 We are going to discuss in the gas phase I will do so you have ln P I suppose I have tried the chemical potential first so the model for the chemical potential is somewhat unique you have different kinds of models in the condensed phase but for the gas phase the model almost always like this where P by convention is in atmospheres because mu I0 is the chemical potential at the temperature T and pressure equal to 1, 1 atmosphere because mu I0 represents 1 atmospheres this P is also an atmospheres actually P by 1 so you will get absurd answers if you use millimetres of mercury for example sometimes the data is given in that form then you will get 760 times the answers that you have to it. So we are doing this calculation Vi bar I told you about equations of state so Vi bar is from experiment or again it is always from experiment or from equation of state in which here again parameters are from experimental data from fit of experimental data. The chemical potentials have to satisfy the Gibbs-Duhem equation but in this case the chemical potential satisfy the Gibbs-Duhem equation if log P I satisfies the Gibbs-Duhem equation because mu I0 is a constant RT log P is a constant RT log Yi will satisfy the Gibbs-Duhem equation anyway right we have shown that is the ideal mixture so the only other term left is RTLN Fi if RTLN Fi satisfies the Gibbs-Duhem equation then you are through RTLN Fi will satisfy the Gibbs-Duhem equation if Vi bar satisfies the Gibbs-Duhem equation the Gibbs-Duhem equation is basically linear equation so I can go term by term and I can move through is the equation is simply sum over Xi partial of mu I with respect to Xj equal to 0 sum over I equals 1 to R this is for all J except that there are only R-1 independent mole fractions. So this is satisfied say GDE is satisfied sum over Xi partial LN Fi this implies Xi partial of Vi bar with respect to Xj is equal to 0 so as far as the gas phase is concerned while the Gibbs-Duhem equation is central to the determination of composition dependence of the chemical potential we would not worry about it because we take Vi bar from experimental data or from equations of state when you take it from equation of state Vi bar again is derived from the expression for V from its definition right it is Vi bar is simply by definition partial with respect to Ni of V at constant T P so the Vi bars will automatically satisfy the Gibbs-Duhem equation in fact when the Gibbs-Duhem equation was first introduced in the universities is it derived by Gibbs very casually it is big monograph 1880s or something took about 1920 before it came into the books 1910 on I so then at that time the experiment researchers were not sure the Gibbs-Duhem equation satisfied so they do experiments extensively calculate this number calculate the right hand side and see if it actually went to 0 so there are many papers in the early up to 30s or even 40s that will say hence the Gibbs-Duhem equation is verified then they got by that time Gibbs had become an icon everybody understood that he was near God as you could get in thermodynamics as we understand it so they said here after the Gibbs-Duhem equation we assume to be correct and if you collect partial experimental data you can complete the data by using the Gibbs-Duhem equation which is what is done very often so the Gibbs-Duhem equation is assumed to hold but here you could get mistakes with Vi bar if your data is wrong if the Gibbs-Duhem equation is not satisfied then now the it is the fault of the data no longer the fault of Gibbs the theory is assumed to be more correct than the experiments you do unless you are extraordinarily careful so as far as the gas phase is concerned we do not use this equation at all we simply go with Vi bar data and calculate this there are hundreds of ways of doing this calculation you can make life simpler by various means let me take for example first of all let me okay let me take the virial equation of state just easy to use it most equations of state are not explicit in the volume it is volume comes out as one of many roots and you have to choose the correct route and then you have to do this differentiation so it is very difficult equations of state will only relate PVT after calculating we have to do a differentiation you can do it numerically trivially but do it analytically is a bit of a nuisance since I am only looking at the concept let me just say the virial equation of state is the following this is also very important Camerling onus said this he said PV by RT can be expanded in a density series he knows it is one for an ideal gas plus B by V plus C by V squared plus etc he said you can write an expansion like this you can show in molecular theory that B corresponds to first of all the first term corresponds to no interaction between molecules the second term corresponds to pair interactions between molecules two at a time and C corresponds to interactions of three molecules at a time without counting the pair interactions that is the total interactions will be the total potential NG will be some over PI this is the external potential suppose you had gravity and counted then you have to take that mg xh into count if you had a box of molecules one molecule is at a higher level than the other so there will be small differences in potential but this is generally negligible unless you have a system with magnetic materials inside and you apply a 500 Tesla magnetic force magnetic field from outside then you can have them dancing then you better take this into count then the molecule on top will have a difference between in properties from the molecule at the bottom next is Phi ij pair interaction some over all pairs some over Phi ijk plus etc this Phi ijk the potential includes up to here all the pair interactions then there is a correction due to triplet interactions the effect of the third molecule on the pair interaction if I have 1 2 3 and these are important if you have electrostatic forces it stops here which is why electrostatics and electrical engineering is so easy it is true that they dazzle you with all kinds of effects but they understand their theory very well because electrostatics stops here all electrical charge-charge interactions do not have a triplet interaction a third charge does not influence the first two charges except in that there are three now that is if you have 1 2 3 you have Phi 1 2 plus Phi 2 3 plus Phi 3 1 you have take all three into count but it stops here in most of chemical engineering and in chemistry we make this assumption that you can stop here this is generally negligible so from here on you say neglected for want of better tools to handle it a slight difference here this is negligible at low densities but at higher densities it is not clear that it is negligible it can make a 10% difference in properties and so on you can go on let me come back here that was shown later the camera link on is actually anticipated this result but he mainly pointed out that each of these terms this these are called virial coefficients and he said these are all only functions of temperature well I should say f of t and composition in a mixture they are not functions of pressure or density if you like f of t as quickly bracket simply means a set of mole fractions y1 y2 etc the argument for that if you take a pure substance for example if you take the limit as v goes to infinity simply goes to 1 when you take v into this is that compressibility factor take z-1 multiply by v and take the limit as v goes to infinity this will give you b right I take z-1 multiply by v the rest of them as at least a v in the denominator so it will all go to 0 so this in the limit as v goes to infinity since all properties in a pure substance are functions of only two variables we are talking one mole of the substance it can only be a function of t you start with everything as a function of t and v and because you are taking the limit as v goes to infinity the property is defined in the limit as v goes to infinity in fact that is how it is determined what you do is plot z-1 z-1 versus 1 by v versus density and take the limit as density goes to 0 then you will get b similarly you can do the same thing next step what you do is take z-1-b by v into v squared and take the limit as v goes to infinity you show that c is a function only of t so the argument the first simplification here is you have a series of coefficients of functions only of temperature they are functions of composition the function of composition again goes back to the same principle that is thermodynamics will not give you the composition dependence you have to guess the composition dependence in this particular case b mix is exactly this sum over i and j because you do not want to count it twice you have to write i less than j no not i less than j sorry I take this back here this is just i sum over i and sum over j because this goes this is a random mixture result but it is exact in the case of the virial equation this can you can only show the exactness in molecular theory simply tells you if you had a random mixture then the probability of finding a i molecule is yi the probability of finding a j molecule is yj the probability of finding an ij pair is yi yj so for example in a binary you will get where b11 b22 are pure results from pure substances in 1 2 is a hypothetical interaction hypothetical second virial coefficient that arises in a system which has only 1 2 interactions in it you cannot realize it by experiment because when you have 1 2 interactions you will have 1 1 and 2 2 with them then you have these are these are the mixing rules this is mixing rules the you have mixing rules these are exact rules you have combinatorial rules or combination rules combination rules will have to tell you what b12 is like now this could be for hard spheres this is an exact result but actually the way these values are derived is through another route through what is called corresponding states so I will come back when I discuss corresponding states I will show you how to do this in practice what you do is to say b12 is equal to this plus ?12 is just a definition in ?12 can be obtained from experimental data by fitting the results you sort of expect that the 1 2 interaction will be intermediate between 1 1 and 2 2 and since we are not very imaginative characters we either use arithmetic mean or geometric mean for temperatures if you want a TC you use an arithmetic mean and for all volumetric properties B is a volumetric property you use the arithmetic mean there are justifications for it you can show in asymptotic conditions that these rules are exact but they are not generally true so let me write that out here so you have for a virial equation you have z is equal to 1 plus b by b so let me use this equation and I am going to approximate this even further by 1 plus b p by rt instead of b I am going to write rt by p this is not a theoretically based equation this is the original virial equation this equation sometimes described as a pressure virial equation pressure explicit virial equation of state but I will use that simply because it is right now it simplifies my life in algebra so if I now calculate v from here is n times v n is the number of modes n into rt by p plus b so I want ?v by ?n i so if I differentiate this n here this n rt by p I will simply get rt by p plus b and then I have to differentiate in what is inside plus or I will write it like this as it is sorry rt by p plus partial with respect to n i of n times b so n times b is equal to because of this rule here called this equation 1 like using equation 1 this is going to be rt by p plus partial with respect to n i I will write this as n squared b by n n squared if I take n squared inside I will get n 1 squared b 1 1 plus 2 n 1 n 2 b 1 2 this is if I am doing do not mind I will change this to 1 I am going to calculate specifically v 1 bar instead of i this will be 1 so I get v 1 bar what I am interested in is v 1 bar minus rt by p I have to differentiate this so if I keep the 1 by n and then differentiate this I will get 2 n 1 b 1 1 because there is an n there it is 2 y 1 b 1 1 then next one I keep n 2 constant remember so I will get again 2 n 1 b 1 2 so plus b 1 2 then the n 2 squared does not contribute then I have to differentiate the n I will get essentially minus 1 by n squared if I get minus n by 1 squared I get back b mix or minus b mix and write out b mix because it may as well cancel terms with this b mix is y 1 squared b 1 1 plus 2 y 1 y 2 b 1 2 plus y 2 squared b 2 okay this should not to simplify this a bit what I will do is so I have I will collect terms together 2 y 1 minus y 1 squared this is 2 y 1 minus this is 1 minus y 2 the whole squared no I think I should rewrite everything in terms of y 2 2 into 1 minus y 2 minus 1 minus y 2 the whole squared times b 1 1 then I have 2 y 1 b 1 2 minus 2 y 1 b 1 2 into y 2 so it is 2 y 1 squared minus 2 y 1 squared plus 2 y 1 squared b 1 2 plus y 2 squared b 2 2 I am going to substitute for this immediately write it 2 into what cancels from here 1 will cancel okay 2 minus 1 is 1 and I have a 2 y 2 here and a 2 y 2 here minus 2 y 2 and a plus 2 y 2 that will cancel minus y 2 squared there is no 2 2 minus 1 1 minus y 2 squared is all you will get into b 1 1 plus y 2 squared b 2 2 plus I will rewrite this keep this as it is into 2 y 1 squared into or y 1 squared this is b 1 1 plus b 2 2 plus 2 delta 1 2 using equation 2 delta 1 2 is b 1 2 minus b 1 1 plus b 2 2 by 2 when you do not have data you just put delta 1 2 equal to 0 you always have to develop these engineering practice you have to have a default option have to have a quantity for which under ignorance under pressure you can either neglect or put equal to 1 because I should have kept this is y 1 itself now let me read it this I get b 1 1 I will tell you what I am really looking at if this was a pure substance this would have been v 1 v 1 minus RT by P would have been exactly b 1 1 so I must separate out b 1 1 I must separate out the pure and then look at what happens and there is a then I have b 1 1 y 1 squared minus y 2 squared this is y 1 minus y 2 into y 1 plus y 2 so y 1 minus y 2 alone I have to keep I do not like it anyway I will keep it choice and say that again here from here I am differentiating this 2 n 1 then 2 n 2 so I made a mistake yeah thanks so this should be 2 y 1 b 1 1 plus 2 y 2 b 1 2 right okay then it does make a difference finally I have got stuck anyway so thanks so 2 into 1 minus y 2 so this is minus 2 y 2 no no what I had written here is correct 1 minus y 2 the whole squared this is alright 2 y 2 into 1 minus y 1 this should have been 2 y 2 squared right I will keep it as it is right now I am simplifying this n sl so this is okay that is all I need b 1 1 minus thanks for bailing me out into b 1 1 must be a minus sign somewhere I am missing a minus sign this b 2 2 is a minus sign here is this is a minus of that so here I get minus y 2 squared into 2 b 1 2 minus b 1 1 minus b 2 2 so essentially this is equal to b 1 1 you know b 1 1 plus b 2 2 minus 2 b 1 2 is ? 1 2 so this is plus ? 1 2 into y 2 squared that is all that is really what I was looking at so you look at log p 1 the integral RT log p 1 is v 1 bar minus RT by p dp from 0 to p this b 1 1 etc all of them are functions of temperature alone so I can integrate directly with respect to pressure p 1 1 p plus ? 1 2 into y 2 squared I am sorry ? y 2 squared into p this is exactly log p 1 pure plus ? 1 2 p into y 2 squared was into RT or log p 1 so notice that p 1 is the same as p 1 pure if ? 1 2 is equal to 0 or effectively if ? v is equal to 0 if the volume change and mixing is 0 then v 1 bar is the same as small v 1 v 2 bar is small same as small v 2 this is called the Lewis and Randall rule part of education is learning jargon you are supposed to know what is the Lewis and Randall rule who is in Randall rule simply means that volume change and mixing is 0 which means instead of v 1 if you want to calculate v 1 in the mixture you need to know only v 1 pure pure component data are adequate for calculation if you simply given the virial equation and no data on b 1 2 again you do the same option anyway so as far as chemical potential of ? 1 is concerned it is ? 1 0 this is gas phase plus RT ln p y 1 into ? 1 this is a few and this is alright okay where log p 1 is simply given by this what we do is define a new quantity because you do not have to carry this ? 1 0 all the time this is a definition f 1 is called the fugacity which is why actually ? 1 was originally introduced as the fugacity coefficient fugacity is corrected pressure in some sense and if you use the same you want 0 as a base in the liquid phase then instead of equating chemical potentials you can equate fugacities and fugacities come out directly in terms of partial pressures and a correction which is calculate which can be calculated from experimental data and for which you now had default options if you do not know anything you put replace fugacities by partial pressures so let me now do the liquid side then we can do vapor liquid equilibrium because all you have to do is calculate the fugacity of a component of a liquid mixture referred to the same you want 0 so I am going to look at liquid phase the liquid phase you do not have any asymptotic behavior you cannot go to v 1 bar v 2 bar and calculate the chemical potentials so you have to use the Gibbs regime equation and you know solve the Gibbs regime equation you have to go to a model model for GXS in the way to model it in some sense in a symmetric form is using the GXS models which we have discussed in class so whenever I give you a liquid mixture I will have to tell you what the model for the GXS is automatically I am going to look at first liquid solvent solvent mixtures so for these cases you remember when we went to the draw this upwards our model is simply mu 1 mu 1 liquid is equal to mu 1 liquid pure plus RTLN ? 1 X 1 where log ? 1 satisfies the Gibbs regime equation that is GXS satisfies the Gibbs or ? 1 satisfies the Gibbs regime equation mu and ? 2 etc similarly log ? 1 satisfy the Gibbs regime equation because this is pure RTLN X 1 satisfies the Gibbs regime equation so is he satisfied but satisfying Gibbs regime equation now simply means assuming a model for GXS that goes to 0 at the right limits any model will do but what will agree with experiment will depend upon how good your physical intuition is having said that I now write mu and liquid pure this is at T and P you are looking at temperature and pressure the solution this is equal to mu and liquid pure T you have to go to a point which you understand the saturation point plus you know partial of mu with respect to P is small V this is rigorous thermodynamics so plus integral V DP from P1 saturation reason you go to P saturation is because at saturation pressure the liquid is in equilibrium with the vapor so I can now go from here I can simply replace L by V and if I know go to V then I can go back to this model I have this model for the vapor pressure vapor phase and if I go to that model I come back at the same mu and 0 which I can then cancel incidentally this is V liquid this whole thing is in the liquid region this is approximately constant so you write this is approximately mu and liquid now instead of liquid I will write it as vapor pure actually I do not have to write pure because if I write T and P saturation it has to be pure as long as I write the saturation pressure here plus V liquid into P – P saturation okay so now I can write mu liquid this quantity I can write in terms of the vapor phase model so I write this as mu10 plus RTLN P1 saturation I am not assuming ideal behavior so I have to get for the pure substance it will be mu1 is equal to mu10 plus RTLN Pf1 pure right y1 is 1 when you get Pf1 pure I write P is P1 saturation because that is the pressure at which I am evaluating it P1 is P1 saturation plus of course this term this is V1 liquid into P – P1 saturation and as soon as you get a mu10 you can define the rest as mu10 plus RTLN f1 this is f1 liquid this is f1 vapor because you have you started with mu1 liquid is equal to mu1 vapor this is the same as f1 liquid equal to f1 vapor remember you are looking at the system at constant T and P so mu10 is a function only of T so if mu this is equal to that mu10 will cancel because the same mu10 then RTLN f1 temperature is the same so LN f1 is the same as same in both phases so f1 liquid so we can now rewrite what exactly f1 liquid is f1 liquid by this comparison here is P1 P1 this side is P1 saturation P1 saturation and then there is to I forgot that ?1 x1 somewhere this is only this part sorry see mu1 liquid is equal to all this plus I must write log ?1 x1 I forgot that or this is mu1 liquid pure so I have to add a log ?1 x1 that is I wanted mu1 liquid mu1 liquid is mu1 liquid pure plus this mu1 liquid pure is this so if you like f1 liquid pure is f1 liquid is f1 liquid pure times ?1 x1 so I have f1 vapor is P y1 times P1 and f1 liquid is f1 liquid pure times ?1 x1 this is valid only for the solvent components only the solvent component I mean that component whose mole fraction can go to unity without change of phase at the same Tnp so this is your this is equal to P y1 times P1 but half the time you lose only this equation for vapor liquid equilibria this is known for a pure substance this is non-ideality correction for the pure substance this is effectively an increase in the saturation pressure when the total pressure is greater than the saturation pressure when there are two components the pressure above the liquid at equilibrium need not is not equal to the saturation pressure but it is different from the saturation pressure it is equal to it there is this difference multiplied by v1 liquid by RT will give you the increase in effective increase in saturation pressure it was first pointed out by a guy not punning for punning sake who was called pointing so it is called pointing effect it is called the pointing except he had a Y in his spelling pointing effect on vapor pressure he said most people measure vapor pressures at room temperature they measure the boiling point we are actually measuring it at atmospheric pressure not at saturation pressure because the easiest measurement is at atmospheric pressure just boil the liquid deeper thermometer measure the temperature that the pressure is one that is the normal boiling point what you get is actually an increase in the saturation pressure if there is a second component present for example if water had something dissolved in it then the actual saturation pressure that you measure would be this saturation pressure times an increase because of what is called the pointing effect. So you have a pointing effect and then you have a non-ideal you have an ideal liquid mixture and then a correction for non-ideality on the right hand side you have partial pressure correction for non-ideal so I will stop there that is the basic equation then you just have to hit that equation to death through the course so many systems so many kinds of behavior so many models for gamma 1 in so many ways of calculating phi 1.