Lecture10

# Lecture10 - EMA 6165 – Polymer Physics Rubber Elasticity...

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Unformatted text preview: EMA 6165 – Polymer Physics Rubber Elasticity Lecture 10 Lecture Dr. Anthony Brennan Dr. University of Florida Department of Materials Science & Department Engineering Engineering EMA 6165 Polymer Physics – AB Brennan EMA 1 Agenda • • • Introduction Thermo-Elastic Behavior Statistical Mechanical Approach – Uniaxial Deformation – Biaxial Deformation • Swollen State • Deviations from Classical Theories – Affine Deformation – Phantom Behavior – Mooney-Rivlin Model EMA 6165 Polymer Physics – AB Brennan EMA 2 Chain Statistics Chain Statistical Mechanics Statistical • Returning to Gibbs Free Energy, we know Returning that: that: dG = dH − TdS − SdT and that: q dH = dE + PdV + VdP q dE = + TdS − PdV + fdl hence, through substitution: dE = fdl + VdP − SdT EMA 6165 Polymer Physics – AB Brennan EMA 3 Thermodynamics Thermodynamics • Again, take the derivative with respect to T: ∂G ∂G = f ∂ P = ∂l T ,l T , P ∂G V ∂T = P ,l −S Now, taking the 2nd Derivative and using Maxwell’s Relationships, one can show that: Relationships, ∂2G ∂l∂T ∂ f = ∂T P ∂2G = ∂l∂T l ,P EMA 6165 Polymer Physics – AB Brennan EMA P 4 Thermodynamics Thermodynamics q From these derivatives, it can be shown From that: that: ∂ H f = ∂l ∂ S −T ∂l T ,P T ,P Equations of State ∂ f ∂ H f= +T ∂T ∂l T ,P q l ,P So now that we have the relationship between So retractive force and enthalpy, lets examine it further. retractive EMA 6165 Polymer Physics – AB Brennan EMA 5 Thermodynamics - Enthalpy Thermodynamics Enthalpy q Begin by taking the derivative with respect to length: ∂ H ∂ E ∂V ∂ P 0 = + V + P ∂ l T,P ∂ l T,P ∂ l T,P ∂ l T,P since: ∂V −4 ≈ 10 ∂l q Hence, we see support for the assumption: ∆ V=0 Hence, V=0 which was stated earlier. which q Since we know that: E ~ f ( l ,V ) EMA 6165 Polymer Physics – AB Brennan EMA 6 Thermodynamics q Lets separate the volume and length effect. dE = q ∂E ∂E dV + dl ∂V l ,T ∂l T ,V If we then take the derivative w/ respect to length If at constant T, P: at ∂E ∂V ∂ E ∂ l 1 ∂ E = + ∂ V l ,T ∂ l T ,P ∂ l T ,P ∂ l T ,V ∂ l T ,P q ∂ H Substitution into the relationship = ∂ l T,P EMA 6165 Polymer Physics – AB Brennan EMA 7 Thermodynamics q Gives the following relationships: ∂ E ∂V ∂ H ∂ E ∂V = + + P ∂ l T ,P ∂V l ,T ∂ l T ,P ∂ l T ,V ∂ l T ,P q Rearrange and factor to get: ∂ H ∂l = T , P ∂ E ∂V + ∂ l T ,V ∂ l T ,P ∂ E + P ∂ V T ,l q What happens to E with T & V at fixed values? q Well, Energy is derived from intramolecular forces, which Well, relates to perturbation of chain dimensions! relates ∂ E ∂V Now, let’s examine the function: l ,T q q First, return to the following: dA = −SdT − PdV + fdl EMA 6165 Polymer Physics – AB Brennan EMA 8 Thermodynamics ∂ P ∂ S = ∂V l ,T ∂ T V , l q q Now, taking the fact that: dE = +TdS − PdV + fdl Differentiate w/ respect to volume at constant T, l Differentiate yields: yields: ∂E ∂S ∂V −P = +T ∂V l ,T ∂V l ,T ∂V l ,T q Substitution of the identity given above, yields: P ∂ E ∂ = +T ∂ − P T V ,l ∂ V l ,T q Which by partial differentiation yields: EMA 6165 Polymer Physics – AB Brennan EMA 9 Thermodynamics ∂ P ∂V ∂ P = − ∂ V l , T ∂ T P ,l ∂ T V ,l q and using the following definitions: 1 ∂ V α= V ∂ P ,l T β= q 1 ∂V − V ∂P l ,T It can be shown that: P ∂ = ∂ V ,l T Coefficient of Thermal Expansion Coefficient of Isothermal Compressibility α β EMA 6165 Polymer Physics – AB Brennan EMA 10 Thermodynamics q q q ∂E α = T − P ∂V l ,T β ∂ H which can be substituted into: ∂ l T ,P to give: And that: to give: ∂H α ∂ V ∂E = +T ∂l T , P ∂l T ,V β ∂ T , P l q Thus, one can evaluate the enthalpy simply by evaluating the Thus, following relationship graphically: following f= ∂ f ∂ H + T ∂T l , P ∂ l T ,P EMA 6165 Polymer Physics – AB Brennan EMA 11 Thermodynamics Force ∂ f slope = + T ∂ T V , l Tg For a good elastomer: Temp ∂S =T f = −T ∂l T ,V ∂ f ∂T EMA 6165 Polymer Physics – AB Brennan EMA V , l 12 Equations of State - Graphical Introduction to Physical Polymer Science, 3rd Edition, Lesley H. Sperling, Wiley Interscience (2001) Introduction ISBN 0-471-32921-5 ISBN EMA 6165 Polymer Physics – AB Brennan EMA 13 Epoxide Network – DMS EMA 6165 Polymer Physics – AB Brennan EMA 14 Epoxide Network – E’ f(kT) EMA 6165 Polymer Physics – AB Brennan EMA 15 Thermodynamics q So, now, let’s evaluate the entropic contributions. So, But, first, we recall that for an “ideal rubber”: But, E ∂ ∂ l H ∂ ∂ l q which allows us to write: f= q S ∂ −T ∂ l T ,P ,V Now, that we have the means to evaluate both Now, enthalpy and internal energy contributions, we will examine entropic contributions: examine EMA 6165 Polymer Physics – AB Brennan EMA 16 Thermodynamics q Consider the behavior of chains confined to a Consider fixed volume with applied stress: fixed Affine Deformation q i.e., deformation for each element is the same as i.e., the total volume, or: the 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 17 Thermodynamics ∆S = q [ Nck ( λ x 2 + λ y 2 + λ z 2 − 3 ) + (ln( λ x λ y λ z )) − 2 Assume bulk deformation, i.e., no swelling, then it can Assume be written that: be ln( λ x λ y λz ) = 0 q So, now how does one obtain a relationship between So, entropy and length, i.e., entropy q Let’s consider uniaxial Let’s deformation first: deformation q Assume for uniaxial deformation in the z direction: ∂ S ∂ l? λx = λy thus: λ xλ yλz = 1 λx = λy = EMA 6165 Polymer Physics – AB Brennan EMA 1 λz 18 Thermodynamics q Furthermore: l0 x * l0 y = Area and q By substitution one can write: σ0 = f = Area z q N c kT 2lx l y by definition: f = σ 0z Area ∂ 2 2 2 * [ ( λ x + λ y + λ z − 3) + ln( λ x λ y λ z )] ∂ lz λ = ll and ∂λ = dl o Hence: d li ∂λi = l0 which which enables: enables: lo dλz = l0 * ∂λz EMA 6165 Polymer Physics – AB Brennan EMA 19 q By substitution yields: σ 0z N c kT * ∂ ( λ 2 + λ 2 + λ 2 − 3) + ln( λ λ λ ) = x y z xyz 2lx l y ∂ λ z * lz q Now: q So: q Nc lx ly lz [ Nc = V = Nv ( ) N v kT ∂ λ x 2 + λ y 2 + λz 2 − 3 σ 0z = 2 * ∂λz 1 λ =λ = As given before: ln( λx λy λz ) = 0 and λ N v kT 2 2 σ 0z = * ∂ λz + − 3 ∂ λz λz 2 x y z EMA 6165 Polymer Physics – AB Brennan EMA 20 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, How i.e., Nv = Mc ?? i.e., EMA 6165 Polymer Physics – AB Brennan EMA 21 Thermodynamics q Consider the plot from our previous lecture: σ0 z G = N v kT (λ z q − λz −2 ) What is the slope? EMA 6165 Polymer Physics – AB Brennan EMA 22 Summary • Equations of State were developed Equations for Gibbs Free Energy for • Uniaxial Deformation was defined Uniaxial for a chains for • Modulus was defined for uniaxial Modulus deformation deformation EMA 6165 Polymer Physics – AB Brennan EMA 23 References • Introduction to Physical Polymer Science, 3rd Introduction Edition, Lesley H. Sperling, Wiley Interscience (2001) ISBN 0-471-32921-5 (2001) • Principles of Polymer Chemistry, P.J. Flory (1953) Principles Cornell University Press, Inc., New York. Cornell • The Physics of Polymers, Gert Strobl (1996) The Springer-Verlag, New York. Springer-Verlag, • Some figures were reproduced from Some • Principles of Polymer Chemistry, P.J. Flory (1953) Principles Cornell University Press, Inc., New York. Cornell EMA 6165 Polymer Physics – AB Brennan EMA 24 ...
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