1
KE-100.3410 Polymer properties
Exercise 5:
Polymer solubility and gas permeation
Exercise 5.1
Polymer solubility in different solvents can be estimated using solubility parameters i.
According to the Hansen model, the overall solubility parameter can be obtained as
   D2   P2   H2
where D, P and H are the dispersive, polar, and hydrogen bonding parameters. In table 3-3
Hansen solubility parameters for some common solvents are listed.
Fried, J.R., Polymer Science and
Technology, 2nd ed., Prentice Hall
2003, pp. 115.
The best solubility is obtained when the solubility parameters of polymer and solvent are
close to each other. For polymers the so called radius of solubility sphere (RAO) can be
calculated. The liquids inside the sphere act as solvents and outside the sphere as nonsolvents.
RA 
 2
 2 D , s    P , P   P , s    H , P   H ,s 
2
D,P
2
where P refers to polymer and S to solvent.
RA
R
 1 for solvent and A  1 for non-solvent.
RAO
RAO
2
2
Estimate using the Hansen model whether PVC is soluble in the monomer vinyl chloride
when the following parameters for PVC are known D=18.2 MPa1/2, P = 7.5 MPa1/2, H = 8.3
MPa1/2 and RAO = 3.5 and for vinyl chloride: D = 15.4 MPa1/2, P = 8.1 MPa1/2, H = 2.4
MPa1/2. Choose the best solvent from the table for PVC.
Solution 5.1
Calculate the radius of the solubility by substituting the solubility parameter values to the
equation:
RA 

 2
 2 D , s    P , P   P ,s    H , P   H ,s 
2
D,P
2
 2*18.2  2*15.4    7.5  8.1  8.3  2.4 
2
2
2
2
 8.2
Compare RA to RA0.
RA 8, 2

 2,3  1 thus PVC does not dissolve in its own monomer.
RAO 3,5
Calculate the solubility parameter for PVC:
   D2   P2   H2  18.22  7.52  8.32  21.4
Choosing from solvent listed in the table the best choice would be 1,4-Dioxane since the
solubility parameter = 20.5 is closest to PVC.
Exercise 5.2
The solubility parameter is related to cohesive energy-density Ecoh or the molar energy of
vaporization of a pure liquid Ev:
 i  Eicoh 
Eiv
Vi
where Ev is defined as the energy change upon isothermal vaporization of the saturated
liquid to the ideal gas state at infinite dilution and Vi is the molar volume of the liquid. The
solubility parameter of a polymer has to be determined indirectly or calculated by groupcontribution methods. Calculation of  by a group-contribution method requires the value of
3
a molar attraction constant Fi, for each chemical group in the polymer repeating unit. Values
of Fi have been obtained by regression analysis of physical property data for a large number of
organic compounds. The solubility parameter of a polymer is then calculated from these
molar attraction constants and the molar volume of the polymer.
i 
F F
i 1
Vi
i

i 1
i
Mi
i
Some values of the molar attraction constants are shown in the table. Calculate the solubility
parameters using Van Krevelen constants for
a) Polyisobutylene, density 0.924 g/cm3
b) Polystyrene, density 1.04 g/cm3
c) Polycarbonate, density 1.20 g/cm3
Fried, J.R., Polymer Science and
Technology, 2nd ed., Prentice Hall
2003, pp. 114.
4
Solution 5.2
a) Based on the structure of polyisobutylene the molar attractions and molecular weight can
be calculated:
Fi
Amount
Fi
Mi (g/mol)
–CH2–
280
1
280
14.027
–C(CH3)2–
840
1
840
42.081
i
1120
56.108
Group
and then the solubility parameter can be calculated with the equation:
i 
F
i
i 1
Mi

i
cm3
mol  18.4MPa1/2
g
56.108
mol
g
0.924 3
cm
1120MPa1/2
b) Polystyrene
Fi
Amount
Fi
Mi (g/mol)
>CH–
140
1
140
13.019
–CH2–
phenyl
280
1517
1
1
280
1517
14.027
77,106
i
1937
104.152
Group
And the solubility parameter can be calculated:
cm3
1937 MPa
mol  19.3MPa1/2
 i  i 1 
Mi
g
104.152
mol
i
g
1.04 3
cm
 Fi
1/2
5
c) Polycarbonate
Group
Fi
Amount
Fi
Mi (g/mol)
–C(CH3)2–
–OCOO–
p-phenylene
840
767
1377
1
1
2
840
767
2754
42.081
60.008
154.212
i
4361
256.301
And the solubility parameter can be calculated
i 
F
i 1
Mi
i

i
4361MPa 1 / 2 cm 3 mol 1
 20.4MPa 1 / 2
g
256.301
mol
g
1.20
cm 3
Exercise 5.3
Polyvinylalcohol film (thickness 0.20mm) is laminated in between two LDPE films (thickness
of each film 0.2 mm). Oxygen transfer coefficient for LDPE is 2.210-13
(cm3(STP)cm)/(cm2sPa) and for PVOH:lla 6,6510-16 cm3(STP)cm)/(cm2sPa).
a)
What is the oxygen transfer coefficient for the laminate at 25°C?
b)
A product is packed in this laminate material. The gas volume of the package is 20
cm3 and surface area is 250 cm2. How long is shelf life of the product when the
oxygen concentration in the packet must not exceed 1.0 mol-%? Oxygen concentration
is 0.0 mol-% just after the packaging.
c)
What would be the shelf life of a product packed in a similar LDPE packaging at room
temperature?
Solution 5.3
a)
6
Gas transfer coefficient in multilayer laminate depends on the properties of the individual
layers in the laminate
l
l
l
l
 1  2  3
P P1 P2 P3
Gas permeation can be calculated from equation
Q
P  A  t  p
l
where
Q
gas flux permeated through
[cm3]
P
Permeation coefficient
[cm3  cm/cm2  s  Pa]
t
A
l
time
surface area of the film
thickness of the film
[s]
[cm2]
[cm]
p
pressure difference
[Pa]
Oxygen permeation coefficient P:
P
l
0.60mm
cm3 ( NTP)  cm

 2.0 1015
0.20mm
0.20mm
0.20mm
l1 l2 l3
cm2  s  Pa


 
P1 P2 P3 2.2 1013 6.65  016 2.2 1013
b)
For ideal gas the volume is equivalent to molar volume (1 mol-% = 1 vol-%). Laminate is ok
for packaging until there is 20 cm3 of oxygen transferred through the material:
Q = 20 cm3  0.01 = 0.20 cm3
Partial pressure of oxygen outside the packet: p1 = 0.21101kPa=21000 Pa
Partial pressure of oxygen in the beginning p2,start = 0 Pa and when the oxygen concentration
in the packet is 1.0 mol-%, p2,end = 0.01101kPa = 1000 Pa.
 Approximation p  constant
Time taken for oxygen transfer
7
Ql
t

P  A  p
0.20cm3  0.060cm
 1.1  106 s  13d
3
cm cm
2.0  10 15 2
 250cm 2  21000 Pa
cm sPa
c)
t
Q l

P  A  p
0.20cm 3  0.060cm
 10400s  2.9h
3
13 cm cm
2
2.2 10
 250cm  21000 Pa
cm 2 sPa
If the packaging material was LDPE-film, the time would be 10400 s which is less than
3hours.
Exercise 5.4
Plastic soft drink bottles are made of poly(ethylene terephthalate) in Finland. Empty 1.5 dm3
bottle is filled to 2.0 bar CO2 pressure at 25°C and the cap is closed tightly. Carbon dioxide
transfer coefficient for PET is P(CO2, 25°C) = 0.11810-13 cm3(STP)cm/(cm2sPa). How
long a time does it take for CO2 pressure to drop one tenth?
Solution 5.4
Assume the bottle is cylinder with wall thickness of 1mm, and diameter of the bottom is 8 cm.
Assume also that gasses are ideal gasses and the CO2 content in air is 0.03%.
Volume of the cylinder: V   r 2 h  h 
V
 r2
Surface area of the cylinder:
 1,5  103 cm3
2
V

A  2 rh  2 r 2  2    r 2   2 
   4cm    850,5cm2
4cm
r



Partial pressure of CO2 outside of the bottle is: po  0.0003 101325Pa  30.4Pa
Pressure difference between inside and outside of the bottle in the beginning (a):
p a  p a  po  2 10 5 Pa  30.4 Pa  199970 Pa
8
Pressure difference at the end (e): pe  pe  po  0.9  2 10 5 Pa  30.4 Pa  179970 Pa
Calculate the average pressure difference:
pavg 
pa  pe 199970 Pa  179970 Pa

 189970 Pa  190000 Pa
2
2
At the end the bottle has 9/10 of the original pressure (10% drop in pressure), so flux of the
CO2 has (ideal gas p1V1 = p2V2):
Q  V2  V1 
 p

p1V1
 1

 V1  V1  1  1  1.5dm3 
 1  0.17dm3
p2
 0.9 
 0.9 p1 
The time this has taken can be calculated:
Q
t
P  A  t  pavg
l
Q l

P  A  pavg
0.17 103 cm 3  0.1cm
 8.92 10 6 s  103d
3
cm cm
0.118 10 13 2
850.5cm 2 190000 Pa
cm sPa
Exercise 5.5
When solvent is added to polymer the change in free energy (GM) reveals whether the
polymer will dissolve:
GM  H M  T SM
Dissolution will happen when GM is negative. Entropy of mixing SM is always positive and
can be expressed with Bolzman relation:
SM  k ln 
where k = 1,3810-23 J/K is Bolzman constant and  describes the different ways that solvent
molecules N1 and polymer molecules N2 can be arranged. Applying Sterling approximation
( ln N !  N ln N  N ) the entropy of mixing can be expressed:
SM  k ( N1 ln v1  N 2 ln v2 )
where v1 is the volume fraction of solvent and v2 volume fraction of polymer. Dimensionless
Flory-Huggins parameter  can be applied to estimate the polymer-solvent interactions. The
9
parameter can be experimentally measured for each polymer-solvent combination. Using
interaction parameter the enthalpy of mixing can be expressed:
H M  kT 1 N1v2
Then the change in free energy follows:
GM  H M  T SM  kT ( N1 ln v1  N 2 ln v2  1 N1v2 )
(Fried, J.R., Polymer Science and Technology, 2nd ed., Prentice Hall 2003, p. 96 – 102).
Calculate the change in free energy of mixing when 10% solution of polystyrene (Mn = 10000
g/mol) in cyclohexane at 34 oC is prepared. Flory-Huggins parameter is 0.50; density of
cyclohexane is 0.7785 g/cm3 and density of styrene 1.06 g/cm3
Solution 5.5
Volume fractions for cyclohexane v1 = 0.9 and for styrene v2 = 0.1. Take volume of solution to
be V = 1 cm3. Calculate the number of solvent molecules (C6H12) N1:
g
Vv 
cm3  6, 025 1023 mol 1  5, 02 1021
N1  n1 N A  1 1 N A 
g
M1
(6 12, 011  12 1, 008)
mol
Number of polystyrene molecules
1cm3  0,9  0, 7785
Vv 
N 2  n2 N A  2 2 N A 
M2
1cm3  0,11, 06
10000
g
mol
g
cm3  6, 025 1023 mol 1  6,39 1018
Calculate the change in free energy of mixing:
GM  H M  T S M  kT ( N1 ln v1  N 2 ln v2  1 N1v2 )
 1,38 1023
 1, 24 J


J
 307,15 K  5, 02 10 21  ln 0,9  6,39 1018  ln 0,1  0,5  5, 02 1021  0,1
K
10
Polymer miscibility
Mixtures of polymers can be treated with the same thermodynamic approach as polymer
solutions. The change in volume during mixing is ignored so they are not quite accurate.
When mixing two polymers the change in free energy (GM) will tell whether polymers are
miscible or there is phase separation:
GM  H M  T SM
Mixture will be miscible when the GM is negative. For polymer mixtures:
V

 2
GM  kT  v1v2 1 1    Nc  v1 ln v1  v2 ln v2  
 z
 Vr

where k = 1,3810-23 J/K is Bolzman constant, T temperature, V volume of mixture, Vr
volume of one polymer (the one with smaller monomer), vi volume fractions of polymers,
Flory-Huggins parameter, z lattice coordination number (usually between 6 and 12), Nc
number of molecule chains per volume unit (Sperling L.H., Introduction to physical polymer
science, 4th ed., Wiley & Sons, 2006, p. 156-157). Flory-Huggins parameter can be calculated
from:
1 
Vm,1
RT
1   2 
2
where Vm,1 is the molar volume of the polymer with higher amount or if equal composition the
one with smaller monomer, T temperature, R = 8.3145 J/(K mol) gas constant, i solubility
parameters of the polymers.
Exercise 5.6*
Is 1000 g of polystyrene miscible with 1000 g polybutadiene at 150oC, when the molecular
weights of both polymers are 1105 g/mol? What is the upper limiting molecular weight that
the polymers are still miscible at that temperature? Would increasing the temperature enhance
the miscibility when the molecular weights are 1105 g/mol? Solubility parameter for PS is
 18.6 (MPa)1/2 and density  = 1.06 g/cm3. Solubility parameter for polybutadiene is 
17.2 (MPa)1/2 and density  = 1.01 g/cm3. Can be assumed that lattice coordination number z
= 6.
11
Solution 5.6*
V

 2
GM  kT  v1v2 1 1    Nc  v1 ln v1  v2 ln v2  
 z
 Vr

Polymer amounts are the same but the molecular weight of butadiene (C4H6) (M0,1= 54.092
g/mol) is smaller than the molecular weight of styrene (C8H8) (M0,2= 104.152 g/mol) and thus
the molar fraction of polybutadiene is bigger. Thus polybutadiene is considered the “solvent”.
Calculate the volume of the mixture and the volume fractions of the polymers:
V  V1  V2 
v1 
m1
1

m2
2

1000 g
1000 g

 1933cm3
g
g
1.06 3 1.01 3
cm
cm
V1
m
1000 g
 1 
 0.512 v2  1  v1  1  0,512  0.488
g
V 1V 1.01
3
1933cm
cm3
V/Vr is the number of molecules:
Vm N
V 1 N A
V
V
V


 1 A 
V
V
Vr
V1M 0,1
M 0,1
1
1
N 0,1 n0,1 N A

g
 6.025 10 23 mol 1
3
cm
 2.18 1025
g
54.092
mol
1933cm3 1.01
Flory-Huggins parameter using the molar volume of butadiene
Vm,1 
n1 M 0,1
:

V1
1
1 
Vm ,1
RT
 1   2 
2

M 0,1
RT 1
 1   2 
2

kg
2
 17.2  18.6  106 Pa
mol

 0.0298
J
kg
8.3145
 423.15K 1010 3
K  mol
m
0.054092

g
 17.2( MPa )1/2  18.6( MPa )1/2
mol
J
g
8.3145
 423.15K 1.01 3
K  mol
cm
54.092

2
12
Nc number of molecule chains:
m
m 
N c  N1  N 2  N A (n1  n2 )  N A  1  2 
 M1 M 2 

 1000 g
1000 g
 6.025 1023 mol 1  

g
 1 105 g
1 105
mol
mol



22
  1.21 10


Calculate the free energy of mixing:
V

 2
GM  kT  v1v2 1 1    N c  v1 ln v1  v2 ln v2  
 z
 Vr



 2
25
 2.18 10  0.512  0.488  0.0298  1  

23 J
 1,38 10
 423.15K  
 6

K
 1.211022   0.512  ln 0.512  0.488ln 0.488  


 582 J
G is positive so the polymers are not miscible. Calculate the upper limiting molecular
weight for miscible system. This is the case when free energy equals to zero i.e. the enthalpy
and entropy of mixing are equal.
GM  H M  T SM  0  H M  T SM
Nc is affected by molecular weights of the polymers. Since the molecular weights are equal,
Nc can be written:
N c  N1  N 2  N A (n1  n2 )  2 N A
m
Mr
Manipulate the equation and solve for Mr:
V

 2
GM  kT  v1v2 1 1    Nc  v1 ln v1  v2 ln v2    0
 z
 Vr


V
 2
v1v2 1 1    Nc  v1 ln v1  v2 ln v2   0
Vr
 z
V
2
m
v1v2 1  2 N A
 v1 ln v1  v2 ln v2   0
Vr
3
Mr
13
 2N A
m
2V
v1v2 1
 v1 ln v1  v2 ln v2   
Mr
3 Vr
 Mr  
N A m  v1 ln v1  v2 ln v2 
1V
v1v2 1
3 Vr
6, 025 1023 mol 1 1000 g   0,512  ln 0,512  0, 488ln 0, 488 
g

 7700
1
mol
 2,18 1025  0,512  0, 488  0, 0298
3
When the molecular weights of PS and polybutadiene are less than 7700g/mol, they form a
miscible system at 150oC.
Solve the limiting T where the polymers with molecular weights of 1105g/mol would be
miscible. As previously the enthalpy and entropy would be equal to have zero free energy of
mixing.
V
 2
v1v2 1 1    Nc  v1 ln v1  v2 ln v2   0
Vr
 z
Increasing temperature affect the Flory-Huggins parameter:
1 
Vm,1
RT
 1   2 
2

M 0,1
RT 1
 1   2 
2
Solving for the limiting temperature Tr:
V
 2
v1v2 1 1    N c  v1 ln v1  v2 ln v2   0
Vr
 z

M 0,1
2V
2
v1v2
1   2    Nc  v1 ln v1  v2 ln v2 
3 Vr
RTr 1
V
2
v1v2 M 0,1 1   2 
Vr
 Tr  
3R 1 N c  v1 ln v1  v2 ln v2 
2

2  2.18 1025  0.512  0.488  54.092
3  8.3145
 5478K

g
 17.2( MPa)1/2  18.6( MPa)1/2
mol

2
J
g
1.01 3 1.211022   0.512  ln 0.512  0.488ln 0.488 
K  mol
cm
14
Temperature is so high that material would decompose before reaching that. Thus increasing
temperature is not a valid method to enhance polymer/polymer miscibility.