VAPOR LIQUID

EQUILIBRIUM

CHE553
CHEMICAL ENGINEERING
THERMODYNAMICS
Part 1
- THE NATURE OF EQUILIBRIUM
- THE PHASE RULE: DUHEM’S THEOREM
- VLE: QUALITATIVE BEHAVIOR
- SIMPLE MODELS FOR VAPOR LIQUID
EQUILIBRUM: RAOULT’S LAW
 3

Learning Outcomes

Derive and simplify equations of VLE

Apply Raoult’s Law to obtain data

for P-x-y and T-x-y diagram

Apply Raoult’s law to solve simple

thermodynamic problems

Caluculate bubble and dew point for

a given mixture
 4

THE NATURE OF EQUILIBRIUM

 Equilibrium is a condition where no changes occur in the properties of a system with
time.
 The temperature, pressure and phase compositions reach final values which
thereafter remain fixed.
 Vapor liquid equilibrium (VLE) is a state where vapor and liquid phase are in
equilibrium with each other.
 Many processes in chemical engineering do not only involve a single phase but a
stream containing both gas and liquid. It is very important to recognize and be able
to calculate the temperature, pressure and composition of each phase at
equilibrium.
 VLE information is useful in separation processes, e.g. distillation, evaporation, liquid-
liquid extraction, adsorption, etc.
 5

Measures of composition
 Mass or mole fraction is defined as the ratio of mass or number of moles of a particular
chemical species in a mixture or solution to the total mass or number of moles of
mixture or solution.
𝑚𝑖 𝑚ሶ 𝑖 𝑛𝑖 𝑛ሶ 𝑖
𝑥𝑖 ≡ = 𝑥𝑖 ≡ =
𝑚 𝑚ሶ 𝑛 𝑛ሶ
 Molar concentration is defined as the ratio of mole fraction of a particular chemical
species in a mixture or solution to molar volume of the mixture or solution.
𝑥𝑖
𝐶𝑖 ≡
𝑉
or multiplying and dividing by molar flow rate, n
𝑥𝑖 × 𝑛ሶ 𝑛𝑖ሶ
𝐶𝑖 ≡ =
𝑉 × 𝑛ሶ 𝑞
 The molar mass of a mixture or solution is the mole fraction-weighted sum of the molar
masses of all species.
𝑀 ≡ ෍ 𝑥𝑖 𝑀𝑖
𝑖
 6

THE PHASE RULE

 For a multiphase system at equilibrium, the number of independent variables
that must be fixed to establish its intensive state is called the number of degrees
of freedom, given as:
F = 2 − + N
F = degree of freedom
 = no. of phases
N = no. of chemical species
 Consider a binary system, F = 4-π
When only a single phase is present, so F = 3, that is; THREE intensive variables
(T, P and xi ) must be specified to fix / determine the state of the system.
 For a binary system with two phases (liquid and vapor) coexist, so F = 2, that is;
TWO intensive variables must be specified to fix/determine the state of the
system.
 7

DUHEM’S THEOREM
For any closed system, the equilibrium state is completely
determined when any two independent variables are fixed.

 The two independent variables subject to specification may in

general be either intensive or extensive. However, the number of
independent intensive variables is given by the phase rule.
 Thus, when F = 1, at least one of the two variables must be extensive,
and when F = 0, both must be extensive.
 8

VLE: QUALITATIVE BEHAVIOR

 Vapor/liquid equilibrium (VLE) is the state of coexistence of liquid and
vapor phases.
 For a system of two chemical species (N=2), the phase rule becomes
F=4-.
 Because there must be at least one phase (=1), the max 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/mass fraction.
 9

 All equilibrium states of the system

can therefore be represented in
3D P-T-composition space as
shown in Fig. 10.1.
 This fig. contains the equilibrium
states of saturated vapor and
saturated liquid for species 1 and
2 of a binary system.
 10

• Line AEDBLA in Fig. 10.1 can be represented by Fig 10.2(a) P-x1-y1 diagram at Ta.
• The horizontal lines are tie lines connecting the compositions of phase in equilibrium.
• The temperatures Tb and Td lie between the two pure species critical temperature
identified by C1 and C2 in Fig. 10.1.
• Line KJIHLK in Fig. 10.1 can be represented by Fig. 10.2(b) T-x1-y1 diagram at Pa.
• Pressure Pb lies between the critical pressures of the two pure species at points C1 and C2
in Fig. 10.1.
• Pressure Pd is above the critical pressures of both pure species, therefore the T-x1-y1
diagram appears as an island.
10
 11

Figure 10.8 shows P-x-y

diagrams at constant T for four
systems (much lower
temperature and pressure).
 12

Figure 10.9 shows t-x-y behavior for

four systems at low pressure – 1 atm.

-Azeotrope is a point/state where x1= y1,

i.e a boiling liquid of this composition
produces a vapor of exact composition.
-Such information is important as it tells us that
no separation (of liquid and vapor phase)can
be performed at this point by distillation.
 13

Figure 10.10: The y1-x1 diagrams at constant P for four systems.

The point at which a curve crosses the diagonal line of the diagram
represents an azeotrope, for such a point x1 = y1 (for (b) and (d)).
 14

SIMPLE MODELS FOR VAPOR/LIQUID

EQUILIBRIUM
 VLE calculation provides information on temperatures, pressures, and compositions of
phases in equilibrium.
 Two simplest models, Raoult’s law and Henry’s law are used to predict the behavior of
systems in vapor liquid equilibrium.

Raoult’s Law
 Assumption
 Vapor phase is an ideal gas (apply for low to moderate pressure)
 Liquid phase is an ideal solution (apply for species that are chemically similar)
 Mathematical expression reflecting the two assumptions above is expressed
quantitatively in Raoult’s law as:
yiP = xiPisat (i = 1,2, ..., N) (10.1)
where xi = mole fraction of liquid phase
yi = mole fraction of vapor phase
Pisat = vapor pressure of pure species i at the temperature of the system
 15

Antoine’s equation is used to calculate Pisat :

Bi
ln Pi sat ( kPa ) = Ai −
T ( K ) + Ci

*Refer Table B.2 for values of Antoine parameters (A, B and C)

Let’s try!
Calculate vapor pressure of methanol(1)/water(2) system at 333.15 K.
 16

Dewpoint and Bubblepoint Calculations with

Raoult’s Law
 Bubble point is the point at which the first drop of a liquid mixture begins to
evaporate (the first bubble of vapor appears).
 Dew point is the point at which the first drop of a gas mixture begins to
condense (the last drops of liquid disappear).

BUBL P DEW P
Calculate {yi} and P Calculate {xi} and P
Given {xi} and T Given {yi} and T

BUBL T DEW T
Calculate {yi} and T Calculate {xi} and T
Given {xi} and P Given {yi} and P
 17

 Application of Raoult’s law:

 Because iyi = 1, eq. (10.1) may be summed over all species to yield
P =  xi Pi sat
i
(10.2)

 This equation applied in bubblepoint calculations, where the vapor

phase compositions are unknown.
 For a binary system with x2 = 1-x1,

( )
P = P2 sat + P1sat − P2sat x1

 A plot of P vs. x1 at constant temperature is a straight line connecting

P2sat at x1 = 0 with P1sat at x1 = 1.
 18

 Equation (10.1) may also be solved for xi and summed over all species.
With ixi = 1, this yields

1
P= (10.3)
y
i
i Pi sat

 This equation applied in dewpoint calculations, where the liquid phase

compositions are unknown.
 19

Example 10.1
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:
2945.47
ln P1 sat = 14.2724 −
T − 49.15
2972.64
ln P2 sat = 14.2043 −
T − 64.15

(a) Prepare a graph showing P vs x1 and P vs y1 for a temperature of 348.15K

(b) Prepare a graph showing T vs x1 and T vs y1 for a pressure of 70kPa
 20

Solution:
(a) Prepare P-x1-y1 diagram (given T = 348.15 K, x1 and y1 in the range 0→1)

BUBL P calculation

1. Calculate P1sat and P2sat Bi Antoine

ln Pi sat ( kPa ) = Ai −
by Antoine eqn. T ( K ) + Ci eqn

2. Calculate P at any x1 (0 to 1) (
P = P2 sat + P1sat − P2sat x1) (10.2)
By eqn. (10.2)

3. Calculate y1 by eqn. (10.1) x1 P1sat

y1 = (10.1)
P
 21

The results of calculations:

P-xy diagram for acetonitrile(1)/nitromethane(2) at 75oC
x1 y1 P
100
0 0 41.98
0.1 0.1805 46.10 90

0.2 0.3313 50.23 P1sat=83.21

80
At 75oC (384.15 K) a 0.3 0.4593 54.35
Subcooled liquid
liquid mixture of 60 0.4 0.5692 58.47 70
mole% acetonitrile and
0.5 0.6647 62.60

P/kPa
40 mole% 60
nitromethane is in 0.6 0.7483 66.72
equilibrium with 50
a vapor containing 0.7 0.8222 70.84
74.83 mole% 0.8 0.8880 74.96 40
acetonitrile at a
pressure of 66.72 kPa 0.9 0.9469 79.09
30
Superheated vapor
1 1 83.21 P2sat=41.98
20

At 348.15 K and x1 = 0.35,

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

P = ?, y1 = ?
x1, y1

At 348.15 K and y1 = 0.6, x1 y1

P = ?, x1 = ?
 22

DEW P calculation (given y1 = 0.6 and T = 348.15K)

1. Calculate P1sat and P2sat Bi Antoine’s

ln Pi sat ( kPa ) = Ai −
by Antoine eqn. T ( K ) + Ci eqn

2. Calculate P by eqn. (10.3) 1 (10.3)

P= sat
y1 P1 sat
+ y2 P2

3. Calculate x1 by eqn. (10.1) y1 P

x1 = (10.1)
P1sat
 23

(b) Prepare T-x1-y1 diagram (given P = 70 kPa, x1 and y1 in the range 0→1)

1. Calculate T1sat and T2sat at the given

Ti sat
(K ) = Bi
− Ci
Ai − ln P(kPa )
pressure by Antoine eqn

2. Select T1sat < T < T2sat to calculate Bi

P1sat and P2sat for these temperature ln Pi sat ( kPa ) = Ai −
by Antoine eqn T ( K ) + Ci

P − P2 sat
3. Evaluate x1 by eqn (10.2) x1 = sat (10.2)
P1 − P2 sat

4. Calculate y1 by eqn (10.1) x1 P1sat

y1 = (10.1)
P
 24

The results of calculations:

x1 y1 T (K) T-xy diagram for acetonitrile(1)/nitromethane(2) at 70 kPa

1.0000 1.0000 342.99

0.8596 0.9247 345.15
365
0.7378 0.8484 347.15
0.6233 0.7656 349.15
0.5156 0.6759 351.15 360

0.4142 0.5789 353.15

0.3184 0.4742 355.15 355

0.2280 0.3614 357.15

T/K
0.1424 0.2401 359.15 350

0.0613 0.1098 361.15

0.0000 0.0000 362.73 345

At x1 = 0.6 and P = 70 kPa, 340

T = ?, y1 = ?
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

At y1 = 0.6 and P = 70 kPa,

x1, y1

T = ?, x1 = ? y1 x1
 25

BUBL T calculation (given x1 = 0.6 and P = 70kPa)

1. Calculate initial T using mole

fraction-weighted average of T1sat 𝑇 = 𝑥1 𝑇1𝑠𝑎𝑡 + 𝑥2 𝑇2𝑠𝑎𝑡
and T2sat

Bi
2. Calculate P1sat and P2sat by ln Pi sat ( kPa ) = Ai −
Antoine eqn. T ( K ) + Ci

3. Calculate    P1 sat P2 sat

P
4. Calculate P2sat from eqn. 10.2 P2 sat = (10.2)
x1 + x2

5. Calculate T from Antoine

T (K ) =
B2
− C2
equation for species 2 A2 − ln P2 (kPa )
sat

NO YES
T (current) = T (previous) STOP
 26

Initial T = 350.89 K

Iteration P1sat (kPa) P2sat (kPa)  P2sat (kPa) T (K)

1 90.99 46.40 1.961 44.40 349.68
2 87.47 44.40 1.970 44.24 349.58
3 87.20 44.24 1.971 44.23 349.57
4 87.18 44.23 0.971 44.23 349.57

The answer is T = 349.57 K

From Antoine eqn., P1sat = 87.17 kPa and by eqn. (10.1), y1 = 0.7472
 27

DEW T calculation (given y1 = 0.6 and P = 70kPa)

1. Calculate initial T using mole

fraction-weighted average of T1sat 𝑇 = 𝑦1 𝑇1𝑠𝑎𝑡 + 𝑦2 𝑇2𝑠𝑎𝑡
and T2sat

Bi
2. Calculate P1sat and P2sat by ln Pi sat ( kPa ) = Ai −
Antoine eqn. T ( K ) + Ci

3. Calculate    P1 sat P2 sat

4. Calculate P1sat from eqn. 10.3 P1 sat = P ( y 1 +y 2 ) (10.3)

5. Calculate T from Antoine

T (K ) =
B1
− C1
equation for species 1 A1 − ln P1 (kPa )
sat

NO YES
T (current) = T (previous) STOP
 28

Initial T = 350.89 K

Iteration P1sat (kPa) P2sat (kPa)  P1sat (kPa) T (K)

1 90.99 46.40 1.961 96.91 352.85
2 96.91 49.78 1.947 96.51 352.72
3 96.51 49.55 1.948 96.53 352.73
4 96.53 49.57 1.948 96.53 352.73

The answer is T = 352.73 K

From Antoine eqn., P1sat = 96.53 kPa and by eqn. (10.1), x1 = 0.4351
 29

REFERENCE
Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to Chemical
Engineering Thermodynamics. Seventh Edition. Mc Graw-Hill.
PREPARED BY:
NORASMAH MOHAMMED MANSHOR
SCHOOL OF CHEMICAL ENGINEERING,
COLLEGE OF ENGINEERING,
UNIVERSITI TEKNOLOGI MARA, SHAH ALAM.
norasmah@uitm.edu.my
03-55436333/019-2368303