 We are doing vapor liquid equilibria, this is the fugacity of the vapor phase is equal to fugacity of the liquid component I which is equal to this is the saturation pressure this is a pointing correction then there is non-ideality correction at saturation up to this is pure this is the fugacity of pure I so looking at each of the terms how one would compute each of these looking at the fugacity coefficients is fairly straight forward this is by definition integral 0 to P bar is partial of respect to N9 of V I think we start we discussed example of the virial equation of state so V is approximately equal to discuss binary and we had delta 12 this is definition default option is delta 12 equal to 0 so this is called the mixing rule for the virial equation of state it is an exact rule can be you cannot derive this mixing rule or the combinatorial rule this is the combined combining rule so you need two rules empirically is regardless of the phase the composition dependence of the chemical potential is not given uniquely by thermodynamics so what you are saying here is instead of the composition dependence of the chemical potential I can use the composition dependence of the partial molar volume to calculate the behavior of the chemical potential but thermodynamics does not give you anything about the partial molar volume either it gives you one equation less than the number of variables so you still have one degree of freedom for which you have to use empiricism when you use the empirical route through the equation of state you have a B mix for which if you have an equation of state that is derived from molecular theory that is derived from mechanistic description then you can get an exact mixing rule otherwise the mixing rules is also empirical the combining rule is always empirical except in limiting cases for example B12 is equal to B11 plus B22 by 2 is a limiting case where if you have take very similar compounds of similar molecular size between methane ethane ethane propane etc you can set ?12 equal to 0 otherwise this is not an exact rule this as far as diameters are concerned that is if you take this diameters the molecule ?12 equal to ?11 plus ?22 by 2 is an exact rule in for hard spheres if you use that rule B will be like a volume therefore B will be like ?3 so ?312 will be equal to ?311 plus ?322 by 2 that would be an exact rule for hard spheres but not for real molecules so the various ways of deriving this but all of them are empirical ultimately only in limiting cases some of these rules are exact so this combining rule is empirical this is this can be exact depending on the kind of equation of state you use as of now for most equations of state both are empirical but if you do this then you can calculate Vi and therefore get the complex independence of the chemical potential then you can solve this equation you still have only one term which is ? I and this comes from we have already seen from G excess by RT expressions these are also empirical so as far as theory is concerned thermodynamics runs out of steam here both this this is also empirical and what is was essentially gives tells you that you will run out of steam forever I mean there is nobody classical thermodynamics does not give you an opportunity to derive expressions for G excess from without see by classical thermodynamics I mean thermodynamics that does not refer to structure of matter so if you go back to molecular theory you can derive expressions for G excess by RT you have to make other assumptions in all of these there is still one empirical assumption that will be at the base of all theories of solution for example even one large equations essentially I got this through a series of assumption again you have to make these assumptions for A and B in one large equation so this is the root let me start with a simple example we will ask an example that has no application at all this is a tree typically now I want let us assume that unlike madras this soil is saturated with water what I mean is the roots are covered with water I want to know how tall this tree will grow in absolutely no direct use but the idea is that thermodynamics can be used for this purpose the assumptions I make in all these assumptions perfectly valid there is a certain partial pressure in the atmosphere take madras for example your saturation the such you can calculate the value of partial pressure of H2O in air simply from the vapor pressure of water at 25 degrees let us say the temperature is 25 degrees at 25 degrees C you can find out what the saturation pressure of water is right now my contention is the following that if there is a leaf here for a leaf to grow at the highest place the water has to go from here to here and this leave the water in the leaf here will be in equilibrium with the water in the vapor in the air so I have P H2O in the leaf is equal to or F H2O if you gas the I will start with chemical potential chemical potential of water in the leaf should be equal to mu H2O in air now if you have water here my let me write this out mu H2O in air is this is a mixture so it is mu 0 plus RTLN PY water in air this is a mixture so you also have to count for non-ideality this we can take as 1 since pressure as P is less than 1 atmosphere less than or equal to 1 atmosphere I am talking of atmospheric pressure and growing trees inside a compressed chamber now the water there is water I am assuming there is water here so mu water liquid pure is simply mu 0 plus write it like this this is equal to this mu water liquid pure is equal to mu water vapor pure which is mu 0 plus RTLN P saturation for water was P saturation water plus a correction the chemical potential is actually as long as pure substance is concerned it is the Gibbs free energy per mole in G is U plus TS minus U plus PV minus TS to remember when we wrote the first law we should not have used the internal energy we should actually have used the total energy so the potential normally potential energy and kinetic energy negligible so you use the internal energy otherwise they should really be replaced by E so in all our derivations wherever you have chemical potential equality this chemical potential should include the potential energy if the potential energy is significant in this case from here to here the potential energy is significant so I will have to add that also so the chemical potential will be this plus I should take into account the height also now this term is negligibly small I can show you that this term is and simply say negligible I told you this is called the pointing effect and basically all of the literature you will have books on vapor liquid equilibria at low pressures at high pressures at the only difference between those two is that at high pressures this pointing effect is important I can make an estimate of this for you for example the pointing effect is exponential of V liquid into P minus P saturation by RT and this V liquid for water is 18 let us keep this as P P in atmospheres then R is 82 cc atmospheres this is 18 cc is per gram mole so this will be 82 cc atmospheres per gram mole into temperature is say 300 degrees for typically I am just making an estimate so this is 2 this is about 9 approximately this is 150 so this is approximately P in atmospheres by 1350 the pointing correction which is the correction for the vapor pressure because the presence of a second substance because the pressure is greater than the saturation pressure is significant only if exponential of this factor exponential of P by or P minus P saturation if you like P minus P saturation P saturation is anyway for water for example at 25 degrees it is 25 millimetres of mercury so it is negligibly small so what counts is P but you can see that unless P is about 50 atmospheres are higher this number is very very small so as a thumb rule you do not worry about the pointing correction at pressures below 25 atmospheres you want to make very accurate calculations you take it into account but otherwise generally below 25 atmospheres you do not have to worry about the pointing correction so that just to give you an idea so I do not have to worry about this so I am equating this is the chemical potential of liquid pure here from here to here it passes through the veins in the stem and get gets there it has to get there if there is chemical potential here this is in contact this chemical potential also be covered by the same law except that the chemical potential is altered by the height so essentially what you get here is mu 0 plus RTLN PW saturation I will throw this out this is one typically PW saturation is about 25 millimetres I told you so there is no fugacity there is no fugacity coefficient correction this can be neglected only that plus mg H so you get mg H is equal to logarithm of P by W water in the air by P saturation by RT there must be an RT by RT so what is the H equal to equal to RTLN this is what you call is essentially the relative humidity in the air if the saturation pressure will represent the maximum water that it will hold and therefore you should get RTLN of R this is a relative humidity this is reported in the papers even the Hindu has it every day all you have to read is the relative humidity there by RT by mg should have picked up a minus sign yeah this the energy here there should be minus because this is the energy that you have had there should be a minus sign here this relative humidity is always less than 1 this is the height to which a tree can grow and still have water supplied from the root to the top if it grows any higher it will not have water from here and so without water supply the leaf would just dry up give you an idea as to what this height is give me the units here or do it in SI units what will the value be or if I start with 8 to 2 cc atmospheres per gram mole degree k into 300 I am looking at only this factor by molecular weight is 18 what is g it is terrible units it is easiest to work with Joules R will be what 8.3 Joules per mole degree k into 300 is degree k so if I am working with Joules this is 9.8 that correct I am making a mistake in unit yeah this is alright point this comes to 0.018 you know yeah you need you know this is 18 kg gram per mole okay you what you want you are saying is you divide by 1000 so it is about take this is 10 this is 0.2 it is about 40,000 200 approximately or 12,000 0.2 into 10 40 into 300 yeah 12,000 so you are talking of 12 km into so h is approximately – 12 h in km and then r so thermodynamics just set a limit this r will have to be the you know you have to take this value of r as close to in the wettest season possible so r can be close to 1 if it is r is 1 then there is no choice no tree can grow there the relative humidity will be less than 1 but if it is less it will vary from season to season in the monsoon season for example in Madras it can be 90% humidity it will be about 0.9 so ln of 0.9 is about 0.1 so in principle it can go to 1.2 km the thing is not interesting by the number is not so fascinating because no tree ever goes to that height before that it gets knocked off by wind or whatever but the fact that you can set a limit to the height is purely from these arguments all these arguments had nothing to do with trees but using those arguments you can finally set a limit on this but let us look at actually what happens where do you use this most in chemical engineering use it actually in a distillation column the most important distillation column in the chemical industry till date is the oil fractionation column you take crude oil distillation column actually somewhat looks like this you know I am bad at pictures but does not matter let me you have a series of trays and show you what each tray looks like give you an idea then this is the feed this is the bottom product this is the top product crude oil is basically a mixture of compounds of different molecular weight and you can actually see this in the dross refinery I promise that I will arrange for you to go and see it we will talk to them what you have is crude oil coming in at some point each of these let me take one tray for example a typical tray now there are much better trays typical tray looks like this drawing the cross section of the tray what you have is from below the vapor comes up we will find out and there is liquid that comes from the bottom from the top explain what this is supposed to be there is a hole in the tray through which the vapor rises this vapor there is a cap on top this tray is called a bubble cap tray the bubble cap is the name of the tray what happens is there is a hole and through which the vapor rises the liquid this there is a small tube attached to the plate and as the vapor rises it will have to bubble through the liquid is up to this level the vapor has to bubble through and escape like this as the vapor bubbles through and escapes like this it has comes into intimate contact to the liquid the contact time may be only a fraction of a minute but during this time the vapor comes to equilibrium with the liquid so this vapor that leaves and at the end of the tray there is a downcomer in the liquid goes through here so because this vapor and liquid are extremely well mixed as the liquid comes from the top flows like this and finally comes down the downcomer there is a downcomer here from the previous there is liquid coming down okay this is liquid coming down from a downcomer it flows on the tray like this it goes across and then goes down to the next tray as it passes over the tray the vapor from below bubbles through it these are devices the various kinds of devices that are made in order to make sure that the gases in intense contact with the liquid typical contact time here will be only few seconds but during that thing because it bubbles through in the form of small bubbles you have large area of contact so you have sufficient contact for it to come to equilibrium as far as thermodynamics is concerned this vapor and this liquid that leave are in equilibrium with one another this is from top this is the vapor from below the vapor coming in and the liquid coming in are not at equilibrium this is coming in from another tray this is coming down from the top tray these two are not in equilibrium but the vapor leaving and the liquid leaving are in equilibrium in practice they are very nearly in equilibrium they reach about it 90% of equilibrium so you can do calculations by finding out how much how close to equilibrium they are so where does thermodynamics come in my box is like this I have liquid vapor so this is the liquid this is the vapor on top so the two compositions are related they are related through this equation so as I go up what happens is normally as a thumb rule if the saturation pressure is higher or if the component is more volatile then why I is greater than Xi right I mean there are other factors that will control but qualitatively this is correct that is the saturation pressure is higher the substance is more volatile it will be more in the vapor phase then if I have only two components one and two component one is more volatile then it will be it will appear the mole fraction in the vapor phase will be more than the liquid phase so at every tray if y n is the vapor leaving the tray n n is the tray number okay look at a binary system for a binary mixture if it is a binary mixture y n is the is in equilibrium with x n x n is the this has a composition x n for each tray and you will find y n plus 1 let us measure the trays from either up or below let me take y n plus 1 is greater than y n then x n plus 1 is less than x n so if I have a large number of trays I can get practically pure one here so if I have many components the most volatile component the lightest component will come out on top in the case of the refinery here what comes out on top is natural gas actually methane effectively lowest hydrocarbon if it comes out in more quantities than you can use you actually burn it there because you cannot afford to pollute the air so you will see the occasionally you get a whiff of it so far away from Manali but most of the time they burn it very effectively now they are trying to pipe it and use it as natural gas at the bottom here you get higher and higher molecular weights at the lowest finally in a tower you will get very high molecular weight compounds which go into lubricating oils so from methane effectively these are not exactly methane of that ratio it is a hydrocarbon mix with carbon to hydrogen ratio of something like 12 is to 4 which is the methane ratio these are all approximate ratios in a distillation column you do not get pure components in a crude oil distillation column this bottom will be some lubricating oil this even this lubricating oil is a liquid only because the high temperature that you maintain here what you this is called the reboiler what you do is send in a mix of a feed which is crude oil at the bottom is where you finally supply energy you supply enough energy to evaporate this blue boil also and this goes up as it goes up you can draw of streams at various points each representing level of purity for a molecular weight range for example the next one would be nafta after that would be petrol then there would be diesel and so on so you will get things like etc all the way to blue boil that you even get waxes but they would not be in the solid form at this stage the temperature will be high enough so that the waxes more so these products you remove depending on you can change the number of trays you can trade change the point at which you draw off depending on the market if you want more diesel or more petrol then you have a large number of trays at the top you will draw off a lot of petrol and so on and depending on that you will have to supply energy on the top the gas that comes off is condensed and is removed or it just burnt as gas so this is a condenser this is a reboiler this is a condenser should realize that higher and higher molecular weight compounds will condense and lower and lower molecular compounds will evaporate as you go up so the latent heats will balance one another so basically you do not need that much of energy in a distillation column the energy of condensation will compensate the energy of evaporation the more volatile component will evaporate and the less volatile component will condense on each tray so this will go on and this is balanced out so you do not really need normally you do not need a detailed enthalpy balance in a distillation column the enthalpy balance simply says that the latent heats of evaporation in condensation approximately equal in that is a thumb rule in thermodynamics that the molar latent heats are approximately the same for hydrocarbons so this is the actual column this is what you would be dealing with but what I need to design this is an input which tells me how the molecular the mole fraction of a component I in the vapor phase is related to its mole fraction in the liquid phase this plus mass balance plus if necessary energy balance will solve all the equations for you will be able to get exact because when you do a mass balance you have the liquid coming in from the previous tray above and the thing coming in for the feed tray there will be a special treatment because there is another stream that is being added for every between two trays if you are taking out a stream that tray will need a special treatment all this is done it is all straight forward it is absolutely the only input you need is this in this input will be given to you in the form of a graph as far as distillation column mass transfer course is concerned so you will think you did not use thermodynamics at all this input comes from thermodynamics you calculate this curve from thermodynamics so in practice what you need to do is do this calculation practice the vapor phase corrections and the pointing correction are not so important most important correction from comes from ? I if there is the non-ideality largest largest non-ideality correction is from the liquid phase that is the most important part of it so we look at for example liquid phase non-ideality in practice in a given problem in order to find out if you understand everything I may ask you to calculate everything may give you a problem at two atmospheres and still ask you to correct calculate the pointing correction just to find out if you know how to calculate not so much for but in actual practice for anything below 10 atmospheres or even 25 atmospheres the pointing correction is neglected this is neglected Phi I saturation Phi I notice also that you get Phi I saturation by Phi I so very often this ratio is one even if individual non-idealities are not one so you can often throw away the corrections for fugacity coefficients so most of the time you would use I will say at low pressures pointing correction is approximately one and you take Phi I by Phi I saturation is approximately one so the equation boils down to P Y I and include treatments we write it as K I x I this P I saturation gamma I by P this is actually a function of composition strictly speaking but in classical treatments in chemical engineering they have used Y I is equal to K I x I or more importantly Y I by Y J by this token is equal to in order to get Y I simply have to do K I x I because sum over Y I Y J is one so sum over this has to be equal to one this is equal to X I by sum over K J by K I x J this alpha J I is defined as K J by K I this K I is called essentially this is called volatility this determines how volatile component I is and K J by K I is called relative volatility it is a rather important engineering quantity because it is much more constant than K I see K I is actually a function composition dependent quantity and can vary quite a lot but usually K I by K J is approximately constant at a given temperature so this is primarily a function of temperature in a typical distillation column you can go from something like 600 degrees here to about 100 degrees here in centigrade you are talking of very wide range of temperatures so very often what they will do is to say in the lower half you use one alpha I J in the upper half use another alpha because it varies with temperature alphas will vary widely so it is convenient divided into sections and use constant alpha over each section but this gives you a representation of equilibrium very nicely for example if you have you have Y I by X I or X I by Y I if I have got alpha I J we write it for a binary so instead of giving you gamma data instead of giving you an expression for G excess from which you calculate gamma I can give you data on K which makes life very simple or data on alpha for example binary makes you get Y 1 is equal to X 1 by 1 plus alpha I I is 1 clearly alpha 21 not this is X 1 plus a very simple relationship between Y 1 and X 1 this is 1- X 1 of course so this is a relationship between Y and X this also gives you a relationship between Y and X you can then therefore calculate an expression for alpha in terms of gamma and if you had a binary this would simply mean Y 1 is equal to P 1 saturation all of that stuff pointing correction for one P 1 saturation gamma 1 into X 1 so you have to compare these two great some connection between alpha 1 this expression and gamma 1 so alpha can be directly related to gamma so the rest of it is just a lot of problems that I have to give you in vapor liquid equilibria and one of the most one interesting way one of the questions is how do you get these values the gamma values you get it through these equations but how do you get the parameters in these equations and let us look at some examples of GXS I have GXS by RT and then log gamma 1 for example already seen this for example you have a X 1 X 2 this is a X 2 squared now we know that gamma 1 goes to 1 as X 1 goes to 1 then as gamma 1 infinity is called infinite dilution activity coefficient come on infinity is limit as X 1 goes to 0 of a X 2 squared so this is simply so if you can measure non-ideality at very low compositions of X 1 then you get this coefficient here log gamma 1 infinity sorry log gamma 1 infinity is simply a so this infinite dilution activity coefficients are often the way in which data are reported that is if you have different correlating equations these constants I have to give you data on these constants for you to do predictions because if you know the GXS expression along with these constants for a particular system then you can calculate gamma and you can solve this set of equations so in order to solve this set of equations I have to give you vapor phase non-ideality which means I have to give you an equation of state or tell you the pressure is low so you throw away all the fees make them one so I have to give you a method of calculating this I have to give you the liquid volume so that you can calculate the pointing correction if necessary and then you have to calculate gammas to calculate gammas I have to give you GXS the way in which the GXS data is given is often you are told that it obeys Porta's equation it obeys some equation and to get the constants any give you the I can give you values for the constants or simply give you limiting activity coefficients so with that actually the vapor liquid equilibrium treatment is over all I have to do is give you a large number of problems and discuss individual problems in particular interest and the other way of getting these is through measurements on azeotropes I will discuss this next class I will discuss getting constants in these equations using data from azeotropic mixtures if the system has an azeotrope okay really that covers vapor liquid equilibrium in the case of liquid liquid mixtures where all the components are solvents we still have to discuss the case where one of the components in the liquid is actually solute in which case you cannot get this PI saturation you will not get PI saturation you will get some other reference value you cannot go to the liquid state okay stop there