 I am going to discuss phase equilibrium again, I am going to discuss the chemical potential of the solute, once I have a model for the chemical potential of the solute I can discuss topics like solubility of solids in liquids, solids in gases, solubility of liquids in liquids and gases in liquids, these are the areas of interest, direct interest. Basically use the chemical potential you know it has to satisfy the Gibbs-Duhem equation, this is the chemical potential in any phase has to satisfy this equation. So if you are talking of a liquid solution, liquid solution consists of solvents and solutes, it may not in particular it may be all solvents in which case for the solvent case we have the nice model, this is within quotes in our context the solvent is defined as a substance which does not change phase if you change the mole fraction to unity without changing the temperature and the pressure, all solvents are essentially components that remain in liquid phase at tp and xi and say components i such that xi is equal to 1, these are components that change phase that gas or solid in pure state at tp of solution. So for solvents we have µi is equal to µi liquid pure at tnp of the solution plus RTln ?i xi, ?i goes to 1 as xi goes to 1, this is automatic because of the way you define this quantity the left hand side when xi and ?i or 1 this term is 0, so this has to be equal to this which is the correct limit, so if you have a solute you have a difficulty here for solutes we still write the same kind of model, this is a function of tnp alone and leave a double dot there I will explain what I mean, it is not a function of composition but it actually has to do with the fact that it depends on the environment I will explain that in a minute. The physically realizable state without change of phase is not xi going to 1, if I have for example carbon dioxide dissolved in water at room temperature as I go to xi going to 1 for carbon dioxide will become a gas, so I can only realize the physical limit of xi going to 0, so the reference point for me I have to therefore what I assume is that xi as xi goes to 1, 0, ?i goes to 1 that is I have given myself an additional degree of freedom I define the ?i star and ?i because I want ?i star to be independent of composition that is when this formulation is useful then I need to fix ?i as equal to unity at some point this point can be either xi going to 1 or xi going to 0, I choose the limit as xi goes to 0 the infinitely dilute solution of the solute in the solvent in that limit I choose ?i equal to 1 that defines ?i star at this stage ?i star is still hypothetical it may not be an experimentally accessible quantity but I will show you how I can get values for it or how I can get rid of it in the equations. So what I do is to make this limit but as a xi goes to 0 this logarithmic function will go to –infinity you have problem of log xi going to –infinity I will circumvent that problem by the following method first of all I look at let us look at individual cases let us look at gas liquid equilibria or gas solubility liquids this is the crux of the fish that you get when you drink your coke you when you drink your coke there is carbon dioxide dissolved in it your temperature you have to drink it cold the temperature grows up and the solubility decreases so the gas comes out you get a fish you know that is the only reason you are able to get that fish is because gas solubility is decreased as temperature increases okay let me get let me equate chemical potential of the let us say solubility let me take example CO2 in water we will call CO2 as component 1 water as component 2 so you have µ1 liquid phase is equal to µ1 vapor phase or gas phase similarly I have µ2 liquid phase is equal to µ2 vapor phase I write G and V to distinguish the fact that one is below the critical temperature the one is above the critical temperature okay this we know how to do because this vapor phase is simply the vapor liquid equilibrium problem this is µ2 liquid µ2 0 plus RTL and PY1 P1 PY2 P2 this is simply your pointing correction in a µ2 0 plus RTLN of the saturation pressure the fugacity coefficient at saturation pointing correction ? 2 X2 so this you have done already so this is the equation I have to worry about even this is known this is simply µ10 plus RTLN PY1 times P1 and P1 of course satisfies the Gibbs equation log P1 satisfies the Gibbs equation but you get it from V1 bar which satisfies the Gibbs equation anyway which is experimentally measured so this part is well known now this part you have this equation here let me write this out I will write µ1 star plus RTLN ?1 X1 I am looking at LN PY1 P1 P1 P1 P1 P1 P1 X1 okay this is equal to µ1 star minus µ10 by RT usual trick in thermodynamics if you do not know a quantity try and produce the same quantity on both sides of the equation which is what we did so far in vapor liquid equilibrium we cancelled µ10 on both sides you had words for it but you did not know the number so you produce µ10 here for for mu 2 for example you produce mu 2 0 on both sides cancelled it so we did not have to talk about it now you have to get rid of this mu 1 star the other thumb rule in thermodynamics is whenever you do not know a free energy divided by T and differentiate with respect to T okay let me do that so you have and also this number will this quantity will keep appearing all the time P y 1 P 1 by ? 1 x 1 let us give it a name h 1 would write h 1 2 because it will turn out see as far as this is concerned I am going to the limit as x 1 goes to 0 and x 1 goes to 0 this can only represent a quantity that is characteristic of the solvent of the medium in which one is dissolved in this case if I have a binary of 1 and 2 then this should depend on the nature of 2 not on composition but on the nature of 2 this quantity is called Henry's law constant normally this phi and ? are omitted because you are looking at dilute solutions in the limit as x 1 goes to 0 I will come to that x 1 goes to 0 ? 1 will go to 1 and normally the pressures that you are talking about are low enough for ideal behavior so this ratio is called the Henry's law constant is just given a name because Henry's suggested at some point that y 1 is proportional to x 1 y 1 is equal to h 2 1 x 1 or P y 1 the partial pressure is h 1 2 x 1 Henry's law constant so Henry's law constant will if I have experimental data for it then I have got a correlation between vapor phase compensation and the liquid phase compensation I have delta log of h 1 2 by delta t as I told you if I differentiate with respect to t mu by t will give you a h by t squared mu 1 0 is pure so you will get minus write this as h 1 bar star minus h 1 0 by RT square this is pure substance it is as an ideal gas so this is simply h 1 h 1 if you like in the vapor phase or in the gas phase mu 1 0 is the chemical potential of pure 1 at temperature t and pressure equal to 1 atmosphere so mu by t for the pure substance will give me h by t squared it is h for the ideal gas for component 1 as an ideal gas at temperature t and pressure equal to 1 because enthalpy is independent of pressure I do not have to write that 0 there for the ideal gas it is simply h 1 gas it is simply the enthalpy of pure 1 this is symbolic when you differentiate mu 1 you will get h 1 bar I put h 1 bar star to indicate that since the same state as the mu 1 that I have there now look at this quantity the whole idea of writing an equation like this is I have all the composition dependence on the left hand side the right hand side is independent of composition therefore the left hand side remains the same if I go to the limit as x 1 goes to 0 right so this is equal to the left hand side in the limit as x 1 goes to 0 and as x 1 goes to 0 gamma 1 x 1 x 1 go to 0 y 1 will also go to 0 I do not have to worry gamma 1 is 1 this is equal to this because because the way I have separated I have written the model for the chemical potential is to dump all the composition dependence in gamma here and in p here other than this x 1 and y 1 the rest of the composition dependence is dumped in gamma 1 and p 1 these are reference quantities the only thing I know about them this I know the only thing I know about this animal is that it is independent of composition because it is independent of composition if I rearrange the equation this quantity becomes independent of composition therefore it is the same this ratio remains the same in the limit as x 1 goes to 0 in the limit as x 1 goes to 0 gamma 1 is 1 that I have the liberty to do because that is what defines me 1 star then y 1 will also go to 0 you do not have to worry about x 1 going to 0 because y 1 by x 1 will remain finite then you have pressure and you have p 1 so h 1 2 is actually Henry's law concepts is usually defined in books as p y 1 by x 1 it is correctly p y 1 p 1 by x 1 p 1 is normally small it is close to 1 so I have this in the limit as x 1 goes to 0 this is written as h 1 bar infinity minus h 1 gas by RT square this h 1 bar infinity is the partial molar enthalpy of component 1 in a solution in the given solvent at infinite dilution so the infinity indicates infinite dilution it is actually a measurable quantity I will show you why it is measurable in a minute actually what you do is measure enthalpy of solution and from the enthalpy of solution you can get h 1 bar and h 2 this is actually been experimentally measured long ago Henry's law concepts have been reported for a long time except you do not actually measure it infinite dilution you measure it in low dilution in an extrapolate to 0 having done this only subject in which you can do this is now integrate this and you get a slightly different set if you integrate this you get log h 1 2 is equal to integral of plus a constant if you like because now you have an interpretation you had mu 1 star by T which you did not know but when you differentiate it you got h 1 bar by T square which you know so you can integrate this only difference is that this integral you can do this integration actually from experimental data on h 1 bar let me also write down the other expression that is differentiation with respect to pressure since this quantity can only depend on temperature and pressure I have to worry only about temperature and pressure dependence of course it depends on the nature of the solvent it will vary from solvent to solvent this is why you will get v 1 bar infinity minus v 1 by RT differentiation of mu with respect to P will give me V or v 1 bar mu 1 will give you v 1 bar this again is measurable and what is more if I integrate this over P I can go to 0 pressure and put down another constant here we will discuss the constants later it does not matter that means I need to access experimentally the Henry's law constant at some sufficiently low pressure if I do it at low pressure then P is 1 so what I do is measure P y 1 for carbon dioxide and plot it against x 1 typically the curve will go like this it will start at 0 0 of course and it will go off vertically because as you increase the pressure at a given temperature you reach the saturation limit this is the x 1 saturation or solubility at temperature T this is typically the experimentally observed curve as you increase the pressure the solubility will increase and then it will stop at a certain value since I said I can this is at given temperature at another temperature may be slightly different this is at T 1 this is at T 2 so this will be at T 1 this will be x 1 saturation in T 2 is less than T 1 I mean this is an experimental observation that the case of gases it also follows from here if you look at the Henry's law constant here this is enthalpy in the solution this is enthalpy in the vapor phase so this enthalpy is always greater than this this number is of the order of the latent heat it is not exactly it is actually normally a little less than the latent heat but it is a positive number so d by dt is negative so the Henry's law constant will decrease as temperature increases this h 1 2 will decrease as T increases if h 1 2 decreases P y 1 decreases the same so what you do is actually you operate at some temperature that is lower you will actually be sorry at some you are not working at saturation you work at some temperature like this at some composition liquid composition like this and the solubility you will find the vapor pressure here of carbon dioxide above the solution at temperature T 2 is this at T 1 is this so you get a difference in the carbon dioxide in the vapor phase that is why you get this in the solution so having said this let me now write out what the chemical potential is therefore mu 1 is equal to mu 1 liquid is equal to mu 1 star plus RTLN gamma 1 x 1 or I write P y 1 P 1 these two are equivalent gamma 1 x 1 it is necessary for you to measure the Henry's law constant Henry's law constant for gases in liquids for example is usually very large but at very high partial pressures you will get very low solubility typical solubility carbon dioxide in water resistant to the power minus 3 solubility for most of the gases very low in terms of mole fraction now let us look at so this is how what you do is then as far as solid vapor liquid equilibria are concerned involving solvents and solutes for solutes will get an equation like this for solvent component this is for solute like components and this is solvents I mean you can have any number of components you can just rewrite these equations anything is H 1 2 if you have many many solvents the 2 will actually represent the solvent environment it may be a mixture of solvents so you have to get Henry's law constant in that particular solvent or mixture of solvents so I can change this one to I and 2 to J I representing all solutes J representing all solvents in which case you will write this as H I S S indicating the solvent in which solvent mixture in which the solute is the Henry's law constant is measured for the solute okay this is one part of it the other thing is solubility of solids in liquids in the game is played exactly the same way so I have now again solvent like substances and solute like substances now let us take example benzoic acid in water it is a classical system in which many many measurements have been done very convenient to measure benzoic acid does not ionize so readily like an ACL if you have highly ionizing solutes the treatment is slightly different it is not different at this stage but it is different because what counts in solution is the number of molecules present and if it ionizes or the ions present sodium and chlorine for example will be treated as two different ions is a sometimes called colligative properties their properties that depend on the number of species in solution number of actual identities whereas benzoic acid will be just benzoic acid it does not ionize so you can treat it as one solute in water if you had an ACL it will be like a three component system effects will be NA plus CL minus and water but let me come back to this let us say benzoic acid is one and water is two for water it is the same equation so we know how to treat this is just a solvent here you are going to get µ2 is equal to µ2 star plus RTLN ? 2 X2 this should be equal to under phase equilibrium conditions if you are looking at solubility of solids you are looking at µ2 solid pure and phase equilibrium demands that in both phases the chemical potential should be the same the solid phase in this case is pure solid this is µ2 solid pure should be equal to µ2 liquid which is equal to this now you do the same thing you ask simply what is ? 2 X2 this is µ2 solid pure – µ2 star by RT squared by RT if you differentiate this will give you H2 solid this will become H2 star and reverse the two in order to take care of the sign I did not reverse it here only because the quantity H1 bar infinity I mean or H1 gas – H1 bar infinity is the positive point here this quantity is positive this is like a liquid enthalpy this is a solid enthalpy this is H2 solid so it is like the latent heat of melting of the order of magnitude of latent heat of melting so let me write it as ? log ? 2 X2 by ? T is equal to H2 bar infinity I am going to go to the limit this side is independent of composition so this side both sides should remain independent when X2 goes to 0 so take limit as X2 goes to 0 it H2 bar infinity – H2 solid by RT squared this is actually your ? H of solution I will come back to that I am going to discuss heats of mixing heats of solution so on in the measurement then you can integrate this and get information you do not do differentiation with respect to P when it comes to the condensed phase because pressure the condensed phase behavior is insensitive to the changes in pressure is insensitive so essentially you lose a degree of freedom when you are dealing with the condensed phase pressure is not a significant variable for the engineer unless you use millions of pounds pressure I mean unless you are talking of enormous changes in pressure we do not deal with normal such changes in pressure normally so in practice you will find effectively one degree of freedom is lost that pressure effectively should be countered out as a variable that does not control anything in the condensed phase so as far as the condensed phase concern this is the equation you will use you will have to make these measurements and then do the integration but let me get back and show you measurements so we have seen that these measurements are important measurements of ? H and calculation of H1 bar ? H of mixing may be solution or in all cases the is simply H after mixing – H before mixing this is this is after mixing for mixing is H1 – H2 small H1 is of course by our notation the enthalpy of the pure substance specific enthalpy of the pure incidentally I forgot to mention this if you look at sufficiently dilute solutions ? 2 is 1 because x 2 goes to 0 ? 2 is 1 so in dilute solutions if this quantity is reasonably independent of temperature I will say two things first ? 2 goes to 1 secondly H2 bar infinity – H2 is approximately constant or approximately independent of temperature so I can rewrite I can integrate this equation and write log x 2 is equal to – by RT plus a constant which means if you plot the solubility of 2 in 1 against 1 by T you should get a straight line but anyway this is a very well known result log x 2 against 1 by T and many physical chemists have actually calculated H2 bar infinity – H2 S from this so data on H2 bar infinity are reported from such calculations so you need to know the solubility at room temperature and then you need to know H2 bar infinity H2 S is usually known so from that you can calculate solubility at any other temperature there is another extreme case of this in if the solution remained ideal if solution is ideal if the mixing process is ideal then H2 bar infinity is the enthalpy of 2 when dissolved in 1 that is in the liquid state if you go to the limit as x 2 goes to 1 and treat this as still being in the liquid phase it is a hypothetical now because you are talking of benzoic acid in water suppose benzoic acid had remained a pure liquid at that temperature then H2 bar infinity would be actually H2 liquid which means the right hand side is actually the latent heat of melting of component 2 that means log x 2 ideal the ideal denotes ideal solution is – L2 of melting by RT plus a constant this was by integrating with respect to temperature I know one condition that the temperature equal to T melting of 2 of component 2 x 2 has to be equal to 1 so I can write log x 2 ideal or ideal solubility is equal to L2 melting by R 1 by Tm – 1 by T the most significant aspect of this result is that the ideal solubility of a solute 2 in a in any solvent is independent of the solvent and means you must have the same solubility in all solvents this has been observed for example for naphthalene in a series of hydro carbons you take propane butane etc you take actually we should start with butane you have to start with even higher molecular weight so that is in the liquid state or go to sufficiently low temperature you take a whole series of hydrocarbon series the homologous series and look at the solubility of say something like naphthalene in that if you take the mole fraction solubility it is independent of the this has been observed it is true for many systems in what actually what Hildebrand and Scott do say that x 2 actual is simply what you do is x 2 ideal is actually gamma 2 x 2 to actual x 2 ideal by gamma 2 so you can have a theory for G excess the excess free energy calculate gamma from that and calculate x 2 ideal divided by gamma 2 fact the whole book on by Hildebrand and Scott on solubility of non electrolytes has a large volume of data that shows you that you can do this calculation very well you calculate the ideal solubility using this formula that is fixed for a solute so you can take a solute and tell me at any temperature what the solubility is then you calculate gamma 2 for that substance from excess free energy information you divide x 2 ideal by gamma 2 you will get the actual it becomes independent of composition therefore the right hand side has to be independent of composition therefore I can go to the limit of x 2 going to 0 then I know this animal as h 2 bar infinity as the partial molal enthalpy at infinite dilution what I will do is to show you that I can actually measure these quantities by measuring heats of mixing I will discuss different units for measurement of for writing the models for chemical potential we will use the molality units for example and we will also talk about polymer solutions separately in polymer solutions when you say dilute solutions you mean parts per million you do not mean mole fractions of 0.21 I do not think those are the only two cases I have discussed the rest of it is just applications so I will stop there.