 separation work is very simple this simplest case would be air I put the air here let us assume here consists of only nitrogen and oxygen for the case of for the sake of simplicity what I really have is a membrane as I said all these are thought experiments let us assume I have a semi permeable membrane that can distinguish between the size of oxygen and size of nitrogen. So nitrogen would go right through you can pull out nitrogen here oxygen will come out here now air is at some pressure p I want nitrogen at pressure pn this at pressure p0 as far as thermodynamics is concerned even this sketch is very complicated drawn envelope like this all I have is air going in and oxygen nitrogen coming out. So if I take one mole of air let us assume 79% is oxygen nitrogen 21% is oxygen all I am saying is I have a mass balance I have ?g you know that Ws dot by m dot is equal to – ?g isothermal the what happens inside is immaterial this is at 25 degrees this is at 25 degrees all I want to do is to calculate how much work I have to do to separate air into oxygen and nitrogen the gaseous mixtures all I have to do is to calculate according to thermodynamics – ?g you can make this if you like n dot usually m is used for mass make this n dot per mole for 1 mole per hour or per second flowing through I want to calculate the rate at which work has to be done for this is very simple so ?g is simply g of nitrogen or µ nitrogen sorry we will call nitrogen as component 1 this will be component 2 so it is – of x1 µ1 plus x2 µ2 – µ air actually sorry this is x1 µ1 pure these are the products this is pure – x1 µ1 plus x2 µ2 this is g before in this case it is – ?g this is not ?g of mixing this ?g for the process so this is equal to if you like ?g mixing because this is ?g separation which is – of ?g mixing this according to the equations the models we have this is gas phase so let me rate gas phase models only one model for the gas phase or µi saturation at temperature T I think I wrote it as not saturation I do not need µi saturation µi 0 sorry µi 0 of T plus RT ln P y1 times ?1 or yi and µi pure is again µi 0 plus RT ln P into ?i pure µi 0 is the chemical potential of pure I at the temperature of the mixture but one atmosphere pressure so normally this is written at one atmosphere therefore P has to be an atmospheric units otherwise you will get the numbers wrong because you are taking log P this is actually log P by 1 which corresponds to one atmosphere pressure at which µi 0 is reckoned so if you use millimetres here you will be off by a large number you have to be careful so this difference therefore ?g mix or ws dot by m dot by n dot is equal to RT into the difference between these two log in the mixture its pressure P in the mixture composition is y1 ?1 by pressure if you are talking of component 1 component 1 is at pressure Pn into ?1 pure you must have another x1 in front sorry we will divide ws dot by RT we will take RT to this side this is x1 this is y1 now sorry plus y2 log P y2 ?2 by P oxygen ?2 pure the simplest case is this if you assume ideal gas and ideal mixing if you assume ideal gas P1 pure and P2 pure R1 and ideal mixing means P1 is equal to P1 pure that is also equal to so the fees will cancel then you have y1 log y1 plus y2 log y2 plus also written slightly differently like this is a small P1 P into y1 is the partial pressure of 1 in the mixture small P2 is partial pressure of 2 in the mixture notice that that work is exactly 0 if P1 is equal to Pn and P2 is equal to PO that means if this is at the partial pressure that nitrogen exerts here and this is at the partial pressure that oxygen exerts here you do not need any work of separation all you need to do is to find this magic membrane through which you get oxygen on one side this is what you do in regular reverse osmosis where you separate solutes and solvent but in this case right now I will discuss that case in a minute you have no work done at all but you actually want this usually at P so normally you get everything at 1 atmosphere at the same pressure normally Pn is equal to PO is equal to P in which case your work done ws dot by n dot is simply RT times y1 log y1 plus y2 log y2 because of pressures if there is non-ideality you have to make these corrections but you can calculate them quite easily you know if you want to calculate non-ideal corrections you know log Vi is simply integral Vi bar by RT minus 1 by P dP from 0 to P you have to take experimental data on V1 bar Vi bar and integrate that quantity you get Vi so you can throw in the Vi and the Vi pure for Vi pure is simply put Vi bar equal to specific volume of I so you need a fair amount of experimental data but these are available for gases not a problem very often in spite of non-ideality P1 and P1 pure are the same because your Vi bar and Vi specific volume Vi pure are really the same in many gaseous systems so the non-ideality here usually cancels out so this is a pretty good estimate of the work required so that in all gaseous substances if you plot log y1 against the cost in the marketplace after all this is the amount of work I put in so if you plot the actual cost which is the plot that I indicated to you if you put cost in marketplace I think I showed you a slide in the very first lecture cost in marketplace versus log y1 log y which is log of the mole fraction as it occurs in nature usually get a straight line for various substances you get an exact straight line in the case of gases even the case of liquids is proportional to this so you can take this data from chemical age of India I do not know if you look at trade journals it is worth your while looking at it will give you some idea as to where the emphasis is in the chemical engineering market place so this take this cost in the marketplace plot against the mole fraction in which the substance occurs naturally you get a straight line in fact in the US market you get a perfect straight line in the Indian market you will have few aberrations petrol for example is much higher than it will be here because that is because of the additional taxation in the government urea in order to please the farmer they give it to him very cheap so urea will be somewhere here sugar will be somewhere here so the cost except for those irrational things where there is market intervention by the government presumably I think they were all introduced with good reasons but now they are going one by one because of that you have a artificial price difference but otherwise 90% of the things it will be perfectly in line in the international market Europe US it is much more even more rational even here in India also the changes are only in about 3 4 substances where the excise duties and other things are artificially either increased or subsidized a nice plot is given in Jetson King so first point about separation work is this is the cost to give you some idea you can look at the numbers notice that W s dot by m dot will come out to be negative means work is done on the system our convention is work done by the system is positive so you have to do work to separate air into oxygen nitrogen okay let me discuss the non-ideality part of it in gas phase basically all non-ideality corrections have to do with calculating this integral vi bar – RT by P this is RT ln Vi so all you need is Vi bar if you look at Vi bar by definition it is partial of V with respect to ni folding TP all other mole numbers constant so all you need an expression for V when you do the differentiation so what I need is equations of state I will tell you the equations of state that I expect you to know one of the problems with equations of state is that they are all usually implicit in the volume that means you will get multiple routes you have to find the correct route and then do this differentiation so normally you do it numerically but let me explain the equation of state that you should know of course ideal gas law ideal gas equation of state second for simply the historical record you have to know the van der Waals equation of state third you have to know the virial equation of state I will discuss this a little more this is the only equation of state that can be derived rigorously from theory if you are looking at molecules you are looking at intermolecular potentials and depending on different intermolecular potentials you will get different equations of state so in principle there is one intermolecular potential that will give you each equation of state ideal gas comes from phi is equal to 0 the intermolecular potential being 0 it is not clear what the van der Waals comes from there is no simple you can get van der Waals like expressions from many potentials but the exact form of the van der Waals equation does not come out from theory the virial equation of state comes out very nicely because this is really a density expansion and this was originally due to a person called cammerling on us I think I told you about cammerling on us in theory it is due to mayor and mayor the husband and wife team who derived this actually it was derived even before but mayor and mayor give exact expressions from statistical mechanics for the virial coefficients this is a very important equation purely from a theoretical point of view when I say van der Waals here you must add here redlich-quang redlich-kister this is k for quang I think this is for kistler redlich-kistler equation of state and then the other sets of you have to know this multi parameter equations of state think you use the BET equation in adsorption isotherms and so on and I will put down one you just have to know at least one so I will choose one Wallace has all these things and Wallace has this integration also will give you expressions for fee for each equation of state all this has been done then you have a set of it equations of state which are really interpolative based on you have to introduce another word which is a centric factor what you do is you ask what is the cause of non-ideality and the cause of non-ideality in molecular terms and this was not necessarily this pictures this thing was derived in classical thermodynamics but it basically has to do with the concept that the center of force in the center of mass do not coincide if I have two molecules at interact I can take the line joining the centers of mass but there is also a line of force if I take even a simple molecule like carbon dioxide I have carbon and oxygen oxygen is bigger than carbon so I am drawing it a little bigger over and over now but if I have two carbon dioxide molecules the total interaction between two molecules is between the carbon carbon carbon oxygen carbon oxygen and so on and oxygen oxygen you have to take an average of all these and find out how the interaction occurs the center of mass may be right in the middle here but the center of force need not be located there because the force between oxygen oxygen may be much larger than the force between carbon and oxygen so on so there can be a distortion so the question is in fact it is better for me to take carbon monoxide because carbon dioxide is quite symmetric but if I take carbon monoxide then you can see that the center of mass will be shifted towards oxygen but the carbon carbon interaction may be stronger than the oxygen-oxygen interaction in which case the center of force will be shifted towards the carbon atom there will be a separation between these two when there is a separation you get eccentricity non-centric forces if you get eccentricity it leads to non-ideal behavior so Pitzer defined an eccentric factor he actually defined it in classical thermodynamics I will tell you how it is defined and what he said was eccentric factors typically vary from 0 to 0.5 so he said find a fluid for which it is 0.5 find a fluid for which it is 0 and then do an interpolative calculation for all other fluids that is what we will do these all derived interpolative ones are derived from the principle of corresponding states so what I will do is discuss equations of state a little more today and I will start with the discussion of corresponding states it is a very appealing idea you I think you already know about the simple principle of corresponding states it was again the first idea was due to Van der Waals let us start with the so I will talk about the principle of corresponding states it is much more general than let us start with the Van der Waals you look at P plus a by V squared if you look at these plots I can do it on the PV diagram if you like I am distorting the scale these are towards this end will go like finally at very low pressures PV is equal to constant so the isotherm will go like a rectangular hyperbola so this part is a rectangular hyperbola and this can be steep it is actually steep because liquids are incompressible at the point is that if you go up towards the critical point these roots come closer and closer till they finally coincide here this equation is cubic if I rearrange this equation I get PV squared plus a into V minus B is equal to RT into V squared you multiply this out let me divide through by P I am looking at the polynomial expression for you get V cube plus V squared into is a minus B here there is so I can write this as minus B plus RT by P minus a V by P is equal to 0 I have got V cube a B by P sorry there is a constant term minus A B by P is equal to 0 this is plus A B by P this is a cubic equation there is three roots what the experimental diagram tells you is two roots are real below the critical point at the critical point all the roots coincide so you will get if two roots are real in a cubic the third root is also real but it is unphysical it is not realizable in the lab so you get three roots at critical point actually let me say this at critical point the three roots are equal that means at critical point I can write this as Tc Pc is equal to 0 because the roots are equal this equation should also be you should be able to write it as V minus Vc whole cube equal to 0 so this is one if you like and this is two these two represent the same polynomial you should be able to equate the coefficients so I get V cube minus Vc cube minus 3 V Vc into V minus Vc minus or plus I think this is correct this is correct so therefore if you have two parameters van der Waals argued that you should be able to find the parameters by making a comparison between these equations so if you look at the comparison you get 3 Vc is B plus RTC B plus RTC by Pc is equal to 3 Vc it is one equation other equation is a by Pc is equal to 3 Vc squared in the last one is a B by Pc is equal to Vc cube so this these two imply that B is equal to Vc by 3 if you look at experimental data on liquids there is a nice book that you should look at it is actually a graduate level book it is called liquids and liquid mixtures I think now it is Rawlinson and somebody the original version we have several editions the first edition was by Rawlinson and he wrote it he was in Imperial College that book has table of data for several substances if you look at the critical that and look at the volume specific volume at the triple point it is almost always one third the wall critical volume they give you both they will give you the critical volume they will give you the volume of the specific volume of the liquid at the triple point the one exception is oxygen unfortunately it is one of the first ones that come because of paramagnetism oxygen alone has a very wide range of liquid behavior so you will find otherwise you will find in almost all other cases this is true the other empirical observations are like this about generalizations I mean there will be exceptions but if you want to generalize some things empirical observations V liquid is approximately constant so V triple point I will say VT is the specific volume at the triple point of the liquid is approximately VC by 3 then the absolute temperature at the boiling point is approximately 0.7 times TC which is why a large amount of data see the boiling point normal boiling point this is normal boiling point normal means at the total pressure of one bar one at most of the data is at one atmosphere so very large amount of data this is the easiest to measure TB is the easiest to measure extensively available TC is somewhat difficult to measure because actual critical point measurements are done by looking at what is called critical opalescence you know the interface disappears and it is very difficult to experimentally observe and detect it exactly I mean to within point one or point even 0 5 degrees it is easy but to get exactly the critical point is very difficult but this is a good thumb rule you can use TB by 0.7 and estimate critical temperatures for engineering purposes in correlations we will need the critical temperature it is very convenient use that estimate you do not have data having said this so this was a B you remember Van der Waals argued B was actually the volume occupied by molecules in liquid phase because he said in the liquid phase essentially you compressed the molecules as much as you can so he said the free space available is V – B so this estimate of B is a I mean this interpretation of B came in very handy Van der Waals thought he had a good equation of state with him of course a is simply 3 PC VC squared and if you combine these here you get PC VC by RTC is equal to 3 by 8 this is well known this is VC by 3 so PC VC by if you divide this by RTC by 3 is equal to no plus 1 and you work it out it is 3 by 8 so any two parametric question of state that is if you have a cubic equation of state and many equations of state are cubic and if you have only two parameters in it it will give you both the parameters in terms of PC VC and PC.