1. Polymer Properties Exercise 2

2. 1. Crystallinity Polyethylene is crystalline polymer which forms orthorhombic unit cell, i.e. a=b=g=90ᵒC, where a, b, a and g are the angles between the b and c, a and c, and a and b axes. a) Calculate the density of each crystal when the values |a|, |b|, and |c| are 736, 492, 254 and 2[-CH2-CH2-] units per cell. b)The crystalline lamella of PE can at its thinnest consist of 22 carbons and at the thickest 150 carbons. Calculate the melting temperature range for PE using Thompson-Gibbs equation:.

3. 1a) The volume for the orthorhombic unit of PE can be calculated: V = 736  492  254  10 -36 m 3 = 9,20  10 -29 m 3 Units are in picometers (pm, 10 -12 ) Weight of the two molecules in the unit is: And the density is:

4. 1b) From lecture notes: T m = melting temperature of a lamellar with thickness L T m 0 = melting temperature of an infinitely thick and complete crystallite (414.2K)  = free surface energy per unit area (79 x 10 -3 J / m 2 )  H m = Enthalpy change per volume (288 x 10 6 J / m 3 ) L = lamellae thickness, ( ) = typical values for PE

5. 1b) Lamella thickness can be calculated from the chain length of the monomer: 22 carbons (11 monomer units): L = 11  254 pm =2.79 nm 150 carbons  19.1 nm Substituting the values gives: The melting temperature range for PE is 57-130 o C.

6. 2. Degree of Crystallinity Experiments were carried out to measure c max, the maximum crystallinity, in a series of random copolymers of ethylene and propylene. Linear PE gave c max = 95 %. Introduction of 4 CH 3 –groups per 100 mainchain carbon atoms reduced c max to 50 % and adding 20 CH 3 –groups per 100 mainchain carbons c max = 0 %. a)Calculate the weight fraction of propylene in copolymer. b)Calculate the glass transition temperatures T g (K) for these amorphous ethylene/propylene copolymers using the equation below. w 1 and w 2 are the weight proportions of comonomers and T g,1 and T g,2 are the glass transition temperatures for equivalent homopolymers. PE T g = -120°C and PP T g = -19°C.

7. 2a) Introducing 4 CH 3 –groups per 100 mainchain carbon atoms means the mainchain contains 50 monomer units. Thus there are 4 propylene groups per 46 ethylene units. Molecular weights of the repeating units: M(C 2 H 4 ) = 28.1 g/mol M(C 3 H 6 ) = 42.1 g/mol

8. 2a) Weight fraction of propene in copolymer where there is 4 CH 3 -groups: Weight fraction of propene in copolymer where there is 20 CH 3 -groups:

9. 2b) Using equation and solving for T g : The glass transition temperature for copolymer containing 11.5 wt.% propene: The glass transition temperature for copolymer containing 50 wt.% propene: T g = -82°C

10. 3. Effect of plasticizer on melting Calculate the molecular weights and molar volume for the components; (C 16 H 28 O 4 ) and (C 3 H 7 NO):

11. 3. Plotting the melting temperature as a function of weight fractions of solvent the cross point of y-axis is and the slope is and the slope is

12. 3. The melting temperature and the melting enthalpy of the pure polymer can then be solved from the plot T m 0 :

13. 3. The slope enables the calculation of  H f for the polymer: