Welcome to this movie on statistical mechanics, my name is your station and I teach statistical mechanics at the Delft University of Technology in the Netherlands. This is the third movie in a series of three on elementary polymer and solution physics, and in particular, this movie deals with polymers in solution and the theory which describes those systems is the so called floryhuggins theory. The first movie, which I made dealt with free chains where the chain doesn't have any interaction. The beats of the chain do not have any interaction. The second one was a movie on polymers, where the beads cannot overlap, and then you get the socalled Florrie exponents. You have a self avoiding chain, and in this movie we are treating a much more complicated problem. Those are chains which cannot overlap, so the beats of change do not want to overlap, but in addition, there is a solution present and then you can have a search that either the polymers collapse into one part of the volume where they form a dense, polymer melt Or they will really dissolve in the solution and that's the problem I want to address in this movie, and I hope that you enjoyed this movie and learned something about it. On floryhuggins theory, in the floryhuggins theory for polymer solutions, we start with polymers, whose beats cannot overlap. That'S similar to the floor, a mean field theory where we had also beats that could not overlap. They had a hard core repulsion. They were moving in the continuum space here we put the polymers on a lattice and the solvent atoms of the solvent molecules are also put on lattice sites and on each letter site that can either be a polymer bead or there can be a solvent molecule, and We can envisage that in that case you have the following two situations: on the lefthand side, we see polymers, they are the colored sites. The color here only serves to distinguish to differ chains, and we have also great circles white insight and they are the solvent molecules and we see that of maalik solvent and the polymer they mix, and so this is a real solution which is characterized by a high Entropy, so we can say that here the entropy is high, it's certainly higher if we compare it to the right hand, side where we see that face has separated. So we have a region in space where the polymers all go, so they form a dense melt and we have, in addition, a solvent which separates from the polymers, and so this is called a separated phase. The polymers here in they form a melt and, on the left hand, side. We have a solution, although we can make a definite statement about the entropy which is higher on the left hand case in the solution than on the right hand, side in a separate two rated face. It'S difficult to tell what the energy in the two cases will be, because that depends on how much the sites of the polymer and the beats of the polymers do like each other and how much they like the solution and also on how much the molecules inside The solution like each other, so we have interaction parameters that we will come back to in the end, describing these three different energies of polymer, polymer, polymer, solvent and solvent, solvent interactions and, depending on the values that these parameters have, the energy can be higher or lower. On the right hand, side and, on the left hand, side. The face which dominates is the one which has the lowest free energy and therefore we need indeed to evaluate the entropy for the two phases. And we need to evaluate the energy. And we start by the entropy, but before we do that, we need to define several quantities that we will use in the calculation and the first one is, and sites and sites is just a number of sites in the system, so it's equivalent to the volume. The second number, which is important, is n and n is the number of beats for each polymer. The next number is n, P and n P is the number of polymers we have in the solution or in the melt. So the number of polymer beats in the system is just n times and P. Finally, we have the number Q and that's the lattice coordination number. That'S the number of nearest neighbor sites for each sites on the lattice. So now we start calculating the entropy by imagining that we build up the system polymer before polymer and we will use some mean field arguments in doing that before we continue. It should be pointed out that the grid that we have on here is a twodimensional grid, but it could also be three dimensional or one dimensional or four dimensional. Whatever we now evaluate how many possibilities we have when we built up the J's polymer of the system, so we assume that there are already J minus one polymers in the system and we are going to build up now. The j'the polymer and we start with the first site for that first beat of the polymer. We can choose any site which is not yet occupied by the J minus one previous polymers. So in total we have how many sites at our disposal we had tons n sites in total and of these J minus 1 times the number of sites per polymer and that was n has been occupied by other polymers by other beats, and therefore we have this. Many probably possibilities to put our first beat for our second beat. We have in principle, we can choose any of the q neighbors of the previous one, but maybe that neighbor is already occupied by one of the previous polymers. So we have to multiply this by a probability that the site is unoccupied and for this probability we use a mean field assumption we just take Q times, then we just take the sites that are still unoccupied. So that's n sides minus J, minus 1 times n minus 1 because we have already placed the first beat and we have to divide that by n sites. This is a meanfield assumption, because we do not. We assume that the average probability of occupation is independent. Of that of the fact that we have placed a first beat here, so it's homogeneous everywhere in the lattice. That assumption is the mean field assumption in order to see how many possibilities we have for the third beat. Let us take first first beat and then we could put a second beat either here or there or there or there. If you are on a square lattice, so suppose we have put it here and then for the second beat we can again in principle. It has four neighbors, so Q is four in this case of a square lattice in the plane, but one has already been occupied. So instead of Q neighbors, we only have Q minus one neighbors. We can where we can place the next site, and we should multiply again by this probability that that site is unoccupied. It could be that some other polymer which started here as occupied, for example this or that or that site, and so what do we obtain? We have aq minus 1 times n sites, minus AJ, minus 1, times N, and now we have minus 2 because we have placed already the second bead. So there's one less site free and we carry on in the same fashion. So for the cave beat you have. The same expression as for the third beat very similar, except for the minus two which we had for the third beat. We have a minus k plus 1 for the cave beat and he carry on until we have exhausted all the N beats of Palawan and the J all in all for polymer J. In order to find the number of possibilities we have for building up that polymer, we should multiply all these numbers, so this number times that number etc etc, and we obtain the following expression so recall that for the first site we just had all the as many Possibilities as there were empty sites for the second one, we could take Q neighbors, and then we should multiply that by the probability that they are unoccupied. That'S this factor, then, for the next one. We have only Q minus 1 because the previous beat was already occupied and we have to multiply that by the unoccupied, probability, P unoccupied, and we carry on like that until we arrive at beat number n now note. When an enumerator, we have a product of terms which decrease. So here we have a certain number. This is that same number, minus 1, minus 2, minus 3, etc. Until we arrive at the last number  and there is a compact way of representing that  namely by the fraction of two factorials  and what we obtain then in the end is that we have n sides minus J, minus 1 times n factorial. So that's what we would have if this would continue all the way down to one, but that's not the case. So we have to divide out all the terms that are smaller than this and and that turns out to be n sites minus J n. So that takes care of all the integers in the denominator. Apart from Q, then we have one factor of Q and in addition, we have n minus two factors of Q minus 1. That'S because we have the terms of Q, they all correspond to links. If you have n sites, we have n minus 1 links the first one of those is Q, and so for the remaining n minus 2. We have Q minus 1 and then we have this factor of n sites in the denominator and that occurs for each beat. Except for the first one, so that should be raised to the power and minus 1. So this is the number of configurations that we have for polymer number J and in order to find the number of configurations for all the polymers, we need to multiply this quantity. Omega J for each polymer soda J runs from 1 up to the number of polymers and then, of course, we need to divide by an P factorial, because when we reorder the polymers, then we have in fact the same configuration and the end doesn't matter in which Order we built up the polymers and that's why we have to compensate here with one of our and P factorial. So when we worked that out, it's wise to start with the prefactor. That'S rather simple, because it's the same for each polymer. So we simply have to raise that to the power RNP, so we have first of all our one over n P factorial, and then we have Q to the power n P. Then we have Q, minus 1 raised to the power n minus 2 times n P, and we divide by n sites raised to the power n P times: n minus 1. So the remaining terms are a bit more complicated for polymer number one. We have here n sites factorial and we need to divide by n sites minus n factorial. So that's this expression for J is 1. So then we obtain the sequence of these terms and sites factorial divided by n science and factorial, and then we have the same term in the numerator and a new term in the denominator that same term in the numerator new term in the denominator, etc, etc. So we see that there are a lot of cancellations. This term cancels against this term. This term cancels against that term, this term councils against the next one, etc, etc. Finally, this term will also cancel and the only terms which remain our n sites factorial and this term n sites minus NP times n factorial. So finally, we have the same term as we had here and then we are left with n sites. Factorial / n sides, n P times n factorial, and so that's the number of ways we can put the polymers in the solution phase. We call that the solution phase was this phase that we had in the left picture. What we want to find out is whether the left or the right situation, so the solution or the separated face is stable, and so we can. We know that when we know the free energies of both  and we are dealing now with the entropy, which is an important part of that free energy, so for the entropies, we need to compare the entropy of this face with that of this, and we are focused Only on the entropy of the polymers and in the right hand, side we can use the the procedure that we are followed here, except that we just assume that there is no solvent present and that we have fewer sites. So here are all the sites that can be occupied by polymer, so it's just n times n P. So that's the number of beats we have available in the polymers and we have no solution for the separated phase that we use. Therefore, the same expression. But now we take n sides to be equal to n times n P. So, for example, this term this P R sub n from the expression and we then obtain 1 over and P factorial. And then we have Q and P Q, minus 1, n minus 2 and P, and now in the denominator, we have n times n P, raised to the power and P times n minus 1, and then we have instead of n sites and times and P factorial. In order to find the entropy difference, we take the K log of the Omega solvent and that of the separated face, and so that means that in the end we have to take the logarithm of the ratio of these two terms and then one thing is simple: We see that these terms in those are exactly the same, and so we then recover the expression only with the different terms and the result then becomes the following expression, which is rather cumbersome to work out and in order to make life a bit easy easier. We define a new parameter, which is the polymer fraction, so we define a parameter that we call Phi and that is n times. N, P and P is the number of polymers and is the number of beats per polymer. So this is the total number of beats divided by the total number of sites. So that's the polymer fraction and in terms of this expert parameter, we can express this entropy difference in a rather straightforward way. First, we recognized this parameter here in the first term. In the logarithm and working it out, we get an PE times, n minus 1 times the logarithm of Phi, and then we use the Stirling approximation for the other terms that occur here. All these factorials here is the result where we have used the following. This is Phi times n sites, and that rightly gives rise to the last term here and sites Phi times the violin and science Phi. Then we have here 1 minus Phi times n sites which gives rise to this term, and the term in the numerator is simple. It just gives me this, and then we realize that the last terms are all proportional to n sites, so we could put n sites here in the denominator. At first sight it doesn't. It seems that this term is not proportional to n sites but and P times n can be written as Phi times n sites and the minus NP can be written as Phi times n sites divided by N, and so we can then write the following for s. Delta s, divided by the number of sites undivided by KD, gives minus Phi over n log Phi minus 1 minus Phi algorithm of 1 minus Phi. This expression follows straight over formally from the previous one. You just need to work out all the terms. One important remark in working out the entropies for the solution and for the separated phase. I have neglected the multiplicity of the solvent itself, but that doesn't matter because that's the same in both case, so that doesn't affect the final difference. The next step is done to calculate the energy both of the solution and of the separated face, and we start by the solution. Now the energy obviously depends on the interaction between solvent molecules, interactions between the polymer beats interactions between polymer beats and the solution, and therefore we now introduce a few interaction parameters where we assume that there are only nearest neighbor parameter interactions and so for the solvent. We have the parameter J SS, which describes us as a solvent, solvent interaction which only interacts between nearest neighbor sites, both occupied by solvent molecules. So we ask ourselves how many such interactions will there be? Well, in total, we have n sites times Q, because the lattice coordination number is Q. So this is the total number of nearest neighbors, but we have come to them doubly. So we need to divide this by 2. We need to multiply by j SS, but now these are. This is the total number of sites and, of course,the we can only allow sites which are not occupied by the polymer and therefore provided. If we multiply now by the occupation probability that the first site is indeed not occupied by a polymer, we have 1 minus Phi. For that, because Phi is the polymer fraction and if we bond the second site to be empty as well. We need to multiply once more with by 1 minus Phi, and therefore this is the interaction energy of the solvent. Then we need to take into account of solvent polymer interactions and we assume that the interaction constant we take for that. The symbol JSP now here is a section of a polymer and that can interact with, in this case two neighbors, because the other two have been occupied by neighboring polymer beads and therefore we start by allowing each polymer beat to F neighbors, and there are n times And P polymer beads, but because the these two sites have already been occupied by other Peet's beats. We have here a Q minus two, because two sites have been occupied and then we need to multiply by the, by the probability that the remaining sites are unoccupied and that probability is 1 minus Phi, and then we multiplied by j SP. Obviously, you could argue that we should also consider the first and the last beat, but we assume that we are in the large and limit, so we have long polymers and then they give a negligible correction to this result. Next, we consider polymer polymer interactions and that's the case which is very similar to the previous one, except that we want to read cross sites now to be occupied by another polymer by another beat, and it's important to emphasize here that, with these polymer polymer interaction, we Do not include the interactions between beads that are neighboring on a chain they're always in the same number of neighbors on a chain, so that gives a constant contribution to the energy, and we don't consider that here. So we take the case where, at this site, the Red Cross or that site is occupied by a beat of another polymer or a bit further on on this polymer, and therefore we have something similar to the previous one. We have n times and P times. Q minus 1, but now we need to divide by 2 because we have both polymer sites. They are indistinguishable, so swapping them gives the same and we do not want to off count. And then we need to multiply by the probability that the neighboring site is indeed occupied, and that gives me a Phi and then I have a J Palmer Palmer and I can add up all these terms. But I've done something in addition and times NP can be written as n sites times Phi and I've used that here. I have also used that for this contribution and therefore, in the final expression, we have an extra factor of Phi included in these two terms and that enables us to write the solvent energy as n sites times a single energy. So we are interested in the end. In the energy per site, we need to compare that solvent energy to the energy of the separated phase, and in that case we have a fraction of the volume which is occupied by the polymer and we have a solvent and because the boundary is a lot smaller. It took it involves much fewer links than the bulky parts we neglect the boundary altogether, so we consider the polymer volume separately from the solvent volume. Then we realized that every polymer in the melt has Q minus 2 neighbors and we need to multiply this by the number of polymer sites and that's n sites times Phi and for the solvent. We have n sites times 1 minus Phi solvent sites. So we subtract polymer fraction from the total and then we have a queue. Each site can have Q neighbors. Now because there is no restriction, you know dealing with the solvent and not with the polymers. We need to divide by 2 for doublecounting and we have a J solvent solvent and the entry are interested in the difference between the free energy. So here we focus on the difference between these two energies. All the terms are proportional to n sites, so we can divide by that and if we subtract these two expressions, then after some calculation we find the following expression and we see that we have extracted now all the terms which depend on the polymer concentration in front Of the square brackets and inside the square brackets, we have three terms solvent, solvent, solvent polymer and polymer polymer, and we see that the terms wherever particles interact among particles among them in themselves. So, with the particles of the same kind, they have a minus sign and four different particles. We have a plus sign. Obviously it did this parameter just as a fixed value, because the QD coordination number is fixed and all the interaction parameters they depend on the chemistry of the solvent and of the polymer. So all these terms are fixed and we lumped all those terms together into KBT. Chi, that's called the floryhuggins parameter guy and we see that, although the original expression here does not depend on temperature at all by this procedure, we have introduced the temperature dependence in Chi such that for low temperature, guys big and for hightemperature guy is small. Now, let us recall the difference between the entropy which we calculated before and which is this expression, and that enables us to find the free energy difference and using the Delta e and using the Delta s. We can find the Delta F just two Delta e minus T Delta s, and then we have a lot of terms that we can work out and the final result is then, this expression that we see here and this term arises from the energy difference. The other terms are just the Delta s, and so this is the energy free energy difference between the solution and the separated face, and this equation determines whether the solution is stable. That'S the case when Delta F is smaller than zero, because then the free energy of the solution is lower than that of this separated face and for positive values of Delta. F, we find that a separated face is stable. Here you see a plot of Delta F as a function of Chi and off I'd two parameters in the formulas given here on top and the color indicates whether the value of Delta F is positive. In that case, the separated face is stable or negative. That'S the blue region and then the solution is stable and then on the right picture. You see a few scans that were taken for fixed values of Chi, so those are cuts in this direction and you see that there are even values of Phi, for which the value for Delta F can be either positive or negative, depending on the concentration Phi. In summary, we have compared two different phases for a collection of polymers and solvent molecules. The left phase is the phase in which the polymers mix with the solute with the solvent molecules, and so they form a solution on the right hand, side they form a separated face, which one of the two is stable depends on the relative free energies and in Order to calculate that, if we are first analyzed the entropy and then the energy, both in a mean field approach  and it turns out that these quantities can be expressed in terms of two parameter. The first of these is the polymer fraction. That'S fine and that's the number of polymer beats divided by the total number of sites, so we have NP times and polymer. Beads and P is number of polymers and and mr. length of the polymer measured in beats and the remaining sites are occupied by the solvent. The second parameter is Chi, that's the floryhuggins parameter and it depends on the interaction constants. These are all nearest neighbor interaction, constants, the first one for two sites, neighboring sites being occupied by solvent molecules, then solvent polymer and then polymer polymer. The interactions were, the occupations of two neighboring sites are the same, set its SS or PP. They both have a negative minus sign and the one in which a solvent talks to a polymer has a positive sign. Furthermore, there is some temperature in introduced because we have divided by KBT. The stability of the face is determined by the sign of Delta F, which depends on those two parameters. Guy and Phi  and here you see exactly for which values of Chi and Phi 2 Delta F is positive, then it's red or when it's negative. In that case, it's blue
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