 Okay we get started I am going to discuss ideal solutions actually one topic I left out is equations of state I will discuss that couple of classes on the line because we have gone on to liquid mixtures and also discuss ideal gaseous mixtures and do equations of state later in the case of ideal gas gaseous mixtures there is one redeeming feature that helps you solve the Gibbs-Duhem equation using experimental data the way this happens is you have ?mu I by ?p is equal to Vi bar is an exact expression in the case of ideal mixtures ideal gaseous mixtures first of all for ideal this is ideal mixing if the process of mixing is ideal there is no volume change on mixing because ideally molecules do not occupy volume so there is no change in volume and ideally molecules do not interact with one another so there is no enthalpy change so these are the two basic definitions of ideal mixing if that is true this is then if it is an ideal gas that is the pressure is sufficiently low Vi is simply RT by P so you get ?i ?i is equal to if you integrate this this is done at constant pressure and composition and temperature and composition so you will get some function of temperature and composition plus RT Lnp in particular you have ?i pure is some function of temperature plus RT Lnp to find this function of composition you separate out ?i and RT Lnp you have f of t and xi you know you have to solve the Gibbs-Duhem equation now let us put down the I will come back to this in a minute but let me put down the Gibbs-Duhem equation we have already seen that this is simply xi partial of ?i with respect to xj we said the simplest solution is ?i varying as log xi right varying as because one left side is energy per mole the right hand side is dimensionless so you have to do some corrections. So I can define this time an ideal solution I will show you that I can also come from another route from the gaseous behavior and come back to the same conclusion first of all define an ideal solution by writing ?i ideal is equal to ?i if you okay I will just write or I will write this as function of function of t in fact I will make it this function of t ?p this is the definition of an ideal solution now clearly as xi goes to 1 f should go to ?i this ideal refers only to the mixture I mean only to this mixing process so in the limit as xi goes to 1 you do not have mixing left hand side will become ?i pure so f of t has to be equal to ?i pure whatever the value of the chemical potential at that temperature and pressure is so I have ?i ideal this will give you all the results that you need to define an ideal solution because ?i ideal by p this side is simply Vi bar ideal on the right hand side I simply get Vi then hi bar ideal by T squared or minus of this is partial of ?i ideal by T with respect to T this is rigorous thermodynamics but if I do the right hand side this comes from the right hand side on the right hand side if I divide by T and differentiate with respect to T at constant pressure and composition I simply get ?i pure with respect to T ?i pure by T with respect to T which is hi sorry minus hi by T squared so you get hi bar is equal to hi Vi bar is equal to Vi which is the definition of ideal mixing ideal mixing says mixing without change in volume and without change in energy and I can go further because of this if I have an ideal gas then I have Mu i is equal to this plus RT Lnp I am sorry for a pure substance again this let me put this in here so if I have Mu i ideal for an ideal gas mixture for a mixture of ideal gases I have to write two lines there two words there for an ideal mixture of ideal gases the ideal mixing process can be in the liquid phase can be in the gas phase but I am taking the special case of ideal mixture of ideal gases for Mu i pure for an ideal gas I got this result all of this is for this is for ideal gas and because I have substituted ideal gas here ideal mix of ideal gases now I get Mu i ideal is equal to some function of T plus RT Lnp xi you have to give an interpretation for f of t simply to say what it represents when xi goes to 1 and p goes to 1 this term goes to 0 when xi goes to 1 you get pure substance pure i so pure i at p is equal to 1 at the p is equal to 1 is sufficiently low pressure we usually measure p in bars so if you use those units then when p is equal to 1 bar you get f of t is simply Mu i the chemical potential of pure i at the temperature of the mixture at pressure equal to unity so we write this as Mu i is equal to Mu i this is Mu i ideal Mu i 0 of t plus RT Lnp xi also from this equation let me give these equations numbers this is 1 and vi bar equal to vi is for ideal mixing this is for ideal gas this is 3 so this will be 4 it will be 5 this is of course one number every day 6 7 then 8 this simply proves that this is hence 8 satisfies conditions of ideal mixing conditions of ideal mixing are simply delta v mix is equal to 0 delta h mix is equal to 0 and then finally this year 8 let us say this is 9 first of all from equations 1 to 3 I have delta Mu i actual minus Mu i ideal by delta p this when I write ideal here it only means ideal mixing on Mu i okay where I have detailed mixing of ideal gases I will write it explicitly but normally when I write Mu i ideal I only mean ideal mixing that is the gases need not be individually ideal but I am assuming the process of mixing is ideal this can happen at very high pressures between say ethane methane mixtures similar substances will mix without change in volume without change in so if I take this difference this is vi bar the first one if I differentiate this I get vi Mu i minus Mu i ideal is therefore equal to I am integrating from 0 to p of vi bar minus vi dp plus some constant now I can determine the constant very simply in the case of gas phase because in the limit as p goes to 0 all gases behave ideally therefore this Mu i also becomes a mixing of ideal gases therefore I have Mu i minus Mu i ideal is 0 at p is equal to 0 so the constant is 0 this is because this is 0 I will write here since all gases behave ideally as p goes to 0 have this integral 0 to p of vi bar minus vi dp I write this as integral 0 to p or minus I am adding and subtracting RT by p inside the integral and splitting the integral into two parts this is defined as these three lines is just a definition right this is RT ln of vi minus RT ln of vi pure when vi bar is replaced by small vi this becomes the specific volume of the pure substance so I recognize this by writing vi pure incidentally Smith and Nannas has a slightly different notation for vi he writes vi cap for this he writes vi so when he writes vi he is talking about pure fugacity the mixture value fugacity coefficient is called fugacity coefficient I will come back to it in a minute vi is called vi cap is the value in the mixture I am too used to this notation to change if I change half way through I will make a mistake again I will stick to this notation so vi for us is vi cap our notation Smith and Nannas our notation is vi pure and this is vi similarly mi cap in Smith and Nannas is chemical potential in the mixture and mi is the chemical potential of pure I for n the reason we give these symbols is because you need default options in chemical engineering you can put vi equal to 1 as an approximation whereas you would know what the default option for chemical potential was so we have not really done any significant at this stage we have not made progress in terms of defining the composition dependence of the chemical potential but we have expressed it in terms of a quantity for example in the gas phase this is completely measurable vi bar is measurable so if I know the composition dependence of the chemical potential if I know the composition dependence of this molar partial molar volume that is all I am saying so if I know one I can get the other I cannot make this connection in the liquid phase because liquids do not have an asymptotic behavior either at a limit of temperature or a limit of pressure because all gases behave ideally at P is equal to 0 this integration constant vanished for you you would not have such a facility in the liquid phase so in the liquid phase you are stuck with solving this equation directly here if you have experimental data on vi bar you have a solution of the Gibbs-Joum equation because the Gibbs-Joum equation simply tells you then that mu i minus mu i ideal is this and mu i ideal is given by this so you got your whole this thing worked out so let me get back and write this equations mu i pure in particular okay let me write this out mu i pure is equal to mu i 0 sorry this is mu i 0 I can just simply write mu i 0 because mu i 0 that I have the notation I have used there is the chemical potential of pure i at the temperature in question and pressure equal to 1 plus this would have been an ideal case times vi pure because if I did the same thing for mu i for the pure case mu i pure minus mu i ideal pure would have been simply vi minus RT by P maybe I should write it here see it follows that mu i pure minus mu i ideal simply equal to 0 to P because it is pure vi bar becomes simply vi the second case because it is ideal it becomes way a minus RT dp this is mu i ideal pure in mu i ideal pure we had already that is mu i 0 you would have mu i ideal pure would have been mu i 0 plus RT L and P X i or RT L and P if I multiply by there is no X i here P Vi pure mu i 0 plus RT L and P would be the value for mu i pure ideal I add on a correction for non-ideality which is Vi pure then I have mu i in the actual mixture is equal to mu i ideal mixture plus RT L and P X i into phi i I have actually mu i is equal to mu i 0 my ideal mixture is already there P X i is there so this is mu i 0 7 to come back these are the unique models for the gas phase phi i is given in terms of vi bar minus RT by P dp in phi i pure is simply vi minus RT by P dp for gases this completes the description remember that the Gibbs-Duhem equation is different written for a homogeneous phase when you write DG is equal to minus STT plus V dp you writing for a homogeneous phase so you have to solve this since luckily we have only three phases of aggregation you have to solve it for the solid state the gas state in the liquid state separately so I have done the gas phase now we will do the liquid phase liquid mixtures for liquid mixtures first no asymptotic behavior if you are able to find an asymptotic behavior you are in for a Nobel Prize again but some of these things are not worth searching in the liquid range has been covered people have looked everywhere so you have to solve need to solve Gibbs-Duhem equation what we do is the simplest solution is mu i ideal again I still have possible simplest solution is however you must notice that in liquid you have all kinds of mixtures I can have gas dissolved in liquid so suppose I referred to carbon dioxide in water and I referred to carbon dioxide then as if you go to the temperature of the solution if you go to the temperature and pressure the solution go to the pure state the pure state is a gaseous state it is not realizable in practice that is I cannot take a mixture of composition Xi where Xi represents the composition of carbon dioxide mole fraction of carbon dioxide in the mixture and take that mole fraction continuously to one without change of phase in the Gibbs-Duhem equation is written for a single phase so I have to solve within that phase so what we do is write this as simplest solution for we give name for mixtures in we classify mixtures into solvent mixtures and solvent solute mixtures so this is valid only for solvent solvent mixtures in solvent solvent mixtures you can take the mole fraction of all the components individually to unity without change of phase that is the definition of a solvent solvent mixtures for every component I can go to mole fraction unity and still get preserve the liquid state so this is the simplest solution this is a solution so what you do in practice is right for non-ideal solutions or real solutions as you have Mu i is equal to Mu i Mu i pure at TP again for solvent solvent mixtures only I will treat the other case separately I write Mu i is equal to that where I have to tell you what gamma i is in this case when Xi goes to 1 left hand side becomes equal to this so gamma i is 1 in the pure state so I you write along with this you notice that gamma i goes to 1 as Xi goes to 1 this quantity gamma i x Xi is called ai it is being called activity and it was term coined by Lewis long ago instantly Lewis is credited with being the first expositor of classical thermodynamics because Gibbs needed an interpreter what Gibbs said was absolutely right but very terse Gibbs would say in the footnote obviously and he will say in the footnote if necessary this can be derived and so on then Lewis will do the derivation so Lewis is booked at all the it was like he took notes from the master and filled them up so in ideal solutions the role of non-ideal solutions the role of mole fraction is played by ai activity and the activity the default option in all chemical engineering design programs as I said gamma i equal to 1 but you can go terribly wrong there are mixtures in which gamma i can be 1500 which means Xi being the being a representative of Mu i is meaningless log Xi does not any longer represent but those are exceptions but typical gamma values of 2 1.5 2 3 etc very very common many many mixtures so we will do lots of examples in which these and that will make a lot of difference as I told you finally you are going to use delta G is the work done delta G is Xi Mu i – Xi Mu i for the Mu i pure so that is the change when you do mixing or unmixing you take air separate it into oxygen and nitrogen you can calculate how much work is done in the process I will do that calculation when you do the difference essentially this RTLN gamma log gamma i Xi represents the work in bringing you will show that it is equal to the work in bringing 1 mole of i into a mixture of J safe you have a binary system and since the work depends on log of Xi if it is log gamma i Xi and gamma i turns out to be 2 you will make a mistake of RTLN 2 it can be a large difference that is all so this is your activity coefficient now you have not really done anything only change words because if I substitute this into the Gibbs-Johem equation now let me go back to the Gibbs-Johem equation it will play you till you solve it in some way or other now substitute this quantity give these finished 9 I think last we will call this 10 this can be 11 so substitute 12 into the Gibbs-Johem equation do this by inspection mu I pure not function of XJ at all so all of it will cancel RTLN Xi satisfies the Gibbs-Johem equation so when I do some over i of Xi partial of log Xi with respect to XJ you know we get 0 right for come I J I equal to J you will get 1 for the last component R you will get minus 1 so you will get 0 so only thing that remains is log gamma i and RT is a constant so this will give you some over i partial of log gamma i with respect to XJ is equal to 0 to solve this you go back to the other formalism remember we wanted to look at delta G you model delta G and you get all your equations for mu I instead of mu I now you will get log gamma i so the equations are very simple you get G excess by RT write this for a binary X1 now log gamma 1 plus X2 log gamma 2 reason is G excesses delta G-delta G ideal delta G is G after mixing- G before mixing- G after mixing ideal- G before mixing G excess by definition is delta G-delta G ideal so this is G after mixing- G after mixing ideal G after mixing is sum over Xi mu I these two will cancel before mixing these two will cancel I am looking at the ideal mixing process now Xi mu I- mu I ideal so mu I ideal is mu I pure plus RT ln Xi mu I non ideal is mu I pure plus RT ln gamma I Xi so the only difference is log gamma I so this is equal to RT sum over I so G excess for a binary this in particular will give you X1 log gamma 1 plus X2 log gamma 2 I suppose I should say here since this okay so then do the same thing that I did before differentiate this partial of delta G excess by RT by delta X1 is log gamma 1- log gamma 2 plus in this term is 0 because of the Gibbs Duhem equation I just have to solve these two equations algebraically equation 12 this is 13 this is 14 and if you recall when we did one large equation we really got an expression for G excess not for delta G the models for solid mixtures are identical we do not often use them in chemical engineering we use them in metallurgical engineering because basically between solids and liquids is not really much difference they both represent the condensed phase molecules reasonably closely packed in the density of the solid is not very different from that of the liquid you have a very well defined crystal structure in solids in amorphous solids you do not have a crystal structure but you have a well defined reasonably well defined structure the case of liquids you have a little more entropy so the models for solids solid mixtures are similar but we do not use them often and chemical engineering I told mean solid mixture problems are in metallurgical thermodynamics so primarily we will deal with liquid mixtures we will of course deal with gaseous mixtures but most of the time your gaseous mixtures will behave themselves you will have to calculate fugacity coefficients but you will know the equation of state and you can calculate for this you will have to go through for liquid mixtures I will write the model for G excess in the corresponding log gamma 1 you can get this from tables this is you will have to write this separately for two classes of mixtures and deal with solvent solute mixtures later for solvent solute mixtures the model for the solvent will remain the same but for the solute I will have to write it differently G excess can be 0 this is ideal solution gamma 1 equal to 1 or I will just write gamma 1 log gamma 1 or G excess can be a x1 x2 G excess by RT this is called the this is ideal you will give a name for the model this is called porters equation if you do that solution that I have given in 14 you take this expression differentiate it and so on with a x2 square for log gamma 1 like for log gamma 1 here you will get 0 gamma 1 is 1 log gamma 1 is 0 so you can build models of this kind with a x1 x2 x1 plus b x1 you can write higher polynomial models this is Margules it is called two suffix model it is as I told you you can write this is a12 and this is b12 so it is called two suffix I think he even writes this is a12 and a21 does not matter two parameters exist and you have to solve that I do not remember the solution but you can solve those equations simultaneously and get the values what I will do is put up a table of this instantly there is a very nice personally do not like the book much but it is well written and it is written by an industrial practitioner so it is written typically with a lot of redundancy it is got everything worked out for an open book examination it is ideal because if I tell you the model you do not have to start solving those equations in the exam and make mistakes in solving simultaneous equations because I will never find out whether you know your term you will make a silly mistake there so the best thing is for you to keep these things Wallace as a table I will try and put that table on the web itself he has this model and he has the solution for log gamma and log gamma 2 he has got it for ternary mixtures for any number of multi component mixtures in case you do not know how to simplify the multi component mixture equation for a binary he will separately give you binary tables also so is got all that it is called I think chemical phase equilibrium I think it is just called phase equilibrium I do not remember exactly but Wallace is a practicing engineer for 15-20 years okay so as I said once you have models for the chemical potential the various models in you also have to go through the model for Van Laar which is an expression for RT by G excess the only thing you have to be careful about is that G excess has to go to 0 in the limits otherwise you can propose your own model I mean but if you have a phase equilibrium if you have vapor liquid equilibrium then you will get Mu I liquid is equal to Mu I vapor this you already shown now the whole purpose of all this exercise was to express Mu I liquid in terms of measurable quantities and we have shown that everything is measurable except the composition dependence is a chemical potential so this alone you have to do modeling so with the model you get it in terms of composition again so what you do is when you solve this phase equilibrium problem for Mu I liquid I have to go through a little more so here you express in terms of P and composition and you will get one term here that represents and Mu I liquid pure because your model is Mu I is equal to Mu I pure plus RT L and gamma I xi when you write it for the liquid phase this is Mu I liquid pure this will come out similarly in terms of me this will come out in terms of Mu I 0 always will come out in terms of TT what we will do here after for composition in the gas phase we use Y for liquid phase we use X that seems to be a convention composition Xi so the only thing that is that still gives you a little difficulty is this Mu I 0 that appears in the gas phase equations in the Mu I liquid pure that appears in the liquid phase equations so you have to relate Mu I liquid to Mu I vapor and do that and then I will go back and do solutes in solute solvent mixtures if you go back to your pure component phase diagram P versus H this is T is equal to TC so you have liquid plus vapor that yeah so if you are talking of liquid pure you are talking of substance below the critical point it has to be that is the way the critical point fine so the way you connect the two is to recognize that at this point the chemical potential of the liquid is the same as the liquid chemical potential of the vapor so if I want to cancel this Mu I 0 I have to express Mu I liquid in terms of Mu I gas if I do that then Mu I gas will have the Mu I 0 in it which will cancel on both sides so having introduced a Mu I 0 which is a hypothetical quantity which is the chemical potential of pure I at the temperature and pressure equal to 1 I do not care what it is it will cancel on both sides so I will simply write the equations down and cancel it so to do this all I do is to take at any temperature T and let us say TP I write Mu I liquid pure this is what I want I know from thermodynamics that this is equal to VI so I integrate this between this point where I can make a connection with the gas along that temperature this is the saturation pressure so I integrate this I get Mu I liquid pure TP minus Mu I liquid pure at T and P saturation equal to integral VI liquid DP from P saturation to P but this is the same as Mu I vapor pure and TP saturation because the pure substance in the pure substance at this saturation pressure and temperature the vapor and liquid are at equilibrium because this is given in terms this is simply Mu I 0 of T plus RT ln PI saturation PI saturation this is referring back to the way I treated the once I have got Mu I 0 on both sides of the equation I can cancel it off I do not have to worry what it is I do not even care if it goes to minus infinity because it is the identical quantity on both sides so to complete the picture let me go ahead and do solvent solute mixtures for the solvent the model is the same Mu I is Mu I pure liquid TP this is a complete specification because gamma I will go to one as XI goes to one for solute mixtures I still have log XI as an ideal solution and simply write Mu I reference but for the solute the physical state that I can realize is between XI equal to 0 and XI equal to saturation value which carbon dioxide dissolved in water the saturation value at room temperature and normal pressure would be something like 10 to the power minus 2 whole fraction whatever the value so XI can vary XI is physically realizable or the physically realizable range of XI is 0 to XI saturation whatever the solubility limit is now there are two constants here one is this Mu I reference in this gamma H is a function of composition I had only one unknown to begin with so I must be able to specify one number uniquely so I will have to choose an XI reference at which I choose gamma I to be one in the case of pure substances I could go all the way to mole fraction one I could use the pure substance as a reference so here I will simply say since XI is going to 0 is realizable I said gamma I equal to 1 you will have this conceptual difficulty that as gamma I goes to 1 XI goes to 0 you get log of 0 so you get minus infinity there but we would not worry about it what I am going to do in thermodynamics we always cheat you on this actually it is very valid it is not to the thing what you do is get an animal that diverges but the same animal on both sides of the equation then you can cancel it so that is one way of doing it the other things in simple way is although this quantity this is an expression as a function of XI so I can divide this by T and differentiate with respect to XI differentiate with respect to T or P if I differentiate if I divide by T and differentiate with respect to T I will get a quantity here which represents minus enthalpy of I bar partial molar enthalpy by T square on this side I will get some reference enthalpy this term will vanish because I am differentiating with respect to temperature so if I do that in the if I this will vanish in the limit as XI goes to 0 if I go to very dilute solutions so what I will do is although I do not know this animal I know its derivative with respect to T or its derivative with respect to P and I will show you in treating phase equilibrium I do not need to know Mu I reference I only need to know its derivatives.