 The fundamental equations of thermodynamics just start with the DG equation you can do a similar treatment with others so we have seen quantities of interest are essentially ?G correspondingly ?A ?U ?H and so on and minimum value of say G constant T and these are the two quantities of interest in order to do this we need G since this is your work calculation this is your equilibrium calculation you need G as a function of temperature pressure and composition put this in squiggly brackets to indicate in I from I equals 1 to the number of components usually you get these relationships through Maxwell relations that is you find out how it depends on T how it depends on various quantities through the Maxwell relations so you have ?G by ?T is this is –S ?G by ?P is V ?G by ? and ?I but therefore what you need really these quantities you know for example S with if you want to know how S varies again you go back to Maxwell relations you got S with respect to P and so on so all that you can do the only thing you cannot get from here is from Maxwell relations writing getting bad to us if I keep all the variables constant except say N1 and N2 I get equations like this or I and J ?mu I with respect to ?NJ both are equal to of course ?Squad G by ?NI ?NJ normally if you want S with respect to Ni for example what you do is to keep S DT and ?I Ni Dni alone all the others you will keep constant then your Maxwell relation will give you the second partial of G with respect to T and Ni is S with respect to Ni that is equal to you will get the derivative of ?I let me do this first of all let me say this has been reduced to an equation in the chemical potential so we will say need ?I as a function of TP XR-1 so from one I get partial of ?I with respect to T is the same as partial of minus the partial of Si with respect to Ni S with respect to Ni this is ?I with respect to T holding P and all NJ constant is a better way of getting this is not a measurable quantity I told you you stop whenever the right hand side is a measurable quantity so if you differentiate ?I by T with respect to T you get this quantity this is the same as Si bar you said all partial molol quantities are obtained by differentiating an extensive variable with respect to Ni holding T and P constant that is important other molol numbers but T and P are the important variables so if you do this you get minus Si bar of course by T I am differentiating Mu I I get minus Si bar minus Mu I by T square this is the same as 1 by T square minus 1 by T square into T Si bar plus GI bar Mu I is the same as GI bar because G we saw that last class G with respect to Ni holding TP and other molol numbers constant is what you call GI bar and this is minus HI bar by T square so I have one equation that says Mu I by T and keep this equation HI bar is simply in the partial molol property with respect to enthalpy changes in enthalpy are measurable so we will show that this is a measurable quantity so this is equation 2 I will show you how it is actually measured little later we will simply note that this is a measurable quantity you can get a lot of data on HI bar so since I am interested in this equation here the chemical potential as a function of TP etc because my chemical equilibrium is governed in phase equilibrium by Mu I alpha is equal to Mu I beta and if I have Mu I as a function of temperature pressure and composition I can solve the phase equilibrium problem in principle so I have got this I need to know Mu I as a function of P this is very straightforward here Mu I with respect to P will be the second partial of G with respect to Ni and then with respect to P that will give you the first partial of V with respect to Ni this is VI bar this is equation 3 and this is also a measurable quantity volume changes enthalpy changes are measurable and from these changes I can actually calculate HI bar and VI bar now when it comes to composition dependence if I take two of these the third equation is the hardest Mu I with respect to XJ or NJ sorry I should finally differentiate with respect to XJ but right now I will do it as it stands here I take Mu I D Ni and Mu J D NJ all others I hold constant I simply get Mu J with respect to Ni all it says is that the chemical potential of component I depends on the number of moles of J exactly like the chemical potential of J depends on the number of moles of I but since I do not know either I am left in square one this is actually the basic problem of mixed thermodynamics or challenge essentially you do not know the chemical composition dependence of the chemical potential uniquely and there probably no unique solution because the different mixtures behave differently the question is how do we describe this composition dependence in some meaningful form how do we model it you have of course we derived last class you know we integrated this and differentiated it and obtained the Gibbs-Duhem equation subject to I will say 0 is equal to you know we got this that way Ni sum over I Ni D Mu I subject to the Gibbs-Duhem equation so these five equations are the fundamental basis of all mixture thermodynamics although you can play the same game with AU or H you can play the equivalent game with all those equations this is the simplest because T and P appear explicitly here and for chemical engineers T and P are the most natural variables so what I have to do out of all this mess is to derive see first of all let me say this having by writing this equation down I am saying all the chemical potentials together the changes in the chemical potential are constrained by one equation and then I have this equation if between my claim is that between these equations I have essentially one independent chemical potential which I cannot whose composition dependence I cannot describe if I fix that I can solve for all the others so I will show that explicitly what I am going to do is derive a set of partial differential equations for the chemical potential that will give you a method of solving for the chemical potential all chemical potentials except one and one will be completely arbitrary and then you actually call on some philosophical ideas in physics you say you cannot model Mu1 why did you choose one why did you choose component 2 why did you choose component 3 so these questions will arise so we will find a common property which is the free energy change of mixing ?g of mixing what we will do is take the pure components form the mixture in a thought experiment and ask what is the change in free energy and if I will show that if I can model that change I can get all the chemical potential so that is what I am going to do in the next probably half an hour or 45 minutes let me start here all I am going to do is I want to simplify this equation to simplify this equation the aim is to get I will say to solve for Mu I because it is an intensive variable it can only be a function of TPX1 through XR-1 there are only R-1 independent composition variables so I should be able to solve for the chemical potential as a function of these in fact if I give you that rest of thermodynamics is trivial it is because you do not know this that you have to make so many extensive measurements in thermodynamics and go through complicated manoeuvres to predict phase equilibria and reaction equilibria so to do this what I do is to go back here and substitute for D Mu I if Mu I is this D Mu I is going to be partial of Mu I with respect to T DT plus partial of Mu I with respect to P D P plus sum over K Mu I with respect to XK TXK where I have written D Mu I with respect to XK this is actually the partial in this equation TP held constant plus XL any arbitrary composition which is not equal to see I am differentiating here Mu I with respect to XK X1 through XR-1 XK cannot be held constant but since mole fractions are not independent I also have to allow one more to vary which is XR I have written up to XR-1 so here L should not be equal to K or R it is this is change in Mu I with respect to change in XK mole fraction of K holding TP in all mole fractions except XK and XR I would not write this explicitly it is sort of understood in the context all the time come back again and again all I do is put this back in here remember that this is Mu I with respect to T at constant P and all mole fractions that means it is the same as what we got here or Mu I this can be written as right now I would not bother whether it is SI bar or this is the original equation or this I can write it in either form this equation I use Mu I with respect to T is holding all the mole numbers constant and P constant it is equivalent to that in a mixture except you are talking of one mole of the mixture so this term is the same as minus SI it is still minus SI bar DT plus VI bar DP this is also the same plus this as it is now you have to recall and write this in general in general this is what we showed in general for any extensive property DE you can write DE is equal to partial of V with respect to T DT plus partial of V with respect to P DP plus sum over EI bar DNI this is going to be a large amount of manipulation but the final form of the equation is so nice it makes the whole process worthwhile where this is really this is what philosophers call an anthropocentric point of view you are stuck with your notions of equality even if it is not realized in real life you are stuck with your ideal notions of equality impartiality and so on so I cannot model one chemical potential Mu I there will be a hue and gray so I have to model a common property of the mixture and then derive values for Mu I Mu II Mu III etc sort of we are stuck with that mode maybe it is easier to model Mu I and be done with it in each mixture you can choose one component for which you can guess its physical behavior and do it but in when deriving a general theory use these principles all the time so I have this equation you know by integrating what you will get integrating assume this is at constant T and P you will integrate exactly like we did before EI bar is an intensive variable so using the assumption or hypothesis or actually using the fact that EI bar is intensive I told you the big difference is between intensive and extensive properties if it is an intensive property if I change only the mole numbers but keep the composition the same EI bar will not change I will get E is equal to sum over Ni EI bar and then by differentiating this or this is the important result you get a Gibbs-Duhem equation for every property EI bar by differentiating this and comparing with this in particular the most useful Gibbs-Duhem equation is this equation here 5 is a particular case of that result and this is what we use repeatedly and therefore this is generally this is referred to as the Gibbs-Duhem equation but you have a whole series of Gibbs-Duhem equation but this result is important whenever I get Ni times EI bar I simply get E back so if you look at this result here if I substitute this D Mu I here if I put Si bar here I get Ni Si bar sum over Ni Si bar is S Vi bar if I put in here Ni Vi bar is V so I will get back SDT and VDP that cancels with the right hand side so what I have left is this here so let me write it out so we will substitute this is 6 this equation is 6 substitute 6 in 5 sum over the left hand side becomes minus SDT and use let us use this we will call this 7 and use 7 you get minus SDT plus VDP plus the last term there which is sum over K coincidentally this sum over K is not over all mole fractions R is not included because the variables are X1 to XR minus 1 so you must write here K normally this is denoted by a prime on the summation that means one mole fraction does not come in the summation usually the last mole fraction by convention so I will put a sum here prime or I will write explicitly K is equal to 1 to R minus 1 and then I have this summation anyway summation over I equals 1 to R of Ni times doh mu I by doh XK DXK this is equal to minus SDT plus VDP right hand side I am copying as it is the left hand side if I substitute this I get minus SDT if I substitute this and some I get minus plus VDP in the last term gives me a double summation so the result simply says sum over E sum over K equals 1 to R minus 1 sum over I equals 1 to R and pull out a total number of moles XI here that is if I divide this by the total number of moles I will get XI but first the argument here is the DXKs are independent thermodynamics demands that this should be 0 the left hand side should be 0 the left hand side should be 0 under all conditions since DXK are arbitrary I can change these mole fractions because this goes only up to R minus 1 I can change any of the mole fractions exactly as I please and this should still be 0 therefore this implies that sum over I equals 1 to R of XI partial of mu I with respect to XK is equal to 0 and you are pulling out the case so for all K for all K equal to 1 to R minus 1 when you derive a theory it has to be proof against all perversities so if you give me any holes I will choose a mixture where I change the mole fraction of only one component and then if you told me this was not 0 for that component then you will get a contradiction here essentially you will get as contradiction of the two laws because you can trace back all the way and go back to the laws from which you derived these. So this is the Gibbs-Johem equation now it looks a little nicer it looks as if all the chemical potentials are treated equally but even this is not good enough notice that the Gibbs-Johem equations now explicitly appear as R minus 1 equations for every K you can write that equation so it is a set of R minus 1 equations which was not obvious from the original Gibbs-Johem equation right this is actually a set of R minus 1 equations for R unknowns µ1 µ2 through µR so you have R unknowns namely µ1 µ2 etc µR and R minus 1 equations because you have what I it is 8 yeah you have equation 8 for K equals 1 to R minus 1 it has to be valid for every K so you have 1 degree of freedom there is no way you can uniquely solve that set of equations if you assume one of the chemical potentials then you have R minus 1 equations for R minus 1 unknowns and again I am assuming given a set of equations you have solutions to it proves they will goes through a painful proof showing that does in fact but having said this I want to set this in a slightly different context let me go back and look at the process of mixing all this trouble arose because of the process of mixing so if I examine that process I can show you that there is one way of getting everything from a combination of chemical potentials that does not specify any one chemical potential uniquely if I look at the process of mixing and I look at G after mixing and G before mixing like AD and BC after mixing and before mixing so G I am looking at delta G for this process delta G is actually G after mixing minus this definition G before mixing I cannot measure delta G but I can measure delta H of mixing delta V of mixing the change in enthalpy change in volume those are measurable quantities but let me derive this here an expression for this this difference is simply sum over I XI Mu I minus Mu I pure if you look at G before mixing all the components were pure components I take X1 moles of 1 X2 moles of 2 etc so I have XI Mu I pure I sum it up I get the free energy before mixing but because Mu I is a partial model property after mixing it simply XI Mu I you can write this for any extensive property you will have to write EI bar and small EI similarly this is GI bar minus Mu I but GI bar has a nomenclature now we call it Mu I symbol okay so then let me write DG again now let me do this sorry let me write D of delta G partial of delta G with respect to say X1 let me do this for a binary and then we will do this for the more complicated it is very straight forward but if I do it for the binary it will be easier let me first do this for a binary I will show you the results and then we will go on to the mixtures let us take a binary mixture so delta G is simply X1 Mu I minus Mu I pure plus X2 into Mu II minus Mu II pure then I want to ask what is partial of delta G with respect to X1 this is done at constant T and P in the case of a binary mixture I can hold only T and P constant as I have to allow X1 to vary so I cannot hold X2 constant if I differentiate this if I differentiate this I get Mu I minus Mu I pure if I differentiate X2 I get a minus sign then I have to differentiate what is inside plus X1 this is a function of composition this is not so I get delta Mu I by delta X1 plus X2 delta Mu II by delta X1 now notice that this term is identically 0 according to the Gibbs-Johm equation the Gibbs-Johm equation there tells you for a binary in a by in the case of binary K can be only one so you get X1 partial of Mu I with respect to X1 plus Mu II with respect to X1 the denominator variable is the same in the Gibbs-Johm equation so you will get X1 XK the same in the denominator numerator will run through all the indices because this is 0 I can now look at these two and solve for one of these for example Mu I minus Mu I pure if I want to solve for Mu I minus Mu I pure all I do is multiply this equation by X2 and add give these numbers again we will call this 9 this is 10 and this is 11 so Mu I minus Mu I pure I will simply say solving 10 and 11 this is simply equal to delta G plus X2 into delta partial of delta G with respect to X1 if I switch the indices 1 and 2 I get the equation for 2 so what this tells you is that I can get the chemical potentials 1 and 2 from delta G itself you can do this in a mixture in a multi-component mixture clearly because you will get this R minus 1 equations and this one you simply solve for the differences Mu I minus Mu II Mu II Mu I minus Mu I pure or Mu II minus Mu II pure and so on but essentially now solving the Gibbs-Johm equations you cannot solve them uniquely so we will simply say the Gibbs-Johm equations represented by 8 have a unique solution they have one degree of freedom regardless of the number of components in fact if I do this in the general case let me just write this down now I take equation 9 differentiate equation 9 with respect to whenever you make these differentiation you choose index other than the dummy index here so with respect to XK then you will get partial of delta G with respect to XK is equal to sum over I XI delta XI with respect to delta XK you hold doing this holding TP XR is not a variable in the problem you are differentiating with respect to XK so you have to allow XK to change so XK XR are not held constant all others are held constant here so you are doing this differentiation here XI with respect to XK holding again XL not equal to K R times Mu I minus Mu I pure plus sum over XI partial of Mu I with respect to XK because Mu I pure is not a function of composition I differentiate this term first and then I differentiate this term and differentiate that time I get luckily the exact thing that I have in the left hand side of the Gibbs Duhem equation so this term is 0 I will copy this here then I can erase that also okay now XI with respect to XK because all others are held constant when XI is equal I is equal to K you will get 1 when I is equal to R this summation is over all components and I is equal to R you will get R XR is 1 minus X1 X2 etc XK so you will get a minus 1 that is the only difference so partial of delta G with respect to XK has only two terms 1 is Mu I minus Mu I pure no sorry Mu K minus Mu K pure when I is equal to K you will get a contribution and then when I is equal to R you will get a minus contribution Mu R minus Mu R pure it is called this equation 12. So if you look at equation 9 and equation 12 this is actually a set of equations this is R minus 1 equations because you have to write one such equation for each value of K other than R so if K equal to 1 K equal to 2 up to R minus 1 so you have R minus 1 equations and I have one additional equation that I have introduced here so I have R equations for R unknowns so you can solve for say Mu K minus Mu K pure I will simply say Mu K for Mu K minus for all K given delta G as a function of TP and X1 through XR minus 1 I have to give you this function if I tell you what delta G is as a function of composition if I give you this as a function of composition then by solving these equations simultaneously I can get every one of the chemical potentials what thermodynamics tells you is that this is not unique so when you say modeling in thermodynamics it invariably boils down to guessing a function F and you can guess any function F you will produce your own functions every one of you can have a model in thermodynamics except that nobody else will use your model unless the results you derive from that model agree with experimental data for a large class of systems no point having a model for each individual systems that means you are only measuring data or you are describing it but if you have the many classes of systems that you can describe with a one function I can have a model for delta G which describes all ideal mixtures never another model that describes all alcohol water and polar non-polar mixtures and so on. So if you can find a model for a class of systems then you are in business so essentially the task of thermodynamics is to guess a delta G the delta G should have some boundary conditions you cannot get a delta G function that is non-zero when you have only a pure substance that is when I do not have a mixing process it should go to zero so when all the mole fractions go to zero except one which is one then delta G should be zero so there is some boundary conditions you have to satisfy for example for a binary mixture you write delta G as X1 X2 times a function arbitrary function that function cannot be infinite it has to be a bounded function that is all. If you write X1 X2 in front automatically when X1 goes to one or X2 goes to one it will go to zero so you are safe so they use some games and then you can say you can use Weierstrass's theorem and say what can f of X1 be whatever it is I can approximate it by a polynomial so I will write a polynomial and you can write a whole series of models like that that is called the Wohl's model Wohl introduced this polynomial expansion so you can get several models initially people derived models in different ways the most interesting model though was the very first one which was quite exciting which was derived by Van Laar, Van Laar was a student of Van der Waals it is one of the most beautiful papers in mixture thermodynamics after that people discovered that you could write arbitrary functions from real analysis they said after all what can this function be it can be a polynomial and so on. So there are different forms of this and what I will do is go through and start with Van Laar's incidentally there are two this whole all these theories the thing that I have derived and the delta G expressions that you get are all all come under what are called physical theories of solutions these come under theories of liquid mixtures they assume that if you know the components in any given mixture in 1890s there was another guy called Dolzaleck in this guy said actually you guys are all mistaken first of all I will show you a concept here of an ideal mixture you can show what is the simplest solution to the Gibbs-Johem equation and that is called the ideal mixture he said Dolzaleck said all mixtures are ideal you know you behave non you observe non-ideal behavior because you are a fool and you are looking at the wrong components actually in a mixture of 1 and 2 the real substance say A and B the real substance may be A 22 B 1 or it may be and A 12 B 13 you do not know they may be they associate that way and if you looked at the right components all mixtures are ideal that is if you look at it as a mixture of A 22 B and A 12 B 13 it is an ideal mixture but if you look at it as a mixture of A and B as some foolish human beings too and this is from Dolzaleck but let me first look at I think the simplest thing is to first start with ideal solutions the word ideal is used in this context as the simplest solution to the Gibbs-Johem equation and this is done by inspection I have this Gibbs-Johem this is sum over 1 into R J is 1 to R – 1 this is the Gibbs-Johem equation simplest solution I will write down a solution I will assume that mu I depends only on the composition of I you can treat it as a pure guess or you can go back and apply everything to an ideal gas mixture and derive this result as well it is possible to do it both ways but as far as I am concerned the Gibbs-Johem equations always have 1 degree of freedom and I can make a guess is just you treat this as an intelligent guess and then we look at the thermodynamic consequences I will show you that this corresponds in fact to your exact notion of an ideal mixture an ideal mixture is one in which it consists of molecules that do not have interactions between them so there should be no energy change when you mix things ideally and I will show you that this result actually agrees with that and I will also derive a result for the entropy of mixing which will be the same as what you would get if you went back to an ideal I will show you but we will come back here just substitute this here if you substitute this here you are going to differentiate with respect to xj again holding all mole numbers all mole fractions constant except j and k right j and r because this one the differentiation is j equals 1 to r minus 1 and you are substituting this here you get xi log xi with respect to xj in the only two variables that can change are i equal to j and i equal to r so you will essentially get when you substitute this here to what was this equation this equation is 8 substitute into 8 you get xj partial of log xj with respect to xj plus xr partial of log xr with respect to xj that is all on the left hand side all others are held constant this is equal to this is 1 by xj into xj which is 1 this is 1 by xr into xr into minus 1 because xr is 1 minus x1 minus x2 etc up to xj so 1 minus 1 so this is identically 0 all I am saying is I have got a set here that satisfies that equation you can actually guess the solutions to this equation by hook or crook I do not care how you get the solution as long as you have a solution it is a valid thermodynamic model for as far as I am concerned so valid thermodynamic model for chemical potentials from this you then derive results in phase equilibria you derive results in reaction equilibria and if it does not agree with any experiment your model will be thrown out the test of the model is simply agreement with experiment finally otherwise it is as far as thermodynamics is concerned supremely indifferent anything that satisfies this set of equations is a valid model in fact you may be able to find for every model you write you may be perverse enough to write all kinds of complicated models involving transcendental functions but you will find one system that satisfies that model you can always find a system but that is no meaning at all I mean what you have to do eventually is to give some physical meaning to the model your parameters in the model you have to relate it to something and so on so this equation this is called an ideal mixture let me show you the consequences of this we have mu i by t with respect to t is h i bar by t squared this is thermodynamics take this model if you differentiate mu i by t I mean if you different first divide by t and then differentiate you get mu i pure by t on this side it is mu i by t right from the model I get mu i pure by t by t and then if I next term is 0 because if I divide by t and then differentiate our log x i with respect to t is 0 because I am differentiating with respect to t holding p and all of all mole fractions constant but this is again from thermodynamics this is thermodynamics here hi by t squared small hi is specific enthalpy of pure I so it says hi bar is equal to hi that means there is no enthalpy change for each for any component per mole when you do a mixture this means delta h is equal to 0 because delta h mixing is h after mixing minus h before mixing h after mixing is x i hi bar some and before mixing it is a hi x i times small hi this you know is ideal behavior this is what you understand by ideal behavior that we know interaction energies if there were interaction energies or equal interaction energies between all pairs right this out between between all pairs of molecules this happens for example in homologous series you mix methane and ethane really no intermolecular force difference between a molecule of ethane a molecule of methane so such mixtures often have delta h equal to 0 no energy changes occur in so the word ideal therefore etymologically is acceptable it describes exactly what you would think of in a mixture of substances that had no intermolecular forces stop there it is only 40 seconds what I do is continue next class and complete the ideal mixture in what I will do is to introduce chemical potential you will write mu i is equal to mu i pure plus rtln ? i x i an activity coefficient the reason engineers need an activity coefficient is because you need a default option it will always be systems about which you are ignorant so in your computer program you must be able to say if I tell you nothing ? i equal to 1 whereas for chemical potentials I did not have a default option but here in the you have to say at the worst case I can I guess ? i equal to 1 or if you do an iterative calculation you will start with ? i equal to 1 so in that sense it is better to use a variable that occurs that tells you the departure from ideality then use a variable that is