 So today's talk is about getting away from the approximation that even when the enthalpy of mixing is finite, we assume that the atoms are randomly distributed to calculate the configurational entropy. And in general that cannot be the case because if the enthalpy of mixing is finite that means some atoms prefer to be next to particular kinds of atoms and therefore the configurational entropy cannot be random. And the sort of model that I'm going to describe has a variety of hierarchies. So the regular solution model that we discussed is known as the zeroth order quasar chemical model because it basically assumes a random mixing for the entropy with a finite enthalpy of mixing. And today I'll discuss the first order quasar chemical model where we will look at pairs of atoms rather than individual atoms and there are even higher order models where you look at clusters of atoms being distributed in your system. Okay, so just to summarize then these are the characteristics of the models that we have discussed so far. The ideal solution has a random mixture of atoms and zero enthalpy of mixing so the entire contribution to the free energy comes from the minus T delta SM term and the system is said to be completely random at all temperatures. In the case of the regular solution model we assume that sorry that should be I'll just change that so that there's no confusion. So this is the regular solution model and this is the quasar chemical model that I'll discuss today. So in the regular solution model we assume that the entropy of mixing is determined by a random mixture of atoms but we admit that the entropy of mixing is not zero and the quasar chemical model deals with this problem where we don't assume that the entropy of mixing is ideal. So in a regular solution model, atoms are treated as independent entities and it's called the zero order quasar chemical model. Okay, so in the quasar chemical approximation first order we say atoms are not distributed at random but pairs of atoms are treated as independent quantities, okay, right. In order to go through the derivation I'm going to go through a very simple case where we have just two energy levels. This is the ground state here and this is a higher energy level by this quantity E1 and G represents the degeneracy of states that means at this state I have two equivalent states with the same energy okay and in the ground state I have three equivalent states with the same energy which is zero in this case for the ground state. So the atoms will distribute themselves amongst these energy levels depending on temperature and the quantity E1 so the number of atoms out of the total number capital N which are in the ground state will be the degeneracies of the ground state the we have these three energy levels say divided by the total number of possibilities where this is the ground state and this is the higher energy state and the chance of occupying the higher energy state E1 over KT. And similarly the number in the higher energy state is given by N1 over N which is the chance of being in the higher energy state times the number of higher energy states with the same energy that exists, okay. So this is straightforward and the importance of this is that it defines what we call a partition function okay so this is very common in thermodynamics partition function is simply the sum of all these terms here where again this is the degeneracy of the higher state here and the energy difference between between the states relative to the ground state. And if I take the logarithm of this term then if you work it out you'll see that that gives us the free energy of the system the Helmholtz free energy of the system. So if I take the Helmholtz free energy of the system minus RT log of the partition function and substitute for the partition function then that is our partition function G0 plus G1 into exponential E1 over KT and expand that and I get this as the Helmholtz free energy and if you want to convert this to Gibbs free energy then you just add a P pressure times a volume change term. So if I want to work out the difference in energy between the state where all the atoms are in the ground state and then at a particular temperature T then this term disappears here and that is our difference in the Helmholtz free energy okay. So that is the meaning of a partition function you know how the atoms are distributed over the different energy states how they are partitioned on the different energy states and the importance of that is that we can take the logarithm of that function multiply by RT and we get the Helmholtz free energy term for that particular system. Okay now if I'm looking at a system of A and B atoms then the number of AA bonds is simply the total number of A atoms minus the number of AB bonds because this fraction of A atoms has been this number of A atoms has been taken up in forming a bond with B and therefore it cannot contribute to AA and since we can't distinguish these two A atoms there's a factor of a half here and this is a coordination number said and the number of BB bonds is similarly given by the number of B atoms minus the B atoms that are occupied in forming AB bonds which are not the same as BB bonds okay this is just a reminder and we have a binding energy between the atoms which was defined as follows so if I take two A atoms from a distance far apart and I bring them close together then I get a curve like this where initially they may attract and then they start to repel as you push them closer and closer together so that the electron codes are interfering and the binding energy is defined as minus two AA or here and the whole basis of the binding energy is that we want to work out the enthalpy change or in this case an internal energy change when we make a mixture of A and B atoms so this is the total energy of our assembly or total internal energy of our assembly where this is the binding energy half the binding energy of the AA atoms multiplied by the number of AA bonds BB number of BB bonds and AB and the number of AB bonds where AB can be distinguished from BA so if I just simplify this then that's minus Z into this term where omega you'll remember from a previous lecture is when I take an air bond and a BB bond break them apart and form two AB bonds so that's nothing new from what we've done in calculating the enthalpy change for the regular solution model okay now let's let's try and make a partition function for a system where we have a number of energy states okay and here is our equation where this is the number of AB bonds the degeneracy of states for that number of AB bonds and the internal energy for that number of AB bonds and you know you can have a different number of AB bonds depending on how the atoms are arranged so there's a whole summation here of the degeneracies of the variety of NAB values and the variety of U NAB values and I substitute for U from the previous slide and this is what we get so G NAB is simply the total number of arrangements for a given value of NAB is the degeneracy of that state how many states do I have which have the exactly the same number of NAB atoms and energy and sigma NAB that means the total number of arrangements I can get in a mixture of NB is simply N factorial or NA factorial into NB factorial as we did in our earlier example now we are going to make an approximation okay and the approximation is that you know to work out the Helmholtz free energy you take the logarithm of the term and that will be dominated by the maximum value of of NAB okay so we'll ignore all other terms in the summation and simply take the maximum number here and therefore the degeneracy for the maximum value of NAB must be equal to this here okay because we've approximated this summation by simply taking the maximum value given that it's the logarithm of omega which gives us the free energy so that's an approximation that we have to make in all of these quasi chemical models to make it make it simpler okay now I can write just as we did for a random solution here where this is N factorial over NA factorial NB factorial I can write the number of possibilities for having a certain number NAB bonds in a mixture as this which corresponds to this term here remember we are dealing with pairs of atoms now okay and that's why we have the coordination number and the half and similarly we divide by the AA bonds the BB bonds and we distinguish between the AB and VA bonds but the important thing I want you to notice here is that while this is an equality for a random solution this I've written as a proportionality and that's because there's a problem in dealing with pairs of atoms that we can't count the number of AB bonds independently okay and I'll illustrate that problem here I've got these four atoms here A, B, B and another B once I've placed this B atom next to this A atom I've got an AB bond and once I've placed a B atom next to this B atom I've got a B B bond and once I've placed this atom here I've got a B B bond but there's no choice here okay once we've made this arrangement this bond VA is there it's not independent of the counting that we are doing so we want to treat these bonds as independent but we can't because there's no way you can do that so how do I change this proportionality here into an equality because you know we need to do that to work out the free energy okay so I hope everyone understands this that once we have fixed these four atoms in position the this bond here VA is not independent of the configuration of the other bonds in other words we are not taking account really of the orientation of the bonds in the lattice in calculating the number of AB bonds okay so this is the number of degeneracies for the maximum value of AB bonds that we described earlier and again we have a proportionality sign giving us the number of possible arrangements which give me this maximum value of an AB but that must also equal this and therefore we can normalize the equation by dividing the top and bottom by these terms here which are equal okay so I can write that the degeneracy for a particular combination of A and B atoms is given by the usual factor and a normalization factor which converts it into an equality right so that's quite a lot to take in but the principles are straightforward so far that in our partition function we have simplified the problem by taking the largest number of NAB bonds okay and ignoring the number of configurations for the largest number of NAB bonds and ignoring all others and in this process here because we can't treat pairs of atoms as being completely independent this if this was an equation it would overestimate okay so what we are doing is we are normalizing the general equation for the number of AB bonds by these these terms here which are equal just just dividing at the bottom by this and multiplying by this okay okay so in our partition function we've already said that we don't need this summation we are simply going to set this to the maximum value of NAB so we don't need this summation here but we have now the complete equation we have here the degeneracies for NAB being the maximum number of AB bonds and given that we know the value of NAB we can work out the internal energy part of the problem so all we have to do is differentiate this equation with respect to the number of AB bonds to find the equilibrium number of AB bonds and when you do that you get an equation like this okay where this is like a regular solution model where we have the coordination number x one minus x being you know the mole fraction of B atoms and mole fraction of A atoms and the total number of atoms this term is special okay so it's taking account of the fact that we have a finite enthalpy of mixing which is given by this term omega the difference in the binding energies and therefore this will not be equal to the number of pairs or a number of AB pairs that you would find in a completely random solution it will depend on this value here so if I take a plot of this this equation this is how the number of AB bonds varies with our enthalpy term okay omega and this is how the number of AA and BB bonds changes as a function of the enthalpy and when we have an ideal solution that means the enthalpy term is zero you know these this quantity is simply given by you know the probability of finding an A atom times the probability of finding a B atom and so on and similarly an AA is simply proportional to one minus x squared and and BB to x squared as usual but when we have a finite enthalpy of mixing that is no longer the case okay so you can do a fairly rigorous calculation of the free energy using that partition function and this is the Helmholtz free energy and if we are working at zero pressure that's equal to the Gibbs free energy we take the logarithm of the partition function this term we've already determined from the omega and the equilibrium number of bonds and this when we expand this we have three terms here this is our normal enthalpy of mixing if you ignore the denominator here okay but this takes account of the fact that there's a finite enthalpy of mixing because the number of AB bonds will not be given by random probabilities this of course is the ideal entropy of mixing and this term here corrects the ideal entropy of mixing for the fact that we don't have a random mixture of atoms okay now you might ask you know what is the importance of this equation you know why can't we just use the empirical thermodynamics which were there in the thermo calc and empty data and backstage and so forth where we have polynomials and so forth there are cases where you cannot do that okay so if you have an ordered system you cannot actually talk about the pairs of atoms or atoms being distributed at random it just doesn't make sense when we come to for example the diffusion of carbon in austenite you know the concentration dependence of the diffusion coefficient is given partly by the fact that the chemical potential is a function of the composition but in part it also comes from the fact that the carbon atoms do not ever want to be in adjacent sites within the lattice because they strongly repel each other and that's that term comes into this this kind of a factor beta q and provides an additional push to the carbon to diffuse down the gradient simply because the carbon atoms don't want to be next to each other so you cannot explain the concentration dependence of carbon in austenite simply by saying that there's a thermodynamic factor which comes from the dependence of the chemical potential on concentration these terms become important and if you go to my website and look for a paper on the diffusion of carbon in austenite Siller and McClellan did a lot of work on this and I derived some coefficients for that work and you will see that it beautifully represents the concentration dependence of the diffusion coefficient of carbon in austenite so there are cases where we cannot simply use empirical terms the other point is that a lot of the thermodynamic data the equilibrium thermodynamic data will be derived at high temperatures because you cannot really achieve equilibrium if you go to sufficiently low temperatures that means that if you try to extrapolate those functions you might get some strange results in using you know thermodynamic databases alone so this method also enables you to extrapolate to low temperatures and by low temperatures in the case of in the case of nickel and iron you're talking about at temperatures below about 600 degrees centigrade or lower so in in doing our calculations on the transformation start temperatures for the non-equilibrium transformations like martensite and bainite and so forth we use quasi-chemical theory to do the extrapolations okay because this is a rigorously founded theory in as far as we've now gone on to look at a non-random distribution of atoms for the configurational entropy if you go to higher approximations of quasi-chemical theory they're generically known as cluster variation methods okay so this is just an introduction to quasi-chemical theory and i'm going to summarize this now so how to derive all these solution models well first of all we categorize and count the variety of atom atom or atom vacancy pairs that are possible in the solution and you know in the case of an inter-sexual solution the number of vacancies is of course much much greater because the vacancies are simply the holes between the atoms so you have to explicitly take that into account you estimate the configurational energy that means the enthalpy term if we are working with Gibbs free energy and internal energy term if you are working with Helmholtz free energy then you write a configurational partition function you know they're basically telling you how the particles are distributed along the different kinds of energies and the degeneracies of those energy levels discover the degeneracy for each configuration and because we cannot strictly count pairs of atoms independently we have to normalize that degeneracy function okay and replace the summation by the largest term in there and then we derive the thermodynamic partition function and the problem is solved of course there might be other terms which come for example from thermal entropy and so on now i've talked for approximately i would say 20 minutes okay or 15 or 20 minutes but the concepts here are actually quite powerful and you actually need to do something with greater chemical theory to understand its full power and i don't need to go into any more complexity beyond what i've taught you because the idea is simply to know that there exists a method which can take account of non-ideal configurational entropy all right so that's the important takeaway from this so someday you come across a problem where you need to treat a solution with a non-ideal configurational entropy of mixing this is the way to do it and to be honest you know all this talk about high entropy alloys which i mentioned in my very first lecture is remarkable that they all assume the ideal entropy of mixing whereas it is simply impossible in that sort of an alloy system to get an ideal entropy of mixing okay so someone needs to do proper work there to see whether these are really high entropy alloys or some wishy washy low level of entropy because the atoms are not distributed at random so i don't want you to worry too much about the mathematics that i've presented i want you to understand that there exist methods which enable you to work out the configurational entropy in a much better way than simply assuming that the atoms are randomly mixed and this is the end of the course so at the end of the course we must have a little celebration so this is how i would like to celebrate if i was there this is a picture that was taken at the indian issue of technology in karagpur and this actually is the first time i had this after emigrating from kinia as a child because in kinia we used to have this as well where you chop off the top of a green coconut you drink the juice and then you scrape off the flesh and eat it and the sad thing is that one of these in britain costs about 450 rupees okay so so it's more or less unaffordable um so thank you all for attending the course i have really enjoyed it and all the videos for this are on my youtube channel if you ever want to look at them again okay harry okay thank you very much for this wonderful series and it's so nice of you to you know have so much of time for our students and other participants we really you know we are really grateful to you and i hope in the future we should be able to do it again sometime