1. A polymer brush in good solvent forms a layer of thickness L that scales with number of
monomers, N and grafting densit...
0.60.4
-2
0.20
0
-3
x
1
-4
0.8
-1
-5
Freeenergyofmixing
mole fraction of solvent
Figure 1: ∆Fm, free energy of mixing, ver...
(a) Predictions of ∆Gm
/(NRT) versus φ2, volume fraction of monomer, for T = 275,
300, 325, 332, 350, 400 K.
The Flory-Hug...
0.60.40.20
0
-0.05
y
-0.1
-0.15
1
-0.2
-0.25
0.8
volume fraction of polymer
Flory-HugginsGibbsFreeEnergy∆G/(NRT)
T=300 K
A...
You might not be able to discern easily any inflexion points, but the next exercise
will show that they are definitely there...
h
0.6
1
0.4
0.5
0
0.2
-0.5
-1
0
y
10.8
SecondDerivativeofFlory-HugginsGibbsFreeEnergy∆G/(NRT),wrt
voluemfractionofpolymer
...
y
0.6
0.2
0.4
0.2
0.15
0
0.10.05
Now what we can do is solve for the two values of polymer volume fraction for which
dgg =...
T
330325320315310305300
0.2
0.15
0.1
0.05
And the result is the envelope of spinodals as a function of T. Notice that the
...
Thus, there are 2 equations, µ†
1 = µ‡
1 and µ†
2 = µ‡
2, and 2 unknowns, φ†
2 and φ‡
2 to
solve. These equations are:
χφ2...
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Polymers Homework Help

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Published on: Mar 4, 2016
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Transcripts - Polymers Homework Help

  • 1. 1. A polymer brush in good solvent forms a layer of thickness L that scales with number of monomers, N and grafting density σ as L ∼ aNσ1/3 . How does the polymer thickness scale in poor solvent? In poor solvent, an isolated chain will collapse to a small sphere of size R ∼ N1/3 a. So if the chains were grafted a distance D apart where D > R, then the surface will be dotted with “mushrooms” of size R ∼ N1/3 a that contain monomer and no solvent. Thus, the thickness of the polymer will not be homogeneous when the polymer is grafted dilutely on the surface in the presence of poor solvent. In that case, L = 0 or L = R ∼ N1/3 a, and is independent of the grafting density when σ = (D/a)−2 is in the range 0 < σ < (R/a)−2 . Notice that the highest grafting density at which the molten globules remain isolated, σ = (R/a)−2 , scales as N−2/3 (this you can see by replacing R with the scaling N1/3 a) so that as the chain gets larger, this upper grafting limit gets smaller. On the other hand, if D < R, then the globules will coalesce so as to form a homogeneous layer which minimise polymer contact with the solvent. In this case, as each chain is grafted to the surface within an area D2 , and the volume of the collapsed chain is the number of monomrs times the voluem of each monomer, then the layer thickness is 1 chain D2 area × Na3 volume chain = L or L = Na3 D2 = Na3 /a2 D2/a2 = aNσ. That is, under poor solvent conditions the layer thickness depends linearly upon the grafting density. 2. A researcher suggests a new model for the phase separation of a binary mixture of solvent (species 1) and solute (species 2). The researcher has written this new solution thery, “the three-hump model”, in a lengthy manuscript for publication. You are given the job of reviewing the manuscript and informing the editor of its correctness or incorrectness. You have not read the researcher’s excessively long description, but one of his figures (reproduced below) rings “alarm bells”, signalling that the model violates fundamen- tal thermodynamics. Using only the researchers figure, explain to the editor why this manuscript should not be published. Describe all incorrrect aspects of this figure. Our online Tutors are available 24*7 to provide Help with Polymers Homework/Assignment or a long term Graduate/Undergraduate Polymers Project. Our Tutors being experienced and proficient in Polymers sensure to provide high quality Polymers Homework Help. Upload your Polymers Assignment at ‘Submit Your Assignment’ button or email it to info@assignmentpedia.com. You can use our ‘Live Chat’ option to schedule an Online Tutoring session with our Polymers Tutors. http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 2. 0.60.4 -2 0.20 0 -3 x 1 -4 0.8 -1 -5 Freeenergyofmixing mole fraction of solvent Figure 1: ∆Fm, free energy of mixing, versus x1, mole fraction of solvent as predicted from the new Three-hump Model of Solution Thermodynamics: ∆Fm/RT = sin (6πx1) − 6πx1(1 − x1). Shown is the model predictions at a given temperature where three phases, whose energy and composition are given by closed circles, coexist in the unstable concentration range (shaded area). The darker shaded regions show the metastable region where you can create a solution containing either 1 homogeneously mixed phase, or 2 partially-mixed phases that exist over short times before they slowly decompose into the three lowest energy phases via the process of spinodal decomposition. • It is not possible to have a two component system separate into more than 2 phases. You can understand this using the phase rule, where F, the number of degrees of freedom of the system is given by the number of components, C, and phases, P: F = C − P + 2. So for a two component system, F = 4 − P. If we had a one phase, there would be F = 3 degrees of freedom required to specify the system: these being temperature, pressure, and mole fraction of solute. If we had two phases, there would be F = 2 degrees of freedom required to specify the system: these being temperature and pressure (You don’t need to specify mole fraction as the equal chemical potentials across phases determines this). For three phases, there would be only F = 1 degree of freedom, an insufficient number of degrees of freedom OR, you can note for this specific model that it is impossible to construct a line, tangent to the ∆Gm curve at three different binodal compositions - which would be necessary for the co-existence of three phases. • Completely unstable regions should be marked by regions of the ∆Gm curve that have negative curvature. Compositions for which ∂2 ∆Gm /∂x2 1 < 0 will spontaneously de- mix. • Metastable regions are compositional regions where ∂2 ∆Gm /∂x2 1 < 0 and where compositions will eventually de-mix, but do so using the slower kinetics of nucleation & growth. • Unstable regions are usually bounded by metastable regions and not the other way around as indicated in the shading of the figure. 3. Using Flory-Huggins solution theory for polymer-solvent solutions containing chains of length N = 100 monomers, and a critical temperature of Tc = 332 K , show (using Maple for needed efficiency) http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 3. (a) Predictions of ∆Gm /(NRT) versus φ2, volume fraction of monomer, for T = 275, 300, 325, 332, 350, 400 K. The Flory-Huggins Solution model is ∆Gm NRT = χφ1φ2 + φ1 ln φ1 + φ2 N ln φ2. You are not given χ, but you are told that the chains are of length N and Tc = 332K. This is sufficient to determine χ over a range of temperatures. The value of χ at the critical temperature, where spinodals and binodals merge and above which the solution is homogeneous, is given by χc ≡ 1 2 + 1 2N + 1 √ N , or χc ≡ χ(Tc) = 0.605. Then, as χ(T) = B RT and χ(Tc) = B RTc , then χ(T) = χ(Tc) × Tc T . Thus, The Flory-Huggins model reduces to: ∆Gm NRT = 0.605 × Tc T (1 − φ2)φ2 + (1 − φ2) ln (1 − φ2) + φ2 N ln φ2. > G300:=y->0.605*332/300*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y); G300 := y → 0.6695333333 y (1 − y) + (1 − y) ln (1 − y) + 1 100 y ln (y) > G325:=y->0.605*332/325*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y); G325 := y → 0.6180307692 y (1 − y) + (1 − y) ln (1 − y) + 1 100 y ln (y) > G332:=y->0.605*332/332*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y); G332 := y → 0.6050000000 y (1 − y) + (1 − y) ln (1 − y) + 1 100 y ln (y) > G350:=y->0.605*332/350*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y); G350 := y → 0.5738857143 y (1 − y) + (1 − y) ln (1 − y) + 1 100 y ln (y) > G400:=y->0.605*332/400*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y); G400 := y → 0.5021500000 y (1 − y) + (1 − y) ln (1 − y) + 1 100 y ln (y) > plot([G300(y),G325(y), G332(y), G350(y), G400(y)], y=0..1); http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 4. 0.60.40.20 0 -0.05 y -0.1 -0.15 1 -0.2 -0.25 0.8 volume fraction of polymer Flory-HugginsGibbsFreeEnergy∆G/(NRT) T=300 K Alternatively, you can express a procedure, which I call g, which takes values of y, volume fraction of polymer, and T, temperature, and generate a procedural function that you can plot, derivitize (if that is a word), or solve: > g:=proc(y, T) description "free energy"; > 0.605*332/T*y*(1-y)+(1-y)*ln(1-y) + y/100*ln(y) > end proc; g := proc(yT )description “free energy′′; 332 ∗ .605 ∗ y ∗ (1 − y) ∗ T −1 + (1 − y) ∗ ln(1 − y) + 1/100 ∗ y ∗ ln(y)end proc; > plot( [g(y, 300),g(y,325), g(y,332), g(y,350), g(y,400)], y=0..1); 0.60.40.20 0 -0.05 y -0.1 -0.15 1 -0.2 -0.25 0.8 http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 5. You might not be able to discern easily any inflexion points, but the next exercise will show that they are definitely there; and indeed you can see them by plotting over a reduced range of y. (b) Spinodal compositions for the temperature range, 275K < T < 400K The spinodal compositions at any T ≤ Tc satisfy ∂2 ∆Gm /(NRT) ∂φ2 2 = 0. The easiest way to demonstrate these spinodal composition is to plot the second derivative of the free energy expression versus composition: spinodal compositions are found as those compositions where the second derivative curve crosses the com- position coordinate. > ddG300:=y->diff(diff(0.605*332/300*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y), y), y); ddG300 := y → −1.339066667 + (1 − y)−1 + 1 100 y−1 > ddG325:=y->diff(diff(0.605*332/325*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y), y), y); ddG325 := y → −1.236061538 + (1 − y)−1 + 1 100 y−1 > ddG332:=y->diff(diff(0.605*332/332*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y), y),y); ddG332 := y → −1.210000000 + (1 − y)−1 + 1 100 y−1 > ddG350:=y->diff(diff(0.605*332/350*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y), y),y); ddG350 := y → −1.147771429 + (1 − y)−1 + 1 100 y−1 > ddG400:=y->diff(diff(0.605*332/400*y*(1-y) + (1-y)*ln(1-y) + y/100*ln(y), y); ddG400 := y → −1.004300000 + (1 − y)−1 + 1 100 y−1 > plot([ddG300(y),ddG325(y), ddG332(y), ddG350(y), ddG400(y)], y=0..1, h=-1 http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 6. h 0.6 1 0.4 0.5 0 0.2 -0.5 -1 0 y 10.8 SecondDerivativeofFlory-HugginsGibbsFreeEnergy∆G/(NRT),wrt voluemfractionofpolymer volume fraction of polymer You can solve for the spinodal compositions at each temperature: > solve( ddG300(y)=0, y); 0.03276645143, 0.2279126287 > solve( ddG325(y)=0, y); 0.05690933409, 0.1421596689 > solve( ddG332(y)=0, y); 0.09090909091, 0.09090909091 Alternatively, you could again resort to formulating procedures, which is much easier and can yield in a few lines the complete spinodal curve as a function of temperature > dg:= proc(y, t); diff(g(y,t), y) end proc; dg := proc(yt) diff (g(y, t), y)end proc; > dgg:= proc(y,t); diff(dg(y,t), y) end proc; dgg := proc(yt) diff (dg(y, t), y)end proc; Here I simply created procedures for calculating the dg ≡ ∂(Gm /NRT)/∂y for values of polymer volume fraction, y, and temperature, as well as dgg ≡ ∂2 (Gm /NRT)/∂y2 . Then I plot the second derivative at T = 300 simply to show that there is indeed 2 inflection or spinodal points. > plot( dgg(y, 300), y=0.01..0.23); http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 7. y 0.6 0.2 0.4 0.2 0.15 0 0.10.05 Now what we can do is solve for the two values of polymer volume fraction for which dgg = 0 at arbitrary temperature: > solve( dgg(y, T) =0, y); 0.5 − 0.001232201533 T + 0.0001369112815 p 13337104.0 − 67064.0 T + 81.0 T 2, 0.5 − 0.001232201533 T − 0.0001369112815 p 13337104.0 − 67064.0 T + 81.0 T 2 These are the 2 spinodals at any T (which correspond to two roots of a polynomial) which I can plot as a function of temperature. I will plot only for temperatures up to Tc as mixtures above Tc are homogenous at all compositions. I simply copy the maple output into the plot function: > plot( [.5-0.1232201533e-2*T+0.1369112815e-3*(13337104.-67064.*T+81.*T^2)^(1/2), .5-0.1232201533e-2*T-0.1369112815e-3*(13337104.-67064.*T+81.*T^2 T=300..332); http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 8. T 330325320315310305300 0.2 0.15 0.1 0.05 And the result is the envelope of spinodals as a function of T. Notice that the spinodals are low polymer volume fractions. (c) Binodal compositions for the same temperature range, and The binodals correspond to where the ∆Gm /(NRT) versus φ2 curve is tangent to the line whose compositional intercepts are the chemical potentials. In other words, if the chemical potentials are given by: µ1 − µ◦ 1 RT = ∆Gm /(NRT) + φ2 ∂(∆Gm /(NRT)) ∂φ1 and µ2 − µ◦ 2 RT = ∆Gm /(NRT) + φ1 ∂(∆Gm /(NRT)) ∂φ2 , then the spinodal fractions of the two phases, which we denote {φ† 1, φ† 2} and {φ‡ 1, φ‡ 2} are those at which µ† 1 − µ◦ 1 RT = µ1 ‡ −µ◦ 1 RT , and µ† 2 − µ◦ 2 RT = µ‡ 2 − µ◦ 2 RT . So, inserting the F-H expression for eq 1 and 2, we get µ1 − µ◦ 1 RT = χφ2 2 + ln (1 − φ2) + φ2 1 − 1 N and µ2 − µ◦ 2 RT = χ(1 − φ2)2 + 1 N ln (φ2) + (1 − φ2) 1 N − 1 . http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215
  • 9. Thus, there are 2 equations, µ† 1 = µ‡ 1 and µ† 2 = µ‡ 2, and 2 unknowns, φ† 2 and φ‡ 2 to solve. These equations are: χφ2† 2 + ln (1 − φ† 2) + φ† 2 1 − 1 N = χφ2‡ 2 + ln (1 − φ‡ 2) + φ‡ 2 1 − 1 N χ(1 − φ† 2)2 + 1 N ln (φ† 2) + (1 − φ† 2) 1 N − 1 = χ(1 − φ‡ 2)2 + 1 N ln (φ‡ 2) + (1 − φ‡ 2) 1 N − 1 Now χ = 0.605 × 332/T, so these above equations provide for each temperature T < Tc = 332, the binodals or the compositions of the two phases, φ† 2 and φ‡ 2. These are also the phase diagram requested below. Note that the binodals are not found as the polyemr volume fractions which minimise ∆Gm , although these minima are excellent first guesses in an iterative solution of the above 2 equations. (d) T-φ2 phase diagram This is the T versus φ† 2 and T versus φ‡ 2 plots.. Discuss how these predictions would be altered for a solution of chains of higher molecular weight. With higher molecular weight chains, the polymer becomes even more difficult to ho- mogenously mix. You need to go to even higher temperatures in order to get the chains to mix homogeneously. That is Tc increases with molecular weight or N . Moreover, the composition at the critical point , φ2c, is smaller for larger N . See figure in Section 5.3.3 http://www.assignmentpedia.com/chemical-engineering-assignment-help.html For further details, visit http://www.assignmentpedia.com/ or email us at info@assignmentpedia.com or call us on +1-520-8371215