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 (R A ) can be calculated  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 R A to R A0. For solvent R A R A0. PVC does not dissolve in its own monomer.

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. 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 g/cm 3 b)Polystyrene, density 1.04 g/cm 3 c)Polycarbonate, density 1.20 g/cm 3

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: GroupFiFi Amount FiFi M i (g/mol) –CH 2 – –C(CH 3 ) 2 – ii

2b) Based on the structure of polystyrene the molar attractions and molecular weight can be calculated: The solubility parameter can be calculated with the equation: GroupFiFi Amount FiFi M i (g/mol) >CH– –CH 2 – phenyl ,106 ii

2c) Based on the structure of polycarbonate the molar attractions and molecular weight can be calculated: The solubility parameter can be calculated with the equation: GroupFiFi Amount FiFi M i (g/mol) –C(CH 3 ) 2 – –OCOO– p-phenylene i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 2.2  (cm 3 (STP)  cm)/(cm 2  s  Pa) and for PVOH:lla 6,65  cm 3 (STP)  cm)/(cm 2  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 cm 3 and surface area is 250 cm 2. 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:

3b) Gas permeation can be calculated from equation Qgas flux permeated through membrane [cm 3 ] PPermeation coefficient [cm 3  cm/cm 2  s  Pa] ttime [s] Asurface area of the membrane [cm 2 ] lthickness of the membrane [cm]  ppressure difference [Pa]

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 cm 3  0.01 = 0.20 cm 3 Partial pressure of oxygen outside the packet: p 1 = 0.21  101kPa=21000 Pa Partial pressure of oxygen in the beginning: p 2,start = 0 Pa When oxygen concentration in the packet is 1.0 mol-%: p 2,end = 0.01  101kPa = 1000 Pa.  Approximation  p  constant = p 1 – p 2,start

3b) Time taken for oxygen transfer

3c) A similar LDPE packaging: If the packaging material was LDPE-film, the time would be 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 dm 3 bottle is filled to 2.0 bar CO 2 pressure at 25°C and the cap is closed tightly. Carbon dioxide transfer coefficient for PET is P(CO 2, 25°C) =  cm 3 (STP)  cm/(cm 2  s  Pa). How long a time does it take for CO 2 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 CO 2 content in air is 0.03%. Volume of the cylinder: Surface area of the cylinder:

4) Partial pressure of CO 2 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:

4) At the end the bottle has 9/10 of the original pressure (10% drop in pressure), so flux of the CO 2 has (ideal gas p 1 V 1 = p 2 V 2 ): The time this has taken can be calculated:

5. Dissolution of polymers Dissolution will happen when  G M is negative: Entropy of mixing  S M is always positive and can be expressed with Bolzman relation: k = 1,38  J/K is Bolzman constant and  describes the different ways that solvent molecules N 1 and polymer molecules N 2 can be arranged. Applying Sterling approximation (ln N! = N ln N – N) the entropy of mixing can be expressed: where v 1 is the volume fraction of solvent and v 2 volume fraction of polymer.

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 Calculate the change in free energy of mixing when 10% solution of polystyrene (M n = g/mol) in cyclohexane at 34 o C is prepared. Flory-Huggins parameter  1 is 0.50; density of cyclohexane is g/cm 3 and density of styrene 1.06 g/cm 3.

5) Volume fractions for cyclohexane v 1 = 0.9 and for styrene v 2 = 0.1. Take volume of solution to be V = 1 cm 3. Calculate the number of solvent molecules (C 6 H 12 ) N 1 : And number of polystyrene molecules:

5) Calculate the change in free energy of mixing: