 I am going to continue talking about the give you an equation I think for quite some time and I am going to talk about we have two equations I had µ1- µ1 pure so binary systems actually the equation itself I do not know if I wrote it in the final form yeah I wrote this we wrote this to get solutions of this form where ?g is any function of composition such that of course f of 0 is equal to f of 1 equal to 0 because when x1 is 0 or x1 is 1 you have got the pure state you do not have a mixing process at all so this is automatically a condition. Now we were discussing the Van Laar equations the Van Laar theory what Van Laar said we started with this this is what I want but this is what I can do luckily thermodynamics tells me if I go around this way I get the same answer if I do it as I if I do it from a to b so this is pressure p this is equal to 0 p equal to constant then this is pure liquids x1 x2 this is mixture liquid mixture of composition x1 this is gas mixture of composition x1 this is pure gases we made some assumptions we talked about excess properties so you have ?g- ?g ideal is equal to is defined as g excess instead we said u excess is equal to or the other way around of this this is set to 0 this is an assumption I told you it is not a bad assumption in the case of molecules of similar size this is also set to 0 this is generally a good assumption when I say sound assumption I mean you can verify using experimental data that volume changes on mixing are generally very small there is no ideal change in volume there is no actual change in volume either so ?v- ?v ideal is actually 0 both are 0 so this is generally a very good assumption except in rare cases in fact you will be hard put to find counter examples where it is significant there are some case this set to 0 I will say this is not in the same class as the thing not bad for mixtures of similar molecular sizes when I say similar you can go up to 5 times roughly it does not matter one molecule can be 5 times as big as the other and still you would not have significant as ?s other than the ideal entropy change in mixing so I am having said that he simply wanted to find you we started with this equation do you where do you is equal to Cv dt plus you is calculating change in you at constant temperature in all the processes so to calculate ?u from one point to the other he said ?u was simply integral of T ?p by ?t-Tdv from p is equal to p to p is equal to 0 this is ?u for a b now this is where the boss stepped in van der Waals he said all fluids obey the van der Waals equation van der Waals equation is p plus a by b squared to b-b is equal to RT because the physical arguments for van der Waals very simple simply said there is a volume occupied by molecules b pb is equal to RT was already known he said the actual volume is the volume minus the volume occupied by the molecules which is b and that van der Waals already predicted that b would be of the order of the specific volume of the liquid because after condensation it says that is the space occupied by the molecules in the balance space is available as for a ideal gas the ideal gas consists of point particles plus free space this is not point particles so he said this is he expected this to be of the order of b liquid and generally densities of liquids are constant so when he said b liquid he took it at the triple point but he could have taken it almost at any condition the argument here was that pressure is change of moment exchange of momentum with the walls it is a rate of change of momentum so when a molecule collides and comes back you take the change in momentum divided by the time between collisions and then you will get an estimate of the momentum exchange you can get the pressure this is in an ideal gas in actual gas he said that the forces are primarily attractive the number of molecules striking the wall is proportional to the density which is proportional to the 1 by the specific volume the number of molecules pulling these molecules away from the wall and therefore reducing the momentum is also proportional to the number density which is also proportional to 1 by v so effectively he said the effect of attractive forces between molecules is proportional to 1 by v squared depending on the molecule you have different a values different strengths so he argued that a by v squared is the reduction in pressure caused by attractive forces so if we had treated the system as an ideal gas the ideal gas would have actually exerted a pressure equal to p plus a by v squared that is this would have been absent in the ideal gas so this is effectively the ideal pressure this is effectively the ideal volume in p v is equal to RT so Van der Waal said you have to use this equation and this equation because it has attractive forces will predict for you condensation will actually do that it can predict you go on compressing the gas below its critical temperature it will condense at some point at some point thermodynamics will tell you what the chemical potential vapor is chemical potential liquid is and you can calculate this in fact one assignment I will give you is to prepare the Van der Waals chart the HP chart for Van der Waals fluid if you do that once then you know practically how the others are but let us look at this because if I do not have these attractive forces you can produce condensation so you would not get and I do the chart here you will get only this region for an ideal gas there will be no condensation at all if you want condensation you have to introduce attractive forces this v-b will still won't give you condensation in fact p into v-b equal to RT is sometimes called a hard sphere model but let me get back here I have got the Van der Waals equation has to be used so if I calculate p so you get p is equal to RT by v-a by v squared so T doh p by doh team is simply RT by v which assumes that a is an absolute constant it is not a function of temperature improvements on the Van der Waals fluid actually show that a is dependent on temperature and therefore you have to use it is rewritten in terms of another constant times a function of temperature it is the redlich kister equation there is some slight changes but there is a very successful equation of state called the redlich kong equation of state not redlich kister when I went as graduate student to US at that time Berkeley had redlich teaching the course teaching thermodynamics apparently it started with the redlich kong equation of state and ended with the redlich kong equation of state he said there is no other fluid so everything was done for the redlich kong equation of state so you get this so this difference is actually a by v squared T doh p by doh t-p is actually plus a by v squared so you get this implies that delta u for a Van der Waals fluid is actually a by v squared dv so it will give you minus a by v if I switch the thing you get a by v at p is equal to 0 v is infinity so you will simply get a by v this I am writing this from p is equal to 0 to p p with a minus sign and p is equal to p the volume is v that is given by this equation you have to solve the cubic and find the correct root but a by v so a has the interpretation of essentially the latent heat of operation delta u is actually the internal energy change when this pure substance goes from liquid to vapor right so this is actually the it is the enthalpy of vaporization is the latent heat but enthalpy internal energy change are approximately equal because p delta v is negligibly small so this has an estimate of the latent heat so a can be calculated as density times the latent heat this v is v liquid because you are talking of at pressure p we started with the liquid and v liquid according to Van der Waals is b is the parameter b in his equations so that approximately equal to a by b so he is got delta u a b is equal to u xs a b because delta u if this was an ideal gas delta u change is 0 because u is a function only of t you are doing the whole process at constant temperature so delta u a b is the same as u xs because delta u a b minus delta u ideal would have been this delta u ideal is 0 in fact the for the whole process delta u ideal is 0 so this is equal to this and it is equal to a by b it is actually it is not a by b exactly it is x1 moles of 1 which you are evaporate which you are evaporating separately per mole it is a11 by b that equation is written per mole so this is delta u is the specific internal energy change per mole for the pure substance I have to write x1 and x1 moles of 1 b1 is the parameter for pure 1 a11 is the parameter because it represents interaction between two molecules you have to write two subscription plus of course x2 moles times a22 by b and then delta u bc is equal to u xs bc this is mixing of ideal gases there is no change in energy when you mix ideal gases to form a mixture is no interaction energy to begin with so there is no change in interaction and then delta u cd is equal to u xs cd instead of this you are now doing condensation so you will get a minus sign right the integral will go from p is equal to 0 to p is equal to p so you will get a minus sign get minus a mix by b mix so I have g xs for the process is equal to u xs for the process this is a d this is equal to u xs for a b plus u xs for bc plus u xs for cb so you get g xs is equal to x1 a11 by b1 plus x2 a22 by b2 minus a mix by b mix the concepts in this equation are basically valid but this particular final result is restricted to the van der Waals fluid so if you have a more complicated equation of state you can use that and do the same calculations not many people have done that and you can do that and get other expressions for g xs yeah because it represents attractive forces in van der Waals recognize that if you mix there will be an a12 a11 is between 1 and 1 a22 is characteristic of the pure fluid 2 it represents interaction between two molecules a12 is if you had one molecule and one two molecule the interaction between of course all these will be multiplied by Avogadro number because you are talking of per mole whereas your interactions are actually between molecules no you can put RT by V minus B if I made that mistake even if you put V minus B as long as B is constant and you differentiate with respect to T it will disappear if I wrote V it is a mistake it is V minus B in fact incidentally when I am on this topic let me say this so you get this this is P is equal to RT by V minus B minus a by V square so this is what he called P hard sphere and then this is P or ? P due to attraction due to attractive forces that is what van der Waals argued much later now we have an exact molecular theory of hard sphere fluids you have a different expression for P hard sphere it is different from the van der Waals expression and you can actually derive an exact theory for hard sphere fluids and you can verify this because now you can do computer experiments you can do Monte Carlo calculations and just describe that briefly because a very interesting set of calculations P hard sphere now we have an exact expression so you can rewrite this equation as P hard sphere exact plus the same thing minus a by V square this equation has been this is also it is called long a Higgins we dome equation of state after three people who wrote it down they borrowed the P hard sphere exact from molecular theory it is been fairly successful theory there is van der Waals as many limitations qualitatively it is always right but in terms of agreement with experiment as a large number of limitations in the long a Higgins we dome equation of state is much better that means van der Waals was able to essentially quantify the effect of attractive forces quite well mistake he made was in the hard sphere estimate you cannot call it a mistake in 1870 that was well ahead of his times so let me get back here so what it what this tells you is that if you have an equation of state you will anyway have to have parameters that are functions of composition because when you write van der Waals equation write P is equal to RT by V minus B minus a by V square it is written for a homogeneous fluid so A and B have to be functions of composition so what the van der Waals suggested and when van der Waals law at least at that time it was law hard spheres HSS hard spheres if you take a mix the probability of in a mixture the probability of finding a 1 1 pair is proportional to x 1 squared so you get x 1 squared a 1 1 probability of finding a 1 2 pair and a 2 1 pair which are both characterized by an interaction 1 2 is 2 x 1 x 2 plus x 2 squared a 2 2 so here he says assume that the mixture was a random mixture even to date molecular models only ask what is the departure from randomness so you can get mixing rules that are slightly different from this you must remember although I called this a theory basically you have 1 degree of freedom there is going to be always guessing at one point what he does is uses a equation of state and finally gets it in terms of these and you have to put in expressions for these these are called mixing rules and till date they are all empirical in classical thermodynamics you simply put in a mixing rule derive results compare with experiment if it agrees then you accept the mixing rule there are hundreds of mixing rules as a whole paper on mixing rules in 1960s I do not think a review has been written after that because there are many new mixing rules I mean you should realize that these coefficients of these coefficients are such that x 1 squared plus 2 x 1 x 2 plus x 2 squared is 1 it is always true all these equations will be of that form then there is one more a 1 2 see I can get a 1 1 and a 2 2 b 1 and b 2 by fitting data for pure substances that is I take this equation of state for pure one I write p is equal to RT by v minus b 1 minus a 1 1 by v squared then I collect a lot of experimental data on PVT and fit the data to the curve so I will get the best values of b 1 and a 1 1 one simple way of getting b 1 is simply to take the density at the triple point triple point is invariant so you can go to that triple point measure the density of the liquid and you will get the value of b it is one way of doing then you can fit a by v squared by taking the data it is one parameter fit is trivial so you get these so I can get a 1 1 and a 2 2 from experiment a 1 2 is the hypothetical fluid in which all the interactions of the R of the form 1 2 that you can get because in any mixture you will have 1 1 interactions 2 2 interactions and 1 2 interactions I cannot isolate 1 2 alone so this is invariably produced from experimental data on mixtures knowing the pure parameters pure substance parameters the other way is to suggest what a 1 2 would be it has to be a mean between 1 1 and 2 2 and we are not very imaginative we only know the arithmetic mean and the geometric mean so you use one of them actually Van der Waals suggested the geometric mean if you like I will write this as K times in view of what has happened afterwards this K 1 2 is approximately 1 but if it if you put 0.98 and if you put 1 you will get a huge difference in calculated the modern properties so subsequently prowess nets and co-workers have actually determined K 1 2 for a very large number of mixtures in the in some limiting cases where the forces the dispersion forces are the dominant the many intermolecular forces and even now our knowledge of intermolecular forces is not complete but for pure dispersion forces you can show that this kind of result is exact so it is exact in a certain limit but as far as Van der Waals is concerned this combining rule was completely empirical this is called come combining rule so one way of getting excess free energies is through the equation of state approach this whole thing is called the equation of state approach because you have a fluid described by an equation of state the equation of state should describe both the gaseous state and the liquid state ideally that is true we still do not have a single equation of state that correctly describes both the liquid in the way but qualitatively the Van der Waals fluid does many many equations of state will predict the correct qualitative shape but it is still difficult to get the exact but every equation of state has parameters that are empirical which are composition dependent according to what are called the mixing rules these can be derived from theory to some extent in special cases in other cases you have to approximate them there is right now we do not have a way of deriving this from theory I mean we have a limiting case limiting cases where we can derive this from theory but no other cases we do not know the rules so what we do is to use usually the geometric approximation is used with a constant in front empirical constant you fit that date fit that to the data but if I write this as a 1 1 a 2 2 then a mix is clearly equal to x 1 if I put that in there it is clearly x 1 square root of a 1 1 plus x 2 square root of a 2 2 so I can now write what g x s is x 1 a 1 1 by b 1 minus a mix I will just expand this as well get 1 by b 1 x 1 plus b 2 x 2 into b 1 b 2 this one will become this b 1 you should write b 2 you see if I can get the algebraic correct okay this is b 2 into b 1 x 1 plus b 2 x 2 into a 1 1 into x 1 that is one term plus b 1 into b 1 x 1 plus b 2 x 2 into a 2 2 into x 2 minus b 1 b 2 into x 1 squared a 1 1 so you can see the cancellations now this b 1 b 2 x 1 squared a 1 1 this term will cancel with this term and similarly this term b 1 b 2 x 2 the second term will cancel with this term what I have left is g x s is equal to there is an x 1 x 2 here x 1 x 2 here x 1 x 2 everywhere so I will pull out x 1 x 2 a 1 1 b 2 squared plus a 2 2 b 1 squared minus 2 b 1 b 2 square root of a 1 1 a 2 2 this is what x 1 x 2 into square root of a 1 1 b 2 minus square root of a 2 2 b 1 the whole square what I am going to do is multiply this by b 1 b 2 and divide by b 1 b 2 so I already have a b 1 b 2 here so it will become b 1 squared b 2 squared if I take it in here it will become b 1 b 2 inside the square sign so this will become a 1 1 by b 1 minus a 2 2 by b 2 the whole square then of course I have the b 1 x 1 plus b 2 x 2 this is the expression that Van Laar finally got see square root of a 1 1 by b 1 minus square root of a 2 2 by b 2 is a constant so what he is saying is g x s is equal to x 1 x 2 by b 1 x 1 plus b 2 x 2 or if you take g x s by x 1 x 2 and take it is reciprocal this is linear in x 1 so if you plot if you actually measure ?g and get g x s you have to subtract of ?g ideal in plot it against x 1 if you get a straight line then you know and the Van Laar equation will fit this data very well it was very important when I was a student this straight line business now it is not you guys use the computer just hit the damn thing on the head it will tell you whether it is a good fit or not it will tell you what the standard deviation is earlier we used to do this visually you have to go on plotting this data and show that is a straight line it is still graphically is a good this thing but there are some significant things that Van der Waals said which is why Van Laar also point of Van Laar pointed out and became therefore he was accepted very widely I do not know have you done principle of corresponding states have you shown these constructs you have done that a bit so you can show that the constant a is proportional to PC VC squared in B is proportional to VC so this quantity a by B square root of a by B is proportional to PC for a Van der Waals fluid is proportional to PC so Van Laar said if the critical pressures are the same the fluids will mix ideally because excess free energy will be 0 see if a by B for 1 is equal to square root of a by B for 1 is equal to square root of a by B for 2 this will be true if critical pressures are equal so he said all fluids is approximately same critical pressure will mix ideally in the limited data they had at that time showed that that was in fact true so it was held at that time as a very successful theory of liquids so this is the final result in Van Laar's theory but nowadays we handle this much more easily we do not do we do not go through equations of state we do not go through mixing rules and combining rules the equation of state approach still remains the same this is the method everywhere you can change things you can change the equation of state from Van der Waals to any other complicated equation of state you like you can take a more modern equation of state substituted there you can change your mixing rules mixing rules simply have to show that in the limits they have to go to the pure component properties but B can be written as any combination that goes to B 1 in the limit as X 1 goes to 1 and B 2 as X 1 goes to 0 similarly for A and then you have to assume a combining rule for A 1 2 if you do all that you will get a theory of mixtures and people have done this repeatedly through simply the ones that survive are the ones that are that agree with experimental data for a large number of systems for each system you cannot have a model the easier way of doing the same thing is you recognize again mu 1 minus mu 1 pure is simply delta G plus X 2 partial of delta G with respect to X 1 so all I need is an expression for delta G what the whole said was what can delta G be in a binary system it has to be X 1 X 2 times some function of X 1 and any function of X 1 according to wirestress can be approximated by a polynomial so he said just use a polynomial for example you write this as X 1 X 2 into A plus B X 1 plus C X 1 squared plus etc you want very accurate data you have up to C X 1 to the power n and so on you go on increasing n you want an accuracy you have to arrive at the in the scene at the right time when people already know the result but haven't stated it then quickly stated then your name is in the books everywhere old didn't do anything new people had already been using the polynomial approximate simply wrote a paper I do not know 30s or 40s I do not remember when it is not an important paper but somehow he managed to get credit for it so everybody uses walls expansions and when too many terms are required some wise fellow had the idea that instead of doing this you make X 1 X 2 by see this is delta G by X 1 X 2 is equal to f of X 1 so if you write this as some other polynomial of X 1 this is equal to A plus B A prime if you like plus B prime X 1 in case this requires too many terms the reciprocal hopefully will require fewer terms now this equation here having a not equal all the other 0 a not equal to 0 and take this one then names because of people who derived various expressions earlier a not equal to 0 that means all others are 0 it is called Forte's equations you get the same equation when a prime is not equal to 0 except a prime will be 1 by a then if a and b are not equal to 0 you get what is called Margules it is called the two suffix equation it is called two suffix because Margules wrote it as a 1 2 you wrote these with subscripts a 1 a 2 and so on so it is called two suffix equation all of these are two suffixes to them it does not matter it is when two are this thing then you can similarly have a b c a b c d etc so you get various all these are called Margules equations because he wrote them down when a prime and b prime are not equal to 0 you get Van Laar's equations the structure is the same because if you look at Van Laar's equations X 1 X 2 by this is B 1 X 1 plus B 2 X 2 so in this case they called Van Laar's equations this would be now compare with Van Laar X 1 X 2 by G is we will call this parameter a just call it a you like a 1 2 so you will get a prime plus b prime X 1 is equal to B 1 X 1 plus B 2 X 2 by B 1 B 2 a 1 2 so what is this is B 1 minus B 2 into X 1 plus B 2 so your a prime is simply B 2 by B 1 or 1 by B 1 a 1 2 and B prime is the coefficient of X 1 except according to Van Laar B 1 B 2 are actually volumes according to Van Laar's right so they are known from pure component properties so this B 1 B 2 all these are known only constant that is unknown is a 1 2 whereas in practice people treat a prime B prime as independent parameters and only if you treat it as independent parameters you get a good fit to the data if you try to use the original Van Laar equation with values for B 1 B 2 from experimental data you do not get a good fit this equation fitted all mixed data even now for alcohol water mixtures probably one of the best correlating equations this works very well porters equations very well for likes of you have a is equal to 0 this one works very well for simple mixtures where molecules are very similar and then the second this thing works a little better if the molecular separation is larger a large number of studies earlier were funded by the petroleum industry is there the ones who had these various hydrocarbons coming out of the distillation column they were similar but they varied in molecular weight from very low to very high you could go all the way from methane molecular weight of what about 15 16 all the way down to tar which could be C 56 if you like you could talk in a very huge molecular weight ranges and if you have these mixtures you could describe them by various Margules equations to ABC not equal to 0 have been used quite widely so this really is the summary the way you do this is simply show that you can get dealt from Delta G an arbitrary model for Delta G you can get all the new ones once you show that all you have to do is guess Delta G there is no restriction on Delta G in thermodynamics except that reasonable restrictions like it has to go to 0 when you go to the pure state it has to be a smooth function you do not go on putting step functions and say Delta G jumps here jumps there it does not do that experiment so you have to have a smooth function that is all in the easiest way to get a smooth function is get a polynomial or it is reciprocal so you do that