Review_2_Spring_2005

Review_2_Spring_2005 - Polymeric Materials Review 2 Spring...

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Unformatted text preview: Polymeric Materials Review 2 Spring 2005 Dr. Anthony Brennan University of Florida Department of Materials Science & Engineering EMA 6165 Polymer Physics – AB Brennan EMA 1 Agenda • Overview • Chain Dimensions – Four Models • Rubber Elasticity – Uniaxial – Biaxial – Swelling • Polymer Solubility – Thermodynamics EMA 6165 Polymer Physics – AB Brennan EMA 2 Properties • • • • • Chemical Physical Electronic Magnetic Optical CERAMICS METALS PROPERTIES ELECTRONIC MATERIALS EMA 6165 Polymer Physics – AB Brennan EMA POLYMERS 3 Importance of Chemistry • Balance of Three Energies – Intermolecular – Intramolecular – Thermal (kT) • Structures – Linear – Branched EMA 6165 Polymer Physics – AB Brennan EMA 4 Freely Rotating Chain Vector Analysis • Freely rotating model: – high temperature – solvated • Fixed bond angle of Fixed 109.5° expands by 2 109.5° • Ignores bond Ignores rotational energy barriers (RIS) barriers r = 2nl 2 2 (Fixed angle = 109.5°) r 2 = 2 Ml EMA 6165 Polymer Physics – AB Brennan EMA 2 5 Hindered Rotating Chain Model 1 0 0 0 1 0 E= 0 0 1 r 2 1 + cos( 180 − τ ) 1 + cos φ = nl 1 − cos( 180 − τ ) 1 − cos φ 2 • Hence, the Hindered Rotating Model accounts for the Hence, torsional angle potentials as well as the valence bond potentials. potentials. • For a simple polyolefin, I.e. PE, this can be readily For applied to determine the effect on the mean square end to end distance. EMA 6165 Polymer Physics – AB Brennan EMA 6 Hindered Rotating Chain Model PE Torsional Energy Potentials PE Torsional EMA 6165 Polymer Physics – AB Brennan EMA 7 Three Dimensional Random Walk Gaussian Distribution Gaussian • Cartesian Coordinates n 2 p( x , y , z , n) = 2π l 3 • −3 ( 2 e −3 x 2 + y 2 + z 2 2 nl 2 ) dxdydz Polar Coordinates: 3-D Random Walk −3( r 2 ) −3 2 2 nl 2 n 2 2 p(r , n) = 2π l 4π r e dr 3 EMA 6165 Polymer Physics – AB Brennan EMA 8 Three Dimensional Random Walk Three Calculated Dimensions Calculated EMA 6165 Polymer Physics – AB Brennan EMA 9 Chain Statistics Chain Statistical Mechanics Statistical Objective: Predict mechanical properties from Objective: first principles, i.e., equilibrium thermodynamics thermodynamics q Two Approaches: q Gibbs Free Energy ∆G = ∆H − T∆S q Helmholtz Free Energy ∆A = ∆E − T∆S EMA 6165 Polymer Physics – AB Brennan EMA 10 Thermodynamics q Since thermal reversibility is a boundary condition, dq = TdS q Which allows the following substitution: ∂ E ∂ S f= − T ∂ l T ,V ∂ l V ,T q This is an Equation of State, which using the identity given previously, one can rewrite: ∂ E ∂ f f= + T ∂ l T ,V ∂ T l ,V q Which gives a second Equation of State. EMA 6165 Polymer Physics – AB Brennan EMA 11 Thermodynamics q Another way to express these relationships are: f E Retractive force due to internal energy = Retractive force due to entropy fS q It can be shown that: 2 f E ∂ ln ro = fS ∂ ln T q Think of the role of chemistry in this function. EMA 6165 Polymer Physics – AB Brennan EMA 12 Thermodynamics q From these derivatives, it can be shown that: ∂ H ∂ S f= − T ∂ l T,P ∂ l T,P Equations of State ∂ H ∂ f f= + T ∂ l T,P ∂ T l,P q So now that we have the relationship between retractive force and enthalpy, lets examine it further. EMA 6165 Polymer Physics – AB Brennan EMA 13 Thermodynamics ∂ P ∂V ∂ P = − ∂ V l , T ∂ T P ,l ∂ T V ,l q And using the following definitions: 1 ∂ V Coefficient of α= V ∂ P ,l Thermal Expansion T β= q 1 − V ∂V ∂P l ,T Coefficient of Isothermal Compressibility It can be shown that: P ∂ = ∂ V ,l T α β EMA 6165 Polymer Physics – AB Brennan EMA 14 Thermodynamics q First, consider the behavior of chains confined to a fixed volume: Affine Deformation q i.e., deformation for each element is the same as the total volume, or: lx ly lz Vi λx * λy * λz = * * = ≈ 1 l0 l0 l0 VT q Now, given that: q It can be shown that: f= ∂S − T ∂l T , P ,V EMA 6165 Polymer Physics – AB Brennan EMA 15 Thermodynamics σ 0z q Thus: 2 N v kT * 2λ z − 2 = 2 λz σ0 z = N v kT * ( λ z −λ z − 2 ) Uniaxial Deformation for an Ideal Elastomer q How does one relate this to molar mass, i.e., Nv = M c ? ? EMA 6165 Polymer Physics – AB Brennan EMA 16 Thermodynamics α=1 3RTρ E= Mc q For q Now let’s evaluate the biaxial oriented elastomer q q Examples are: – Garbage bag – Packaging film λy λx z Deform along x and y plane: EMA 6165 Polymer Physics – AB Brennan EMA λy λx 17 Thermodynamics q Then through simple algebra: σ0 = N kT ( λ − λ λ −3 v x x x y σ0 = N v kT ( λx − λx λy −2 y q ) ) −2 −3 True stress is calculated as follows: fz f z A0 A0 σt = = = * = σ0 * A A0 A Area z 1 1 σ0 * * = σ 0z * * lx l y λx λ y l0 x l0 y z EMA 6165 Polymer Physics – AB Brennan EMA 18 Thermodynamics Biaxial Stress in x direction q ( σ t x = N v kT λ x + 2 − λ x − 2 λ y − 2 ) Biaxial Stress in y direction q ( σ t y = N v kT λ y + 2 − λ x − 2 λ y − 2 q ) Differences in Principal Stresses: σt x − σt y = [ N v kT λ x λ y 2 2 EMA 6165 Polymer Physics – AB Brennan EMA 19 Thermodynamics q Substitution gives: P= 4d 0 N v kT 1 * λ − 7 3 R0 λ λ for a spherical balloon, where the thin film exhibits: P versus λ q What does a plot of q Where is the maximum ∆V = 0 look like? P? EMA 6165 Polymer Physics – AB Brennan EMA 20 Rubber Elasticity Effect of chain ends Effect q A front factor is added to the stress relationship for uniaxial deformation: f * − 2 −2 σ= N v RT ( λ − λ ) f* q Thus, for a functionality of 4, the stress is reduced by a factor of 2 compared to the standard form. q Next, lets evaluate the effect of chain ends: 2N A N v ( total ) = N v ( ideal ) − V 2N A = ( free ends per chain) where: V EMA 6165 Polymer Physics – AB Brennan EMA 21 Rubber Elasticity Graphical Analysis of Effect of Chain Ends q q Which is also written as: ρ R T 2 MC σo = 1 − MC Mo ( λ − λ−2 ) Thus, Mo is critical in determining the stress and furthermore: 2 M C >> M o q σ ≤0 Graphically, this relationship can be shown as: N V kT G or ρ R T MC q Other y-axis values used are: − M O1 EMA 6165 Polymer Physics – AB Brennan EMA 22 Rubber Elasticity Phantom Networks q Another common correction used to correlate with end to end distance is: ρ R T 2 MC r −2 σo = (λ − λ ) 1 − Mo r MC Now, let’s evaluate the Phantom Network: 2 i 2 0 q σo = 1− 2 ρ R T f MC σO 2 M C ri 2 1 − M o r02 λ−1 = 1 Gaussian Chain N v kT 2 N v kT 3 ( λ − λ−2 ) Phantom Chain λ−1 EMA 6165 Polymer Physics – AB Brennan EMA fA =4 fA =3 fP =3 23 Rubber Elasticity Swelling Behavior Consider the following: Increase solvent to increase volume, but what happens to stress? EMA 6165 Polymer Physics – AB Brennan EMA 24 Rubber Elasticity Swelling Behavior q Since it is an isotropic deformation, one can write 1/ 3 V −1/ 3 λx = λy = λz = = ϕ B V0 q Thus, we have: N v kT −2 / 3 −1 ∆Gelastic = ( 3ϕ B − 3) − ( ln ϕ B ) 2 [ q The material stops swelling. Why? q Elastic Term! What is the factor of interest? ∆Gelastic = ρRT −2 / 3 −1 ( 3ϕ B − 3) − ( ln ϕ B ) 2 Mc [ EMA 6165 Polymer Physics – AB Brennan EMA 25 Rubber Elasticity Rubber Deformation of Swollen Networks Deformation • Which ultimately leads to: σ0x q 1 1/ 3 = N v kTϕ λ x − 2 λx Lets examine some other theories that deal with active chains σ 0i = N v RT −2 σ c = ( N c + N p ) RT ( λ − λ ) EMA 6165 Polymer Physics – AB Brennan EMA 26 Polymer Solubility Entropy of Mixing ∆Gmix = N 1 ∆G1 + N 2 ∆G2 Where N1 = # of molecules and n1 = mole fraction ∆ =0 H = − kT ( N 1 ln n1 + N 2 ln n2 ) then, because: ∆Gmix Where n1 =mole fraction of solvent n2 =mole fraction of solute EMA 6165 Polymer Physics – AB Brennan EMA 27 Polymer Solubility Entropy of Mixing Or by summing all the sites of the lattice, Or one can define the Configurational Entropy or Combinatorial Entropy: Entropy ∆Smix = −k ∑ N i ln N i i =1 or in terms of moles: ∆Smix = − R ∑ ni ln ni i =1 EMA 6165 Polymer Physics – AB Brennan EMA 28 Polymer Solubility Entropy of Mixing Hence the actual value for Gibbs Free Energy of Mixing: ∆Gmix = kT ( N 1 ln υ + 1 Where N1 = # N 2 ln υ ) 2 of molecules N 2 = # of segments υ= volume fraction EMA 6165 Polymer Physics – AB Brennan EMA 29 Polymer Solubility Entropy of Mixing ∆ S R n=500 xi = mole fraction of segments n=100 n=50 x2 x1 Why? ∆ ≠0 H EMA 6165 Polymer Physics – AB Brennan EMA Consider: London van der Waals Dipole H-bonding Acid-Base, etc. 30 Polymer Solubility Enthalpy of Mixing Since we are interested in 1,2 pairwise interactions, a new term is defined 1∆ w χ= z 2 kT Chi Parameter ∆ mix = N 1υ χk T H 2 Flory-Huggins Pairwise interaction EMA 6165 Polymer Physics – AB Brennan EMA 31 Polymer Solubility Enthalpy of Mixing Substitute in for ∆Hmix ∆ mix H ∆E ∆E 2 1 − = Vm υ1υ2 V2 V1 1 2 ∆ 1 E V1 1 2 Hildebrand and Scott Cohesive Energy Density EMA 6165 Polymer Physics – AB Brennan EMA 32 Polymer Solubility Enthalpy of Mixing Which provides a definition for the Solubility Parameter: δ= ∆E1 V1 1 2 cal 3 cm 1 2 Which for an Athermal Process: δ1 = δ2 EMA 6165 Polymer Physics – AB Brennan EMA 33 Polymer Solubility Osmotic Pressure by definition: ∆ G π= − υ 1 RT ln 1 −υ + [( π =− 2) V1 Vol. fraction Molar volume of solvent 1 1 − υ + 2 x x = # of segments EMA 6165 Polymer Physics – AB Brennan EMA χ υ 1 2 2 Chi Parameter 34 Polymer Solubility Solution gives Flory θ temperature where : Which gives A2 = 0 And an alternate expression for χ A2 1 1 = 2 as N AU A2 = 2 2M EMA 6165 Polymer Physics – AB Brennan EMA 35 Polymer Solubility Thus... χ C 1 1 = = 2 2 2( 1 − υ 2 ) C 2 υ 1,C Concentration Dependence at critical point Substitute back into first derivative to get..... χ C 1 1 = 1 + 1 2 2 X 2 Molecular Weight Dependence at Critical Point EMA 6165 Polymer Physics – AB Brennan EMA 36 Polymer Solubility Finally, consider Chain Expansion Factor. θ α − α = 2Cm Ψ1 1 − M T 5 3 constant as M→∞ then α −α ≈ M 5 3 EMA 6165 Polymer Physics – AB Brennan EMA 1 2 When T = Θ α=1 0 .1 37 Polymer Solubility Universal Calibration Mark Houwink ( η] M = φ ro 2 [ α 3 } ) 3 2 φ ≈ 2.5* 10 dl mol 21 dl ≈ 2.1* 10 mol 21 Hydrodynamic Volume GPC ∝ Hydrodynamic Volume EMA 6165 Polymer Physics – AB Brennan EMA 38 Estimating Solubility Parameters m1 −mo 1 Q= mo ρ s Where m =mass ρ =density of solvent s EMA 6165 Polymer Physics – AB Brennan EMA 39 Osmometry Osmotic pressure: arises due to concentration gradients i.e. chemical potential: µsolvent ≠ µsolute ∂ ∆Gmix = ∂ Xi ∆G 1 = 1 2 RT ln( 1 − υ2 ) + 1 − υ2 + χ1υ2 x EMA 6165 Polymer Physics – AB Brennan EMA 40 Distribution MV = ∑ Ni Mi ∑ Ni Mi 1+ a M 1+ a Where a relates to α, the Chain Expansion Factor 0.5 < a < 0.8 Good n Poor MV wi ≈M W M Z Consider the influence on Mn and MW massi EMA 6165 Polymer Physics – AB Brennan EMA 41 Polymer-Polymer Compatibility UCST/LCST Behavior EMA 6165 Polymer Physics – AB Brennan EMA 42 Glass Transition Characteristics Characteristics of a glass • Modulus (1 Hz) q EG ~ 109 Pa > ER~ 105-7 Pa q Viscosity q η G ~1013 P > η R~ 103-4 P q Amorphous by X-Ray techniques EMA 6165 Polymer Physics – AB Brennan EMA 43 Glass Transition Characteristics Characteristics of a glass q Coefficient of Thermal Expansion q α G ∼ 10 to 100 ppm < α R = ~500 ppm q DG < DR ∂E ≠0 ∂l and ∆v ≠0 EMA 6165 Polymer Physics – AB Brennan EMA 44 Vol Spec (cc/g) Glass Transition Characteristics α glass α liquid V f’ fast V0,G Flory Fox WLF slow V’ Free Volume Approach T Temperature (K) EMA 6165 Polymer Physics – AB Brennan EMA 45 Glass Transition Characteristics ∆ ln t = − ∆E ACT kT ∆T Consider as t T ↑↓ 2 In terms of molecular motion, f is a necessary factor EMA 6165 Polymer Physics – AB Brennan EMA 46 Summary q Relationships between molecular structure and bulk properties developed q Equilibrium thermodynamics basis for theories of chain dimensions, rubber elasticity and polymer solubility (Ideal Behavior). q Glass formation is a spinodal decomposition process. EMA 6165 Polymer Physics – AB Brennan EMA 47 ...
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