Solution Behavior
Figure 8.7 d
95
105
115
Not to Scale
85
Bubble
Point
75
Temperature, C
 If we heat a
mixture of benzene
and toluene, the
temperature where
it starts to boil is
called the bubble
point
 The combined
partial pressures
equal 1 atm
 1
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of benzene
in the liquid or vapor
phase, xa
Ideal Solution Behavior
Figure 8.7 d
115
Not to Scale
95
105
Dew Point
85
Bubble
Point
75
Temperature, C
 If we cool a
mixture of benzene
and toluene vapor,
the temperature
where it starts to
condense is called
the dew point
 The combined
partial pressures
equal 1 atm
 1
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of benzene
in the liquid or vapor
phase, xa
115
Bubble Point
Temperature at
the bubble point
85
Temperature, C
95
105
Dew Point
75
Liquid Phase
composition at
the dew point
Vapor Phase
composition at
the bubble point
Heat a liquid
mixture until it
starts to boil
0
0.2
0.4
0.6
0.8
Mole fraction of benzene in the liquid or vapor phase, xa
1.0
Now let’s look at non-ideal solution
behavior
 First let’s attack positive deviations
from ideal solution behavior (Type II
behavior)
 The activity coefficients of both (or
all) species is greater than one.
 The acetone water system is an
example of this behavior
 So is the isopropanol-water system
shown in Figure 8.8
10
Not to Scale
P=1 atm
isopropanol
water
1
Activity Coefficient, 
 Notice that the
behavior of each
species approaches
ideal (=1) as it’s
concentration
increases
15
Type II behavior of activity
coefficients – always greater than 1
Figure 8.8 b
Pure
0.8 1.0
isopropanol
Mole fraction of isopropanol
in the liquid phase, xa
Pure water
0 0.2
0.4
0.6
i 1
Type II behavior
Not to Scale
Pi   i xi Pi
0
1000
Pressure, torr – at 84 C
 Activity coefficients
greater than 1
mean that the
partial pressure of
the vapor is
greater than that
predicted with
Raoult’s law
800
Actual isopropanol
partial pressure
600
400
Predicted with =1
200
0
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of isopropanol in
the liquid or vapor phase, xa
i 1
Type II behavior
Not to Scale
Pi   i xi Pi
0
1000
Pressure, torr – at 84 C
 Activity coefficients
greater than 1
mean that the
partial pressure of
the vapor is
greater than that
predicted with
Raoult’s law
800
600
Actual water partial
pressure
400
200
Predicted with =1
0
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of isopropanol in
the liquid or vapor phase, xa
i 1
Type II behavior
Figure 8.8
a Scale
Not to
1000
Pressure, torr – at 84 C
 In this case (but not
every type II case)
the total pressure
curve (at constant
temperature)
displays a maximum,
which produces a
minimum boiling
azeotrope.
800
isopropanol partial
pressure
600
400
water partial
pressure
200
0
0
Pi   i xi Pi
0
Total pressure
0.2
0.4
0.6
0.8
1.0
Mole fraction of isopropanol in
the liquid or vapor phase, xa
i 1
Type II Solution Behavior
Figure 8.8c
0.8
1.0
Azeotrope
0.4
0.6
Equilibrium
curve
0.2
Reference
curve
0
Mole fraction of isopropanol
in the vapor phase
 When the total
pressure
displays a
maximum, the
equilibrium
curve crosses
the reference
curve
Not to Scale
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of isopropanol
in the liquid phase, xa
i 1
Azeotropes
Azeotrope
Mole fraction of
isopropanol in the liquid
phase
0 0.2 0.4 0.6 0.8 1.0
 At an azeotrope,
the liquid and
vapor have the
same
concentration
Not to Scale
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of
isopropanol in the liquid
phase, xa
100
Dew Point Curve
85
90
Azeotrope
95
Not to Scale
80
Bubble Point
Curve
75
 At mole fractions
below the
azeotrope, almost
pure water, and the
azeotropic
composition can be
distilled
 At mole fractions
above the
azeotrope, almost
pure isopropanol
and the azeotropic
composition can be
distilled
Temperature , C
Type II Solution Behavior
i 1
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of isopropanol
in the liquid phase, xa
Type II Solution Behavior
i 1
 Type II solutions do not necessarily
exhibit an azeotrope
 For example, the acetone-water
system does not
 Two factors contribute to this
behavior
 Difference in boiling point
 Deviation from ideal behavior
Type III Behavior
  for both species is less than 1
 Similar to Type II – just insert
maximum for minimum etc
 Consider the acetone-chloroform
system
Figure 8.9 b
1.0
chloroform
acetone
0.4
0.6
Not to Scale
P=1 atm
0.3
Activity Coefficient, 
 Notice that the
behavior of each
species approaches
ideal (=1) as it’s
concentration
increases
1.5
Type III behavior of activity
coefficients – always less than 1
Pure 0 0.2 0.4 0.6 0.8
1.0
Pure
chloroform
Mole fraction of acetoneacetone
in
the liquid phase, xa
i 1
Type III behavior
Not to Scale
1000
Pressure, torr – at 60 C
 Activity coefficients
less than 1 mean
that the partial
pressure of the
vapor is less than
that predicted with
Raoult’s law
800
600
Predicted
with =1
400
Actual acetone
partial pressure
200
0
Pi   i xi Pi
0
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of acetone in the
liquid or vapor phase, xa
i 1
Type III behavior
Not to Scale
1000
Pressure, torr – at60 C
 Activity coefficients
less than 1 mean
that the partial
pressure of the
vapor is less than
that predicted with
Raoult’s law
800
Predicted with =1
600
400
200 Actual chloroform
partial pressure
0
Pi   i xi Pi
0
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of acetone in the
liquid or vapor phase, xa
i 1
Type III behavior
Figure 8.9
a Scale
Not to
1000
Pressure, torr – at 60 C
 In this case (but not
every type III case)
the total pressure
curve (at constant
temperature)
displays a minimum,
which produces a
maximum boiling
azeotrope.
800 Total pressure
600
Actual acetone
partial pressure
400
200
Actual chloroform
partial pressure
0
0
Pi   i xi Pi
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of acetone in the
liquid or vapor phase, xa
i 1
Not to Scale
0.8
Equilibrium
curve
Azeotrope
0.2
0.4
0.6
Reference
curve
0
Mole fraction of acetone iin
the vapor phase
 When the total
pressure
displays a
minimum, the
equilibrium
curve crosses
the reference
curve
1.0
Type III Solution Behavior
Figure 8.9c
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of acetone in
the liquid phase, xa
i 1
Azeotropes
1.0
0.8
Equilibrium
curve
0.2
0.4
0.6
Azeotrope
0
Mole fraction of acetone iin
the vapor phase
 At an azeotrope,
the liquid and
vapor have the
same
concentration
Not to Scale
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of acetone in
the liquid phase, xa
75
Dew Point Curve
60
65
Azeotrope
70
Not to Scale
55
Bubble Point
Curve
50
 At mole fractions
below the
azeotrope, almost
pure chloroform,
and the azeotropic
composition can be
distilled
 At mole fractions
above the
azeotrope, almost
pure chloroform and
the azeotropic
composition can be
distilled
Temperature , C
Type II Solution Behavior
i 1
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of acetone in the
liquid phase, xa
What controls whether the activity
coefficients are greater or less than one?
 When >1 it indicates that the
solution species are repelled by each
other, or at least are more attracted
to themselves than to the other
species
 When <1 it indicates that the
solution species are more attracted to
each other than to their own kind
More about azeotropes
 For a given degree of mutual
attraction or repulsion, an azeotrope
is more likely for species with a small
difference in boiling point
 For pairs of compounds with the same
difference in boiling point, an
azeotrope is more likely for the pair
whose activity coefficients deviate
from one by the largest amounts
 Binary azeotropes between
compounds with wide differences in
boiling point are rare
 Most azeotropes are of the minimum
boiling type (over 90%)
Type IV Behavior
Two Liquid Phase --Heteroazeotropes
 Type II behavior occurs when two
species repel each other
 For a moderately strong repulsion an
azeotrope forms
 If the repulsion is strong enough, the
two liquids actually separate into two
separate phases
For example
 Consider the water and butanol
system
 Between 65% water and 98% water,
two phases occur
 At less than 65% water only a single
phase exists
 At greater than 98% water only a
single phase occurs
Water-butanol system
 From 0 to 65%
water one phase
exists
 From 98% to
100% water one
phase exists
 Between these
values, two phases
exist – one of 65%
water and one of
98% water
2
phases
1
phase
0
0.2
0.4
0.6
Mole fraction water
0.8
1
phase
1.0
In the single phase region…
water
20.
10.
1
5.0
1
1
2.0

butanol
1.0
1
Activity Coefficient, 

30
 The solution exhibits Type II behavior
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of water in the
liquid phase, xa

2

2
butanol
water
1
1
800
Total Pressure
400
Partial Pressure of
water
Partial Pressure of nbutanol
0
Pressure, torr – at 90 C
 In the two phase
region the total
pressure is the
sum of the
pressure from each
phase
 However, both
phases must be
exposed to the gas
phase
1200
Type IV Behavior
Pressure behavior
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of water in
the liquid phase, xa
1.0
0.8
0.4
0.6
Equilibrium
Curve
0.2
Reference
Line
0
 The composition
of the gas phase
stays constant
when two liquid
phases are
present
Mole fraction of water in the
vapor phase
Type IV
Vapor-Liquid Equilibrium
0
0.2
0.4
0.6
0.8
1.0
Mole fraction of water in the
liquid phase(s), xa
120
Vapor Phase
110
Dew Point Curve
100
Vapor
+ L1
Vapor
+ L2
L2
90
L1
Bubble Point Curve L1 + L2
80
Temperature , C
 Call the n-butanol
rich phase L1
 Call the water rich
phase L2
130
Type IV
Bubble Points and Dew Points
0
0.2
0.4
0.6
0.8
Mole fraction of water in the
liquid phase, xa
1.0
Boiling is necessary for this
behavior
 The predicted behavior
will only occur if both
phases are in contact
with the gas phase
Gas Phase
 Realistically this only
occurs during boiling
Liquid Phase II
 This type of behavior
is common in
petroleum refining
Liquid Phase I
Steam Distillation
 If we take type IV behavior to the
extreme, we consider two liquids that
are essentially completely immiscible
 For example water and mercury
 In this case we are always in the two
phase region
P   yi P   Pi  P
0
water
P
0
Mercury
The total pressure is the sum of the
pure component vapor pressures
 Consider example 8.7
 If n-butanol and water were
completely insoluable, what would
the boiling point be at one atm?
 What would the composition of the
vapor be?
P   yi P   Pi  P 0 water  P 0 n  butanol
Estimate the partial pressures of each
component with the Antoine Equation
log Pwater
B
1657.462
 A
 7.94917 
C T
227.02  T
1558.190
log Pn -butanol  7.838 
196.8812  T
Pn -butanol  Pwater  760 mmHg
Solve for T
iteratively
T=89 0C
At the boiling point of 89 C
 PH20 = 0.67 atm
 Pn-butanol= 0.33 atm
 Compare these results to Figure 8.12
 Remember that butanol and water
actually do exhibit considerable solubility
Distillation of the Four Types of
Behavior
 For systems that do not have an
azeotrope, distillation columns can
produce practically pure products
 The highest vapor pressure component is
separated into the overhead component
 The lower vapor pressure component is
separated into the bottoms
Gas-Liquid Equilibrium – Henry’s
Law
 In the previous discussions, both
components could exist as a pure
liquid at the temperatures of interest
 In other words we were interested in
vapor-liquid equilibria
 How can we extend this discussion to
include gas-liquid equilibria, for
species that do not condense
 Use Henry’s Law
Our discussions so far have been
about systems at low pressure
 At low pressures, the gas phase obeys the
ideal gas law
 At higher pressures we’ll need to consider
the fugacity coefficient in our calculations
 At both low and moderate pressures (up to
~1/2 the critical pressure) we won’t need
to adjust our liquid phase calculations
Low Pressure VLE Calculations
 Graphs are great to get a general
idea of how systems behave, but they
aren’t very accurate
 We need to develop a standard
approach to calculate vapor-liquid
equilibrium properties
At low pressures the fugacity
coefficients are 1 – but the liquid
phase activity coefficients aren’t
 Estimate them using the Van Laar
equation
log(  a ) 
Axb2
A

 xa  xb 
B

2
log(  b ) 
Bx a2
B 

 xa  xb 
A 

There are other estimating equations – this
one is just easy to use -- the basis for the
Van Laar equation is developed in chapter 9
2
The 6 Most Common VLE
Calculations
 Find the dew point, for a known
temperature
 Find the dew point, for a known pressure
 Find the bubble point for a known
temperature
 Find the bubble point for a known pressure
 Isothermal flash calculations
 Adiabatic flash calculations
These are all examples problems that can
be formulated as equilibrium flash
calculations
Flash Calculations
V
Vapour
F
T, P
Liquid
L
F can be a liquid,
a gas or a two
phase mixture
This is called a
flash calculation,
because if the
pressure is
reduced enough,
the liquid
changes to vapor
in a “flash”
To solve any of these problems we
need to identify our equations and
unknowns -- Table 8.6
 Material Balances
F V  L
xi F  yiV  xi L
Remember from Process
Engineering that you can
write n independent
material balances, if you
have n components
 Summation of mole fractions
x
i
1
y
i
1
More equations to be solved
 Equilibrium
yi  i Pi
Ki  
xi
P
Simplified version assuming ideal
gas
 Vapor Pressure equations
 Antoine equation is probably the most
accurate of the simple equations
More equations to be solved
 Calculate the activity coefficients
 Van Laar equation is simplest analytical
approach
 Energy balance
 Adiabatic flashes
H F  Hv  H L
We also need the inlet conditions
 Feed specification
 xi
 Temperature
 Pressure
Let’s do some example calculations
Dew Point – Example 8.9
 Estimate the boiling pressure and the
composition of the vapor in
equilibrium with a liquid that is:
 0.1238 mol fraction ethanol
 remainder water
 85.3 oC
First – Use Antoine’s equation to
find the pure component partial
pressures at this temperature
B
log Pi  A 
C T
At 85.3 C,
Pwater=0.5772 atm
PEtOH=1.3088
1657.462
log Pwater  7.94917 
227.02  T
1554.3
log PEtOH  8.04494 
222.65  T
Second
Find Activity Coefficients
 Use the Van Laar Equation
log(  water ) 
log(  EtOH ) 
Ax
2
EtOH
A

 xwater  xEtOH 
B

2
Axwater
2
A

 xEtOH  xwater 
B

2
Find the values
of A and B for
water and
ethanol in
Table A.7
water=1.0388
EtOH=2.9235
Next
Calculate the partial pressures
Pwater  x

0
water water water
P
Pwater  0.8762 * 2.9235 * 0.5772
Pwater  0.5254 atm
PEtOH  0.4737 atm
PTotal  0.9991 atm
Similarly for
Ethanol…
and…
Finally
Find the vapor phase mole fractions
Pi
yi 
PTotal
Pwater 0.5254
y water 

 0.5259
PTotal 0.9991
PEtOH 0.4737
y EtOH 

 0.4741
PEtOH 0.9991
Pressure Specified Calculations are
Similar
 They are harder, because the
Temperature appears in the Antoine
Equation
 They need to be solved interatively
Calculational Steps
 Guess a Temperature
 Calculate the Pure Component Vapor
Pressures
 Calculate the activity coefficients
 Calculate the partial pressures
 Calculate the total pressure, and
compare to the given pressure
Graphical Solution
 Add Figure 8.17
Bubble Point Calculations
 Mirror image of Dew Point
Calculations
 Temperature specified calculations are
easier
 Pressure specified calculations require an
iterative approach
Isothermal Flash Calculations
 Both T and P are specified
 The division of mass between liquid and
vapor is unknown
 Consider Example 8.13
 An ethanol-water mixture
 xaFeed=0.126
 is brought to equilibrium at
 1 atm
 91.8 C
 Estimate the vapor fraction and the mole
fraction of each species in the vapor phase
Let’s Solve it Graphically First
 Add figure 8.19
Analytical Solution
 Find the pure component vapor
pressures
 Assume a value for V/F
 Estimate the activity coefficients
 Calculate the K factors
 Use a material balance to find the
mole fractions in the vapor and liquid
phase
xaF F  yaV  xa L
xaF F  yaV  xa L
V
L
xaF  ya  xa
F
F
V
 V
 y a  x a 1  
F
 F
Divide by F
Equation 8.12
Adiabatic Flash Calculations
 In addition to the equations from the
previous example, you need an energy
balance
H F  HV  H L
 You must “guess” a temperature, then
perform the calculations, and finally check
to see if the energy balance requirements
are met
Solutions using K factor
approximating tools
 Activity coefficients are functions of
T,P and x (the liquid composition)
 Thus, K is also a function of T,P and
the liquid composition
 By ignoring the contribution of
composition, DePreister created
Figure 8.20 – an estimating tool for
finding K
K factor chart
Colligative Properties – Another
application of Raoult’s Law
 Boiling Point elevation
 Freezing Point depression
 Osmosis (Chapter 14)
 Predictable using Raoult’s Law
What happens as the concentration
of the solvent approaches 1?
 The solvent activity coefficient, ,
approaches 1
 Thus for dilute solutions of anything
the solution obeys Raoult’s law
 The identity of the solute doesn’t
matter, as long as it’s a dilute solution
and it’s not volatile!!
Boiling Point Elevation
 When the solute has a very high
boiling point, the solution vapor
pressure only depends on the solvent
0
PT  Psolvent  Psolute
PT  xsolvent P
0
solvent
Boiling Occurs when the vapor
pressure equals the pressure of the
surroundings
 Example 8.15
 One mole of sucrose (MW=342.3
g/mole) is dissolved in 1000 g of
water
 What is the vapor pressure of this
Boiling
solution at 100 0C?
PT  0.982 *1 atm  0.982 atm
won’t occur,
unless
you’re at an
altitude
above sea
level
What temperature will cause
boiling?
First – determine what pure component
vapor pressure is required
PT  xsolvent P
0
solvent
1atm  0.982 P
P
0
solvent
0
solvent
 1.018atm
From the
steam tables
we find
T=100.51 0C
– or a boiling
point
elevation of
0.51 0C
We can approximate the behavior of
this solution at low concentrations of
solute
 The change in boiling point is directly
proportional to the molality of the
solution
Tboiling  K b m
Figure 8.22
Freezing Point Depression
 Just as adding a non-volatile solute
increases the boiling point, it also
decreases the freezing point
 See Example 8.16
T freezing  K f m