This lecture: Crystallization and Melting. Next Lecture: The Glass Transition Temperature

Transcription

1 Thermal Transitions: Crystallization, Melting and the Glass Transition This lecture: Crystallization and Melting Next Lecture: The Glass Transition Temperature Today: Why do polymers crystallize in a chain folded fashion? Why do polymers melt over a range of temperatures? What are the factors that affect the Tm? Chapter 8 in CD (Polymer Science and Engineering)

2 Thermal Transitions Temperature Viscoelastic liquid Crystallization Melting Glass Transition Semicrystalline Solid Glassy Solid

3 Crystallization and the Glass Transition Volume Glassy Solid Cool Liquid or Melt Crystalline Solid Temperature T g T c

4 Kinetics, Crystallization and the Glass Transition Volume Glassy Solid Cool Quickly Liquid or Melt Volume Cool Quickly Does not Crystallize! Cool Slowly T g Temperature Crystallizable Polymer T g T g Temperature NonCrystallizable Polymer

5 Polymer Crystallization WHAT DO WE KNOW FROM EXPERIMENTAL OBSERVATION? Crystallization occurs relatively slowly At high undercoolings And results in the formation of chain folded lamellae

6 Polymer Crystallization Melt SMALL MOLECULES Specific Volume Heat Cool What is undercooling? T c T m Temp Specific Volume POLYMERS Cool Semicrystalline Solid Heat Tc T m Temp

7 Crystallization Kinetics General Features Induction period formation of primary nuclei Degree of Crystallinity Secondary Crystallization Primary crystallization a period of fast spherulitic growth Primary Crystallization Secondary crystallization a period of slower crystallization that occurs once the spherulites have impinged on one another Induction Period (Nucleation Step) Time

8 Thermodynamic Considerations The free energy of this primary nucleus is given by G cryst = (4xl)σ + (2x 2 )σ e (x 2 l) g The last term represents the free energy that we would obtain if all the segments were in the bulk. The first two terms are the excess free energy that must be added in to account for those segments at the surface. l σ e x σ x

9 Extended Chain Crystals and Annealing Thermodynamically most stable form Courtesy of I.R. Harrison, Penn State Crystals irreversibly thicken on annealing so they would like to get to the extended form

10 Critical Nucleus Size So why don t polymers crystallize in the extended chain form to begin with? To answer this let s calculate the critical nucleus size. 0 T m= Tm

11 Critical Nucleus Size The critical nucleus size is given by the values of l and x that minimize G cryst ; G l = G x = 0 Solving the two simultaneous equations l = 4σe g Nucleation at temperature T

12 Critical Nucleus Size Substituting for g; l = 4σeTm0 h f T l * where T is the undercooling; 0 T = Tm T c Note the inverse dependence of fold period on T. Critical Nucleus Size Also, at the equilibrium melting temperature, T c = T m0, and only lamellae with infinite fold periods could grow.

13 The Rate of Primary Nucleation An expression for the rate of nucleation can also be obtained: ν nuc f c l = 4σeTm 0 h f ( Tm 0 T) = 4σeTm h f T l * For l* large, T ~ exp const.σ 2 0 σ e [ T m ] 2 v nuc has to be small h 2 kt[ T] 2 Note that the rate of nucleation is vanishingly small at low undercoolings. Only nuclei with relatively short fold periods (compared to chain length) form at an appreciable rate. At small T V nuc is small T 0

14 Primary Crystallization: Once a nucleus has been formed growth is predominantly in the lateral direction. There is a considerable increase in the fold period behind the lamellar front during crystallization from the melt Crystallization

15 The Crystalline Melting Temperature Semicrystalline Solid Heat Melt

16 Characteristics: Melting Temperatures Temperature ( 0 C) Melting is Complete Melting Starts 20 Crystallization Temperature of Crystallization

17 The Melting Temperature of Polymer Crystals Temperature 0 Tm = Tm 1 2σ e l hf T m 0 l

18 Factors that Affect the Melting Temperature of Polymer Crystals T m ~ C Why? O= C H N C = O T m ~ C N H

19 The Effect of Chemical Structure G f = H f T S f And at Equilibrium G f = 0 Equilibrium at temperature T m Τ m = H f / S f Hence Τ m = H f / S f The subscript f stands for fusion

20 Intermolecular Interactions T m ~ C T m ~ C O = C N H Is this nylon 6 or nylon 66?

21 The Effect of Intermolecular Interactions OK, now it s your turn! Supposing you could synthesize such beasts, would you expect syndiotactic poly(propylene) or syndiotactic PVC to have the higher melting point? A. B. CH Cl CH CH 3

22 Entropy and Chain Flexibility Flexible Τ m = H f / S f Stiff

23 Entropy and Chain Flexibility S f S = k lnω S f = k (lnω melt lnω crystal ) = Large

24 Conformational Entropy and the Melting Point S f = k (ln( lnω melt lnω crystal ) = Small T m = H f / S f

25 Entropy and the Melting Point Polyethylene T m ~ C O Poly (ethylene oxide) T m ~ 65 0 C Poly (pxylene xylene) T m ~ C

26 Entropy and the Melting Point Polyethylene CH CH 3 Isotactic Polypropylene T m ~ C T m ~? CH T m ~? Isotactic Polystyrene

27 S 1 The Effect of Diluents S 2 > S 1 S 2

28 The Effect of Copolymerization and Molecular Weight