May 11, 2015 - Tentukan transformasi laplace dari fungsi berikut ini : f = 0,03(1-cos2t). Jawab : Command windows : >> f = sym('1-cos2*t)'). >> F = laplace(f). Atau pada M-file : Syms t % Untuk inisialisasi variabel. f = 1-cos2*t. laplace(g). b). Inv
Summary of Ideal Chains Ideal chains: no interactions between monomers separated by many bonds Mean square end-to-end distance of ideal linear polymer
R 2 Nb 2
2 Nb Mean square radius of gyration of ideal linear polymer Rg2 6 3 3 / 2 3R 2 exp Probability distribution function P3d N , R 2 2 2Nb 2 Nb 2 3 R Free energy of an ideal chain F kT 2 Nb 2 3kT R Entropic Hooke’s Law f 2 Nb 3 1 g ( r ) Pair correlation function rb2
Real Chains Include interactions between all monomers Short-range (in space) interactions
Probability of a monomer to be in contact with another monomer for d-dimensional chain d N * b d R If chains are ideal R bN 1/ 2
* N 1d / 2
small for d>2
Number of contacts between pairs of monomers that are far along the chain, but close in space N * N 2d / 2 small for d>4 For d 2GN/2 because ring sections reinforce each other
f f 2 f 2
Biaxial Compression Ideal chain R||
Real chain D
R||
D
On length scales smaller than compression blob of size D chain is almost 5/3 2 unperturbed D D g g b b Occupied part of the tube 1/ 2 2/3 N b N R|| D Nb R|| D bN 1/ 2 g D g Free energy of confinement 5/3 2 N b N b Fconf kT kTN Fconf kT kTN g D g D
R Fconf kT 0 D
2
Fconf
RF kT D
5/3
Challenge Problem 2: Biaxial Confinement of a Semiflexible Chain Consider a semiflexible polymer – e.g. doublestranded DNA with Kuhn length b=100nm and contour length L=16mm. Assume that excluded volume diameter of double helix is d=3nm (larger than its actual diameter 2nm due to electrostatic repulsions). R|| D
RF Calculate the size RF of this lambda phage DNA in dilute solution and the length R|| occupied by this DNA in a cylindrical channel of diameter D (for db where |U(r)|>1, NB = 1
Polymer blend: NA >>1, NB >> 1
Flory-Huggins Free Energy of Mixing Fmix 1 kT ln ln 1 1 n NB NA
At lower T for >0 repulsive interactions are important and there is a composition range ' '' with thermodynamically stable phase separated state. This composition range, called miscibility gap, is determined by the common tangent line.
Fmix/(kn)
At high T entropy of mixing dominates, Fmix is convex and homogeneous 0 mixture is stable at all compositions. -0.1
250o K
-0.2 -0.3
300o K
-0.4 -0.5 -0.6 -0.7
350o K 0 ’ 0.2
0.4
0.6
0.8
” 5o K NA = 200 NB = 100 T
1
Phase Diagrams Fmix 1 kT ln ln 1 1 n NB NA
Critical composition
cr
NB N A NB
For a symmetric blend NA=NB=N cr=2/N cr=1/2
For polymer solutions NA = N, NB = 1 1 1 1 1 cr cr N 2 N 2N
0
N=2.7
NFmix kTn
there is a miscibility gap for ' ' '
NA = NB
’ sp1
-0.1
sp2
4 3 Ncr2
Two Phases
1 0
Single Phase
N
For cr mixture is stable at all compositions. 2 1 1 1 > For cr 2 NA NB
0
0.2 0.4 0.6 0.8
cr
”
1
Phase Diagram of Polymer Solutions Polymer solutions phase separate upon decreasing solvent quality below q-temperature
B A T
Upper critical solution temperature B>0 binodal
Solution phase separates below the binodal in poor solvent regime into a dilute supernatant of isolated globules at ` and concentrated sediment at ``. dilute supernatant concentrated sediment
` ``
Single Phase
`
spinodal
”
unstable (spinodal decomposition)
metastable Two Phases (nucleation and growth) M=53.3kg/mole Polyisoprene in dioxane Takano et al., Polym J. 17, 1123, 1985
Intermolecular Interactions Osmotic pressure c 2 P RT A2c ... Mn A2 – second virial coefficient
h
membrane solvent
Osmometer
Poly(a-methylstyrene) in toluene at 25 oC Noda et al, Macromol. 16, 668, 1981
30
106 P/cRT (moles/g)
solution
25 20
Mn= 70800 g/mole
15 10
Mn= 200000 g/mole
5
0 0.00
Mn= 506000 g/mole 0.01
c (g/ml)
0.02
Mixtures at Low Compositions
Fmix 1 kT ln ln 1 1 n NB NA Expand ln(1-) in powers of composition 2 1 3 Fmix 1 kT ln 2 ... n NB 2 NB 6N B NA Osmotic pressure Fmix P
Fmix V
2 3 n kT 1 3 3 2 ... b b N A NB 2 3N B 2
nA
cn v 2 Virial expansion in powers 3 P kT c wc ... n n of number density cn = /b3 NA 2
1 3 b6 2 b Excluded volume v 3-body interaction w 3N B NB v T q 2 A2 M 02 1 2 3 In polymer solutions NB = 1 3 T b b N Av
Polymer Melts Consider a blend with a small concentration of NA chains in a melt of chemically identical NB chains. No energetic contribution to mixing = 0. 1 3 b3 is very small for NB >> 1 Excluded volume v 2 b NB NB Flory Theorem 6 b 2 xT b gT bN B g N Thermal blob T B 2 v Chains smaller than thermal blob NA < NB2 are nearly ideal. In monodisperse NA = NB and weakly polydisperse melts chains are almost ideal. RA 3/5 2 In strongly asymmetric blends NA > NB bNA long chains are swollen 1/2 3/ 5 1 / 10 3/ 5 NA NA NA 1 / 2 b R A xT bN B 2 bN A 2 2 N 1 N N N gT B A B B
Challenge Problem 3:
RA2/(b2NA)
Long NA-mer in a 3-d Melt of NB-mers NB NB NB NB NB NB NB
1/ 5
NA R 2 2 b N A NB 2 A
NA/NB2
M. Lang
Why doesn’t Flory Theorem work?
Challenge Problem 4: Mixing of Polymers with Asymmetric Monomers NA – monomers per A chain
NB – monomers per B chain
v0 – volume of a lattice site = volume of a small B monomer mv0 – volume of an A monomer m times larger than B monomer Derive the free energy of mixing Fmix of A and B polymers.
Calculate the size RA of dilute Achains in a 2-d of B-chains for m>>1 in the case of =0.
Summary of Thermodynamics of Mixtures Free energy of mixing consists of entropic and energetic parts. Entropic part per unit volume (translational entropy of mixing Smix )
TSmix 1 kT ln ln 1 V vB vA Energetic part per unit volume U mix kT 1 V v0
vA – volume of A chain vB – volume of B chain v0 – volume of a lattice site
B Flory interaction parameter A T Many low molecular weight liquids are miscible Some polymer – solvent pairs are miscible Very few polymer blends are miscible
Chains are almost ideal in polymer melts as long as they are shorter than square of the average degree of polymeriztion NA