Polymer properties Exercise 5. 1. Solubility parameters Solubility can be estimated using solubility parameters. According to Hansen model the overall.

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

• 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

2 2 2D P H

2 2 2

, , , , , ,2 2A D P D s P P P s H P H sR

D, P and H are the dispersive, polar, and hydrogen bonding parameters

1)• Calculate the radius of the solubility by substituting the

solubility parameter values to the equation:

• Compare RA to RA0. For solvent RA < RA0 and non-solvent RA > RA0.

• PVC does not dissolve in its own monomer.

2 2 2

, , , , , ,

2 2 2

2 2

2*18.2 2*15.4 7.5 8.1 8.3 2.4 8.2

A D P D s P P P s H P H sR

8,22,3 1

3,5A

AO

R

R

1)

• Calculate the solubility parameter for PVC:

• 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 2 2 2 2 218.2 7.5 8.3 21.4D P H

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:

a) Polyisobutylene, density 0.924 g/cm3

b) Polystyrene, density 1.04 g/cm3

c) Polycarbonate, density 1.20 g/cm3

1 1i i

i ii

ii

i

F F

MV

2a)

• Based on the structure of polyisobutylene the molar attractions and molecular weight can be calculated:

• The solubility parameter can be calculated with the equation:

CH2 C

CH3

CH3n

Group Fi Amount Fi Mi (g/mol)

–CH2– 280 1 280 14.027–C(CH3)2– 840 1 840 42.081

i 1120 56.108

31/2

1/21

3

112018.4

56.108

0.924

ii

ii

i

cmF MPamol MPa

M gmolg

cm

2b)• Based on the structure of polystyrene the molar attractions

and molecular weight can be calculated:


CH2 CHn Group Fi Amount Fi Mi (g/mol)

>CH– 140 1 140 13.019–CH2– 280 1 280 14.027phenyl 1517 1 1517 77,106

i 1937 104.152

31/2

1/21

3

193719.3

104.152

1.04

ii

ii

i

cmF MPamol MPa

M gmolg

cm

2c)

• Based on the structure of polycarbonate the molar attractions and molecular weight can be calculated:


O C O

O

C

CH3

CH3n

Group Fi Amount Fi Mi (g/mol)–C(CH3)2– 840 1 840 42.081–OCOO– 767 1 767 60.008p-phenylene 1377 2 2754 154.212

i 4361 256.301

2/1

3

132/11 4.20

20.1

301.526

4361MPa

cmg

molg

molcmMPaM

F

i

i

ii

i

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 2.210-13 (cm3(STP)cm)/(cm2sPa) and for PVOH:lla 6,6510-16 cm3(STP)cm)/(cm2sPa).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:

• Oxygen permeation coefficient P:

31 2

1 2 3

ll ll

P P P P

315

231 2

13 16 131 2 3

0.60 ( )2.0 10

0.20 0.20 0.20

2.2 10 6.65 0 2.2 10

l mm cm NTP cmP

l mm mm mml l cm s PaP P P

3b)

• Gas permeation can be calculated from equation

• Q gas flux permeated through membrane [cm3]• P Permeation coefficient [cm3 cm/cm2 s Pa]• t time [s]• A surface area of the membrane [cm2]• l thickness of the membrane [cm]• p pressure difference [Pa]

P A t pQ

l

3b)• 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 = 0.21101kPa=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 = 0.01101kPa = 1000 Pa.

Approximation p constant = p1 – p2,start

P A t pQ

l

3b)

• Time taken for oxygen transfer

P A t pQ

l

pAP

lQt

PacmsPacmcmcm

cmcm

21000250100.2

060.020.0

22

315

3

ds 13101.1 6

3c)

• A similar LDPE packaging:

• 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.

PacmsPacmcmcm

cmcm

pAP

lQt

21000250102.2

060.020.0

22

313

3

hs 9.210400

P A t pQ

l

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) = 0.11810-13 cm3(STP)cm/(cm2sPa).

• How long a time does it take for CO2 pressure to drop one tenth?

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:

• Surface area of the cylinder:

22

VV r h h

r

3 3

22 2 21,5 102 2 2 2 4 850,5

4

V cmA rh r r cm cm

r cm

P A t pQ

l

4)

• Partial pressure of CO2 outside is:

• Pressure difference between inside and outside of the bottle in the beginning (a):

• Pressure difference at the end (e):

• The average pressure difference:

PaPapo 4.301013250003.0

PaPaPappp oaa 1999704.30102 5

PaPaPappp oee 1799704.301029.0 5

PaPaPaPapp

p eaavg 190000189970

2

179970199970

2

P A t pQ

l

4)

• 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):

• The time this has taken can be calculated:

P A t pQ

l

33

1

111

2

1112 17.01

9.0

15.11

9.0dmdm

p

pVV

p

VpVVQ

avgpAP

lQt

ds

PacmsPacm

cmcm

cmcm

pAP

lQt

avg

1031092.8

1900005.85010118.0

1.01017.0 6

22

313

33

5. Dissolution of polymers• Dissolution will happen when GM is negative:

• Entropy of mixing SM is always positive and can be expressed with Bolzman relation:

k = 1,3810-23 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:

where v1 is the volume fraction of solvent and v2 volume fraction of polymer.

M M MG H T S

lnMS k

1 1 2 2( ln ln )MS k N v N v

• 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:

• Then the change in free energy follows:

5. Dissolution of polymers

1 1 2MH kT N v

1 1 2 2 1 1 2( ln ln )M M MG H T S kT N v N v N v

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)• 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:

• And number of polystyrene molecules:


33

23 1 211 11 1

1

1 0,9 0,77856,025 10 5,02 10

(6 12,011 12 1,008)A A

gcmVv cmN n N N mol

gMmol

33

23 1 182 22 2

2

1 0,1 1,066,025 10 6,39 10

10000A A

gcmVv cmN n N N mol

gMmol

5)

• Calculate the change in free energy of mixing:


1 1 2 2 1 1 2

23 21 18 21

( ln ln )

1,38 10 307,15 5,02 10 ln 0,9 6,39 10 ln 0,1 0,5 5,02 10 0,1

1,24

M M MG H T S kT N v N v N v

JK

KJ

Polymer properties Exercise 5. 1. Solubility parameters Solubility can be estimated using solubility parameters. According to Hansen model the overall.

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