VAPOR – LIQUID
EQUILIBRIUM
• The Nature of Equilibrium
• The Phase rule.
• VLE Quantitative Behavior
• Raoult’s Law
• VLE by Modified Raoult’s Law
2
In this chapter we first discuss the nature of equilibrium, and then
consider two rules that give the number of independent variables required
to determine equilibrium states.
Then a qualitative discussion of vapor/liquid phase behavior.
Introduce the simplest formulations
temperatures,
pressures,
and
that
allow
phase compositions
for
calculation
of
systems
in
vapor/liquid equilibrium. The first, known as Raoult's law, is valid only
for systems at low to moderate pressures and in general only for systems
comprised of chemically similar species.
A modification of Raoult's law that removes the restriction to
chemically similar species is treated.
Finally the calculations based on equilibrium ratios or K-values are
considered.
3
Phase equilibrium
• Application
– Distillation, absorption, and extraction bring phases of
different composition into contact.
• Both the extent of change and the rate of transfer
depend on the departure of the system from
equilibrium.
• Quantitative treatment of mass transfer the
equilibrium T, P, and phase compositions must be
known.
4
The nature of equilibrium
• A static condition in which no changes occur in the
macroscopic properties of a system with time.
• At the microscopic level, conditions are not static.
– The average rate of passage of molecules is the same in
both directions, and no net interphase transfer of material
occurs.
• An isolated system consisting of liquid and vapor
phases in intimate contact eventually reaches a final
state wherein no tendency exists for change to occur
within the system.
– Fixed temperature, pressure, and phase composition
5
Phase rule vs. Duhem’s theorem
• (The number of variables that is independently fixed in a
system at equilibrium) = (the number of variables that
characterize the intensive state of the system) - (the number
of independent equations connecting the variable):
• Phase rule: F = 2 + ( N − 1)( ) − ( − 1)( N ) = 2 −  + N
• Duhem’s rule:
F = 2 + ( N −1)( ) +   − ( −1)(N ) + N  = 2
– for any closed system formed initially from given masses of prescribed
chemical species, the equilibrium state is completely determined
when any two independent variables are fixed.
– Two ? When phase rule F = 1, at least one of the two variables must be
extensive, and when F = 0, both must be extensive.
6
VLE: QUALITATIVE BEHAVIOR
Vapor/Liquid equilibrium (VLE) is the state of coexistence of liquid and
vapor phases. In this qualitative discussion, we limit consideration to systems
comprised of two chemical species, because systems of greater complexity
cannot be adequately represented graphically.
When N = 2, the phase rule becomes F = 4 - π. Since there must be at
least one phase (π = 1), the maximum number of phase-rule variables which
must be specified to fix the intensive state of the system is three: namely, P, T,
and one mole (or mass) fraction. All equilibrium states of the system can
therefore be represented in three dimensional P-T- composition space. Within this
space, the states of pairs of phases coexisting at equilibrium (F = 4 - 2 = 2)
define surfaces. A schematic three-dimensional diagram illustrating these
surfaces for VLE is shown in Fig. below:
7
8
This figure shows schematically the P-T-composition surfaces
which contain the equilibrium states of saturated vapor and
saturated liquid for a binary system:
The under surface contains the saturated-vapor states;
it is the P-T-yl surface.
The upper surface contains the saturated-liquid states;
it is the P-T-xl surface.
These surfaces intersect along the lines UBHC2 and
RKAC1, which represent the vapor pressure-vs.-T curves for pure
species 1 and 2.
9
Because of the complexity of Fig. 10.1, the detailed characteristics
of binary VLE are usually depicted by two-dimensional graphs. The
three principal planes, each perpendicular to one of the coordinate
axes.
A vertical plane perpendicular to the temperature axis is
outlined as ALBDEA. The lines on this plane form a P-x1-y1 phase
diagram at constant T.
A horizontal plane perpendicular to the P axis is identified by
HIJKLH. Viewed from the top, the lines on this plane represent a
T-x1-y1 diagram at constant P.
10
11
The third plane, vertical and perpendicular to the
composition axis, is indicated by MNQRSLM, this is the P-T
diagram.
At points A and B in Fig. 10.3 saturated-liquid and
saturated-vapor lines intersect. At such points a saturated
liquid of one composition and a saturated vapor of another
composition have the same T and P, and the two phases are
therefore in equilibrium.
12
13
14
• Within this space, the states of pairs of phases coexisting at
equilibrium define surfaces.
– The subcooled-liquid region lies above the upper surface; the
superheated-vapor region lies below the under surface.
– UBHC1 and KAC2 represent the vapor pressure-vs.-T curves for pure
species 1 and 2.
– C1 and C2 are the critical points of pure species 1 and 2.
– L is a bubble point and the upper surface is the bubblepoint
surface.
– Line VL is an example of a tie line, which connects points
representing phases in equilibrium.
– W is a dewpoint and the lower surface is the dewpoint surface.
• Pxy diagram at constant T
• Txy diagram at constant P
• PT diagram at constant composition
15
Fig 10.8
16
Fig 10.9
17
Simple models for VLE
When thermodynamics is applied to vapor liquid
equilibrium, the goal is to find by calculation the temperatures,
pressures, and compositions of phases in equilibrium.
18
• Raoult’s law:
– the vapor phase is an ideal gas (apply for low to moderate
pressure)
– the liquid phase is an ideal solution (apply when the species
that are chemically similar)
–
yi P = xi Pi sat
(i = 1, 2, ..., N )
• where xi is a liquid-phase mole fraction, yi is a vapor-phase mole
fraction, and Pi sat is the vapor pressure of pure species i at the
temperature of the system. The product yi P on the left side of Eq. is
known as the partial pressure of species i
19
Ideal-solution behavior is often approximated by
liquid phases wherein the molecular species are not too
different in size and are of the same chemical nature
although it provides a realistic description of actual
behavior for a small class of systems, it is valid for any
species present at a mole fraction approaching unity,
provided that the vapor phase is an ideal gas.
20
Dewpoint and Bubblepoint Calculations with
Raoult's Law
Engineering interest centers on dewpoint and bubblepoint calculations;
there are four classes:
In each case the name suggests the quantities to be calculated:
either a BUBL (vapor) or a DEW (liquid) composition and either P or T.
Thus, one must specify either the liquid-phase or the vapor-phase
composition and either T or P.
21
yi P = xi Pi
sat
(i = 1, 2, ..., N )
y
i
=1
i
P=
x
i
i
=1
1
sat
y
/
P
 i i
(i = 1, 2, ..., N )
i
For dewpoint calculation
P =  xi Pi sat (i = 1, 2, ..., N ) For bubblepoint calculation
i
Binary system
P = P2sat + ( P1sat − P2sat ) x1
22
23
Binary system acetonitrile (1)/nitromethane(2) conforms closely to Raoult’s law. Vapor
pressures for the pure species are given by the following Antoine equations:
ln P1sat / kPa = 14.2724 −
2945 .47
2972 .64
sat
ln
P
/
kPa
=
14
.
2043
−
2
t /  C + 224.00
t /  C + 209.00
(a) Prepare a graph showing P vs. x1 and P vs. y1 for a temperature of 75°C.
(b) Prepare a graph showing t vs. x1 and t vs. y1 for a pressure of 70 kPa.
(a) BUBL P
P = P2sat + ( P1sat − P2sat ) x1
At 75°C
P1sat = 83.21 P2sat = 41.98
P = 41.98 + (83.21 − 41.98) x1
e.g. x1 = 0.6
P = 66.72
x1 P1sat (0.6)(83.21)
y1 =
=
= 0.7483
P
66.72
At 75°C, a liquid mixture of 60 mol-% (1) and 40 mol-% (2) is in equilibrium with a vapor
containing 74.83 mol-% (1) at pressure of 66.72 kPa.
Fig. 10.11
24
Fig. 10.11
25
(b) BUBL T, having P = 70 kPa
ln P1sat / kPa = 14.2724 −
Select t
P1sat
P2sat
2945 .47
2972 .64
sat
ln
P
/
kPa
=
14
.
2043
−
2
t /  C + 224.00
t /  C + 209.00
P − P2sat
x1 = sat
P1 − P2sat
t vs. x1
y1 =
sat
1 1
t vs. y1
xP
P
Fig. 10.12
26
VLE modified Raoult’s law
• Account is taken of deviation from solution ideality in
the liquid phase by a factor inserted into Raoult’s
law:
yi P = xi i Pi sat
(i = 1, 2, 3, ...N )
The activity coefficient, f (T, xi)
P =  xi i Pi sat
i
1
P=
 yi /  i Pi sat
i
27
28
For the system methanol (1)/methyl acetate (2), the following equations provide a reasonable
correlation for the activity coefficients:
ln  1 = (2.771 − 0.00523T ) x22 ln  2 = (2.771 − 0.00523T ) x12
The Antoine equations provide vapor pressures:
ln P1sat / kPa = 16.59158 −
3643 .31
T ( K ) − 33.424
ln P2sat / kPa = 14.25326 −
2665 .54
T ( K ) − 53.424
Calculate
(a): P and {yi} for T = 318.15 K and x1 = 0.25
(b): P and {xi} for T = 318.15 K and y1 = 0.60
(c): T and {yi} for P = 101.33 kPa and x1 = 0.85
(d): T and {xi} for P = 101.33 kPa and y1 = 0.40
(e): the azeotropic pressure and the azeotropic composition for T = 318.15 K
(a) for T = 318.15, and x1 = 0.25
P1sat = 44.51 P2sat = 65.64  1 = 1.864  2 = 1.072
P =  xi i Pi sat = (0.25)(1.864)(44.51) + (0.75)(1.072)(65.64) = 73.50
i
yi P = xi i Pi sat
y1 = 0.282 y2 = 0.718
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(b): for T = 318.15 K and y1 = 0.60
P1sat = 44.51 P2sat = 65.64
An iterative process is applied, with
1
P=
 yi /  i Pi sat
1 = 1  2 = 1
i
x1 =
Converges at:
P = 62.89 kPa  1 = 1.0378
y1 P
 1 P1sat
x2 = 1 − x1
 2 = 2.0935 x1 = 0.8169
(c): for P = 101.33 kPa and x1 = 0.85
T1sat = 337.71 T2sat = 330.08
A iterative process is applied, with
ln P1sat / kPa = 16.59158 −
T = (0.85)T1sat + (0.15)T2sat = 336.57
3643 .31
T ( K ) − 33.424
 1 = ...  2 = ...
P =  xi i Pi sat
i
P1sat = ...
Converges at:
T = 331.20 K
 1 = 1.0236  2 = 2.1182
y1 = 0.670
y2 = 0.330
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(d): for P = 101.33 kPa and y1 = 0.40
T1sat = 337.71 T2sat = 330.08
A iterative process is applied, with
T = (0.40)T1sat + (0.60)T2sat = 333.13
1 = 1  2 = 1
ln Pi sat / kPa = Ai −
ln P1sat / kPa = 16.59158 −
3643 .31
T ( K ) − 33.424
Bi
T ( K ) − Ci
P1sat = ... P2sat = ...
P =  xi i Pi sat
i
x1 = ... x2 = ...
 1 = ...  2 = ...
P1sat = ... P2sat = ...
Converges at:
T = 326.70 K
 1 = 1.3629  2 = 1.2523
x1 = 0.4602 x2 = 0.5398
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(e): the azeotropic pressure and the azeotropic composition for T = 318.15 K
y1
12 
Define the relative volatility:
Azeotrope
12
x1 =0
y2
y1 = x1 y2 = x2
yi P = xi i Pi sat
x1
x2
 1P1sat
12 =
 2 P2sat
12 = 1
P1sat exp(2.771 − 0.00523T )
=
= 2.052 12
sat
P2
P1sat
= 0.224
x1 =1 =
sat
P2 exp(2.771 − 0.00523T )
Since α12 is a continuous function of x1: from 2.052 to 0.224, α12 = 1 at some point
There exists the azeotrope!
 1P1sat
12 =
=1
sat
 2 P2
 1az P2sat
= sat = 1.4747
az
2
P1
ln  1 = (2.771 − 0.00523T ) x22
ln  2 = (2.771 − 0.00523T ) x12
ln
1
= (2.771 − 0.00523T )( x2 − x1 ) = (2.771 − 0.00523T )(1 − 2 x1 )
2
x = 0.325 = y
az
1
az
1
 1az = 1.657
P az =  1az P1sat = 73.3276kPa
VLE from K-value correlations
• A convenient measure, the K-value:
Ki 
yi
xi
– the “lightness” of a constituent species, i.e., of its
tendency to favor the vapor phase.
– When Ki is greater than unity, species i exhibits a higher
concentration in the vapor phase; when less, a higher
concentration in the liquid phase, and is considered a
"heavy" constituent.
33
– The Raoult’s law:
Pi sat
Ki =
P
– The modified Raoult’s law: K i =
 i Pi sat
P
34
Fig 10.13
35
Fig 10.14
36
For a mixture of 10 mol-% methane, 20 mol-% ethane, and 70 mol-% propane at 50°F,
determine: (a) the dewpoint pressure, (b) the bubblepoint pressure. The K-values are given by
Fig. 10.13.
(a) at its dewpoint, only an insignificant amount of liquid is present:
Species
Methane
Ethane
Propane
yi
0.10
0.20
0.70
P = 100 (psia)
Ki
yi /Ki
20.0
0.005
3.25
0.062
0.92
0.761
Σ (yi /Ki) = 0.828
P = 150 (psia)
Ki
yi /Ki
13.2
0.008
2.25
0.089
0.65
1.077
Σ (yi /Ki) = 1.174
P = 126 (psia)
Ki
yi /Ki
16.0
0.006
2.65
0.075
0.762
0.919
Σ (yi /Ki) = 1.000
(b) at bubblepoint, the system is almost completely condensed:
Species
Methane
Ethane
Propane
xi
0.10
0.20
0.70
P = 380 (psia)
Ki
xi Ki
5.60
0.560
1.11
0.222
0.335
0.235
Σ (xi Ki) = 1.017
P = 400 (psia)
Ki
xi Ki
5.25
0.525
1.07
0.214
0.32
0.224
Σ (xi Ki) = 0.963
P = 385 (psia)
Ki
xi Ki
5.49
0.549
1.10
0.220
0.33
0.231
Σ (xi Ki) = 1.000
37
Flash calculations
• A liquid at a pressure equal to or greater than its bubblepoint
pressure “flashes” or partially evaporates when the pressure
is reduced, producing a two-phase system of vapor and liquid
in equilibrium.
• Consider a system containing one mole of nonreacting
chemical species:
L +V = 1
The moles of vapor
The moles of liquid
zi = xi L + yiV
zi = xi (1 − V ) + yiV
The vapor mole fraction
The liquid mole fraction
yi =
zi K i
1 + V ( K i − 1)
zi K i
 1 + V ( K − 1)38= 1
i
The system acetone (1)/acetonitrile (2)/nitromethane(3) at 80°C and 110 kPa has the overall
composition, z1 = 0.45, z2 = 0.35, z3 = 0.20, Assuming that Raoult’s law is appropriate to this
system, determine L, V, {xi}, and {yi}. The vapor pressures of the pure species are given.
Do a BUBL P calculation, with {zi} = {xi} :
Pbubl = x1P1sat + x2 P2sat + x3 P3sat = (0.45)(195.75) + (0.35)(97.84) + (0.20)(50.32) = 132.40 kPa
Do a DEW P calculation, with {zi} = {yi} :
Pdew =
y1 / P1sat
1
= 101 .52 kPa
sat
sat
+ y2 / P2 + y3 / P3
L = 1 − V = 0.2636 mol
Since Pdew < P = 110 kPa < Pbubl, the system is in the two-phase region,
Pi sat
Ki =
P
zi K i
 1 + V ( K − 1) = 1
i
K1 = 1.7795 K2 = 0.8895 K3 = 0.4575
yi =
x1 = 0.2859
x2 = 0.3810
x3 = 0.3331
Ki 
yi
xi
V = 0.7364 mol
zi K i
1 + V ( K i − 1)
y1 = 0.5087
y2 = 0.3389
39
y3 = 0.1524