WILL A POLYMER DISSOLVE IN A GIVEN SOLVENT? WILL A PARTICULAR POLYMER MIX WITH ANOTHER POLYMER?

Size: px

Start display at page:

Download "WILL A POLYMER DISSOLVE IN A GIVEN SOLVENT? WILL A PARTICULAR POLYMER MIX WITH ANOTHER POLYMER?"

John Shelton
6 years ago
Views:

1 MIXING WILL POLYMER DISSOLVE IN GIVEN SOLVENT? WILL PRTICULR POLYMER MIX WITH NOTHER POLYMER?

2 POLYMER SOLUTIONS ND LENDS MISCILE IMMISCILE lends Single phase Phase separated Solvent (or polymer) molecules Solvent rich phase Phase separated solution Solvent poor phase CN WE PREDICT PHSE EHVIOUR?

3 THE SECOND LW OF THERMODYNMICS ND FREE ENERGY For a process to occur spontaneously,s must increase; eg mixing Stot = S sys + Ssurr ut,if heat is released by the system Define free energy S surr = - Q sys T = - H sys T G = -T S tot = H -T S

4 WHY OTHER TO CLCULTE THE FREE ENERGY? 1. ECUSE IF WE DON'T THE GHOSTS OF OLTZMNN ND EHRENFEST WILL COME CK TO HUNT US 2. ONE CONDITION FOR FORMING SINGLE PHSE IS THT THE FREE ENERGY MUST E NEGTIVE 3. THE OTHER CONDITION IS THT THE SECOND DERIVTIVE OF THE FREE ENERGY MUST E POSITIVE

5 THEORIES OF MIXING G m = H m - T S m ssume the interactions between the molecules Can be approximated by a mean field.this allows The enthalpy and entropy to be treated seperately

6 OLTZMNN S TOM Entropy is associated with the distribution of energy and matter in a system.this can be expressed formally in terms of the equation carved on oltzmann s tomb S = k lnw Which,today,is normally written S = k ln Ω Where Ω is the number of arrangements vailable to the system[t a given V, E, N]

7 LTTICE MODELS - Summary NUMER OF CONFIGURTIONS Equal-size small molecules n o! n! n! Flory model for polymer solutions n o! n! N! ( z - 1 ) N ( m - 1 ) σ N 1 n o N ( m - 1 ) Solvent Polymer chain segment How are these expresions obtained?

8 REGULR SOLUTION THEORY NUMER OF CONFIGURTIONS Ω = N + N! N! N! HENCE - S m R = n lnx +n lnx

9 SOME NOTES CONCERNING THE ENTROPY OF MIXING THE QUNTITIES N OR N CN EITHER E THE NUMER OF MOLECULES OF ND, RESPECTIVELY, OR THE NUMER OF MOLES THE CONSTNT CHNGES DEPENDING ON CHOICE MOLES MOLECULES R (or RT) k (or kt) IN CLCULTING S WE USED STIRLING'S m PPROXIMTION ln(n!) = N ln N - N THE FINL NSWER IS WRITTEN IN TERMS OF MOLE FRCTIONS X,X N X = N + N N X = N + N

10 POLYMER SOLUTIONS ND LENDS COUNTING THE CONFIGURTIONS IS MORE DIFFICULT, UT FLORY ND HUGGINS OTINED; Solvent Polymer chain segment - S m R = n lnφ +n lnφ

11 DEFINITION OF VOLUME FRCTIONS - S m = n s ln φ s + n p ln φ p R n φ s = s n s + Mn p 75 / 100 Solvent Polymer chain segment Mn φ p p = n + Mn s p 25/100

12 THE ENTROPY OF MIXING POLYMERS RELTIVE TO SMLL MOLECULES - S R =n lnφ + n lnφ n = 75 n = 25 - S R =n lnφ + n lnφ n = 75 n = 1 Solvent Polymer chain segment

13 THE ENTHLPY OF MIXING The enthalpy of the system depends upon the interactions etween the molecules - we need an expression for the Change in enthalpy on going from the pure components To the mixture There are two related approaches that are most commonly Used Hildebrand - solubility parameters χ Flory - parameter oth depend upon considering interactions etween pairs of molecules and both assume Dispersion and weak polar forces only

14 INTERCTIONS DISPERSION FORCES DIPOLE / DIPOLE (freely rotating) } Interaction energy ~ _ 1 r 6 HYDROGEN ONDS and STRONG POLR FORCES (charge transfer) Interaction energy complicated,but short range COULOMIC (eg ionomers) Interaction energy longer range ~ 1/r

15 COHESIVE ENERGY DENSITY For the pure components,define the cohesive energyper pair of molecules as; C = E vap V -P.E. C = E vap V -P.E.

16 MIXTURES Change in energy (per pair) related to - 2C -C -C ssume Obtain 1/2 1/2 C =C C 1/2 C -2C C 1/2 +C 1/2 1/2 2 C -C The molar energy difference per pair on forming contacts from and contacts.now multiply y the number of contacts in The mixture.ut first,simplify!

17 SOLUILITY PRMETERS ND THE ENTHLPY OF MIXING 1/2 δ = C 1/2 δ = C (δ - δ ) 2 Define solubility parameters The cohesive energy density change upon forming pairs from and pairs Remember that this is per mole.to get the heat of mixing we must now multiply by the molar volume(to get an energy from the energy density),then the number of moles,then a factor that gives the number of contacts H m = (n + n ) V m φ φ (δ - δ ) 2

18 FREE ENERGY OF MIXING- REGULR SOLUTION THEORY G m = H m - T S m G m RT = n ln X + n ln X + (n + n ) V m φ φ Β (δ Α δ Β ) 2

+ kt φ P N lnφ + φ lnφ

+ kt φ P N lnφ + φ lnφ 3.01 practice problems thermo solutions 3.01 Issued: 1.08.04 Fall 004 Not due THERODYNAICS 1. Flory-Huggins Theory. We introduced a simple lattice model for polymer solutions in lectures 4 and 5. The Flory-Huggins