Polymer properties
Exercise 5
1. Solubility parameters
• Solubility can be estimated using solubility parameters.
According to Hansen model the overall solubility parameter
can be obtained as
D, P and H are the
    
2
D
2
P
2
H
dispersive, polar, and
hydrogen bonding
parameters
• 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 (RA) can be calculated
RA 
 2
 2 D , s    P , P   P , s    H , P   H ,s 
2
D,P
2
2
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
 2*18.2  2*15.4    7.5  8.1  8.3  2.4   8.2
2
2
2
• Compare RA to RA0. For solvent RA < RA0 and
non-solvent RA > RA0.
RA 8, 2

 2,3  1
RAO 3,5
• PVC does not dissolve in its own monomer.
1)
• 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 d = 20.5 is closest to
PVC.
2. Solubility parameters from molar
attraction constants
• 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
a) Polyisobutylene, density 0.924 g/cm3
b) Polystyrene, density 1.04 g/cm3
c) Polycarbonate, density 1.20 g/cm3
2a)
• 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
• The solubility parameter can be calculated with the equation:
3
cm
Fi 1120MPa

mol  18.4MPa1/2
 i  i 1 
Mi
g
56.108
mol
i
g
0.924 3
cm
1/2
2b)
• Based on the structure of polystyrene the molar attractions
and molecular weight can be calculated:
Fi
Amount
Fi
Mi (g/mol)
>CH–
140
1
140
13.019
–CH2–
280
1
280
14.027
phenyl
1517
1
1517
77,106
i
1937
104.152
Group
• The solubility parameter can be calculated with the equation:
3
cm
Fi 1937 MPa1/2

mol  19.3MPa1/2
 i  i 1 
Mi
g
104.152
mol
i
g
1.04 3
cm
2c)
• Based on the structure of polycarbonate the molar attractions
and molecular weight can be calculated:
Fi
Amount
Fi
Mi (g/mol)
–C(CH3)2–
840
1
840
42.081
–OCOO–
767
1
767
60.008
p-phenylene 1377
2
2754
154.212
i
4361
256.301
Group
• The solubility parameter can be calculated with the equation:
i 
F
i 1
Mi
i
i
4361MPa1/ 2 cm 3 mol 1

 20.4MPa1/ 2
526.301 g
mol
1.20 g 3
cm
3. Gas permeability
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?
3a)
• Gas transfer coefficient in multilayer laminate depends on the
properties of the individual layers in the laminate:
l1 l2 l3
l
  
P P1 P2 P3
• Oxygen permeation coefficient P:
3
l
0.60mm
15 cm ( NTP )  cm
P

 2.0 10
2
0.20mm
0.20mm
0.20mm
l1 l2 l3
cm
 s  Pa


 
P1 P2 P3 2.2 1013 6.65  016 2.2 1013
3b)
• Gas permeation can be calculated from equation
P  A  t  p
Q
l
•
•
•
•
•
•
Q
P
t
A
l
p
gas flux permeated through membrane [cm3]
Permeation coefficient [cm3  cm/cm2  s  Pa]
time [s]
surface area of the membrane [cm2]
thickness of the membrane [cm]
pressure difference [Pa]
3b)
P  A  t  p
Q
l
• For ideal gas the volume is equivalent to molar volume (1 mol% = 1 vol-%). Laminate is ok for packaging until there is 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
• When oxygen concentration in the packet is 1.0 mol-%:
p2,end = 0.01101kPa = 1000 Pa.
 Approximation p  constant = p1 – p2,start
3b)
P  A  t  p
Q
l
• Time taken for oxygen transfer
Q l
t
P  A  p
0.20cm3  0.060cm
6

1
.
1

10
s  13d

3
15 cm cm
2
2.0 10

250
cm
 21000 Pa
2
cm sPa
P  A  t  p
Q
l
3c)
• A similar LDPE packaging:
Q l
t

P  A  p
0.20cm3  0.060cm
3
13 cm cm
2
2.2 10

250
cm
 21000 Pa
2
cm sPa
 10400s  2.9h
• If the packaging material was LDPE-film, the time would be
10400 s which is less than 3 hours when with PVOH barrier
layer, time was 13 d.
4. Gas permeation
• 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?
P  A  t  p
Q
l
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
2
V r h  h  2
r
• Surface area of the cylinder:
 1,5  103 cm3
2
V
2
A  2 rh  2 r  2    r   2 
   4cm    850,5cm2
4cm
r



2
P  A  t  p
Q
l
4)
• Partial pressure of CO2 outside is:
po  0.0003 101325 Pa  30.4 Pa
• 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
• Pressure difference at the end (e):
pe  pe  po  0.9  2 10 5 Pa  30.4 Pa  179970 Pa
• The average pressure difference:
pavg 
pa  pe 199970 Pa  179970 Pa

 189970 Pa  190000 Pa
2
2
4)
P  A  t  p
Q
l
• 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):
 p1

p1V1

3 1


Q  V2  V1 
 V1  V1 
 1  1.5dm 
 1  0.17dm3
p2
 0.9 
 0.9 p1 
• The time this has taken can be calculated: t 
Q l
t

P  A  pavg
Q l
P  A  pavg
0.17 103 cm 3  0.1cm
6

8
.
92

10
s  103d
3
cm cm
0.118 10 13 2
850.5cm 2 190000 Pa
cm sPa
5. Dissolution of polymers
• Dissolution will happen when GM is negative:
GM  H M  T SM
• Entropy of mixing SM is always positive and can be expressed with
Bolzman relation:
SM  k ln 
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.
5. Dissolution of polymers
• Dimensionless Flory-Huggins parameter 1 can be applied to estimate the
polymer-solvent interactions.
• The 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 )
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
1 is 0.50; density of cyclohexane is 0.7785 g/cm3 and density of styrene 1.06
g/cm3.
5)
GM  H M  T SM  kT ( N1 ln v1  N 2 ln v2  1 N1v2 )
• 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
3
Vv1 1
cm
N1  n1 N A 
NA 
 6, 025 1023 mol 1  5, 02 1021
g
M1
(6 12, 011  12 1, 008)
mol
1cm3  0,9  0, 7785
• And number of polystyrene molecules:
Vv2 2
N 2  n2 N A 
NA 
M2
1cm3  0,11, 06
10000
g
mol
g
cm3  6, 025 1023 mol 1  6,39 1018
5)
GM  H M  T SM  kT ( N1 ln v1  N 2 ln v2  1 N1v2 )
• 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