Chapter 4. Activity Coefficient Models. Part 1. Random-mixing
based models
Problem 1. Azeotropes with the one-parameter Margules equation
The molar excess Gibbs energy for some simple (symmetric) binary systems can be
represented by the one-parameter Margules equation, where B is a parameter
characteristic of the mixture:
4.48
g E  Bx1 x2
i. Show that the activity coefficient of component 1 is given as:
B 2
ln  1 
x2
4.49
RT
What is the value of the activity coefficient at infinite dilution of the component 1?
ii. Assume that the vapor phase is ideal and that the system is at low pressures. Show,
using the one-parameter Margules equation, that at every temperature for which an
azeotrope exists (concentration where y=x for all components), the azeotropic
composition of compound 1 x1az and the azeotropic pressure Paz are related by the
following equation:
 ln( P az / P1sat ) 
1

1


az
sat 
x1az
 ln( P / P2 ) 
1/ 2
4.50
Problem 2. Henry’s law constants and infinite dilution activity coefficients
At 300 K, some experimental data are available for dilute liquid mixtures of components
1 and 2. When 1 is dilute in an excess of 2, the Henry’s law constant is H1,2=2 bar. When
2 is dilute in an excess of 1, the Henry’s law constant is H2,1=1.60 bar. Estimate the
vapour composition, which is in equilibrium with an equimolar liquid mixture of 1 and 2
at 300 K. Assume that the vapour is an ideal gas. At 300 K, the pure-component vapour
pressures are 1.07 bar for pure liquid 1 and 1.33 for pure liquid 2.
Why can’t we use the one-parameter Margules of the previous problem?
Hint: Assume the validity of a simple two-parameter activity coefficient model e.g. van
Laar or Margules two-parameter.
Problem 3. Comparison with experimental data
The vapour-liquid equilibrium data reported in Table 4.6 (Gmehling et al., Vapor-liquid
equilibrium data collection, Dechema Chemistry Data Series, Dechema, Frankfurt, 1979)
are for the system acetone (1)/cyclohexane (2) at 25 oC. Assume for simplicity that, at the
low pressures involved, we have ideal vapour behaviour.
1
Table 4.6. VLE of acetone (1)/cyclohexane (2) at 25 oC.
Pressure
x1
y1
Activity
Activity
(mmHg)
Coefficient of Coefficient of
compound (1) compound (2)
97.45
0.0
0.0
118.05
0.0115
0.1810
172.90
0.0575
0.4580
253.90
0.4235
0.6800
259.40
0.5760
0.7050
252.00
0.9250
0.8580
243.80
0.9625
0.9160
230.40
1.0000
1.0000
i. Calculate the experimental activity coefficients of the two compounds and complete
the table above. Do we have positive or negative deviations from Raoult’s law?
ii. Use a single experimental data point (at P=253.9 mm Hg) to estimate the parameters
of an activity coefficient model and then calculate the equilibrium pressure and
vapour compositions at liquid acetone concentrations: 0.0115, 0.5760 and 0.9625.
Prepare a table showing for these three compositions the percentage deviations
between calculated and experimental activity coefficients (for both components),
equilibrium pressures and vapour phase mole fractions. Compare the results to the
experimental data and comment on the results.
Hint. For the van Laar equation, show that the parameters A and B for a binary system
can be obtained using the activity coefficient values for the two compounds using the
equations:
 ln  1 x1 
B  ln  2 1 

 ln  2 x 2 
AB
ln  1
ln  2
 x2

 x1



2
4.51
2
Problem 4. The van Laar equation
i.
Show that the two forms of the van Laar equation [equations 4.9 and 4.10] are
equivalent. What is the value of ln  1 using the two forms of the van Laar
equation ?
ii.
Show that when the van der Waals parameters are obtained from the critical
point, then   Pc (where  is defined as in equation 4.9).
iii.
Chlorine ( Pc  7.71 MPa) and carbon tetrachloride ( Pc  4.56 MPa) form an
ideal solution. On the other hand, carbon disulfide ( Pc  7.9 MPa) and
2
iv.
methanol ( Pc  8.10 MPa) split into two liquid phases. Can the van Laar
equation explain these observations ?
2
Starting from g E  b 1 2  1   2  derive the expression for the activity
coefficient (eq. 4.9) for a binary mixture (b is the mixture co-volume
parameter and  is defined as in equation 4.9). The “segment” fractions are
defined in equation 4.9 and Table 4.4.
Hint: Show first that the excess Gibbs energy for the van Laar equation can be
written as:
x x bb
2
4.52
g E  1 2 1 2  1   2 
b
Problem 5. K-factors for distillations using the Regular Solution Theory (Modified
from Prausnitz et al.6)
For distillation-column design, we need K factors (Ki=yi/xi). A liquid mixture at 50 oC
contains 30 mol% n-hexane and 70 mol% benzene. At 50 oC, the pure component vapour
pressure of n-hexane is 0.533 bar and of benzene is equal to 0.380 bar.
1. Calculate the K factors of n-hexane and benzene in this mixture. Assume that the
pressure is sufficiently low to neglect gas-phase corrections and Poynting factors.
Then calculate the relative volatility of hexane/benzene.
2. What would be the relative volatility assuming an ideal liquid solution?
The molar volumes and solubility parameters are presented in Table 4.7.
Table 4.7. Molar volumes and solubility parameters for n-hexane and benzene (all values
at 25 oC).
Compound
Volume (cm3/mol)
n-hexane
Benzene
132
89
Solubility parameter
(J cm-3)1/2
14.9
18.8
Problem 6. Azeotropes with the Regular Solution Theory (Modified from Prausnitz
et al.6)
A binary liquid mixture contains the non-polar components 1 and 2. The mixture is to be
separated by ordinary distillation. To determine if this is feasible, it is necessary to know
whether or not the mixture has an azeotrope. The pure component vapour pressures at
300 K and the solubility parameters are given for the two compounds in Table 4.8. The
pure-component molar volumes are for both compounds equal to 160 cm3/mol. At 300 K,
does this mixture have an azeotrope? If so, what is its composition? Assume that the
vapour phase is ideal. What would the result be if both liquid and vapor phases are
considered ideal ?
3
Table 4.8. Pure component vapour pressures and solubility parameters at 300 K for the
two components.
Compound
Vapor Pressure (kPa)
(1)
(2)
53.3
80
Solubility parameter
(J cm-3)1/2
14.3
17.4
Problem 7. Using azeotropic data to predict VLE for a binary mixture (From
Sandler, S.I., Chemical and Engineering Thermodynamics (3rd edition). John Wiley &
Sons, 1999)
Benzene and cyclohexane form an azeotrope at 0.525 mole fraction benzene, temperature
of 77.6 oC and a total pressure of 1.013 bar. At this temperature the vapour pressure of
pure benzene is 0.993 bar and that or pure cyclohexane is 0.980 bar.
i. Using the van Laar model, estimate the activity coefficients of benzene and
cyclohexane at liquid mole fractions (benzene) 0.0, 0.2, 0.5, 0.8 and 1.0. Using this
activity coefficient information, compute the equilibrium pressure versus liquid and
vapour composition at the same temperature.
ii. Make predictions for the activity coefficients of benzene and cyclohexane using the
regular solution theory and compare these with the results obtained in question (i).
Volume and solubility parameters of benzene are given in problem 5, while for
cyclohexane they are 109 cm3/mol and 16.8 (J/cm3)1/2.
Problem 8. Interrelations between models – Derivations
1. Starting from the expression of the regular solution theory for multicomponent
systems using lij=0, derive the expression for binary systems (equation 4.11).
2. The excess Gibbs free energy equation of the van Laar equation is prior to the
assumption of any mixing rules given as:


g E    xi  i   
 i

4.53
a

b
Consider a binary mixture. Derive the expression for the activity coefficient of
compound (1) in the following cases (types of mixing rules):
i.    xi  i
4.54
i
ii.    xi x j  ij
i
 ij   i  j
4.55
j
iii. the classical van der Waals one fluid mixing rules for the energy and co-volume
parameters (a,b) using also the well-known geometric mean combining rule for the
cross energy parameter and the arithmetic mean rule for the cross co-volume
parameter (equations 3.2 and 3.3 in Chapter 3). Assume that all interaction
parameters (kij and lij) are equal to zero.
4
Comment on the results and mention examples of mixtures your three models can
potentially be applied to.
Problem 9. Permeability of pharmaceuticals through polymers using the Regular
Solution Theory
dC
Starting from Fick’s law: J   D
4.56
dx
Derive the equation for the permeability of pharmaceuticals through polymers using the
Flory-Huggins/Regular solution theory, according to Michaels et al.25 (equation 4.25):
S f  Tm 
4.57
ln  J max  l  exp(1   )  
 1  ln (  D)
R  T

where the maximum attainable flux of a pharmaceutical through the polymer is given by:
D.C *
J max 
4.58
l
D is the diffusion coefficient, l is the thickness of the film and C* is the concentration of
the pharmaceutical in the polymer at equilibrium with crystalline solid steroid.
Hint: Show first that the FH equation for the activity of the pharmaceutical at very low
solubilities can be given by the equation: ln a1  ln 1  (1  12 ) .
Problem 10. Screening of solvents for pharmaceuticals using the regular solution
theory (RST)
i. Starting from the expression for the activity coefficient of RST (lij=0) for
multicomponent systems:
2
 V
ln  i   i  i   av  
 RT


m
av
   i i
4.11
i 1
derive the expression for a binary mixture (components 1 and 2)
ii. What are the expressions for the infinite dilution activity coefficients of compounds 1
and 2 ?
iii. Show that in the case of solid-liquid equilibria, the (real) solubility of a solid (2) in a
liquid (1) is given by the equation:
x 2real 
x 2ideal
4.4
2
5
iv. Kolar et al., (Fluid Phase Equilibria, 2002, 194-197, 771-782) and Abildskov and
O’Connell (Ind. Eng. Chem. Res., 2003, 42, 5622-5634) have illustrated how the RST can
be used for screening solvents for pharmaceuticals. The key concept derived from RST is
that (at low solute concentrations) the solubility x2 is a parabola with respect to the
solubility parameter of the solvent 1 with a maximum located at: 1   2 [  2 is the
solubility parameter of the pharmaceutical]. Illustrate this by showing that for the RST
and for low solute (2) concentrations, the following equation is derived:
 VL 



VL 
VL
4.59
ln x2    2  12   2 2 2  1   ln x2id  2  22 
RT 
RT
 RT 



Figure 4.10 shows the lnx2 - 1 plot of hydrocortisone at 298.15 K. The enthalpy of
fusion, liquid volume and melting point of this solid are 34.87 kJ/mol, 0.293 m3/kmol and
485.5 K, respectively.
Figure 4.10. Plot of lnx2 - 1 of hydrocortisone at 298.15 K. From Abildskov and
O’Connell (Ind. Eng. Chem. Res., 2003, 42, 5622-5634). Reprinted with permission from
American Chemical Society (2003).
i.
ii.
iii.
calculate the ideal solubility at 298.15 K and provide, based on the plot, an
estimate of the solubility parameter of the solid.
what is the solubility of hydrocortisone in carbon tetrachloride, based on the
plot and on the equation shown in this problem?
Explain, with reference to figure 4.4, why RST performs well for morphine
but not for the other pharmaceutical.
6
Problem 11. LLE with the regular solution theory
Show for the RST that a mixture of two components (1and 2) will give two liquid phases
if ( Tc = critical solution temperature):
RTc 
x1 
2 x1 x 2V12V22
x1V1  x2V2 
3

1
2

2
4.60
V12  V22  V1V2  V1
V2  V1
What are the LLE conditions if the two components have equal liquid volumes?
4.61
Problem 12. Infinite dilution conditions
i. For polymers, weight fraction activity coefficients are often used:
1 
1
4.62
w1
Show that, for a binary solvent(1)/polymer(2) solution, the infinite dilution activity
coefficients of the solvent based on molar and weight fractions are related via the
equation:
M 
4.63
1   1  2 
 M1 
where M1 is the molecular weight of the solvent and M2 is the molecular weight of
the polymer.
ii. For the benzene/PIB(50000 g/mole) solution with solvent's weight fraction equal to
0.0564 and molar activity coefficient  1 =0.3396, show that:
- the mole fraction of the solvent is equal to 0.9745.
- the activity of the solvent is equal to 0.331.
- the weight fraction activity coefficient of the solvent is equal to 5.868.
Problem 13. Solubility Parameters
i. Mixed Solvents
Cellulose nitrate (10.27 H) is insoluble in both ethyl ether (7.4 H) and ethanol (12.7
H). Can this polymer be soluble in a mixture of the two ? Why/Why not ? At which
composition? (the values in parentheses are the solubility parameters in Hildebrand
units, i.e. cal1/2/cm3/2).
ii. Hansen solubility parameters
A polymer has a solubility parameter equal to 9.95 (with dispersion, polar and
hydrogen bonding parts equal to 7.0, 5.0 and 5.0) and a solubility Radius equal to 3.0.
Is it certain that a solvent with a solubility parameter equal to 10 (8.0, 6.0, 0.0 are the
dispersion, polar and hydrogen bonding parts, respectively) will dissolve it?
Why/Why not?
The solubility parameter and radius values are in Hildebrand units, i.e. cal1/2/cm3/2.
7
Problem 14. Solubility of mixed solvents in commercial paint resins - An industrial
case
Mixed solvents are often used in the paints and coatings industry. The choice of suitable
solvents is often based on the three-dimensional Hansen theory. For a mixed solvent, the
solubility parameter is given by the equation:
4.64
    i i
i
where  i is the volume fraction of the solvent i and  i are its solubility parameters
(either total solubility parameter or the Hansen solubility parameters:  d ,  p ,  h ).
For two resins, a polyester resin known as Desmophen (product of Bayer) and an epoxy
resin known as Epikote 1001 (product of Shell) and for two possible solvents, n-butanol
and xylene, the Hansen solubility parameters are given in Table 4.9.
The solubility radius is for the polyester resin equal to R=8.2 and for the epoxy resin is
R=6.20.
i. Are the two resins soluble in xylene?
ii. Is the polyester resin soluble in a mixed solvent composed of n-butanol/xylene with
80% volume fraction of xylene?
iii. Is the epoxy resin soluble in a mixed solvent composed of n-butanol/xylene with 55%
volume fraction of xylene?
iv. Explain briefly the results.
The solubility parameter and radii values are in Hildebrand units, i.e. cal1/2/cm3/2.
Hint: The basic equation which determines the solubility range according to Hansen’s
method is:
4 d 1   d 2    p1   p 2    h1   h 2   R
2
2
2
4.65
Table 4.9. Hansen solubility parameters for Desmophen and Epikote 1001 for two
possible solvents, n-butanol and xylene.
Compound
d
p
n-butanol
Xylene
Epikote 1001
Desmophen
7.8
8.7
9.95
10.53
2.8
0.5
5.88
7.30
h
7.7
1.5
5.61
6.0
Problem 15. Evaporation of isopropanol from a PVAC film (Modified from
Prausnitz et al.6)
A film of poly(vinyl acetate), PVAC, contains traces of isopropanol. For health reasons,
the alcohol content of the film must be reduced to a very low value: government
regulations require that that the volume fraction of isopropanol should obey the
relationship: 1  104 . To remove the alcohol, it is proposed to evaporate it at 125 oC. At
this temperature, chromatographic experiments, which are carried at infinite dilution of
8
alcohol, give a value of the Flory-Huggins parameter equal to 0.44. The vapor pressure of
alcohol is 4.49 bar. Assuming that the PVAC is of very high molecular weight, calculate
the low pressure, which must be maintained in the evaporator to achieve the required
purity of the film. What would the value of pressure be if i) the Flory-Huggins parameter
is assumed equal to zero (near-athermal solution), ii) if the solution is assumed ideal.
Comment on the results.
Hint: Under the conditions prevailed here, the activity of the solvent is simply given as
the ratio of the partial pressure to the solvent vapor pressure and the polymer is assumed
to be completely non-volatile.
Problem 16. The Flory-Huggins model
i. Show that, for the Flory-Huggins model, one of the most widely-used models for
polymers, the molar-based activity coefficient at infinite dilution is, for a binary
solution, given by the equation:
V1
V

4.66
 1  1  12
V2
V2
Derive also the infinite dilution activity coefficient expression for compound (1) in a
binary solution for the Entropic-FV combinatorial-Free Volume term (equation 4.19).
ii. At the critical solution temperature, the following conditions apply:
 ln 1
4.67a
0
 2
ln  1  ln
 2 ln  1
0
4.67b
 22
Derive the following equations for the critical Flory-Huggins interaction parameter and
the critical volume fraction:
1
4.68a
 crit

2
1 r
2
1
1 
4.68b
  1 

2
r
What are the values of the critical volume fraction and FH parameters for “high”
polymers (i.e. polymers of high molecular weight)?
crit
1
Hint: Use the FH model written for the solvent activity as:
 1
ln  1  ln  1  1   2   12  22
 r
4.69
Problem 17. Are phthalates completely miscible in PVC?
There are, especially in the recent years, many debates regarding the use of phthalates
like DOP (dioctyl phthalate) as plasticizers in several polymers and especially PVC
9
(polyvinyl chloride). Plasticizers are added in PVC because they can lower the glass
transition temperature of the polymer, but it has now been established that DOP and other
plasticizers sometimes migrate to the surface. This phenomenon results to the destruction
of the plastic material and may cause health problems as well, since certain phthalates are
considered to be dangerous chemicals (carcinogenic). Tables 4.10 and 4.11 present values
of the Hildebrand and Hansen solubility parameters, the Hansen solubility radius as well
as the Flory-Huggins parameters for PVC and DOP. Figure 4.11 shows how the FloryHuggins parameter for PVC/DOP depends on concentration, according to Su et al., (J.
Appl. Polymer Sci., 1976, 20, 1025-1034).
On basis of this information, can we conclude that DOP is completely soluble (miscible)
in PVC? (If yes, this would imply that the cause of DOP migration cannot be fully
explained by thermodynamics). Justify your answer.
Table 4.10. Solubility parameters of PVC and DOP (all values are given in Hildebrand
units = (cal/cm3)1/2) from various sources: A.F.M. Barton. CRC Handbook of polymerliquid interaction parameters and solubility parameters. CRC Press, 1990; C.M. Hansen.
Hansen Solubility Parameters. A User’s Handbook. CRC Press, 2000.
Compound total
value
DOP-1
7.94
DOP-2
8.92
PVC-1
9.314
PVC-2
11.03
PVC-3
10.49
dispersion
polar
hydrogen
bonding
8.14
3.2
1.52
9.4
8.91
4.5
3.68
3.5
4.08
R-Hansen
3.2
1.7
Table 4.11. Flory-Huggins (FH) parameters for the PVC/DOP system obtained from
various measurements.
Anagnostopoulos et al.
T (oC)
53
76
116-118
Bigg
Patel and Gilbert
113-114
Reference
Doty and Zable
FH parameter
0.01
0.03
-0.03
0.06 - extrapolated from
swelling data
0.05
-0.13
P.M. Doty, H.S. Zable, J. Polym. Sci., 1946, 1, 90; C.E. Anaglostopoulos, A.Y. Coran, H.R.
Gamrath, J. Appl. Polym. Sci., 1960, 4, 181; D.C.H. Bigg, J. Appl. Polym. Sci., 1975, 19, 3119;
S.V. Patel, M. Gilbert, Plast. Rubber Process Appl., 1986, 6, 321.
10
Problem 18. The Flory-Huggins model
For the polymer solution cyclohexane (1) / polystyrene (2) [PS] the FH parameter is
available at solvent volume fraction  1 =0.60 and it is 12 =0.72. The density of
cyclohexane is 0.7575x103 kg/m3 and the density of the polymer is 0.10482x104 kg/m3.
The average molecular weight of PS is 25900 g/mol. All data are reported at the same
temperature (=373.15 K).
i. Estimate the activity of the solvent as well as the activity coefficient of the solvent
(based on weight fractions) at solvent weight fraction w1=0.55. What is the
percentage deviation as compared to the experimental value 1 =1.802 reported by
Krigbaum and Geymer, J. Amer. Chem. Society, 1959, 81(8), 1859-1959?
ii. Estimate the FH-parameter value so that the experimental activity coefficient value
(at w1=0.55) is reproduced.
Figure 4.11. The Flory-Huggins interaction parameter for the PVC/DOP system (  23 ) as
a function of concentration (volume fraction of PVC). From Su, Patterson and Schreiber
J. Appl. Polymer Sci., 1976, 20, 1025-1034. Reprinted with permission from Wiley
(1976).
Problem 19. Pressure of a polymer solution using Flory-Huggins (From Prausnitz et
al.6).
Estimate using the Flory-Huggins model, the total pressure of a liquid solution containing
50% (weight-based) polyvinyl acetate (with molecular weight equal to 83400 g/mol) and
50% vinyl acetate (molecular weight = 84.1 g/mol) at 125 oC. At this temperature, the
density of the polymer is 1.11 g/cm3 and that of the solvent is 0.783 g/cm3. The vapor
11
pressure of the solvent is 3340 Torr (mmHg). From experiments, we know that the ratio
of the pressure to the solvent weight fraction at infinite dilution, i.e. the so-called Henry’s
P
law constant is 18.3 bar: lim( w1  0)    18.3bar
 w1 
What is the most important assumption involved in the calculations?
Problem 20. Infinite dilution activity coefficients via Flory-Huggins
Ashworth and Price, (Macromolecules, 1986, 19(2), 358-263) published the FloryHuggins interaction parameter values for the system benzene (1) / polydimethylsiloxane
(2) at various concentrations and at temperature T = 30 oC.
i. Estimate the activity and the weight-fraction activity coefficient of the solvent at
volume fraction 1 =0.0165. At this concentration the value of the FH parameter is
0.7317.
ii. Prove that, for the Flory-Huggins model, the molar-based and weight-based activity
coefficients at infinite dilution are given by the equations:
V
V
4.70
ln  1  ln 1  1  1  12
V2
V2
and
1 M2
1

4.71
1 
exp  12  1  
r M1
r

iii. Estimate the values of the above two activity coefficients at infinite dilution of the
solvent (at molar basis and weight-fraction basis)
Data:
solvent: density = 11.11 g/cm3, molecular weight=78.11 g/mol.
polymer: density=0.9523 g/cm3, average molecular weight=3350 g/mol.
Problem 21. Selective dissolution
Selective dissolution is a separation process which has been proposed for recycling of
polymers. The separation is possible because the individual polymers in a mixture of
different polymers are dissolved in different solvents. A typical mixture of polymers is
composed of polyethylene (PE), polystyrene (PS) and polyvinylchloride (PVC). Choose
based on the data available in tables 4.12 and 4.13 three suitable solvents for each
polymer and fulfill table 4.14 indicating also the method used for the solvent selection.
12
Table 4.12. Solubility parameters for various polymers and solvents.
Polymer
Polyethylene (PE)
Polystyrene (PS)
Polyvinylchloride (PVC)
Solubility parameter
(J/cm3)1/2
15.8-17.1
17.4-19.0
19.2-22.1
13
Solvent
Acetone
Acetonitrile
Benzene
Butanol
Carbondisulfide
Carbontetrachloride
Chlorobenzene
Chloroform
Cyclohexane
Cyclohexanone
Decalin
Dioxan
Ethanol
Ethylacetate
Hexane
Methanol
Methylethylketone
Nitrobenzene
Nonane
Octene-1
Phenol
Tetrahydrofuran
Toluene
Water
Xylene
Solubility parameter
(J/cm3)1/2
20.3
24.8
18.8
28.7
20.5
17.6
21.3
19.0
16.8
21.3
17.6
20.5
26.1
18.2
14.9
29.7
19.0
22.5
15.6
15.5
24.6
18.5
18.3
48.0
18.1
Table 4.13. Infinite dilution activity coefficients of solvents in polymers.
Infinite Dilution Activity
Coefficient Values ( 1 )
Acetone
Acenonitrile
Benzene
Butanol
Carbondisulfide
Carbon tetrachloride
Chlorobenzene
Chloroform
Cyclohexane
Cyclohexanone
Decalin
Dioxane
Ethanol
Ethylacetate
Hexane
Methanol
Methylethylketone
Nitrobenzene
Nonane
Octene-1
Phenol
Tetrahydrofuran
Toluene
Water
Xylene
Polyethylene
(PE)
Polystyrene
(PS)
Polyvinylchloride
(PVC)
10-16
3.0-5.0
25-35
2.0
4.0
2.6
4.0
2.5
30-140
5.4
30-100
6.0-11
4.7
10
6.0-16
3.0
-
16
14-33
4.0-6.0
10-20
4.5
2.5-3.0
3.0-6.0
4.5-5
8.0-10
3.4
5.0-6.0
4.5
9.5-28
8.0-13
10-20
10-50
7.3
6.0-8.0
8.0-9.0
8.0-9.0
4.0-7.0
4.0-6.0
76-250
3.5
9.0-20
14-26
7.0-9.0
22-32
8.0
6.0
6.0-9.0
16-18
5.0
45.5
12-17
54-764
19-21
8,0-10
35
11
7.3
7.5-12
-
14
Table 4.14. Choose three suitable solvents for each polymer indicating also the method
for solvent selection.
Solvent
Method for solvent selection
PE
PVC
PS
Problem 22. Deriving the Flory-Huggins activity expressions
For a binary solvent(1)-polymer(2) system, the total Gibbs energy change of mixing is
given for the Flory-Huggins model by the equation:
G mix
 n1 ln 1  n2 ln  2  n1 2 12
4.72
RT
Assuming that the FH interaction parameter is independent of concentration, show that
the activities of solvent and polymer are given by the equations:
 1
ln a1  ln  1  1   2   12  22
4.73a
 r
4.73b
ln a2  ln  2  1  r 1  12 r12
where r=V2/V1
Hint: From basic thermodynamics (eq.1.13): ln ai 
15
  nGmix 


ni  RT  T , P ,n j  i
4.74
Problem 23. Features from the FH equation combined with RST
The FH interaction parameter can be related to the RST-based solubility parameters (of
solvent, 1 and polymer, 2) as (including also its entropic part):
V
4.75
   S   h   S  1  1   2 2
RT
Flory himself used the following notation for the entropic and enthalpic parts of the FH
parameters and the theta temperature,  (=critical solution temperature, CST, at infinite
polymer molecular weight):
1
 S   1
2
h  
4.76

T
Show that the FH parameter is a linear function of the inverse temperature
(   a  b / T ) and determine the two constants as a function of the  , ,  .
Derive the Schultz-Flory plot equation for the relationship between CST and theta
temperature (starting from the critical value of the FH parameter, see problem 16):
1
1
1  1
1 
4.77
 1  
 
CST    1  r 2r 
Derive the following equation:
  12    2 2   ( 2 ) 2  S 

4.78
   
1   


RT
V1   RT   RT
V1 

How do you think the above two equations (4.77 and 4.78) can be used in data
analysis?
 1
i.
ii.
iii.
iv.
Problem 24. Rules of thumb from the Flory-Huggins model
The rules of thumb based on infinite dilution activity coefficients shown in Appendix 4.B
can be (approximately) derived from the rule of thumb based on the Flory-Huggins
model i.e.
4.79
12  0.5  1  6
Starting from the Flory-Huggins model, and assuming constant densities for the solvent
and the polymer, show that miscible mixtures are obtained if 1  4.5 .
Hint: Prove that for very high molecular weight polymers, the solvent (1) infinite
dilution activity coefficient obtained from the FH model is given by the equation:
 
4.80
ln 1  1  12  ln  2 
 1 
where  i are the densities of the solvent and the polymer (1=solvent, 2=polymer).
16
Problem 25. About the Flory-Huggins model
i. Show that the two forms of the Flory-Huggins (FH) model, here written for the
solvent (1) activity coefficient of a binary solvent(1)/polymer(2)solution:
 
 
ln  1  ln  1   1   1   12  22
 x1 
 x1 
4.81a
   1
ln  1  ln  1   1   2  12  22
 x1   r 
4.81b
are equivalent, if r is the ratio of the polymer (2)/solvent(1) volume i.e. r=V2/V1
ii. It is often stated that the FH parameter is a function of concentration and sometimes a
simple linear function can be used: 12  a  b 2 . Is it appropriate to use this expression
for the FH parameter in the above-mentioned expressions for the FH equation (equations
4.81a and 4.81b) ? Why/why not? Justify your answer.
iii. Assume that the FH parameter is zero for nearly athermal solutions such as mixtures
for alkanes. Then, using the FH model estimate the n-pentane activity coefficient at
infinite dilution in mixtures with n-decane, n-eicosane and n-C44. Assume that T=100
o
C. Use two versions of the concentration fractions,  i , both based on molar volumes
and on the van der Waals volumes as estimated using the Bondi R-values given in
Table 4.15. What do you observe? Compare the results obtained with the two
versions of the FH model.
Table 4.15. Values of R and Q for the groups of problem 25. The van der Waals volumes
can be calculated from the R values by multiplying them with 15.17 (thus expressing the
van der Waals volumes in cm3/mole).
Group
CH3
CH2
R
0.9011
0.6744
Q
0.848
0.540
Hint. The molar volumes can be estimated e.g. using the GC-VOL method (Elbro, H.S.,
Fredenslund, Aa., and Rasmussen, P., 1991. Ind. Eng. Chem. Res., 30(12): 2576-2582) or
the van Krevelen method, which is based on the van der Waals volumes.
17