Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

1. REFRIGERATION AND LIQUEFACTION

1.1. REFRIGERATION CONCEPT

The word “refrigeration” means maintaining a temperature lower than ambient
temperature. To achieve this, it is necessary to continuously absorb heat at a
low temperature level, which is achieved by evaporating a liquid in a
continuous flow process in a uniform state; the vapor that is formed returns to
its original state (liquid) by compression and condensation of the same.
Alternatively, it may be absorbed by a low volatility liquid, from which it is
subsequently evaporated at higher pressure.

1.2. TYPES OF REFRIGERATION PROCESSES

Two systems stand out for their diffusion:
- Vapor compression refrigeration.
- Absorption refrigeration.
Vapor compression systems are used in most commercial, industrial and
domestic systems. But both the vapor compression and absorption systems
produce a cold region by evaporation of a refrigerant fluid at low temperature
and pressure.

In compression refrigeration, mechanical energy is consumed in a compressor

that compresses evaporated working fluid coming from the evaporator (cold
chamber), so that the heat taken by the refrigerant fluid in the evaporator can
be dissipated to a higher thermal level in the condenser.
In absorption refrigeration, the heat taken by the refrigerant fluid at low
temperature and pressure is transferred at intermediate temperature and high
pressure after having been evaporated from a solution by means of heating. It
differs from the previous one in that it does not require mechanical energy; any
economical heat source can be used.

1.3. VAPOR COMPRESSION REFRIGERATION

Carnot cycle

The Carnot thermal cycle can be operated in a reverse direction. It is possible

to absorb heat from a low temperature (TC) source and discharge it into a high
temperature (TH) source, if work is administered to the cycle, the Carnot
refrigeration cycle represents the most efficient cycle (KJ of cooling / KJ of
work supplied).

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The Carnot thermobomb consists of two isothermal and two isentropic

processes and uses the same components as the thermal one.
Stage 1-2: The fluid absorbs heat isothermally in a low-temperature
exchanger (evaporator), with a corresponding increase in entropy.
Stage 2-3: The temperature of the fluid increases by means of isentropic
compression.
Stage 3-4: The fluid then discharges heat isothermally to a high temperature
source and experiences a reduction in entropy, which compensates for its
increase in lower temperature.
Stage 4-1: The fluid is then expanded isentropically to the pressure and
temperature at which the cycle began and reversible expansion work is
extracted in this stage.

Ranking Cycle
Describe the following graphs as a practical and technically feasible cycle
applied to commonly used refrigeration systems.

W=QH - QC 1.1
w = qH - qC 1.2 per unit mass

These graphs represent compression-expansion cooling acts.

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Fig 1.2(a), and isenthalpic compression-throttling cooling.

Alternatively to the Ts chart, Ph charts of Fig. 1.3

Figure No 1.3. R-12 refrigerant. Source Handbook J. H. Perry

For troubleshooting purposes, pH charts for some refrigeration systems must

be available.
Entropy balance; per unit mass of refrigerant.

qH qC
− = 0( 1. 3 )
TH TC

Since stage 3-4 occurs isothermally and reversibly:

qH
+ Se − S S = 0( 1 . 4 )
TH
q H =T H ( S S − Se ) = T H ΔS H ( 1 .5 )

COP, defined as coefficient of behavior or coefficient of performance, is a

common measure of the effectiveness of a refrigerator.

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calor absorbido a la temperatura más baja

COP =
trabajo neto
qC TC
COP = = ( 1 .6 )
W neto TH − TC
The coefficient of performance of a Carnot refrigerator can be expressed
only in terms of the temperatures of the hot and cold sources (absolute
temperatures).
From equation (1.6)
W T H −T C
= ( 1. 7 )
QC TC

Examination of Fig. 1.1, indicates that the compression and expansion stages
of the Carnot refrigerator occur within the two-phase region, which is
technically not applicable. The problem can be eliminated by allowing the
refrigerant to evaporate completely in the evaporator producing a saturated
vapor.
Isentropic compression of the refrigerant to the condenser pressure then
produces a superheated value at point 3, in Fig 1.2 (a) and Fig 1.2 (b). The
fluid leaving the condenser is mainly liquid, its specific volume is relatively
low, so the work produced in the expander is low or negligible; which can be
replaced by isenthalpic throttling, which are much cheaper than expanders.

In the Rankine cycle with compression and superheat: with expansion

T H
qH = qC ( 1 .8 )
TC
Where:
q C = h2 − h1 q H = h3 − h 4 ( 1. 9 )
W = q H − q C = ( h3 − h4 ) − ( h2 − h1 )( 1. 10 )

h 2 − h1
COP = ( 1 .11 )
( h3 − h4 ) − ( h2 − h1 )
o o
Q C =m qC = m ( h2 − h1 )(1 . 12 )
o QC
m = ( 1 . 13 )
h2 − h1

Compression cycle – throttling with superheat

q C =h2 − h1 ¿ } ¿¿donde h 4 = h1 (1.14)¿

W neto= ( h3 − h 4 ) − (h2 − h 1) ¿ } ¿¿(1.15)¿

( h 2− h1 ) h2 − h 1
COP = = (1 . 16)
h3 − h2 h3 − h 2

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o QC
m = ( 1 . 17 )
h2 − h1

1.4. REFRIGERATION UNITS

The normal or standard unit of cooling capacity is called a “Ton” and
represents the rate of cooling that would be supplied by 907.2 Kg or (1 short
ton of 2000 lbm) of melted ice per day.

1 Ton = 12,000 BTU/h

= 11 376 KJ/h
= 3024 Kcal / h

1.5. REFRIGERANT SELECTION

The irreversibilities inherent in the vapor compression cycle cause the
coefficient of performance of practical refrigerators to depend on some grade
of refrigerant, such as:
- Economic
- Not being toxic
- Not be corrosive
- Ease of handling
- Boiling vapor pressure is greater than the atmosphere so that the
equipment does not operate in a vacuum.
- The freezing temperature of the refrigerant fluid must be much lower
than the minimum reached by the system.
- It is advisable for the system to operate at T and P well below the critical
values; otherwise, it would be difficult to liquefy the refrigerant.
- Small specific volume and others.

SOME COOLANT FLUIDS

The following table includes commercially used refrigerants. Industrial.

INITIALS CHEMICAL NAME FORMULA
R – 11 Tetrachlorofluoromethane C Cl3 F
R – 12 Dichlorodifluoromethane C Cl2 F2
R – 13 Chlorotrifluoromethane C Cl F3
R – 13 B1 Bromotrifluoromethane Br CF3
R – 14 Tetrafluoromethane C F4
R – 21 Dichlorofluoromethane CHCl2F
R – 22 Chlorodifluoromethane CHCl F2
R – 23 Trifluoromethane CH F3
R – 134a Tetrafluoroethane CHF2 – CHF2
R – 4D Methane CH4
R – 4D Methyl chloride CH3Cl
R – 170 Ethane CH3 - CH3
R – 290 Propane CH3 = CH2 – CH3
R – 500 Azeotrope R-12 and R-152a -

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R – 504 Azeotrope R-32 and R-115a -

R – 600 n-butane -
R - 600a Isobutane -
R – 747 Carbon dioxide -

1.6. CASCADE CYCLES

To achieve low temperatures of the order of -185°C or lower in a single vapour
compression cycle is not possible with commonly used refrigerants, this is
achieved using staged cycles or so-called “cascade cycles” which theoretically
can have any number of stages, the practical limit possible is three. It is
specifically intended for use in the liquefaction of natural gas used in the three-
stage cascade cycle. Fig 1.4 (2) and Fig.1.4 (b)
Three-stage cycle in cascades
Cycle I should be a refrigerant with a normal boiling point around a lower
temperature than desired. Heat is absorbed in the cycle I evaporator, the cycle
I condenser discharges its heat into the cycle II evaporator.
The cycle II refrigerant must be one having a normal boiling point close to that
which comprises approximately 2/3 of the difference between the ambient
temperature and that of the cycle I evaporator, the cycle II condenser
discharges its heat to the cycle II evaporator, to which it finally transfers the
heat to the cooling water at some convenient heat sink. The refrigerant for
cycle III must be one with a normal boiling point close to ½ between the
temperature of the cooling water and that of the evaporator of cycle II.

In a three-stage cascade cycle designed to provide refrigeration for natural gas

liquefaction the refrigerants should be in this order:
I Methane
II Ethylene
III Propane

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1.7. ABSORPTION REFRIGERATION

Absorption refrigeration machine is based on the use of heat as a source of
energy for refrigeration, coming at a high temperature level (such as water
vapor).
The work required by a Carnot refrigerator which absorbs heat at a temperature
TC and expels it at the surrounding ambient temperature TS and then:
T S − TC
W = ( q C )( 1 . 18 )
TC
qC heat absorbed
Also the heat required to produce the work: (W)

W TS
η = = 1− ( 1 . 19 )
qH T H

( )
TH
qH = W ( 1 . 20 )
T H −T S
TH T S −T C
Q H = qC . ( 1. 21)
TH − TS TC

1.8. LIQUEFACTION OF GASES

The first permanent liquefaction of gas was achieved by George Claude
(French) at the beginning of the 20th century. Claude used air as the working
fluid and consisted of an isothermal compression of the gas to a high pressure,
followed by an isentropic expansion to a low pressure. During isentropic
expansion the temperature of the gas decreases and finally it condenses.

This Claude process is not a practical process in large scale applications.

Compression pressures would be extremely high, and there are no expansion
devices available that can efficiently meet the required degree of expansion.

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Linde process, a practical scheme for liquefaction was developed in Germany

by Carl Von Linde at the beginning of the 20th century. The Linde process
where the incoming gas is mixed with the returning non-liquefied gas and the
mixture is compressed to around 7000 kPa. The compressed gas is pre-cooled
and this cooled gas enters countercurrent with non-liquefied gas. The gas
leaving the exchanger is throttled where T4 gas is reduced to T5 and partially
liquefied.

Liquefaction applications have industrial importance in obtaining liquefied

natural gas (LNG), which can be transported in large quantities by ship under
strict safety conditions.
In the Linde process, assuming Z as the liquefied gas fraction, the balance is
o

carried out around the separator, valve and cooler, where Δ (m h )4 =0

h6 Z + h8 (1 − Z ) = h3
Where: 4
h =h5

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LINDE PROCESS

REFRIGERATION AND LIQUEFACTION PROBLEMS

1. A conventional refrigerator that uses R-134a as a refrigerant, to keep the

refrigerated space at -30oC, disposing of its waste heat to the cooling water that
enters the condenser at 18oC at a flow rate of 0.25 kg/s and exits at 26oC. The
refrigerant enters the condenser at 1.2 MPa and 65oC and exits at 42oC. The
compressor inlet state is 60 kPa and - 34oC, and a net heat gain from the
compressor is estimated to be 450W from the surroundings. Determine: (a) The
quality of the refrigerant at the evaporator inlet, (b) The refrigeration load, (c)
The COP of the refrigerator.
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SOLUTION

(a) From the R-134a refrigerant table

(b) To determine the mass flow, the balance is performed on the condenser

( c) To calculate the net work, balance the compressor considering the Qo input

Finally, the system performance coefficient is calculated.

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2. A cooling system requires 1.5 kW of energy for a cooling rate of 4 kJ/s.

Calculate: (a) The COP coefficient of performance, (b) The heat that is
discarded in the condenser (c) If the heat waste is 40 °C, what is the lowest
temperature that the system can possibly maintain?

Solution
o

a) q C =4 kJ / s W =1 .5 kW
qC 4
COP= o
= =2. 667
1.5
W

b) q H =qC +W
q H =(4 +1. 5 )
q H =5 . 5 kJ /s

TC
COP=
c) T H −T C

T C =T H ( COP+1
COP
)
T H = 40+273 . 15=315 . 15 ° K

T C =315 .15 (
2 .667
3 . 667 )
T C =227 .753 ° K
T C =−45 . 40 °C

3. A refrigerator with tetrafluoroethane as a refrigerant operates with an operating

(evaporating) temperature of -15 °F and a condensing temperature of 80 °F.
Saturated liquid refrigerant from the condenser flows through a throttling valve
to the evaporator from which it emerges as saturated vapor (a) For a cooling rate
of 5 BTU/s, what is the circulation rate of the refrigerant? (b) By how much
should the flow rate be varied if the throttling valve is replaced by the turbine in
which the coolant expands isentropically? (c) Suppose the cycle in (a) is
modified by the inclusion of a counterflow heat exchanger between the
condenser and the throttling valve, in which heat is transferred to the vapor
returning from the evaporator. If the condenser liquid enters the exchanger at 80
°F and if the evaporator vapor enters the exchanger at -15 °F and leaves at 70 °F,
what is the velocity of the refrigerant? (d) For each of (a), (b), and (c), determine
COP for isentropic compression of the vapor.

SOLUTION

At point 2: saturated vapor region, of the saturated vapor table

Tsat
= 15 °F Table R – 134th

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h2 =h g =100 .799 BTU /lbm S 2=S g =0 . 22714 BTU /lbm−R

a. Using Figure No 1.2 (b)
h1 =h 4
o
o QC 5 BTU / s
m= =
h2 −h1 ( 100. 799−37 .978 ) BTU / lbm
o
m=0 . 0796 lbm/s
b. Using Figure No 1.2(a)

S 4 =S 1
S 4 =S 1=S 1 f +x 1 ( S1 g −S 1f )
S 4 −S 1 f 0 . 07892−0 . 0733
x 1= = =0. 2935
S1 g−S 1 f 0 . 22714−0 .01733

h1 =h1 f +x 1 (h 1 g−h1 f )=7 .505+0. 2935 (100 .799−7 .505 )=34 .887 BTU /lbm

o
5
m= =0 . 0759 lbm /s
100 . 799−34 . 887

c. To solve this part you must use the following figures

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The temperature difference between: h1=h4

T 2 '=T H −T C =70 ° F
T 2 '=70 ° F P=14 . 667 psia from the figure for R-134 to
h2 '=117 .5 BTU /lbm
S2 '=0 . 62 BTU /lbm−° R
o
o QC 5 lbm
m= = =0 . 063
h'2 −h1 117 . 5−37 . 978 5

d. For the solution the following figure must be taken into account:
In an isentropic process:
S3 =S2 =0 . 22714=Sg
For P = 101.37 psia:
S3 =0 . 22714 BTU /lbm−° R=S 2=Sg
Figure:
h3 =118. 3 BTU /lbm
h2 −h4 100 .799−37 . 978
COP( a)= = =3. 59
h3 −h 2 118−100 .799
h −h 100. 799−34 .887
COP( b)= 2 1 = =3 . 766
h3 −h2 118−100 . 799

COP( c ) para S=0. 62 P=101 .37 psia ( T sat )=80 ° F

h3 =138 BTU /lbm .( Fig . R−134 a)
o o
W =(h 3−h2' ) M = (138−11705) x 0 . 063=1 .2914 BTU /s
o
QC 5
COP ( c )= o
= =3 . 87
1. 2915
W

4. It is required to cool 33,000 gallons of a solution whose relative density is 1.2

and specific color 1.05 BTU/lbm - °F, from 10 °F to 0 °F, using a direct
expansion with ammonia. Cooling water is available at 65 °F under extreme
conditions it can be raised to 15 °F in the condenser. Estimate the size of the
refrigeration unit and the power expressed in Kw.

Solution: Use Figure No 1.2(b)

o o
Q C =m Cp( T C −T 2 ) at point 1 for the solution
1 g /cc=8 . 337 lbm /gal
o
QC =3000 x 8. 337 x 1 . 2 x 1. 05(10−0 )° F
o
gal lbm BTU
QC =3000 x 8 .337 x 1 .2 x 1.05 x 10 ° F
h gal 1b−° F
o
Q C =315 139 BTU /h

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o
315 139
QC = =26 . 26 TON
12 000 (Refrigeration units)

For ammonia at point 2

T 2= 0 ° Fpsat =30 . 42 psia
h2 =hg=611. 8 BTU / lbm S2 =S g =4 . 3357 BTU / lbm−R

At point 3
S2 =S3 =1 .3352 BTU / lbm−RP=158 psia vapor recalentado
T 3=300 ° F
h3 =709 BTU /lbm ammonia graph
Point 4
T 4 =80 ° F Psat
4 =153 psia
h 4 =h f =132 BTU /lbm
S 4 =S f =0. 2749 BTU /lbm−° R

PointT 1=0° F : it is h1 =h 4 =132 BTU /lbm

132−42 . 9
h 4 =h f +x 1 h fg x 1= =0. 1566
611.8−42 . 9
o
QC h2 −h1 611 . 8−132
COP= = =
W h3 −h2 709−611 . 8
o
o QC315 139 BTU /h
W= = =17 .9 BTU /s
COP 4 . 89 x 3600
o
BTU 1 Kw
W =17 . 90 x o
s 0. 947831 BTU / s W =18 .89 Kw
(*) 1 BTU/h = 0.293 W

5. A quick freezing plant requires the cold rooms to be maintained at -160°F. A

two-cycle cascade refrigeration unit is being designed to provide the required
cooling. The low temperature evaporator must operate at -180 °F, while the
high temperature condenser will operate at 102 °F. The low temperature cycle
uses FREON 503 as a refrigerant and will discharge a heat of condensation at -
34°F. The high temperature cycle will use FREON 22 as the working fluid and
absorb heat at -50.8°F. If the refrigerator must absorb 20000 BTU/h from the
freezing chamber. Determine the power of the compressors of both cycles. It
can be assumed that compressions are adiabatic and reversible and that
expansions are isenthalpic.

SOLUTION

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R- 503 (I) R- 22 (II)

Point 2: T= 180oF, h= 50 BTU/lbm Point 2: T= -50.8oF, h= 98 BTU/lbm

Point 3: T= 115oF, h= 88 BTU/lbm Point 3: T= 180oF, h=132

BTU/lbm

Point 4: T= -34oF, h=0.25 BTU/lbm Point 4: T= 102oF, h=30 BTU/lbm

R – 503 (I)
q C =h2 −h1 =50−0 .25
q C =49 . 75 BTU /lbm
o
o QC 20000 BTU / h
m= =
qC 49 . 75 BTU / lbm
o
m =402 lbm/h
W =( h 2−h 1 )+( h4 −h3 ): h4 =h 1
W =h2 −h3 =50−88=−38 BTU /lbm
o o
W Ι =m W =−15 276 BTU / h
q H =h4 −h 3=−87 .75 BTU /lbm
o
Q H =m q H =35 275 BTU / h heat removed in the low temperature condenser
o
Q C =20000 BTU / h heat absorbed in the evaporator

R – 22 (II)

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o o
o QC 35 275
m= = =518 . 75lbm / h
Q C =35 275 BTU / h=QH ( Ι ) qC 98−30
o o o
W =m (h2 −h3 )=518 . 75 (98−132 ) W ΙΙ=−17 638 BTU / h

W total =W Ι +W ΙΙ=−15276−17638=−32 914 BTU /h

6.
A two-stage compression refrigeration system operates with 134A refrigerant
between the pressure limits of 0.8 and 0.14 MPa. Each stage operates in an ideal
refrigeration cycle. Heat rejection from the lower cycle to the upper cycle occurs
in counterflow in an adiabatic heat exchanger, where both streams enter at 0.4
MPa. If the mass flow of the refrigerant in the upper cycle is 0.24 Kg/s,
determine: (a) The mass of the refrigerant in the lower cycle, (b) The rate of heat
removal from the refrigerated space and the power of the compressor and (c)
The COP of the system.

Pure methane gas is liquefied in a simple Linde process, compression is 6 MPa and
pre-cooling is 300 °K, the separator is maintained at a pressure of 100 kPa and the
non-liquefied gas in the process at this pressure stops cooling at 295 °K. What
fraction of gas is liquefied in the process and what is the temperature of the high
pressure gas entering the throttling valve?

From the superheated methane table, the following data is obtained:

h3 = 1140 KJ / Kg T = 300 K y P = 6 MPa
h 8 = 1188 .9 KJ / Kg T =295 k y 100 kPa
Tsat
For a P = 100 kPa = 111.45 °K
h16 = 285 . 4 KJ / Kg h7g = 796 . 7 KJ / Kg

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h8 −−h 3 1188 .9−1140

Z= =
h8 −h 6 1188 .9−285 . 4

Z=0 . 0541

Then:

5.41% is liquid entering the throttle valve.

Balance in the cooler

(h 4 −h3 ) + (1 − Z ) ( h8 − h7 ) = 0

h 4 =1140−0 . 9459 (1188 . 9 −796 . 9)

h 4 =769 .2 KJ / Kg

Interpolation from the superheated steam table

a P = 6 MPa y h = 769 .2 KJ / Kg
T =206 . 5 K

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2. MULTICOMPONENT SYSTEMS AND BALANCE

THERMODYNAMICS.

2.1. GENERAL CONSIDERATIONS

The multicomponent system includes thermodynamic relationships, so it can
describe systems of two or more chemical components of variable
composition and with two or more homogeneous phases or multiphase
systems, excluding chemical reactions, whose subject is considered in
equilibrium systems of chemical reactions.
For the study of which it is considered as follows:

PROPERTIES OF MIXTURES
A. Mixed in ideal behavior
 Ideal mixtures are additive.
 Dalton's law of partial pressures is considered.
 Amagat's law of partial volumes.
B. Royal mix.
 Mixing rule using equations of state
 Using generalized correlations.
 Molecular properties inherent to each component are
taken into account.

SEPARATION PROCESSES
TO. IDEAL SYSTEMS.
For its treatment, Raoult's law is considered.
y i P=x i Psat i (2.1)
B. REAL SYSTEMS
 In the vapor phase, the fugacity coefficient is considered
as a function of the temperature and the molar fraction of
each component.
 In the liquid phase, the activity coefficient is considered
as a function of the temperature and quantity of each
component; appropriate models are used for its
calculations, according to the nature of the mixtures.
 Also in the real behavior, the working pressures of the
system must be considered, whether they are moderate or
high, if the liquid phase is compressed, the Poyting factor
is considered.

P v
i exp ∫P 0
Φ . y i P=γ i x i PSat dP
i RT (2.2)

Equation (2.2) is a general equation, which is considered in

separation processes, in vapor-liquid equilibrium systems, which is
an equation developed based on equation (2.1). and is used according

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to the cases presented, where the exponential part represents the

Poyting factor.

C. IN BALANCE OF CHEMICAL REACTIONS

For its treatment the following must be taken into account:
o
 Heat of reaction. ΔH
 Equilibrium constant of chemical reactions K(T)
 The molar fraction of the components participating in the
process y(∈) , on which the conversion depends.

2.2. CHEMICAL POTENTIAL OF THE COMPONENTS OF A MIXTURE.

In the following relationship for n moles of substances:
d (ng )=nvdP−nsdT (2.3)
In a closed system, where the composition is considered constant, taking into
account a multiphase system, each phase is considered a system, that is, an
open system.

[ d (ng )
dP ]
= nv

(2.4)

[ ]d (ng )
dT
=−ns

(2.4)
Considering each phase as an open system, with respect to another phase, the
Gibbs free energy is a function of pressure, temperature and the number of
moles of each component.
G=g (P , T ,n1 ,n 2 , n3 , .. . .. .. . .. .. . ni ) (2.5)

d (ng )= [ ]
∂(ng )
∂P T , ni
dP− [ ]
∂(ng )
dT P , ni
dT + ∑
i
[ ]
∂(ng)
∂ni P ,T ,n j
dni
(2.6)

d (ng )=nvdP−nsdT + ∑
i [ ]
∂(ng )
∂ ni P ,T , n j
dni
(2.7)

After equations (2.6) and (2.7), the chemical potential is defined as follows:

μi =
[ ] d (ng )
dni P,T ,nj
(2.8)

Equation (2.7) in simplified form is expressed:

d (ng )=nvdP−nsdT + ∑ μ i dni

i (2.9)

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2.3. PHYSICAL-CHEMICAL EQUILIBRIUM CONDITIONS IN SYSTEMS

MULTICOMPONENT
A heterogeneous system is a system where there is more than one phase,
regardless of the number of chemical components that comprise them. Let us
assume a simple case where a two-phase component, for example, liquid
water and water vapor with number of moles nl and nv respectively, with
Gibbs free energy ng.
ng=nl gl + nv g v (2.10)
For the isothermal and isobaric transformation, with an infinitesimal change
we have:
d (ng )=d (nl gl )+d (nv g v ) (2.11)

Considering that there is no expansion work on the system, the following can
be deduced:
−dg=0
dnl +dn v=0
dnl =−dn v
dn v ( g v−gl )=0
gl =gv (2.12)
From the last relation of (2.12), it expresses the condition of phase
equilibrium, when a system is in a state of equilibrium, all its properties
remain in a state of equilibrium, all its properties remain constant, and the
most stable states are those with the lowest energy content, that is, of
maximum entropy, at constant entropy the energy is minimal indicating
stable equilibrium.

2.4. CHEMICAL POTENTIAL AND MULTICOMPONENT SYSTEM

In a system where there are several phases and several components, it
requires a series of analyses, one of the very important thermodynamic
μ
properties proposed by Willard Gibbs is the chemical potential i defined in
equation (2.8), where each phase is considered homogeneous, that is, it
behaves like a pure substance, and from equation (2.9), we have:

d (ng )l =nvdP−nsdT + ∑ μ l dnli

i i
(2.13)
d (ng ) =nvdP−nsdT + ∑ μ
v v
v dni
i i
Under equilibrium conditions the net flow is:
dnl =−dnv

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∑ ( μli−μ vi )=0
i

μli =μvi =−−−−−−−−−−−−μiF

μ11 =μ21 =μ31 −−−−−−−−−−−μ F1

(2.14)
μ12 =μ22 =μ32 −−−−−−−−−−−μF2

μ13 =μ23 =μ33 −−−−−−−−−−−μ3F

μ1i =μ2i =μ3i −−−−−−−−−−−μ Fi

Equation (2.14) allows us to establish the variance V in a thermodynamic

phase process, where the Gibbs phase rule is applied.

V=C–F+2

TABLE OF APPLICATIONS OF SEPARATION PROCESSES WITH PHASE

CHANGE

PROCESS phase phase comp. Variance variables

s type no. C V
F
Distillation of binary 2 L,V 2 2 P,X
miscibles
Binary immiscible 3 L,L,V 2 1 P
distillation
Gas absorption 2 L,V 3 3 P,T,X
Liquid-liquid extraction 2 L,L 3 3 P,T,X
Crystallization from a 2 S,L 2 2 P,T
solution
Crystallization from a 2 S,L 1 1 P
liquid

2.5. MULTICOMPONENT SYSTEMS AND IDEAL MIXING

IDEAL BEHAVIOR AND RELATIONS OF SOME THERMODYNAMICS

PROPERTIES.
Some relationships of specific thermodynamic properties, which are linked with
directly measurable thermodynamic properties, such as pressure, temperature and
specific volume, are necessary for all thermodynamic processes.

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h=u+Pv
g=h−Ts
a=u−Ts (2.14)
These relationships can be expressed in the form of differentials as:
du=Tds−Pdv
dh=Tds+vdP
dg=sdT+vdP
da=−sdT−Pdv (2.15)

MIXING OF IDEAL GASES

The mixing of gases is considered the application of additive properties, assuming the
ideal behavior of gases at low pressures, where the forces of repulsion and attraction
of the molecules are almost zero, assuming that they are at considerable distances, so
the ideal gas equation is applied:
Pv=nRT (to)
(2.16)
Pi v =ni RT (b)
From relations (a) and (b) of equation (3.1) we obtain:
P i ni
= = yi
P n (2.17)
where yi is the molar fraction of component i of the ideal mixture., then
Pi = y i P (2.18)

This last relation expresses the partial pressure of component i of the ideal system.
Let M be an extensive thermodynamic property and m an intensive property expressed
per unit mass or mole of substance.
nm( P ,T )=M (P , T )=∑ ni mi ( P ,T )
i (2.19)
m( P ,T )=∑ y i mi ( P ,T )
(2.20)
Equation (2.19) represents the value of a thermodynamic property of the mixture as a
function of the number of moles at a given pressure and temperature and the
relationship (2.20) the value of an intensive property in terms of the molar fraction yi.
They can then be expressed for different thermodynamic properties as:

h( P , T )=∑ y i h i ( P ,T )
(2.21)
u( P , T )=∑ y i ui (P , T )
(2.22)
These last relations express the values of the enthalpy and the internal energy of a
mixture.
Then for the variation of the thermodynamic properties of mixing we can generalize:
Δm=m( P ,T )−∑ y i mi (P ,T )
(2.23)

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From the above relationships Δm=0 , the system is found to be in equilibrium, and is
assumed to be the behavior of a pure substance.
That is Δh=0 Δu=0 Δv =0 , , the variations for ideal mixing.

The entropy variation in an ideal mixture is not equal to the previous properties as we
will see:
Tds=dh−vdP (2.24)

where dh = 0, for ideal mixture

ds=−Rd (ln P) (2.25)

Integrating this equation from a pure state until after a mixing process we have:

1
s( P , T )−si ( P , T )=−R ln( )
yi
(2.27)
1
si ( P , T )=s( P , T )+R ln( )
yi
(2.28)
s( P , T )=∑ y i si ( P , T )
(2.29)

[
s( P , T )=∑ y i s( P , T )+R ln(
1
yi
)
] (2.30)

s( P , T )=∑ y i s( P .T )+R ∑ y i ln( 1y )

i
(2.31)
Δs=s( P ,T )−∑ y i s( P. T )=R ∑ y i ln( 1
yi
)
(2.32)
From the sequence of operation of the relations (2.24) to (2.31) the relation (2.32) is
obtained which represents the variation of the entropy, this shows that an ideal mixing
process is irreversible.
Similarly, the Gibbs free energy of an ideal mixture is obtained as follows:

g=h−Ts (2.33)

g( P ;T )=h (P , T )−Ts( P , T ) (3.18)

g( P , T )=∑ y i h( P ;T )−T
[ 1
∑ y i s( P ,T )+R ∑ yi ln( y )
i ] (2.34)
1
g( P , T )=∑ y i h( P ;T )−T ∑ y i s( P;T )−RT ∑ y i ln( )
yi (2.35)

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1
g( P , T )=∑ y i gi ( P ;T )−RT ∑ y i ln( )
yi (2.36)

For the calculation of chemical potential of an ideal mixture:

μi =
[ ]d (ng )
dni P,T ,nj

From equation (3.21) we can deduce:

1
ng (P , T )=∑ ni gi ( P ;T )−RT ∑ ni ln( )
yi (2.37)
By deriving equation (3.22), we obtain the chemical potential of a mixture:
μi =g i + RT ln y i (2.38)
For the liquid phase, the same equations as for the gas phase are considered, except
that the molar fraction yi becomes xi.

APPLICATION PROBLEMS
1. A rigid container divided into two compartments by a separator. One compartment
contains 7 kg of oxygen at 40oC and 100 kPa, and the other contains 4 kg of nitrogen
gas at 20oC and 150 kPa. The separator is then removed and the gases are allowed to
mix. Calculate:
a. The temperature of the mixture and
b. The pressure of the mixture after equilibrium has been established.

2. An insulated rigid container is divided into two compartments by a divider. One of

them contains 3 Kmol of oxygen and the other contains 5 Kmol of carbon dioxide,
at the beginning both gases are at 25oC and 200 kPa. The separator is then removed
and the two gases are allowed to mix. Assuming that the surroundings of the vessel
are at 25oC, and the system has ideal behavior during mixing. Calculate the entropy
change of the process.

3 A volume of 0.3 m3 of oxygen at 200 K and 8 MPa is mixed with 0.4 m3 of nitrogen
at the same temperature and pressure, forming a mixture at 200 K and 8 MPa.
Calculate the volume of the mixture using the ideal gas equation and Amagat's
law.

4 A cylinder/piston system containing a mixture of 0.5 kg of hydrogen and 1.6 kg of

nitrogen at 300 K and 100 kPa. Heat is then transferred to the mixture at constant
pressure until the volume doubles. Assuming that the specific heats remain
constant at the average temperature, calculate: (a) The heat transfer, (b) The
entropy change of the mixture.

5 An insulated vessel containing 1 kg of oxygen at 15oC and 300 kPa is connected to a

non-insulated 2 m3 vessel containing nitrogen at 50oC and 500 kPa. The valve
connecting the two containers opens and the two gases form a homogeneous
mixture at 25oC. Calculate: (a) The final pressure of the vessel, (b) The heat transfer,

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and (c) The entropy generated during mixing, assuming ambient temperature of
25oC
.
3. RAOULT'S LAW

3.1. GENERAL CONSIDERATIONS

It is applied to liquid-vapor equilibrium, for which the ideal gas model is attributed
to the vapor phase and the ideal solution model to the liquid phase.
The relationships are established based on the phase equilibrium criterion.
L V F
μi =μ i =μi (3.1)
V V
μi =gi ( T , P ) + RTln y i (3.2)
L L
μi =g i ( T , P ) + RTln x i (3.3)
L V
gi ( T , P )−gi ( T , P )=RTln
( )
yi
xi
(3.4)

dg=RTd(lnP) (3.5)

The pressure has no considerable effect on the liquid phase; we will only consider
the vapor phase.
Integrating equation (3.4) from P to PSat

( )
sat
Pi
gi ( T , P )−gi ( T , P ) =RTln
V sat V
(3.6)
P

By equilibrium condition:
gVi ( T , P sat )=giL ( T , P )

Then the same is done (3.4) and (3.6) and we obtain:

( ) ( )
sat
yi Pi
RTln =RTln
xi P

sat
y i P=x i Pi (3.7)

For the determination of vapor pressure, appropriate correlations are used according
to the nature of the substance in question and the limitations of the correlation or
equation, for example:
Antoine's equation:
B
ln P Sat= A− ( 3 .7 )
t +C
It has limitations on the temperature where the coefficients are valid

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0≤t ≤101° C and should not be used if PSat =1500 mmHg as the error
increases.
The Gómez-Nieto and Thodos equation has a very wide range of validity and can
be applied to estimate vapor pressures of polar, non-polar and associated
substances.
(*Leave the student to find out about other equations and correlations)

3.2. DETERMINATION OF BUBBLE POINTS AND DEW POINT

Analytical methods or graphical methods can be used, for binary systems the
determination of bubble point and dew point is less tedious than for systems
with several components (multicomponent), if we add to this the coexistence of
a number of phases greater than two. In any case, for Rauolt's law applied to
separation processes, it is very common to use these graphs that are presented,
one at constant pressure and the other at constant temperature, to illustrate them
in a simpler way.

3.3. CALCULATION OF BUBBLE POINT AND DEW POINT

CONSTANT TEMPERATURE, for a binary process:
Dado : T , { X 1 }
Calcular : P , { Y 1 } P de burbuja
Y 1 composición de burbuja
P=P1 + P2 =X 1 Psat sat
1 + X 2 P2
X 2=1− X 1

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

2 + X 1 ( P1 −P2 )
P=P sat sat sat
( 3 .8 )
X1 P
Y 1= ( 3 . 9)
P
Dado T , { Y 1 }
Calcular : P , { X 1 } P de rocío
X 1 composición de rocio 1
X 1 + X 2 =1
Y P Y P
X 1 = 1sat X 2= 2sat ( 3 .10 )
P1 P2
1
P= ( 3. 11 )
Y1 Y2
+
P1sat P2sat

CALCULATION AT CONSTANT PRESSURE

Dado: P , { X 1 }
Calcular : T , { Y 1 }
P= X 1 Psat sat
1 + X 2 P2
Psat
1
Haciendo : α 12= (3 . 12)
Psat
2

P=P 2
sat
[
P
X2 + X1
P1sat
P2sat ] ( 3. 13 )

Psat sat
2 =P j = ( 3 .14 )
X 2 +α 12 X 1
B1
ln P1sat = A1 − ( 3. 15 )
t +C1
B2
ln P2sat = A 2− ( 3. 16 )
t +C 2

ln
( )
Psat
1
Psat
2
=( A1 − A2 )−
(
B1 B
− 2
t + C1 t +C2 )
( )
B1 B2
ln α 12=( A 1− A 2 ) − − ( 3 .17 )
t +C 1 t +C 2
Bi
t sar
i = −C i ( 3 .18 )
A i −ln Psat
j
The operational solution technique is:
t sat
1 + t2
sat
t 0=
1. Calculate 2 using equation (3.18)
2. Determine Pjsat, using equation (3.14)
α
3. It is determined 12 by equation (3.17)
4. With equation (3.18) t is calculated.

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5. They are compared with 0

t yt
, if the values differ, then continue the
calculations with the sequence (3), with the last value of t until the values of t
do not differ.

X 1 P1sat
Y 1=
P

Dado P , {Y 1 }
Calcular T , { X 1 } Punto de rocio
1
P= ( 3 .18 )
Y 1 Y2
+
P1sat P2sat
P sat
1
sat
Multiplying the second member of equation (3.18) by P 2 , we obtain:
1 = P [ Y 1 +Y 2 α 12 ]
Psat ( 3 . 19)
P sat
1
Donde : α 12=
P sat
2
Y1P
X 1= sat
P1
The iterative calculation procedure is the same as above except that it rotates

based on
Psat
1

3.3. FOR MULTICOMPONENT SYSTEMS

AT CONSTANT TEMPERATURE
BUBBLE POINT

Dado : T y { X i }
Calcular : P y {Y i }
De la ecuación: Y i =X i Psat
i

∑ Y i P= ∑ X i Psat
i
i i
P=∑ X i Pisat , Donde: i =1, 2 , 3 , 4 , .. . n ( 3 .20 )
i
sat
X i Pi
Y i=
P

DEW POINT
Dado: T , {Y i }
Calcular : P , { X i }
De la ecuación: X i Pisat=Y i P

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YiP
∑ Xi= ∑ Psat
=1
i i i
1
P= ( 3 .21 ) Yi P
∑ ( Y i / Pisat ) X i=
Pisat
I

AT CONSTANT PRESSURE
BUBBLE POINT

Dado : P , { X i }
Calcular : T , {Y i }
Pisat
Donde : α ij= ( 3. 22 )
P sat
J

P=
[∑ i

P
Xi
Pisat
PJsat ] . P sat
J

Psat
J = ( 3. 23 )
∑ X i α ij
i
De la ecuación ( 3. 17 )

( )
Bi Bj
ln ( α ij )=( A i− A j )− − ( 3 . 24 )
t + Ci t + C j

The iterative calculation procedure

sat sat
1. Assuming
P=P i t
and calculating i , using the Antoine Equation.
t 0 =∑ i X i t sat
i
2. It is estimated
sat
3.
α
It is calculated ij
P
with equation (3.24), it is calculated j
with equation (3.23).
4. Using the Antoine equation, the new value of t is calculated.
5. The values of 0
t y t
are compared, if they are different return to point (3), the
operation is repeated until the last two temperatures are equal.
X i P sat
i
Y i=
P
DEW POINT
Dado : P , {Y i }
Calcular : T , { X i }
By analogy with equation (3.22)

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Yi
j =P ∑
Psat ( 3 . 25)
α ij
Yi P
Xi=
Psat
i
The above procedure is similar to the previous one (bubble), using equation
(3.20)

FLASH DISTILLATION OR INSTANT EVAPORATION

Flashing or instant evaporation is a very common operation in practice and
consists of the decomposition of a liquid through a restriction (orifice or valve)
to pass through a pipe to a flash chamber, where the liquid and vapor phases are
separated. If the liquid has two or more components, the composition of the
liquid and vapour that coexist in the chamber is different from that of the
compressed liquid, because flashing produces a considerable variation in
pressure and therefore also in temperature.

Si hacemos : F=1
Z i=X i L+Y i V
Z i=X i L+Y i ( 1−L)

The equilibrium relationship

Y i =K i X i
Y i Psat i
ó = =K i
Xi P
Zi
X i=
L+ K i (1−L)
Zi K i
Y i=
L+V ( K i −1)
∑ X i=1 ∑ Y i=1
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It will depend on the number of components in each phase, so the equation is

solved iteratively.

ISSUES
1) Prepare a diagram P− { X } , {Y } , for a temperature of 90°C and a diagram
T −{ X } , { Y } , for a pressure of 90 KPa, assuming that Raoult's law is valid
for the following systems:
Benzene (1) / ethylbenzene (2) and
1-chlorobutane (1) / Chlorobenzene (2).
From the tables for the Antoine equation the parameters are:

SUBSTANC TO B C
E
Benzene 13.8594 2773.78 220.07
Ethylbenzene 14.0025 3279.47 213.20
1-clobutane 13.9600 2826.26 224.10
Chlorobenze 13.9926 3295.12 217.55
ne

Para t=90 °C
Psat [
1 =exp 13 . 8594−
2773 .78
90+220 . 07 ]
=136 . 148 KPa

Psat
[
2 =exp 14 .0025−
3279 . 47
90+213 . 20 ]
=24 . 20 KPa

2 + ( P 1 −P2 ) X 1
P=P sat sat sat

P( X )=24 . 20+111 . 95 X 1
1
tabulando P ( X 1 ) X 1 =0 . 0 , 0 .2 , .. . ,1 . 0
P= X 1 Psat sat
1 + X 2 P2
X i1 Psat i
Y ( X1 ) =
P= X 1 Psat sat
1 +(1−X 1 ) P2
P i ( Xi )

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Cálculo a T = 90°C

CALCULO A T= 90°C

160

140

120
PRESION (kPa)

100

80

60
Las composiciones x e y se
40 refieren a x1 e y1
respectivamente.
20

0
0.00 0.20 0.40 0.60 0.80 1.00
COMPOSICION

Para la P=90 KPa

t sat =
B
−C
[
P1 (t 1 )=exp A 1−
B1
t +C 1 ]
t sat
A−ln P
1 =76 .293 ° C t 2sat =131 . 917 °C [
P2 (t 1 )=exp A 2−
B2
t+C 2 ]
P−P sat
2 P−P2 ( t 1 )
X 1= =
P1sat −P2 sat P1 ( t 1 )−P 2 ( t 1 )
X 1 P1 ( t 1 )
Y 1=
P

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P = 90 kPa

140

130

120
Tem peratura

110

100

90

80

70
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
com posicion

T=76.293, 9…. 131.917

t1 P1 (t1) P2 (t2) X1 Y1
131.917 395.05 90.00 0.0000 0.000
121.00 307.05 66.02 0.0995 0.339
2. In the 112.00 246.16 50.40 0.2023 0.553 previous
problem, 103.00 195.27 37.71 0.3320 0.720
94.00 152.63 27.94 0.4980 0.844
85.00 117.53 20.29 0.7170 0.936
76.293 90.00 14.59 1.0000 1.000

determine the bubble point and dew point in the following cases:
to). t = 90oC, x1=0.50
t = 90oC, y1=0.50

b). P = 90 kPa, x1=0.50

P = 90 kPa, y1=0.50

SOLUTION
a ) Dado T =90° C y { X 1 }=0 . 50 P y {Y 1 }
, calculate
From the previous problem:

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

Psat
1 =136 . 148 KPa
Psat
2 =24 . 20 KPa
P=P sat sat sat
2 +(P 1 −P 2 )X 1
P=24 .20+111. 95×0 . 50
P=80. 175 KPa Punto de burbuja
X 1 P1sat 0. 50×136. 148
Y 1= =
P 80. 175
Y 1 =0 . 849

This means that at 90°C a liquid mixture of 0.50 mol of benzene and 0.450 mol
of ethylbenzene, in the vapor phase, contains 0.849 mol of benzene.

Dado T =90 °C y {Y 1 }=0 . 50 Calcular : P , { X 1 }

1 1
P= =
Y1 Y2 0 .50 0 .50
+ +
P1 P2sat
sat 136 .148 24 .20
P=41. 1641 .16 KPa presión de rocío
X 1 =Y 1 P/ P1sat =0 . 50×41. 16 /36 . 183
X 1 =0 . 1511
The composition of benzene is 0.1511 mol in liquid phase which corresponds to
the dew point.

b) Given: P=90 kPa, X1=0.50

Find: t, Y1 Bubble point

SOLUTION
Bi
t sat
i = −C i
With the equation: A i−ln P determine
t sat o
1 =76 .29 C y t2sat =131 . 917o C
The sequence of operations is as follows:
sat sat
t + t2
1. t 0= 1 o
=104.104 C
2

[
2. α 12=exp ( A 1−A 2 )−
B1
+
B2
C 1+t C 2+ t ]
=5.104

sat P
3. P2 =P j= =29.567 kPadonde : j=2
x 2+ α 12 x1

sat B2 o
4. t 2 =t n+1 = −C 2=95.67 C
A 2−lnP

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5. Si,|t n−t n+1|≠ 0 , entonces , entonces volver al (2)

Si ,|t n−t n+1|≅ 0 , entonces Fin

6. t n=94.08 o C , cuando n=3 y t n+1=94.08o C
o
t=94.08 C es temperatura de burbuja
sat
P2 =27.90 kPa
sat
x 1 P1 0.50 ×152.93
y 1= = =0.8496 , es la composici on de burbuja
P 90

Given: P=90 kPa, Y1=0.50

Find: t, X1 Dew Point

SOLUTION
Bi
t sat
i = −C i
With the equation: A i− ln P determine
t sat o
1 =76 .29 C y t2sat =131 . 917o C
The sequence of operations is as follows:
sat sat
t + t2
1. t 0= 1 o
=104.104 C
2

2. α 12=exp ( A 1−A 2 )− [ B1
+
B2
C 1+t C 2+ t ]
=5.104

3. P1 =P j=P [ y 1 + y 2 α 12 ] =274.68 kPa , donde : j=1

sat

sat B2 o
4. t 2 =t n+1 = −C 2=116.42 C
A 2−lnP
5. Si,|t n−t n+1|≠ 0 , entonces , entonces volver al (2)

Si ,|t n−t n+1|≅ 0 , entonces Fin

6. t n=114.02 o C , cuando n=2 y t n +1=114.54 o C
o
t=114.54 C es temperatura de Rocio
sat
P2 =263.47 kPa

y1 P 0.50 ×90
x 1= sat
= =0.1714 , es la composicion de Rocio
P 1
263.47

PROBLEM
In the system water (1), benzene (2), Chlorobenzene (3), the vapor pressure is
given:

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

3799 . 89
ln P1sat =16 . 2620− (1 )
t +226 . 35
2773 .78
ln P2sat =13 . 8594− (2)
t+220 . 07
3295. 12
ln P3sat =13 . 9926− (3 )
t+217 . 55

Applying Raoult's law for equilibrium states, calculate:

a ) P , {Y i } para t=90 °C , X 1=0 . 15 , X 2=0. 45 , X 3 =0 . 40
b ) P , { X i } para t=110° C , Y 1=0. 24 , Y 2 =0 .26 , Y 3=0. 50
c ) t , {Y i } para P=85 KPa , X 1 =0 . 32 , X 2 =0 . 0 . 47 , X 3 =0 . 21
d ) t , {Y i } para P=90 KPa , Y 1 =0 . 54 , Y 2=0 . 15 , Y 3 =0 .31
e ) Para t=90 °C y P=80 KPa: Z 1 =0 .60 , Z 2 =0 . 25 , Z 3 =0 .15

Determine:
V j, X j, Y j

SOLUTION
a) Using equation (1), (2), (3) determine the vapor pressures.
Psat
1 =70 .129 KPa
Psat
2 =136 . 148 KPa
Psat
3 =26 . 536 KPa
P= X 1 Psat sat sat
1 + X 2 P2 + X 3 P3 =82 . 40 KPa
X 1 P1sat X 2 Psat
2 X 3 P sat
3
Y 1= , Y 2= , Y 3=
P P P
Y 1 =0 .1277 Y 2 =0 .7435 Y 3 =0 . 1288

b) Calculation for t=110°C Y1=0.24, Y2=0.26, Y3=0.50

Yo Psat
1 / KPa . Y i / Pisat
1 143.25 1.6754x10-3
2 234.109 1.1106x10-3
3 51.045 9.7953x10-3

( )
Yi
∑ Pi
=0 .0125766
i

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1
P=

( )
Yi
∑ Pisat
i

P=79 . 512 KPa .

corresponde a la presión de rocío
Y P Y P Y P
X 1 = 1sat =0 .1332 , X 2 = 2sat =0 .0883 , X 3 = 3sat =0 . 7785
P1 P2 P3

t , {Y i } ,
c) Calculate : for P=85KPa: X1=0.32, X2=0.47, X3=0.21
B
t ( ° C )= i −C i
A i−ln P
P=P1sat =P sat
2 =P 3
sat
en ELV

Yo t sat
i (° C )
Xi ti
1 95.15 30.448
2 74.49 35.010
3 127.49 26.773
t 0 =exp ∑ X i t i=92. 230 ° C

1st ITERATION

[
α ij=exp ( Ai − A j ) −
(
Bi
−
Bj
t +Ci t +C j )]
ij α ij Xi X i α ij
13 2.6612 0.32 0.8516
23 5.0676 0.47 2.3799
33 1.0000 0.21 0.2100

∑ X i α ij=3 . 4415
i
sat sat P 85
P3 =P j = = =24 . 7 KPa
∑ X i αij 3 . 4415
3295 .12
t1= −217 . 7 °C
13 .9926−ln24 .7
t 1=87 .96 ° C (1)

2nd ITERATION

ij α ij Xi X i α ij
13 2.6220 0.32 0.8390
23 5.2083 0.47 2.4479
33 1.0000 0.21 0.2100

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

∑ X i α ij=3 . 4969
i
Psat
3 =24 . 3 KPa
t 1=87. 49° C
t 2≠t 1

3rd ITERATION

ij α ij Xi X i α ij
13 2.6220 0.32 0.8390
23 5.2083 0.47 2.4479
33 1.0000 0.21 0.2100

Psat sat
3 =24 . 3 KPa=P j
t 1=87. 49° C=t 2
t=87 . 49 ° C

P = 85kPa
X i P sat
Yo Xi
α ij×P sat sat
j =P i Y i=
i
P
1 0.32 63.7146 0.24
2 0.47 126.5617 0.70
3 0.21 24.300 0.06
∑ Y i=1 .00
i
d) Calculate
t , { X i } , dado : P=90 KPa ; Y 1 =0 .54 Y 2=0. 15 Y 3 =0 .31

Yo Yi t sat
i Y i t sat
i
1 0.54 96.71 52.222
2 0.15 76.27 11.443
3 0.31 129.57 40.166

t 0 =∑ Y i t i =103 . 83° C
Bi
t sat
i = −C i
A i−ln P
P=90 KPa , t 0 =103 .83 ° C

ij α ij Y i / α ij
13 2.7563 0.1959
23 4.7405 0.0316
33 1.0000 0.3100

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∑ ( Y i /α ij ) =0 .5375

[
α ij= exp ( Ai − A j ) −

2nd ITERATION
(
Bi
t +Ci
−
Bj
t +C j )]
ij α ij Y i / α ij
13 2.792 0.1934
23 4.629 0.0324
33 1.0000 0.3100
∑ ( Y i / α ij ) =0 .5358
j =P ∑ ( Y i /α ij )
Psat
Psatj =90×0 .5358=48 . 32 KPa
t j =t 2=108 . 16≈t 1
t=108 .16 ° C j=3
Y1P 0 .549×90
X 1= = =0 .36
α 13 P3sat 2. 792×48. 38
Y2P 0 .15×90
X 2= sat
= =0 . 06
α 23 P 3 4 . 629×48 . 38
Y3P 0 .31×90
X 3= = =0. 58
α 33 P sat
3
1. 00×48 . 38

and) FLASHING: Given t=90°C, P=80KPa: Z1=0.60, Z2=0.25, Z3=0.15

Hallar :V , X i , Y i
donde : X i=Y i=Z i
de la pregunta(a )
Psat
1 =70 .129 KPa
Psat
2 =136 . 148 KPa
Psat
3 =26 . 536 KPa

Pb =∑ X i P sat
1 =80 .09 KPa
1
Pr = =62 .33 KPa.

( )
Yi
∑ P1sat
Pr < P< Pb si hay flasheo : 62. 33<80 <80 . 09

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

Psat
i
K i=
P
K 1 =0 . 8766
K 2 =1. 7018
K 3 =0 . 3320
Zi K i Zi
X 1= Y 1=
V ( K i −1 ) +1 V ( K i −1 ) +1
n Z i ( K i −1 )
∑V =0
i ( K i −1 ) +1

0 .60 (0. 8766−1) 0. 25(1 .7018−1) 0. 15(0 . 332−1 )

+ + =0
V (0 . 8766−1 )+1 V (1. 7018−1)+1 V (0 .332−1)+1
0 . 07404 0 .17545 0 .1002
+ + =0
0 .1234 V +1 0 .7018 V +1 0 . 668 V −1
V =0. 0062 L=0 . 9938 F=1

Z ( K +1 )
n n
Z i ( K i +1 )2
f (v )=∑ i i f ( v )=−∑
'

i V ( K i +1)+1 i ( V ( K i +1 )+1 )2

'
f (v )+f (v ) ΔV =0 f (v )
V n+1 =V n − Método de Newton
ΔV =V n+1 −V n f '( v )
V 0 =0

DETERMINATION OF SATURATION PRESSURES

MODIFIED HILLER EQUATION

a
lnPr sat =
Tr
[ 1−Tr 2 + B ( 3+Tr ) ( 1−Tr )3 ]

revalued in the reduced state, two parameters A and B

WAGNER EQUATION

Aτ + Bτ 1. 5 +Cτ 3 + Dτ 6
lnPr sat =
Tr
Donde : τ =1−Tr

Experimental data on constants A, B, C, and D exist for a wide range of substances.

FROST-KALWARF-THODOS EQUATION
B DP sat
lnPr sat = A− + C ln T +
T T2

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

LEE-KESLER EQUATION
It is based on the generalized correlation of Pitzer et al.
lnPr sat =f ∘ (Tr )+W f ' (Tr )
6 . 09648
f ∘(Tr )=5. 97214− −1. 28862 ln Tr +0 .169347 Tr 6
Tr
15 . 6875
f ' (Tr )=15 . 2518− −13 . 4721 ln Tr +0 . 43577 Tr 6
Tr

For the acentric factor W =α / β

6 . 09648
α=−ln Pc−5 . 97241+ +1 .28862 ln(T br )−0 .169347 T 6
T br br

15 . 6875
β=15 . 2518− −13 . 4721 ln(T br )+0 . 43577 T 6
T br br

 Tb Normal boiling point

 In the range from Tb to Tc error is 1-2% ifT ¿ ¿ the calculated vapor pressure is
lower than the actual one and the error is greater.

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4. MULTICOMPONENT SYSTEMS AND REAL MIXING

4.1. VOLUMETRIC PROPERTIES

The study of mixture behavior has been oriented along two main paths; one from the
state equations and another from the generalized correlation methods.
The most commonly used equations of state in mixture property correlations have been:
The Van Der Waals equation, Redlich-Kwong-Soave, Peng-Robinson, the virial
equation and others. Many mixing rules have been proposed. In general, a mixing rule
can be described by an equation that describes any parameter or state property of the
mixture Qm in terms of the composition and the individual parameters of the pure
components Qi, which can be presented in this way.
c c
Qm=∑ ∑ y i⋅y j⋅Qij ⋯( 4 .1 )
i j
For example, for a binary mixture, applying this relationship results (1)

Qm=Y 2 Q11 +Y 1 Y 2 Q 12+Y 2 Y 1 Q 21+Y 2 Q22

1 2

The terms Qij in the calculations take into account the way in which substances interact
with each other, it is necessary to differentiate the ways of calculating Qij
Qi +Q j c
Qij = equivale a Q m=∑ y i⋅Qi ⋯( 4 . 2 )
2 j

[∑ √ ]
c 2

Qij =√Q i +Q j resulta Q m= y i⋅ Qi ⋯( 4 . 3)

j

Including interaction parameters Kij, we have:

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

K ij⋅( Qi +Q j )
Qij = ⋯( 4 . 4 )
2
The parameters of Kii, Kij, are weighted through regression of experimental data, or other
ways in ELV.

Qij =K ij⋅√ Qi +Q j ( K ii =1 )
( 1−K ij )⋅( Q i +Q j )
Qij =
2
( K ii =0 )
Qij =( 1−K ij )⋅√ Qi +Q j ( K ii =0 )
The virial equation truncated in two terms can be used for moderate conditions and
when the substances involved are not very polar or associate with each other.

La Ecución Virial P⋅V

=1+
B
R⋅T V
Where:

R⋅T c 0 . 422 0 .172

B= ( B °+ ω B') B °=0 .083- B'=0. 139-
Pc T 1. 6 T 4. 2
r r

This equation should not be used if Vrm< 2 , so you must calculate:

c c T
T r m=
Tc m=∑ yi⋅¿Tc ¿Pc m=∑ yi⋅¿ Pc ¿ Tcm
P
i i i i Pr m =
Pc m
The coefficient B for the mixture can be calculated as follows:
c c
Bm=∑ ∑ y i⋅y j⋅Bij ⋯(4 . 5)
i i
For example for a ternary mixture:

Bm=Y 2 B 1 +Y 2 B2 +Y 2 B3 +2Y 1 Y 2 B12 +2 Y 1 Y 3 B13 +2Y 2 Y 3 B 23

1 2 3

Since the individual coefficients BK are a function of TCk, PCk and ω k, the binary
coefficients Bjk can be calculated; this requires binary pseudo-critical values.

Pseudo-critical relations, for normal fluids

[ ]}
3
V 1 +V 1
3 3
C C ¿
√
i j
T C =( 1−K ij )⋅ T C ⋅T C VC = ¿ ¿ ⋯( 4 .6 ) ¿
ij i j ij 2
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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

For molecules that do not differ much from each other in size or chemical structure, the
binary interaction parameters can be considered null, such as: hydrocarbons, rare gases,
permanent gases, carbon monoxide, perhalocarbons, the following relationship can be
used:

K ij =1−
√
8⋅ V C ⋅V C
i j
⋯( 4 . 7)

( )
3
V 1 +V 1
C3 C3
i j

The equations of two constants are treated separately, since the mixing rules that
correspond to it are (P – R, R – K, VDW YS – R – K) the following relationships can be
calculated:

a m=∑ ∑ y i⋅y j⋅√ ai⋅a j⋅( 1−K ij )

i j
b m=∑ y i⋅bi
i } ⋯(4 . 8)

These relations (4.8) are valid for the cubic equations mentioned above.

4.2 FUGACITY AND FUGACITY COEFFICIENT

In phase equilibrium calculations, chemical potentials are used, alternatively the concept
of fugacity is introduced which takes the place of μi.
The relationship of the Gibas equation to constant T is known

d g idi = R T d ( ln P ) ⋯ ( 4 . 10 )
Whose integration is:
gid
i = βi ( T ) +R T ln P ⋯( 4 .11 )
Where:
βi(T) integration constant as a function of T
give indicates ideal gas
The concept of fugacity lies in equation (4.10), an equation valid only for pure
substances i in the ideal gas state.
For a real fluid an analogous equation is written:
gi =β i ( T ) +R T ln f i ⋯ ( 4 . 12 )
Subtracting equation (2) less (3):

()
gi -g idi = R T ln i
f
P
⋯ ( 4 . 13 )
It is necessary to define the concept of “residual properties” which is given by:
m iR =m i−mid i
By analogy
giR=gi -g id i ⋯( 4 .14 )

Equation (5) defines the residual Gibbs energy

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

gi -g id
i = R T ln ()
fi
P
giR= R T ln ( φi ) ⋯( 4 . 15 )

giR
ln ( φi )=
RT

fi
φ i= ⋯( 4 .16 )
P
Equation (7) expresses the fugacity coefficient
fi = fugacity of species i
The fugacity of the ideal gas state of a pure substance i is equal to its pressure
f gi
i =P

For ideal gas: giR = 0

φi , = 1

It is also known that:

P
dP
ln φ i=∫ ( Z i −1 ) a T constante ⋯(4 . 17 )
0 P
4.4 SOME APPLICATIONS WITH STATE EQUATIONS OF
φi
1. With virial equation truncated with two terms
BP
Z=1+
RT
BP
( Z-1 )=
RT (4.18)
Replacing in equation (8) and integrating:
P
BP dP
ln φ i=∫
0 RT P

φ i=exp
BP
RT ( ) ⋯(4 .19 )
2. Using the general equation of state
Doing
PV
Z=
RT

From equation (8) we obtain:

dV
ln φ i=Z−1−ln Z +ln V −∫ Z ⋯ ( 4 . 20 )
V
4.4. GENERALIZED CORRELATIONS FOR FUGACITY
COEFFICIENTS
The Lee/Kesler method:
P=P c P r
dP=Pc dPr
Equation (8) becomes:
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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

Pr
dPr
ln φ i=∫ ( Z i −1 ) ⋯( 4 . 21 )
0 Pr
°
Where the integration is at constant Tr and using the equation Z=Z +ωZ ' , we
obtain:
Pr Pr
dPr dPr
ln φ i=∫ ( Z −1 ) +ω ∫ Z'
°
⋯ ( 4 . 22 )
0 P r 0 P r
For simplicity this last equation can be written:
ln φ i=ln φ °+ωln φ ' ⋯( 4 . 23 )
Where:
Pr
dP r
ln φ =∫ ( Z −1 )
° °
⋯ ( 4 . 24 )
0 Pr
Pr
dP r
ln φ '=∫ Z ' ⋯ ( 4 . 25 )
0 Pr

φ=( φ ° ) ( φ ' ) ω ⋯ ( 4 . 26 )
For evaluation there are tables or graphs such as tables E – 13 to E – 16 of Van
Ness – Abott. (5th. Edition)

From the Pitzer/Abott correlation, from equation (4.18), we have the following:

BP BP c Pr
( Z-1 )= =( )
RT RT c T r

BP c
=B0 +ωB 1
RT c

Pr
i
ln φ i= ( Bi0 +ωB 1i )
T ri

4.5 FUGACITY OF LIQUID AND SATURATED VAPOR

Equation (2) defining the fugacity of pure component i, can be written for substances i
as a saturated vapor:
V
gV
i = βi ( T ) +R T ln f
i
⋯ ( 4 . 27 )
For saturated liquid
L
giL= β i ( T )+ R T ln f i
⋯ ( 4 . 28 )
V
i
V
i
L f
g -g i = R T ln L
i
f
According to the equilibrium condition giV = giL then:
V L Sat
f i= f i = f i
⋯ ( 4 . 29 )

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

Sat
i
Sat
i f
φ = Sat
⋯( 4 .30 )
i
P
By equilibrium condition:
V L Sat
φ i=φ i=φ i
⋯ ( 4 . 31 )
From equations (19) and (20):
L Sat Sat Sat
f i= f i
=φ i
P i
⋯( 4 .32 )
For isothermal change of state from saturated liquid to compressed liquid at a pressure
P, a new relation can be obtained:
dG=Vd P-S dT
P

i = ∫ V i dP
gi -g sat ⋯ ( 4 . 33 )
Sat
i
P
gi =β i ( T ) +R T ln f i ⋯ ( 4 . 34 )
Sat
gsat
i =βi (T )+ R T ln f i
⋯( 4 .35 )

Equation (3) is doubled for gi and gisat by subtracting the two equations we have:

fi
gi -g sat
i =RT ln Sat
⋯ ( 4 . 36 )
i
f

Equating the expressions gi and gisat (equations (23) and (24)) gives:
P
fi 1
ln Sat =
i
∫ V dP
RT Sati i
f P
Sat Sat Sat
i i i
f =φ P
V i ( P-P )
L Sat
fi i
ln Sat
= ⋯( 4 .37 )
f i RT

By substituting equation (25) into the equation:

V i ( P-P )
L Sat
Sat Sat i
i i
f i=φ P exp ⋯( 4 .38 )
RT

The exponential is known as the Poynting factor.

FUGACITY PROBLEM
1. For a binary gas mixture, nitrogen (1), isobutane (2), whose composition in the
gas phase is: y1 = 0.35, y2 = 0.65, at a temperature of 150oC and a pressure of 60
bar. Determine: the molar volume of the mixture, the fugacity coefficient of each
of the pure components and the fugacity coefficient of component i of the
solution.
a) Considering as ideal behavior.

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

b) Actual behavior, using the Redlich-Kwong cubic equation, the Pitzer-

Abott correlation and the Lee/Kesler correlation. Assume for calculations
K ij =0.11

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

SOLUTION
a) Considering as ideal behavior.

Molar volume of the mixture.

RT 83.14 × 423.15
v 1=v 2= = =586.35 cc /mol
P 60
❑
v=∑ y i v i=0.35 × 586.35+0.65 ×586.35=586.35 cc /mol
i

Fugacityϕ i= ϕ^ i=1 coefficient, of the fugacity coefficient of the pure component

and of component i of the solution respectively is 1, in the ideal state.

b) Real behavior.
Using the Redlich-Kwong cubic equation, the Pitzer-Abott
correlation and the Lee/Kesler correlation. Assume for calculations
K ij =0.11

Molar volume of the mixture.

Pure component data

Yo Tci (K) PCI ωi Zci Vc i

(bar)

1 126.2 34 0.038 0.289 89.2

2 408.1 36.48 0.181 0.282 262.7

cc∗¯¿
R=83.14 ¿ , P=60 ,̄ T =150o C=423.15 K , y 1=0.35 y 2=0.65
mol∗K

K 12=0.11

Calculation of critical properties of the mixture

[ ]
1 1 3
3 3
Tc12= √ Tc1 ×Tc 2 ( 1−K 12) =201.98 K
❑ Vc + Vc
1 2
Vc 12= =160.65 cc /mol
2

Zc 1 +Zc 2 Zc 12 R Tc12
Zc12= =0.2855 Pc12= =29.843 ¯
¿
2 Vc 12

The RK parameters of the pure components i=1,2

2 2.5
R Tci R Tci
a i=0.42748 bi=0.08664
Pc i Pci
a 1=15 549101.96 b 1=26.7368
a 2=272519 446.6 b2=80.5824

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

The interaction parameter ofa ij=a12 ,donde aii =a jj , corresponds to the pure
component.
2 2.5
R Tc12
a 12=0.42748 =57 381 736.13
Pc 12
The RK parameters of the mixture are calculated as follows:
❑ ❑
a=∑ ∑ y 1 y 2 a12= y 1 a11 +2 y 1 y 2 a12 + y 2 a22=143 152921.1
2 2

i=1 j=1
❑
b=∑ y i bi= y 1 b1 + y 2 b2=61.7364
i=1
RT a
The Redlich-Kwong equation is: P= v−b − 0.5
T v ( v +b )
a' P bP a
A= =0.3373 B= =0.1053 , donde a '= 0.5
( RT ) 2
RT T
Z −Z + ( A−B−B ) Z− A × B=0
3 2 2

Z=0.7738
v=Z × v o=0.7738 × 586.35=453.72 cc /mol
Determination of the fugacity of each pureϕ i component, using the RK equation.
a 1=15 549101.96 b 1=26.7368
a 2=272519 446.6 b2=80.5824
a'1 P b1 P a
A 1= 2
=0.0366 B1= =0.0456 , donde a' = 0.5
( RT ) RT T
a'2 P b2 P a
A 2= 2
=0.6422 B2= =0.1374 , donde a' = 0.5
( RT ) RT T
Zi −Z i + ( A i−Bi−Bi ) Z i− A i × Bi =0
3 2 2

Z1 =1.01257 Z 2=0.3356

[
ϕ i=exp Z−1−ln ( Z−B )−
A
B
ln (
Z+B
Z )]
ϕ 1=1.011 ϕ2 =0.522
Calculation of the fugacity coefficient of component i of the gas mixtureϕ^ i
RK parameters of the mixture
a=143 152 921.1 A=0.3373
b=61.7364 B=0.1053
Z=0.7738
RK parameters of pure components
a 1=15 549101.96 b 1=26.7368
a 2=272519 446.6 b2=80.5824
With this data we replace in the equation:

[ (
A 2 √ ai bi
)] ( )
❑
bi Z +B
ϕ^ i=exp ( Z−1 ) −ln ( Z−B )− − ln
b B √a b
❑
Z
ϕ^ 1=1.2366 ϕ^ 2 =0.6145

Molar volume of the mixtureWrite the equation here.

Using the Pitzer-Abbot correlation

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

From the truncated virial equation in two terms:

Tc12=201.98 K Pc12=29.843 T̄ =423.15 k P=60 ¯¿
T P
Tr 12= =2.0943 Pc 12= =2.0105
Tc12 Pc12
ω 1+ ω2
ω 12= =0.1095
2
From the Pitzer-Abbot correlation for the mixture
0 0.422 1 0.172
B12=0.083− 1.6 =−0.04632 B12=0.139− 4.2 =0.1313
Tr 12 Tr 12
For pure substances, it is calculated in a similar way:
0 1
B1=0.0221 B1=0.1379

0 1
B2=−0.3152 B2=−0.00866
Calculation of the second virial coefficient Bij =B 12 y B1 , B 2
R Tc12 0
B12=
Pc 12
( B12 +ω 12 B 12) =−17.9581
1

Similarly for each pure component:

R Tc 1 0 R Tc2 0
B 1=
Pc 1
( B1 +ω 1 B1 )=8.4371 B2=
1
( B +ω B1 )=−294.62
Pc 2 2 2 2
y 1=0.35 y 2=0.65
❑ ❑
B=∑ ∑ y 1 y 2 B12= y 1 B 11 +2 y 1 y 2 B12 + y 2 B22=¿−131.614 ¿
2 2

i=1 j=1
BP
Z=1+ =0.7755
RT
v=Z v o=454.73 cc /mol is the molar volume of the mixture.
Calculation of fugacity coefficient of pure components:ϕ 1 y ϕ2

ϕ 1=exp
[Pr 1 0
Tr 1
( ]
B1 +ω 1 B1 ) =1.0145
1

ϕ 2=exp
[Pr 2 0
]
( B + ω B1 ) =0.605
Tr 2 2 2 2
Calculation of the fugacity coefficient of components i of the solution or mixture.
δ 12=2 B12−B11 −B 22=250.2667 donde :δ ii =δ i
ϕ^ 1=exp
[P
RT
( ]
B11 + y 22 δ 12 ) =1.2175

ϕ^ 2=exp
[P
RT
( ]
B22+ y 21 δ 12) = 0.6377

Using the Lee/Kesler correlation

The reduced pressures and temperatures of each component are calculated respectively,
since this correlation is a function of (Tr, Pr, ω )
Pr 1=1.765 , Pr 2=1.645 y Tr 1=3.352 , Tr 2=1.037 ω 1=0.038 , ω2=0.181

0 1
Z1 Z1
Pr Pr

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

Tr 1.500 1.765 2.000 1.500 1.765 2.000

1.015 0.082
3.000 1.0101 3 8 0.1076
1.015 1.018 0.077 0.089
3.352 1.0128 7 7 8 5 0.1012
1.022 0.072
3.500 1.0156 1 8 0.0949

0 1
ϕ1 ϕ1
Pr Pr
Tr 1.500 1.765 2.000 1.500 1.765 2.000
3.000 1.0069 1.0116 1.0889 1.1194
3.352 1.0104 1.0127 1.0151 1.0839 1.0978 1.1117
3.500 1.0139 1.0186 1.0789 1.1041

0 1
Z2 Z2
Pr Pr
Tr 1.500 1.645 2.000 1.500 1.645
2.000
- -
1.020 0.2715 0.3297 0.0524 0.0722
- -
1.037 0.2923 0.3148 0.3374 0.0038 -0.0307 0.0577
-
1.050 0.3131 0.3452 0.0451 0.0432

0 1
ϕ2 ϕ2
Pr Pr
Tr 1.500 1.645 2.000 1.500 1.645 2.000
1.020 0.5224 0.4266 0.9594 0.9419
1.037 0.5476 0.4982 0.4488 0.989 0.984 0.979
1.050 0.5728 0.471 1.0186 1.0162

0 1
Zi =Z i +ω i Z i
Z1 =1.018 Z 2=0.3092
v i=Z i v 1
v 1=596.90 cc /mol v 2=181.30 cc /mol

v= y 1 v 1+ y 2 v 2=326.76 cc /mol , is the molar volume of the mixture or solution.

The fugacity coefficient of pure components is:

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Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

1 ωi
ϕ i=( ϕ i )( ϕi )
0

ϕ 1=1.0163 ϕ2=0.4967
This correlation is only useful for predicting the fugacity coefficient of the pure
components, but in any case it must be very close to the fugacity coefficients of the
components i of the solution. In some applications it can be very useful with the
limitations of the case.

1. Estimate the fugacity of liquid 1-Butene at its normal boiling point temperature
and 150 Bar.
Solution

Tc = 420.0 K
Pc = 40.43 bar
Zc = 0.277
Vc = 239.3 cc/mol
R = 83.14 cc-bar/mol-K Tb = 266.9 K at PSat = 1.01325 bar
ω = 0.191
P = 150 bar

PSat Tb
Pr = =0 . 0251 T r= =0 .6355
Pc Tc

Calculation of saturated liquid volume:

L
v i =v c Z =91 . 4273 cc / mol
c
( 1−T r )0 . 2857

φ Sat =exp
Pr
Tr [
( B °+ ω B' )

0 . 422
] 0 .172
B °= 0 .083- =−0 . 788621 B'=0. 139- =−1 . 015631
T 1. 6 T 4. 2
r r
Sat
φ =0 . 961934

V i ( P-P )
L Sat
Sat Sat i
i i
f i=φ P exp
RT

fi = 1.80075 bar

2. For SO2 at 600 K and 300 Bar, determine a good estimate of the fugacity
and gR/RT.

Solution

Tc = 430.8 K Pc = 78.84 bar ω =0.245 R = 83.14 cc-bar/mol-K

T=600K P=300bar

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T P
T r= =1 . 393 Pr = =3 . 805
Tc Pc
Using the equation:

Fig 3.15 Page. 102 5th Ed. Van Ness or Table E – 15 and E – 16 Page. 756 and
757 Van Ness 5th Ed.

φ=( φ ° ) ( φ ' ) ω
φ º = 0.672φ ' = 1.354 φ=( 0 . 672 )( 1 .354 )0 . 245=0 . 724
f i=φi Pi=217 . 14 bar
g Ri
=−0 . 323
RT

3. For methane, ethane, and propane, determine the fugacity and fugacity
coefficient for P = 35 bar and T 0 373.15 K.
Redlich-Kwong equation

RT a
P= − ⋯( 1 )
v−b v ( v+b )
Where:
R 2⋅T 2 .5
c
a=0 . 42748 0.5
Pc⋅T
R⋅T c
b=0 . 08664
Pc
Solution
dv
ln φ i=( Z−1 )−ln Z +ln v−∫ Z
v
ln φ i=( Z−1 )−ln Z +ln v−∫
v
−
a
( dv
v−b RT ( v +b ) v )
ln φ i=( Z−1 )−ln Z +ln
v
( )
+
a
v−b RT b
ln
v
v −b ( ) ⋯( 2 )

f i= φ i P i ⋯( 3)

Data

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Compound Tc(K) (bar)

Pc Zc ω Vc (cc/mol) M (g/mol)
Methane 190.6 45.99 0.286 0.012 98.6 16.043
Ethane 305.3 48.72 0.279 0.100 145.5 30.070
Propane 369.8 42.48 0.276 0.152 200.0 44.097

P = 35 bar T = 373.15 KR = 83.14 cc-bar/mol-K

PV
Z= ⋯( 4 )
RT
ln φ i=( Z−1 )−ln Z +ln ( Z A
)
+ ln
Z−B B
Z
Z−B ( )
(a) Methane
a = 1668160.945
b = 29.85299
v = 864.8097 cc/mol
Z = 0.975652
Replacing in (2φ ): = 0.974631
¿
Replacing in (3 f ): = 34.11209 bar
(b) Ethane
a = 5113330.323
b = 45.13859
v = 761.0635 cc/mol
Z = 0.858609
Replacing in (2φ ): = 0.870929
¿
Replacing in (3 f ): = 30.48253 bar

(c) Propane
a = 9469496.265
b = 62.70625
v = 566.9129 cc/mol
Z = 0.639574
Replacing in (2φ ): = 0.735709
¿
Replacing in (3 f ): = 25.74983 bar

4.6. FUGACITY AND FUGACITY COEFFICIENT FOR A

COMPONENT IN SOLUTION

The relationships that describe for a component i in a mixture of gases or a solution of liquids,
with an equation analogous to that of ideal behavior.
¿
μi =β i ( T ) +R T ln f i ⋯ ( 4 . 37 )

[ ]
¿
f i ⇔ X i P reemplaza ⋯ ( 4.38 )
Where:
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Relation (2) expresses the fugacity of species i in solution

According to the phase equilibrium criterion
μ α=μ β =…=μ F ( i=1,2,… ,N ) ⋯( 4 .39 )
i i i
The same way:
¿ ¿ ¿
f α =f β =…=f ( i=1,2,…,N ) ⋯( 4 . 40 )
i i iF
Multiple phases at the same T and P are in equilibrium when the fugacity of each constituent
species is the same in all phases; which is used to solve phase equilibrium problems.
For multicomponent liquid vapor equation (4) can be expressed:
¿V ¿L
i i
f =f ⋯ ( 4 .41 )
Equation (5), is similar to the pure components equation.
From the definition of Gibbs energy, the definition of a partial property provides an equation for
the partial residual Gibbs energy.
g
ng R =n g- ng i ⋯ ( 4 . 42 )
Equation (6) applies to moles of mixture, its differentiation with respect to ni at T, P and nj
constant.

[ ] [ ] [ ] ∂ ( ng i )
g
∂ ( ng R ) ∂ ( ng )
= − ⋯ ( 4 . 43 )
∂n i T , P , nj ∂n i T , P, n ∂ ni T , P ,n
j j

Equation (4.43) by similarity defines the partial molar property, as:

giR=gi −gidi ⋯ ( 4 . 44 )

An equation defining the Gibbs Partial Residual Energy, i

ḡ R
From the equation:
g
μ i ¿ β i (T ) +R T ln Xi P
i
⋯ ( 4 . 45 )
Subtracting equation (4.43) from equation (4.37), both for the same T and P, we have:
¿
g fi
μi −μ ii ¿ R T ln ⋯( 14 . 46 )
XiP
This result can be combined with equation (8) with the identity
μi =ḡ R ⋯(14 . 47)
You get:
R ¿
ḡ i=RT ln φi ⋯( 4 . 48)
By definition:
¿
¿ fi
φ i= ⋯(4 . 49 )
Xi P

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¿
φ i Expresses the fugacity coefficient of component i in solution. How similar it is φ i with gR for
¿g
ideal ḡ =0 gas, and φ =1
R ii

gi
Where M molar value or unit mass of the thermodynamic property of a fluid and M (ideal) aty
P from which gives:
R
i
m̄ =m̄ i− m̄iid
Whose definition expresses equation (6) and (7), in addition:
d ( ng )=( nv ) d P- ( ns ) dT + ∑ μi dn i (α )
i
μi =ḡ i ' (β)

Using the mathematical property, the term

d ( )
ng
RT can be presented:

d ( )
ng
=
RT RT
1 ng
d ( ng )− 2 d T
RT
⋯( 4 . 50 )

The relationship g=h−T⋅S ⋯(γ )

Substituting (α), (β) and (γ) in (4.50) we have:
d ( )
ng
=
RT RT
1
[
ng
nvd P- nSd T + ∑ Ḡ i dni − 2 dT
i RT ]
ḡ
d
ng
( )
RT RT
=
nv nh
d P- 2 dT +∑ i dni
RT i RT
⋯( 4 .51 )

Equation (4.51) expresses a g/RT relationship.

Equation (4.51) can be analogously expressed for an ideal solution:

( )
g g g ḡ i
ng i nv i nh i g

d = d P- 2 dT + ∑ i dni ⋯ ( 4 . 52 )
RT RT RT i RT
Subtracting equation (4.51) and (4.52)
ḡ R
d
RT( )
ng R nv R
=
RT
nh R
d P- 2 dT + ∑
RT i RT
i
dn i ⋯( 4 . 53 )

( )
ng R nv R nh R ¿
d = d P- 2 dT + ∑ ln φ i dni ⋯ ( 4 . 54 )
RT RT RT i
Dividing equation (17) and (18) at constant T and constant composition

[ ]
v ∂ ( g /RT )
R R
= T = Cte ⋯( 4.55)

[ ]
RT ∂P T,X ∂ (ngR /RT )

[ ]
= ⋯(4 .57 ) ¿
hR ∂ ( gR /RT ) ∂n
= P = Cte ⋯( 4.56) i P,T,n j
RT ∂T P,X
ln {^φ i

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[ ]
R
∂ ln { φ^ i i
¿ = v̄ ⋯ ( 4 . 58 )
∂P T , X RT

[ ∂ ln { φ^ i
∂T
¿
] P, X
=
h̄ R
RT 2
⋯( 4 .59 )
Then:
gR
=∑ X i ln { φ^ i ⋯( 4 . 60 ) ¿
RT
The Gibbs/Duhem equation expresses the derivative of equation (24):
∑ X i d ln { φ^ i =0 (T y P constantes )¿
i

APPLICATION EXAMPLE:
DETERMINATION OF
φ^ i IN GAS MIXTURE; WITH THE VIRIAL EQUATION:
BP
Z=1+ ⋯( 1 )
RT
B=∑ ∑ Y i Y j Bij ⋯( 2 )
i j
For a binary mixture i = 1, 2 and J = 1, 2, then equation (2):
B=Y 2 B11 +2Y 1 Y 2 B12 +Y 2 B 22 ⋯( 3 )
i 2
In equation (1):
n BP
nZ=n+ ⋯( 4 )
RT
Differentiating with respect to n1:

Z̄i =
[ ] ∂ ( nZ )
∂ n1 P , T , n2
=1+
[ ]
P ∂ ( nB )
RT ∂n1 T ,n2
⋯( 5 )

An expression
φ^ i for n moles of a mixture from the compressibility factor data is:

RP P
ng dP dP
ln {^φi= =∫(nZ-n) ⋯(6) ¿ln {^φ¿i=∫ (nZ-n) ⋯( 7) ¿ ¿
RT 0 P 0 P
Differentiating equation (7) with respect to ni, holding P,T constant, nj constant:

[ ]
P
∂ ( nZ-n ) dP
ln { φ^ i=∫ ⋯( 8 ) ¿
0 ∂ni P,T ,nj P
Doing:

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Z̄=
[] ∂ ( nZ )
∂ni P ,T ,nj
⋯(9)
P
=∫ (Z̄i−1)
dP
⋯( 11 ) ¿
∂n P
=1 ⋯( 10 ) 0
∂ni
ln { φ^ i
From (5) and (11):

[ ] [ ]
P
1 ∂ ( nB ) P ∂ ( nB )
ln { φ^ i= ∫
RT 0 ∂ ni
dP =
RT ∂ n1
⋯( 12 ) ¿
T , n2 T,n2

B=Y 1 ( 1−Y 2 ) B11 +2Y 1 Y 2 B12 +Y 2 ( 1−Y 1 ) B22

B=Y 1 B11 −Y 1 Y 2 B11 +2 Y 1 Y 2 B12+Y 2 B 22−Y 1 Y 2 B 22
B=Y 1 B11 +Y 2 B 22+Y 1 Y 2 δ 12
δ 12=2 B12−B11 −B22
Doing:
n
Y i= i
n
n1 n2
nB=n 1 B11 + n2 B22+ δ 12
n

[ ]
∂ ( nB )
∂n1 T , n 2
1 n
=B11 + − 1 n2 δ 12
n n2
∂ ( nB )
∂ n1 ( ) [ ]T ,n
2
=B 11 + ( 1−Y 1 ) Y 2 δ 12 =B 11 +Y 2 δ 12
2

Then:

P P
^ln { φ1= (B11+Y 2δ12) (⋯ 13) ¿ ln {φ^ ¿2= (B22+Y 2δ12) ⋯ ( 14 ) ¿ ¿
RT 2 RT 1
Generalizing:

[ ]
P 1
ln { φ^ i= BKK+ ∑ ∑ (2δiK−δij) ¿δiK=2BiK−Bi −BKK ¿δij=2Bij−Bi −Bj ¿¿
Where:
RT 2 i j
δ ii=0 δ jj =0 δ iK =δ Ki

For calculations: RT c
ij
Bij=
Pc ( Bº +ω ij B ' ) ⋯( 15 )
ij
ωi +ω 2
ω ij= ⋯( 16 )
2
1
T c = T c 59
ij i
Tc
j ( )
2
( 1−K ij ) ⋯( 17 )
RT c
ij
Pc =Z c ⋯ (18 )
ij ij V c
ij
 Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

ZC + ZC
i j
Zij = ⋯ (19 )
2

[ ]
3
V 1 +V 1
3 3
c c
i j
Vc = ⋯( 20 )
ij 2

PROBLEM

^ ^
For the ethylene (1) / propylene (2) system as a gas, estimate f 1 , f 2 , φ^ 1 , φ^ 2 T = 150ºC, P = 30 bar
and Y1 = 0.35
a. By applying equations (8) and (9)
b. Assuming the mixture is an ideal solution.

Solution (a)

Tc Pc
ij ij
( ºK )
ij
( bar )
Vc
ij ( mol
cc
) ω ij Tr
ij
Bº B' B

11 282.3 50.40 131.0 0.087 1.499 -0.138 0.108 -59.892

22 366.6 46.65 188.4 0.140 1.157 -0.251 0.046 -159.43
12 321.261 48.189 157.97 0.114 1.317 -0.189 0.085 -99.181

cc
δ 12=2 B12− B11 − B22=20 . 96
mol
0. 422 0 .172
Bºij =0 . 083− B ' ij =0 .139−
T 1. 6 T 4. 2
r r
ij ij
RT c
ij
Bij=
Pc ( Bº +ω ij B ' )
ij

φ^ 1=0.957 { ^f 1=28.722 bar ¿φ^2=0.875 {f¿^ 2=26.244 bar ¿¿

For an ideal solution:

P
Pr = =3 . 805
Pc

( )
Pr
k
φ^ i=exp
Tr ( Bº kk +ω kk B ' kk )
k ,k

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φ^ 1=0. 950 { ^f id=28. 508 bar¿φ^2=0. 875 { f¿^ id=26. 188 bar¿¿
1 2

FUGACITY PROBLEM
^ ^ ^
For the system: methane (1), ethane (2), propane (3), as a gas determine f 1 , f 2 , f 3 φ^ 1 , φ^ 2 , φ^ 3 at ;
temperature of 100ºC, pressure of 35 bar for Y1 = 0.21, Y2 =0.43

a. Using the two-term truncated virial equation

( mol )
Tc Pc cc
Substance Yi ω ij Zij i i Vc
( ºK ) ( bar ) i

1 0.21 0.012 0.286 190.6 45.99 98.6

2 0.43 0.100 0.279 305.3 48.72 145.5
3 0.36 0.152 0.276 369.8 42.48 200.0

Tc Pc
ij ij
( ºK )
ij
( bar )
Vc
ij ( mol
cc
) Zc
ij
ω ij Tr
ij

11 190.6 45.964 98.600 0.286 0.012 1.958

22 305.3 48.672 145.500 0.279 0.100 1.222
1.2 241.226 47.005 120.533 0.282 0.056 1.547
13 265.488 43.259 143.378 0.281 0.082 1.406
23 336.006 45.253 171.308 0.278 0.126 1.111
33 369.800 42.428 200.000 0.276 0.152 1.009

For mixing calculations the following relationships will be used:

ωi +ω 2 ZC + ZC
√
i j
ω ij= T c = T c T c Z ij =
2 ij i j 2

[ ]
3
V 1 +V 1
c
3
c
3 RT c
i j ij T
Vc = Pc = Z c Tr =
ij 2 ij ij V c ij Tc
ij ij
Pitzer/Abott coefficient calculations:
0. 422 RT c
Bºij =0 . 083− ij
T 1. 6
r
Bij=
Pc ( Bº +ω ij B ' )
ij
ij
0 .172 δ ij=2 Bij−Bii −B jj
B' ij=0. 139−
T 4.2
r
ij

ij Bºij B' ij Bij δ ij

11 -0.061016 0.12880 -20.5029 0
22 -0.22320 0.06490 -113.0151 0
1.2 -0.12696 0.11150 -16.5946 30.442
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13 -0.161645 0.097886 -51.5056 107.809

23 -0.27360 0.028500 -166.6817 23.482
33 -0.33299 -0.026648 -244.2371 0

φ^ i=exp
P
RT
1
{ [
BKK + ∑ ∑ Y i Y j ( 2 δ ik −δ ij )
2 i j ]}
1
k=1 ∑ ∑ Y Y (2δ −δ )
2 i j i j ik ij
1
⋅¿ {Y Y (2 δ −δ )+Y Y (2 δ −δ )+Y Y (2 δ −δ )+ ¿ }{Y Y (2 δ −δ )+Y Y ( 2 δ −δ )+Y Y (2 δ −δ )+ ¿}¿ {}
2 1 2 11 11 1 2 11 12 1 3 11 13 2 1 21 21 2 2 21 22 2 3 21 23
{RTP [ B +Y
φ^ 1=exp 11
2
2 δ 12 +Y 2 Y 3 ( δ12 +δ 13−δ 23 ) + ] Y 32 δ 13 }
φ^ =exp { [ B +Y δ }
P
2
RT 22
1
2 δ 12+Y 1 Y 2 ( δ 12+δ23−δ13 ) + ] Y 32 23

φ^ =exp { [ B +Y δ }
P
3
RT 33 δ +Y 1 Y 2
12 13
( δ13 +δ 23−δ 12) + ] Y 22 23

^f =φ^ P
k k

φ^1=1.02 {f^1=35.67 bar¿φ^2=0.8 0 {f^¿2=30.82 bar¿φ^3=0.769 {^f¿3=27.123 bar¿

For an ideal solution, it is treated as a pure substance:
P
Pr =
k Pc
k

( )
Pr
k
φ^ i =exp
dk Tr
( Bº kk + ωkk B ' kk )
k
^f = φ
^ k ⋅P
i
dk dk

φ^1=0.978 {^f1=34.0 bar¿φ^2=0.870 {^f¿2=31.0 bar¿φ^3=0.750 {f^¿3=27.0 bar¿

CALCULATION OF FUGACITY COEFFICIENT OF A COMPONENT IN A MIXTURE,
USING CUBIC EQUATIONS

From the equation:

( ) ng R nv R nh R ¿
d = d P- 2 dT + ∑ ln φ i dni ⋯( 1)
RT RT RT i
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For constant T, v, nj where ni = nj, equation (1) makes:

[ ] ( )
ng R
∂
( )
RT n ( Z-1 ) ∂ P
ln { φ^ i= − ⋯ (2 ) ¿
∂n i T , v,n j P ∂ni T , v ,n

Where:
PV R =( Z−1 ) RT=( Z id −Z ) RT
nZRT
P= ⋯ (3 )
nV

Deriving with respect to ni:

( )
∂P
∂n i T , V ,n j
=
[ ]
P ∂ ( nZ )
nZ ∂ ni
T ,V ,nj
⋯( 4 )

Equation (4) in (2):

[ ] ( )
ng R
∂
ln { φ^ i=
RT
∂n i T , v,n j
− ( )[
Z−1 ∂ ( nZ )
Z ∂ni ] T , v ,n
⋯ (5 ) ¿

In addition to the equation:

V
gR dV
=Z −1−ln Z−∫ ( Z-1 ) ⋯( 6 )
RT ∞ V

Multiplying this equation by n and differentiating with respect to ni:

[ ]( )
gR
∂

[ ] [ ] [( ) ]
V
RT ∂ ( nZ ) ∂ ( ln Z ) ∂ nZ dV
= − −∫ −1 ⋯( 7 )
∂ ni T ,V ,n j ∂ni T ,V ,n ∂ ni T , V , nj ∞ ∂n i T ,V , n j V
j
Replacing this equation (7) in (5) and operating we have:

{[ ] }
V
∂ ( nZ ) dV
ln { φ^ i=∫ -1 −ln Z ⋯( 8 ) ¿
∞ ∂ ni T ,v,n V

For example for a Redlich – Kwong equation

V a ' (T )
Z= −
V- b RT ( V +b )
V na' (T )
nZ= − , luego en la ecuación ( 8 )
V- nb RT ( V + nb )

b̄i
ln { φ^ i= ( Z −1 )−ln
b
( V −b ) Z a
V
63+
b RT
āi b̄ i
1+ − ln
a b
V +b ¿
V [ ]( )
 Offprint of Thermodynamics for Chemical Engineering Policarpo Suero I.

Where:
b=∑ Y i b i
i
1
a=∑ ∑ Y i Y j aij o a=( ai a j ) 2
i j
R2⋅T R⋅T c
c 2. 5
a=0 . 42748 0 .5
b=0. 08664
Pc⋅T Pc

ā i=
[ ]
∂ ( na )
∂ni T ,nj
b̄i =
[ ] ∂ ( nb )
∂ ni T ,n j

Generalizing:
Vi ai ( T )
Zi = −
V i−b i RT ( V i −εb i ) ( V i−σb i )

Where:
(
Ωa⋅α T r , ωi ⋅R 2⋅T
i ) c
2
i b i=
Ωb⋅R⋅T c
i
a i (T )= Pc
Pc
i i

For SRK:

[ ( √ )]
2
( i )
α T r , ωi = 1+ ( 0. 480+ 1. 574 ω i−0 . 176 ω 2 )⋅ 1− T c
i i

For Peng-Robison:

[ ( √ )]
2
( i )
α T r , ωi = 1+ ( 0. 37464+1 . 5226 ωi −0 .26992 ω 2 )⋅ 1− T c
i i

Then:

Equation SRK PR
ε 0 -0.414214
σ 1.0 2.414214
Ωa 0.42748 0.457235
Ωb 0.08664 0.077796
C 0.69315 0.62323

This equation represents a general cubic equation:

a
b̄i
ln { φ^ i= ( Z −1 )−ln
b
( V −b ) Z b RT
V
+
ε−σ
āi b̄i
1+ − ln
a b
V +σb
V +εV
¿ [ ]( )
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5. EXCESS PROPERTIES AND ACTIVITY COEFFICIENT

5.1. EXCESS PROPERTIES
To determine fugacity, the residual properties have first been defined, to measure the deviations
from ideality of the gases as:

m R = m − mid
id
Where g , is the extensive thermodynamic property of the ideal gas.
E
A mathematical expression analogous to the residual property is the thermodynamic excess m
property ().
m E = m − mSi ( 5. 1 )
Where: If indicates ideal solution.

To study the ELV, the property of practical interest is the Gibbs free energy.

G = G − G ( a ) ¿} ¿¿ ( 5.2 ) ¿
E Si

E
WhereG : is called “excess Gibbs free energy”.
Deriving equation (5.2 b), with respect to i for constant nj at T and P.

[ ∂ ( ng E )
∂ni ] P, T , nj
=
∂ni [ ]
∂ ( ng )
P, T , n j
−[∂ ( ngSi )
∂ni ] P , T , nj
( 5 .3 )

ḡ E = ḡ i − ḡ Si .. .. . .. .. . .. .. ( 5 . 4 )
Similarly: i

Ḡ E : defined as “Gibbs partial molar excess free energy”

Also:
d ḡi = RT d ( ln f^ i ) ( 5 .5 )
Integrating equation (5.5) from the initial state of component i to its solution state:

ḡi − ḡ id
i = RT ln ( )f^ i
fi
(5 . 6 )

Equation (5.6) expresses a real relationship where:

^f = xi f i
i Si
f i =fugacity of the real pure component.
An expression that represents an ideal solution in a similar way to equation (5.6) is:

( )
xi f i
ḡ Si − g i = RT ln
i fi
ḡ Si − g i = RT ln x i ( 5 .7 )
i

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Subtracting equation (5.6) and (5.7) we have:

ḡi − ḡ
i Si = RT ln ( ) f^ i
xi f i

ḡ E = = RT ln
( )f^ i
xi f i
( 5. 8 )

Where:  is defined as the activity coefficient of component i of the solution, expressed by the
following equation:
^
fi
γ i = (5. 9 )
xi f i

Then an expression representing the excess Gibbs free energy:

ḡ E = RT ln γ i
E
ḡ
ln γ i = ( 5. 10 )
RT

ḡ E
RT
=
∂ni [
∂ ( ng E /RT )
] P,T , ηj

[
∂ ( ng / RT )
]
E
ln γ i = donde n i ¿ n j ( 5. 11 )
∂ni P ,T , η j
E γ i =1
For an ideal ḡ =0 solution,
^
f = x i f i ( 5 . 12 )
i Si
Equation 5.12 represents an ideal solution relationship, and the Lewis/Randall rule is also known.
ni
= xi
Solving equation (5.11), dividing the second term by , then n .
gE
= ∑ x i ln γ i (5 . 13 )
RT
Deriving this equation:

∑ x i d ( ln γ i ) = 0 ( Ec .Gibbs / Duhem )( 5 .14 )

gE
Where RT : is a constant whose derivative is equal to zero, and the expression for a binary system
can be given by the following relation:

x1
[ d ( ln γ i )
d x1 ] [ + x2
d ( ln γ 2 )
d x2 ] =0

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E
Another way to express ḡ is from other relationships:
ḡi = β i ( T ) + RT ln f^ i
ḡ Si = β i ( T ) + RT ln x i f i
i

Subtracting these equations:

ḡi − ḡ Si = RT ln
i ( ) f^ i
xi f i
ḡ E = RT ln γ i
i

d ( n g ) = nv dP − ns dt

d
ng E
RT
= ( )
n vE
RT
n hE
d P − 2 dT + ∑ ln γ i dni ( 5 . 15 )
RT
For a practical application of equation (5.15)

[
∂ ( n gE /RT )
]
E
v (5.16)
=
RT ∂P T,x

[
∂ ( n g E / RT )
]
E
h (5.17)
=−T
RT ∂T P, x

ln γ i=
[∂ ( n g E / RT )
∂ ni ] P ,T ,nj
(5.18)

Where :
ln γ i is a partial property with respect toG E / RT , and then:

[ ]
∂ ( ln γ i ) v̄ E
i
= ( 5 .19 )
∂P T, x RT

[ ]
∂ ( ln γ i ) h̄ E
i
= ( 5. 20 )
∂T P, x RT 2
By analogy:

gE
= ∑ x i ln γ i ( Ecuación de Gibbs / Duhem )
RT

∑ x i d ( ln γ i ) = 0 P , T , cte

PROBLEM
The excess Gibbs energy of a binary liquid mixture at T and P is given by:
gE
= ( −2.6 x 1 − 1.8 x 2 ) x1 x 2
RT
For a given temperature and pressure:

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a) Find the expression for ln γ 1 y ln γ 2

gE
= ∑ x i ln γ i
b) Show that when these expressions are combined according to the RT
gE
given equation for RT is recovered.

∑ xi d ( ln γ i ) = 0
c) Show that these expressions satisfy the Gibbs/Duhem equation

( ) ( )
d ln γ 1 d ln γ 2
= =0
dx 1 x 1=1 dx 1 x =0
d) Prove that 1

gE
e) Plot a graph against X1 with the values of RT ln γ 1 y ln γ 2 , , calculated from the equation for
gE
RT and by the equations developed in (a), identify the points ln γ 1 ∞ y ln γ 2 ∞ and show their
values.

Solution:
a) We know that:
gE
= ( −2.6 x 1 − 1.8 x 2 ) x1 x 2
RT
gE
= (−2. 6 x 1 − 1 .8 ( 1−x 1 ) ) x 1 ( 1−x 1 )
RT
gE
= ( −0 .8 x 1 − 1 . 8 ) x 1 ( 1− x1 )
RT
gE
= −1.8 x 1 + x 2 + 0. 8 x 3
RT 1 1
From the equations:
d M d M
M̄ 1 = M + x 2 M̄ 2 = M x 1
d x1 dx 1
gE
d
gE RT
ln γ 1 = + ( 1−x 1 )
RT d x1
E
g
E d
g RT
ln γ 2 = + x1
RT d x1
but:
gE
d
RT
= −1 . 8 + 2 x 1 + 2 . 4 x 2
d x1 1

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Then:
ln γ 1 =−1.8 + 2 x 1 + 1.4 x 2 − 1.6 x 3
1 1
ln γ2 =−x 2 − 1.6 x 3
1 1
E
g
= ∑ x i ln γ 1
b) From the equation RT
gE
RT 1 (
=x −1.8 + 2x 1 + 1.4 x 2 − 1.6 x 3 ) + ( 1−x 1 ) (−x 2 − 1.6 x 3 )
1 1 1 1
then this equation is reversed to

the initial equation

c) Dividing the Gibbs/Duhem equation by dx1

∑ x i d ln γ i = 0
x 1 d ln γ 1 + x 2 d ln γ 2 = 0
Now we divide by dx1:
d ln γ 1
= 2+2.8 x 1− 4.8 x 2 ( 1 )
dx 1 1

d ln γ 2
= −2 x 1 − 4.8 x 2 ( 2 )
dx 1 1

d ln γ 1 ( b) ¿
x1 = 2 x 1 +2.8 x 2 − 4.8x 3 ( a )
dx 1 1 1

¿
x2
d ln γ 2
dx 1
= ( 1−x1 ) (−2x 1 − 4 .8 x 2 ) ¿ ¿
1 }
d ln γ 1
=0
d) For x1 = 1 in the ec (1) dx 1
d ln γ 2
=0
For x1 = 0 in Ec (2) dx 1

e) We know that:

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gE
=−1 .8 x 1 + x 2 + 0. 8 x 3
RT 1 1
ln γ 1 = −1 . 8 + 2 x 1 + 1 . 4 x 2 − 1 . 6 x
1 13
ln γ 2 =−x 2 − 1 .6 x
1 13
ln γ 1 ∞ ( 0 ) = −1 .8 ln γ 2 ∞ ( 1 ) = −2. 6

Tabular x1 = 0.0, 0.1, 0.2, .........., 1.0

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6. ELV ACTIVITY COEFFICIENT

6.1. BASIC CONCEPTS

For the i components of a vapor mixture the fugacity can be expressed as follows:
^f v = y φ ^ i P ( 6 .1 )
i i
^f L = y φ
i i ^ i P ( 6 .2 )
In liquid/vapor equilibrium:
^f L = f^ v
i i
For an ideal
φ i gas: = 1, then equation (7.2) will be:
^f = y P
i i
P Sat
In the limit where xi = yi = 1, the total pressure is equal to i
^f L = f^ iv = P Sat
i i

For the VAPOR phase, the fugacity of the components i are calculated from the fugacity
φ^ i
^
f vi
^
φ v = ( 6 .3 )
i yi P
For the liquid phase, the fugacity is calculated from the activity coefficient
^
fLi
γ i = (6 . 4)
xi f L
i
^f L = x γ f
i i i iL

^f L = f^ v
Applying the ELV criterion: i i

From equations (7.3) and (7.4)

xi γ i f L = yi φ
^ v P
i i
^
yi φ P
iv
γ i = ( 6 . 5)
xi f
iL

For compressed liquids:

V
^f L = φ
Pi iL
i ^
i
Sat P
i
Sat exp ∫P Sat RT
dP ( 6 . 6 )
i

Replacing equation (6.6) in equation (6.5)

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^
yi φ P
iv
γ i = ( 6 . 7)
Pi
V L
exp ∫P
i
x i φ^ Sat P Sat dP
i i i Sat RT

A new coefficient Φ i is introduced to simplify equation (6.7)

^
φ
iv
Φi = ( 6 .8 )
Pi
V L
^
φ Sat exp ∫P i
dP
i i
Sat RT
Then equation (6.7) is expressed in terms of this new coefficient already defined:
yi P
γ i = Φi ( 6 . 9)
xi P
i Sat

Φ i is obtained from liquid-vapor equilibrium (LVE) calculations

φ^ = φ^ v = 1 , Φ i → 1
At low pressures, the behavior of the vapor phase is considered as ideal i Sat i gas, so
it is assumed as Φ i= 1, then equation (7.9) is expressed as follows:
yi P
γ i = ( 6 . 10 )
xi P
i Sat

The evaluation of
γ i a binary system from the excess Gibbs free energy equation,

gE
= ∑ x i ln γ 1
RT (6.11 a)
for i = 1, 2
E
g
=x 1 ln γ 1 + x 2 ln γ 2 (6.11 b)
RT

To graph
ln γ 1 y ln γ 2 vs and x2, divide this equation by x1x2, you get a table for a given
x1
temperature (constant)
E
g g
E
P(kPi) xi yi ln γ 1 ln γ 2 RT RT x 1 x 2
P 0.00 0.00 - 0.00 0.00 -
2 Sat

P 1.00 1.00 0.00 0.00 -

1 Sat

gE gE
ln γ 1 , ln γ 2 , ,
The four thermodynamic functions for which we have experimental RT x 1 x 2 RT values
are properties of the liquid phase, where γ 1 ≥ 1 y ln γ 1 ≥ 0 , i = 1, 2
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Chart 6.1
From equation (6.11), as x1  0,
E
g
=0
RT
E
g
lim =0
x1 →0 RT
When x2  0, x1 = 1.
gE
The value of RT  0, then x1 = 0 and x1 = 1
gE
The magnitude of x1 x 2 RT becomes indeterminate for both x1= = 0 and x1 = 1 because in both limits gE
= 0, as x1.x2 = 0.
For x1  0, using L'Hospital's Rule

( )
E E
g g
E
d
g RT RT
lim = lim = lim (a )
x1 →0 x 1 x 2 RT x1 →0 x 1 x1 →0 d x1

lim
d ( )
gE
RT
The value of x1 →0
d x 1 can be obtained by deriving equation (6.11 b) between dx1
E
g
=x 1 ln γ 1 + x 2 ln γ 2
RT
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You get:

d ( )
gE
RT
= x1
d ln γ 1
+ ln γ 1 + x 2
d ln γ 2
−ln γ 2 ( b )
dx 1 dx1 dx 1

Where x 1 + x 2 = 1 : and
d x2
= −1
d x1 To justify the negative sign of the derivative of the equation
From the Gibbs/Duhem equation

∑ x i d ln γ i = 0
x 1 d ln γ 1 + x 2 d ln γ 2 = 0
For a binary system dividing this equation by dx1 gives:
d ln γ 1 d ln γ 2
x1 + x2 =0
dx 1 dx 1 a constant P and T (6.12)
The activity coefficients for the liquid phases at low pressure are almost independent of P, so
combining equation (b) and (7.10) gives:

d
gE
RT( )
= ln
γ1
dx 1 γ2
in the limit, as x1  0, x2  1, then

lim
d ( )
gE
RT γ1
= lim ln = ln γ 1 ∞

x1 →0 dx 1 x 1 →0 γ2
From equation (a):
E
g
lim = ln γ 1 ∞

x1 →0 x1 x 2 RT

In analogous form: x1  0, x2  0
gE gE
lim = ln γ 2 ∞

x1 →1 x 1 x 2 RT
. Where the values of x1 x 2 RT : are equal at the limits of dilution

infinite ln γ 1 , ln γ 2 .
The Gibbs/Duhem equation can be expressed as follows;

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dln γ 1 −x 2 dln γ 2
= (6.13)
d x1 x1 d x1

From graph (6.1) a mathematical expression can be given by the equation:

E
g
= A 21 x 1+ A 12 x 2
x1 x 2 RT
E
g
=( A ¿ ¿ 21 x 1 + A12 x 2 )x 1 x 2 ¿ (6.14)
RT

The equations for and are derived from this last equation by applying the equation:
ln γ
[ ] ( ngRT )
E
∂ (6.15)
i=¿ ¿
∂ ni P,T , nj

Doing: for x1 , and for x2 , and , then:

[ ]
E
g n 1 n2
=(A 21 x 1 + A12 x 2 ) 2
x1 x 2 RT ( n 1+ n2 )

Multiplying both sides of the above equation and differentiating with respect to n1:
ln γ
[ ] ( ngRT )
E

[ ]
∂
1=¿
∂ n1 P, T , n2
=n2 ( A 21 x1 + A12 x 2 )
[ 1
( n1 +n2)
2
−
2 n1
( n1 +n2)
3
]
+
n1 A21
( n1+n2)
2
¿

Reconverting n1 to x1

Making x2 = 1 – x1

ln γ 1=x 22 [ A 12+ 2 ( A 21−a12 ) x 1 ] (6.16 a)

ln γ 2=x 1 [ A 21+ 2 ( A 12−A 21 ) x 2 ]
2
(6.16 b)

In an analogous way, the same is obtained for component 2. These are the equations of Max Margules
(1856 – 1920), Austrian meteorologist and physicist.
For limiting conclusions:

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With Margules's conclusions for and one can easily construct a correlation of the original
data set.

The addition:

6.2. MODELING FROM EXPERIMENTAL DATA

The following is an ELV data set for the methanol (1)/water (2) system at the temperature of
333.15ºK (taken from K . Kurihara, et, al J. Chem Eng. Data Vol 40, pp 679-687, 1955)

P(kPa) xi yi P(kPa) xi yi
19.9530 0.0000 0.0000 60.6140 0.5282 0.8085
39.2230 0.1686 0.5714 63.9980 0.6044 0.8383
42.9840 0.2167 0.6268 67.9240 0.6804 0.8733
48.8520 0.3039 0.6943 70.2290 0.7255 0.8922
52.7840 0.3681 0.7345 72.8320 0.7776 0.9141
56.6520 0.4461 0.7742 84.5620 1.0000 1.0000

a) Based on the calculations, find the parameters for the Margules

E
g
equation that provide the best fit to the data and prepare a P – X – Y diagram in which you
RT
compare the experimental points with the curves determined from the correlation, establish the
error.
b) Repeat (a) for the Van Laar and Wilson equation
Solution

Doing:

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gE
= x 1 ln γ 1 + x 2 ln γ 2
RT
The results box shows the tabulation of the data based on the relationships.
Least squares: y = ax + b

By deriving this equation and then equating it to zero, the values of a and b are obtained:

E E
g g
Yo RT RT x 1 x 2

1
2 1.572 1.013 0.452 0.013 0.087 0.6207
3 1.470 1.026 0.385 0.026 0.104 0.6127
4 1.320 1.075 0.278 0.073 0.135 0.6282
5 1.246 1.112 0.220 0.106 0.148 0.6363
6 1.163 1.157 0.151 0.015 0.148 0.5990
7 1.097 1.233 0.093 0.209 0.148 0.5939
8 1.050 1.311 0.049 0.271 0.136 0.5688
9 1.031 1.350 0.031 0.300 0.117 0.5381
10 41.021 1.382 0.021 0.324 0.104 0.5222
11 1.012 1.410 0.012 0.343 0.086 0.4973
12

Equation (5) is differentiated twice first with respect to “a” and then with respect to “b”.

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Earring:

Intersection:
E
g
= A 12+( A ¿ ¿ 21+ A12) x 1 ¿
x1 x 2 RT

Tabulating A12 = 0.683 A21 = 0.475

So the new equation is:
E
g
=0.683+ 0.208 x1
x1 x 2 RT

Then, with this data, it is tabulated again:

For the Van Laar Equation

x 1i x 2i E
E g
Tabular g :, that is, the inverse of:
x1 x 2 RT
x 1 x 2 RT
Where:

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x1 x2
y= 1 A 12− A21
g
E
y=ax+ b , whereb= : and a=
A 12 A 12 × A 21
RT

A12 = 0.705 A21 = 0.485

Operating
From this data obtain the calculated values using the following equations:

b) For the Wilson equation

Using the least squares we have obtained:

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With the following experimental data, extracted from the Journal Eng. Chemical, for the binary system
diethyl ether (1) / Ethyl – 2 dimethyl ethyl ether (2), for the temperature of 303.15 K, represent in it the
models: Margules, Van Laar and Wilson for the liquid phase, considering the vapor phase as ideal
behavior where i = 1.

P (kpa) X1 Y1

19.29 0 0
29.97 0.1293 0.4755
31.44 0.1455 0.5080
39.53 0.2516 0.6518
46.33 0.3322 0.7186
52.5 0.4044 0.7667
58.84 0.4971 0.8094
70.29 0.6929 0.8889
77.61 0.8133 0.9388
83.73 0.945 0.9804

By the equation:

We found :

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E
g
Then with the equation: =x 1 ln γ 1 + x 2 ln γ 2
RT

E
g
E
g
P x1 y1 1 2 ln1 ln 2 RT RT x 1 x 2

19.29 0 0 … 1 … 0 … …
29.967 0.1293 0.4755 0.12860502 0.93590434 0.25157565 -0.066242 -0.02514818 -0.22337754
31.44 0.1455 0.508 1.28086227 0.93843316 0.2475335 -0.06354365 -0.01828192 -0.14704381
39.53 0.2516 0.6518 1.19494994 0.95343167 0.17810429 -0.04768751 0.0091217 0.04844305
46.33 0.3322 0.7186 1.16941609 1.01206347 0.15650455 0.01199129 0.0599986 0.27045502
52.5 0.4044 0.7667 1.16142935 1.06607345 0.14965145 0.06398222 0.09862686 0.04094769
58.84 0.4971 0.8094 1.11792142 1.15606352 0.11147109 0.14502072 0.1283432 0.51339006
70.29 0.6929 0.8889 1.05219217 1.31182433 0.05087577 0.27629998 0.12010355 0.56442397
77.61 0.8133 0.9388 1.04534385 1.31884136 0.04434588 0.2767536 0.0877364 0.57780955
83 0.945 0.9804 1.01361205 1.33374523 0.01352023 0.28799095 0.02861612 0.55057477
85.7 1 1 1 … 0 … … …

E
g
=( A 21−A 12 ) x 1 + A12
x1 x 2 RT
y=m x 1+ b
E
g
Then we graph vs x1
RT x 1 x 2
Fitting to a straight line:
A12 = -0.193 r = 0.88092754
A21 - A12 = 1.062  A21 = 0.869
A12 = ln 1 = - 0.193 A21 = ln 2 = 0.869
We find the corrected 1 and 2:

 To correct 1 and 2:

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 To correct the pressure:

P = y1 P + y2 P = x1 1 P1sat + x2 2 P2sat
E
g
 To correct: =x 1 ln γ 1 + x 2 ln γ 2
RT

 To correct y1:

sat sat
x 1 γ 1 P1 x 1 γ 1 P1
y 1= =
P ( x1 γ 1 P1sat + x 2 γ 2 Psat
2 )

x 1 γ 1 × 85.7
y=
( x 1 γ 1 ×85.7+ x2 γ 2 × 19.29 )

E
g
E
g
P x1 y1 1 2 ln1 ln2 RT RT x 1 x 2

19.29 0 0 0.8244 2.3845 -0.193 0.869 -3

28.2951 0.1293 0.41662425 1.0638 0.9838 0.0647433 -0.014506 -4.259179x10-3 -0.037832
29.7104 0.1455 0.4560004 1.0883 0.9802 0.0849046 -0.017715 -2.783884x10-3 -0.0232391
39.8705 0.2516 0.6501607 1.2107 0.9554 0.178409 -0.040168 0.048259 0.0787367
47.8732 0.3322 0.7420503 1.256 0.9412 0.210567 -0.0530707 0.0345098 0.15556
54.6279 0.4044 0.7984377 1.2664 0.9373 0.216319 -0.056138 0.0540434 0.224376
62.1947 0.4971 0.848095 1.2438 0.9520 0.198913 -0.04116 0.0781803 0.3127319
73.5468 0.6929 0.910122 1.1274 1.1096 0.109375 0.099224 0.106257 0.49935616
78.4401 0.8133 0.938357 1.0548 1.3344 0.0484247 0.288511 0.0932489 0.6141139
83.50487 0.945 0.976486 1.0058 2.3845 0.004962 0.613707 0.0384429 0.739642
85.7 1 1 1 1

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VAN LAAR MODEL:

We also work with the experimental data given above and obtain the following data in order to find the
van Laar parameters:

For ELV experimental data

E
g
Q= =x 1 ln γ 1+ x 2 ln γ 2
RT

It is shown that:

Where:
1 A 12− A21
a A 12 = b = A12∗A 21
a = 2.2723 b = 0.46722

The parameters obtained are according to the Van Laar model.

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A12 : 0.44023 A21: 0.36647

Obtaining the Van Laar parameters we proceed to calculate the following corrected data for the VAN
LAAR model by applying the following equation:

A12 A21
ln (γ 1 )= ln ( γ 2 )=
A X A X
(1+ 12 1 )2 (1+ 21 2 )2
A21 X 2 A12 X 1
With the new values the following table is generated with the corrected values

GE/
X1 Y1 calc. P calc. lnY1 calc lnY2 calc GË/RT X1X2RT
0 0 19.29 0.4423 0 0 0.4423
0.1293 0.4614 32.97 0.317 0.055 0.0483 0.429
0.1455 0.4907 34.443 0.3043 0.0622 0.0531 0.4277
0.2516 0.627 43.009 0.2238 0.1054 0.07889 0.41
0.3322 0.6962 48.5977 0.1726 0.137 0.09154 0.4126
0.4044 0.7451 53.1594 0.1336 0.1646 0.098 0.407
0.4971 0.797 58.5403 0.0919 0.1989 0.1 0.4
0.6929 0.887 69.0486 0.0319 0.2677 0.0822 0.386
0.8133 0.959 75.3898 0.011 0.3076 0.0574 0.3783
0.945 0.987 82.5807 0.000936 0.3495 0.01922 0.3698
1 1 85.7 0 0.3664 0 0.36647

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WILSON MODEL:

The constants a and b are the same as for the Margules model. (a=−0 . 039122 and
b= 0 . 401937 )

g
E
− x1 x2
= ln ( x 1 + x 2 Λ12) + ln ( x 2 + x 1 Λ 21)
RT x 1 x 2 x 1 x 2 x1 x2

ln γ 1=−ln ( x1 + Λ 12 x 2) + x2
( Λ12
−
Λ 21
x 1+ Λ12 x 2 x 2 + Λ21 Λ21 x 2 )
ln γ 2=−ln ( x2 + Λ 21 x 1 )−x 1
( Λ 12
−
Λ 21
x 1 + Λ12 x2 x 2+ Λ 21 Λ21 x2 )
Wilson's equations expressed at infinite dilution, that is, at the limits of the graph are:

=0 : ln γ ∞1 =−ln Λ12− Λ 21+1

x1

=1 : ln γ ∞2 =−ln Λ21− Λ 12+1

x2

LNY1 LNY2 GË/

X1 Y1 calc. P calc. calc. calc. GË/RT X1X2RT
0 0 19,29 -0,1942 0 0 -0,1942
0,1293 0,3263 26,3 -0,1469 -0,0034 -0,0219 -0,1945
0,1455 0,397 27,24 -0,1401 -0,0043 -0,0241 0,1938
0,2516 0,5776 33,705 -0,1019 -0,014 -0,0361 -0,1917
0,3322 0,6782 38,995 -0,0726 -0,026 -0,0414 -0,1866
0,4044 0,7496 44,057 -0,0472 -0,0406 -0,043 -0,1785
0,4971 0,822 51,031 -0,0149 -0,0678 -0,041 -0,164
0,6929 0,9176 62,68 -0,0319 -0,1374 -0,0202 -0,0949
0,8133 0,956 69,896 -0,0409 -0,1639 0,0317 -0,0208
0,945 0,988 79,259 -0,0327 -0,02 0,0336 0,646
1 1 85,7 0 -0,095 0 -0,0034

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