 This is the first lecture of this course on chemical engineering thermodynamics I thought I would give you an overview of thermodynamics in the chemical industry the outline of this lecture and I am going to use power point only for this lecture and the outline of this lecture is as follows I talk about the chemical industry in general then I tell you a bit about thermodynamics then I am going to tell you about the masters in their ideas that is the people who actually develop the subject of thermodynamics. Tell you about the important aspects of chemical engineering thermodynamics that we will discuss mainly work and equilibrium in particular I will talk about work of separation and show you that thermodynamics actually predicts market prices reasonably well and then we will discuss a bit of reaction equilibrium and finally I will close with a molecular picture of mixing I must say that classical thermodynamics does not depend upon the molecular structure of matter and it is actually independent of that but the molecular picture helps give you a handle on understanding what the properties mean in molecular terms. So I will start off by telling you something about the chemical industry the first thing that strikes you about the world of chemicals is you think there are infinite chemicals but basically there are 4 groups they classified to 4 groups and the raw materials that we discussed are usually oil natural gas, coal, biomass, rock salt, sulphur, air and water and the major fuels are LPG, gasoline, diesel, kerosene the bulk chemicals are divided into base chemicals and intermediates about 300 intermediates and the specialty chemicals are those that are made from bulk chemicals these numbers will increase but certainly not infinite it is reasonable number that you can handle. The chemical industry basically consists of the following consists of a reactor usually at the core whole lot of separation and recycle systems heat exchanger networks and utilities this is broadly the picture of the chemical industry and the after the reactor there are separation processes and I have tell you something about separation process because industrial separation process can be divided as shown in this picture roughly you have equilibrium separation processes which constitute about 55% of the total and then you have 30% rate governed separation process and you have mechanical separation process in the balance the equilibrium separation process can be basically designed on the basis of thermodynamics itself thermodynamics deals with equilibrium and many of these processor work operate so close to equilibrium that you can make quantitative predictions with thermodynamics. The point is that separations are expensive most chemicals do not occur in the pure state in nature so separation process contribute to major part 70% of capital investment it is estimated in the petroleum and the petrochemical industry is on separation equipment alone now that is about the chemical industry about thermodynamics thermodynamics is subject with extraordinary with the applications you can ask questions that you may not think are significant but as diverse as how tall can trees grow how old is the universe and so on thermodynamics as a supervisory of an intangible role in engineering it sets the boundaries of design fundamentally it will tell you what is the maximum possible thing for example what is the maximum work you can extract from a system or what is the minimum work you have to put in order to produce a change of state and so on. And in chemical engineering the applications are in phase and reaction equilibria in calculating separation work and in design of equipment in near equilibrium processes the 55% separation process are practically designed on the basis of mass balance energy balance and just thermodynamics now the perfection of classical thermodynamics if you I do not know if you have seen this book by Lewis and Randall on thermodynamics one of the best books it is a well mark it is an old book it is been reprinted I think the revised edition is by Pitzer and Kurt but if you look at the preface even Lewis says when you visit an ancient monument you see only the perfection of the completed whole enter such an edifice that is still under construction you realize that these great structures are but the result of giving to ordinary human effort a direction and a purpose I think the point that Lewis makes particularly is that thermodynamics as a subject appears so complete and the completion itself gives it a perfection sometimes a little frustrating for the novice because it seems as if you are drawing conclusions out of nothing but I think it is nice to go back to the masters who gave direction and purpose to the edifice called thermodynamics and ask what did they contribute some sense it is nice to have a sense of history of the subject. So let me start with a few of these people Galileo Galilee must be recognized as the person who introduced in effect the scientific spirit he said you not only have to collect data you have to then use your right brain to arrive at conclusions for reaching conclusions and then when you hypothesis you have to also verify experimentally because a lot of your imagination of the right brain is often wrong because Galileo did not talk about the right brain the left brain but in our context we can explain what he said in these terms Isaac Newton of course was the one who essentially as far as thermodynamics is concerned introduced the concept of work and told you how work is calculated. Sadik Arnaud is the case of a rare case of an engineer contributing to fundamental science the second law actually came before the first law and the Sadik Arnaud introduced the idea of absolute temperature then there was Joule who introduced the conservation of energy in a very rigorous form I will say a few words about each of these people and Rudolf Clausius who introduced the word entropy. So this is I cannot get a picture of Galileo as a younger person fortunately but Galileo was the father of modern science and all effectively Newton discovered the laws as you can see there of calculus gravity motion and optics and then turned 626. So this was an Englishman whose contribution well known then with Carnot and Clausius the French engineer scientist you are moving a couple of centuries after Newton and Galileo the Carnot is the was a bachelor who died young essentially came up with the second law. The German physicist and mathematician Clausius was also a neologist he was had a knack of coining words that remain in use even today. So he introduced the word entropy let me talk a little bit about their ideas the heat to work conversion essentially Carnot pointed out as asymmetric and he introduced the concept of a heat reservoir that was so large that little addition of heat and little taking a little heat away from it does not change its thermodynamic state and he introduced the concept of a reversible heat engine he realized that in order to produce work you have to have a cyclic operation otherwise the amount of work produced would not be significant and he introduced the concept of a heat engine that would work both ways and the thermodynamic concept of absolute temperature came from Carnot again you also talked about the efficiency of the reversible the Carnot now called Carnot engine operating between two heat reservoirs at T1 and T2 and said that the efficiency must be 1-T2 by T1 he said no heat engine can be more efficient in the Carnot engine. The entropy definition was actually given by Clausius although it was Carnot who pointed out the delta Q by T while delta Q was a function of path delta Q by T defined a quantity that was a change in a property and he used this to define entropy as a function of state and the conclusion was that the entropy of the universe which is the only isolated system we really know can never decrease. So Joule and Kelvin these are pictures of Joule and Kelvin Joule was characterized by dogged persistence and say a few words about what he did then talk about Lord Kelvin but what Joule said it was even before Joule was known that there were two ways of exchanging energy with the system and between the system and the surroundings and it was Joule who essentially asserted that there are only two ways as far as a closed system is concerned the interaction energy interaction was either an exchange of heat or exchange of work and he did several experiments to show that this is true he supposed to have told the Royal society again and again demonstrated this in front of the Royal society till the members of the Royal society were absolutely convinced they did not want any more persuasion but Joule insisted on doing it in several ways. But he essentially showed that heat and work are equivalent and the internal energy is a function of state the internal energy of the universe a constant of course internal energy he defined as the difference between heat and work will give you a more precise definition as we go along and he pointed out that the internal energy of an isolated system is a constant Lord Kelvin was one of the greatest supporters of Joule, Lord Kelvin came from the aristocracy whereas Joule was an ordinary citizen and this support is very valuable but Lord Kelvin known to be very embarrassingly self-confident the story that he wrote the tripods and he sent one of the attendants asking him to go look up the results of the tripods and said find out who is second and he suppose the attendant seems to have come back and said Lord you are second so he was actually he was very confident that he was going to be the first anchor somebody else had to be second but he provided a lot of support to Joule in the Royal society in those days. Then a little later this van der Waals and van Laar van der Waals again from Netherlands both of them and van der Waals is a Nobel laureate one of the first in the area of thermodynamics and he also an opportunist he apparently got his son to succeed him and he was the first one to produce an equation of state that is even now used as one of the best equations in conceptual terms it has no conflict in any fundamental way with thermodynamics the laws of thermodynamics he talked about copied capillarity and he also introduced the idea of corresponding states van laar was a student he was socially awkward and but he was a great scientist and he is the first one to produce a classical thermodynamic theory of solutions and still one of the best. Then there were Boltzmann and Gibbs Boltzmann was a moody German he actually finally committed suicide but his ideas were far ahead of time and introduced kinetic theory essentially and then there was Gibbs Boltzmann on the tomb of Boltzmann this formula of S for an isolated system is given S is equal to K log omega where the omega is the number of microscopic states in which the macroscopic system can exist and Gibbs was one of the most remarkable people the best of the early expositors he also developed the statistical mechanical principles he was a remarkable man he was in Yale for many years without any payment and subsequently he was finally recognized before he finally recognized another great scientist and his book on elementary principles of statistical mechanics still reads beautifully and after about 150 pages that he had written all of this work was done in isolation all by himself he finally calculates a specific heat of argon and he makes a comment that the our the specific heat that we predict does not agree with experiment therefore we must consider our methods as tentative actually couple of years later the specific heat is actually remeasured it was found that Gibbs's prediction was right in the old experimental data was wrong. So it is an absolutely remarkable person and a lot of results a lot of things you can do classical thermodynamics in many ways is a very complete subject and a lot of things you think you can derive on your own you will find in a footnote to Gibbs's notes it can be shown that in the results recorded there. So basically let me just say that classical thermodynamics is essentially two laws it says the energy of an isolated system is a constant and the entropy of an isolated system is ever increasing the equivalent statement of the laws as far as we are concerned is that the internal energy u and the entropy s are functions of state you know other functions of state like pressure, temperature, composition, volume and so on but these two were introduced because of the two laws of thermodynamics once these are introduced the rest of it is practically calculus you can derive all the results in thermodynamics through calculus essentially chemical engineers are interested in two kinds of applications one is criteria of equilibrium and the other is extreme mind work. Let me talk a little bit about closed systems the combined form of the two laws and we will see these in some more detail as the course unfolds but I am trying to give you an overview here essentially we are going to draw the conclusion that the change in internal energy is less than or equal to Tds-delta w the convention used here for work is that for work and heat the conventions are heat is positive if it is added to the system work is positive if it is work done by the system the modern convention is actually the other way around so in terms of delta w and delta w the science in this presentation in most of the course would actually be the opposite of what is presented in most books today's books but it is a trivial difference because finally in terms of actually measurable quantities the equations will be the same. So in isentropic processes you can show that the maximum work you can get when a process when a system changes from one state to another is actually equal to minus the change in internal energy and minus w which is the work done on the system the minimum work you have to do to produce a change of state is equal to delta u in isothermal processes we will show that this is actually equal to the free energy change delta a for PVT systems delta w is PDV and the equilibrium criteria and becomes simply u is equal to minimum at constant SNV because you can rearrange that first equation you put du is equal to delta w is equal to PDV so you get du is less than or equal to 0 in SNV are constant which means u has to be a minimum at equilibrium but this is a constraint of constant SNV so if you subtract of minus Ts on from both sides you get the result that a should be a minimum at constant T and V then at constant SNP similarly h should be a minimum and g will be a minimum at constant T and V it is clear that because temperature and pressure temperature and volume are the pairs of variables that can be held constant through easily you can monitor these values so a being a minimum and g being a minimum are the important equations. If you look at open systems you have to subtract of flow work so you define redefine shaft work or useful work is delta w plus what I have written there and if you put that in you can ask what is the minimum work you have to do in a flow process in a steady adiabatic process and you get the result that my minus delta Ws by delta m is equal to delta h similar in steady isothermal operation it is equal to delta g. Now this is also valid for systems with negligible hold up and a chemical plant for example if you take a whole plant where raw material center and the products leave at ambient temperature and pressure no matter what happens inside the plant you can calculate simply the delta g between the product and their products and the raw materials and this is the minimum work you have to do because the actual work will be much more it is remarkable that thermodynamics is able to make such predictions while not paying attention to the details. So let me show you an example is an example of an undersea portable device I have a tank one and tank two this can be used for example for working on the hull of ships and this is a portable device is carried on the back of a diver and you can ask how much work can this do what should be the what you have is air in tank one at very high pressure which expands into tank two because you are talking of pressures under the sea and the pressure can be quite high even at 10 meters can be quite high. So what you do is always have an evacuated tank which is much larger into which the air expands and in doing so it does work WS you can ask what is the amount of work done. So I have given you some example tank one you started 5 atmospheres 50 atmospheres and go tank two is at one atmosphere at t is equal to 0 and you allow the air from tank one to go into tank two the volume of tank one is 5 litres and tank two is 50 litres and you can ask when comes to equilibrium what will be the final pressure you can show that the pressure is 5.43 the operation is isothermal because the whole thing occurs under water into particular depth in the temperature is constant effectively. So what you can do is calculate therefore the maximum work maximum work according to thermodynamics is delta A and you are talking of M1 the mass of air in tank one having a change delta A1 that is a specific Helmholtz free energy small a and M2 delta A2 that is the gas in tank two and the second part is compressed the first part expands. So you have a network done of 46.9 kilojoules. So if you are looking at what we are really looking at is mixtures in the in chemical engineering thermodynamics and you have to start to the basic laws for DG for example because the Gibbs free energy is the most important variable because typically chemists deal with close systems and they use A as a very important variable the Helmholtz free energy and chemical engineers deal mostly with closed systems. So we are looking at G as the most important variable the Gibbs free energy DG is simply minus SDT plus VDP plus some or mu I DNI you are looking at the natural function of temperature and pressure and mole numbers and the change in G with respect to Ni at constant Tp etc and other mole numbers is a variable that will appear again and again it is called partial molal property GI bar it is given a name chemical potential mu I mu I has other definitions but basically you can take the first equation DG equal to minus SDT plus VDP plus mu I some over I of mu I DNI and because if you are looking at a process that at constant T and P and because mu I is an intensive variable if you look at changing G from a system with n moles to K times n moles you actually get this result by integrating the first equation you get G is equal to some over Ni mu I this is true of any partial molal property. So in particular for the chemical for the Gibbs free energy the chemical potential is the partial molal property and G is simply equal to some over Ni mu I and you can write this you would by differentiating this again and comparing with the first equation you get what is called the Gibbs-Duhem equation minus SDT plus VDP plus some over Ni mu I gives you pointed out that chemical potential is actually the natural variable for discussing thermodynamic theory not composition but like the composition variables the mole fractions for example out of n only n minus 1 are independent because there is a constraint that some over Xi is equal to 1 or some over DXI is equal to 0 an equivalent equation can be derived for the chemical potential and that is this equation here because chemical potential is given by this equation from calculus and you can use the Gibbs-Duhem equation re-express the Gibbs-Duhem equation by substituting this for D mu I and you get a simple result that some over I of Xi dou mu I by dou XA the sum over all I is equal to 0 for every J equal to 1 to R minus 1. So you have a total of R minus 1 equations for R chemical potentials in any given phase therefore there is one degree of freedom now you can therefore guess the composition dependence of the chemical potential of one species then all the others are determined by the Gibbs-Duhem equation the simplest of these solutions is called mu I is equal to mu I reference RTL and Xi and if it is valid all the way for the whole range of compositions then when Xi is equal to 1 mu I reference becomes the chemical potential of pure I. So as I said the Gibbs-Duhem equation has R unknowns and R minus 1 variables and R minus 1 equations sorry so it has no unique solution the central problem then has to build models for any one chemical potential instead of doing it for one chemical potential you try and do it for delta G mix because delta G mix is simply the free energy change due to mixing of components. So after mixing the Gibbs free energy is Xi mu I sum over all I and before mixing it was all pure substances so it is Xi mu I pure and delta G minus delta G ideal is called G excess. So if you can make a guess for G excess then you can calculate the chemical potentials and in given G excess for example in a binary the chemical potential mu 1 is simply delta G plus X2 times partial of delta G with respect to X1 and this is formally written as mu I pure plus RT log gamma I Xi where gamma I represents the non-ideality and RT log gamma 1 is G excess plus X2 delta of G excess by delta X1. So this way all you have to do is guess what G excess is or delta G of mixing is the various models for this because you have complete freedom only thing is the model that you propose has to give results that agree with experiment for a large number of systems then it stays in the literature otherwise it is forgotten. So the models for example a one law model doles are like these two are classical thermodynamic models the others are all molecular models you have various models like the Wilson the Unique Wacke the Scachard Hildebrand the Flory Higgins model and so on. So basically what you are looking at is bounds for work you have to calculate the change in the thermodynamic properties either U or A or the specific enthalpy or the specific Gibbs free energy particular interest in separation process is the free energy of mixing as I pointed out already delta G mix is delta H mix minus T delta S mix all of the models normally calculate try and calculate H delta H or delta S and then set the other have a default option for the other for example you can say delta H is 0 and I will calculate the entropy of mixing or you can say delta S mix is the same as the delta S of ideal mixture and calculate delta model delta H delta H ideal delta G ideal is simply RT times XI log XI just to give you an illustration take work of separation for example you have let us say a distillation column in which you are separating ethyl benzene from the three xylenes para meta and ortho and I have a certain feed I have a certain overhead composition in a distillation column I have a middle composition the bottom composition the question is how much work is required for this. Now the minimum work that you have to put into the system is simply G of the products minus G of the feed and in this case there are three product streams so delta G of P1 plus delta G of P2 plus delta G of P3 minus delta GF you can write it in terms of delta G for each stream because the pure component free energies will cancel in any case. So delta G ideal in this case turns out to be 2.86 kilo joules per mole if you use a regular solution model we will learn all these models later in the course you can show that that gives you an excess free energy of minus 1.88. So net delta G is 0.98 which is the minimum work that you have to do it should be minus W within parenthesis then minimum because it is work done on the system. Now delta H is 0.81 so you can the enthalpy contribution is approximately 80% this particular case that we discuss is an example of a dominant enthalpy effect. Now if you look at work per mole minus RT times X1 log gamma on X1 plus X2 log gamma 2 X2 represents delta G and it is the minimum isothermal work to be done to separate one mole of a binary mixture into X1 moles of component 1 and X2 moles of component 2 of this minus RT ln of gamma on X1 is the work needed to produce one mole of the substance from its naturally occurring state of composition X1. Now in a free market the price you would expect the price per mole be determined by the work needed to produce it and I will give you an example of a plot this is from Jetson King the classic book of 1970 and he plots log of the price in dollars per pound against log of the weight fraction it should be mole fraction times the activity coefficient but as an approximate crude approximation you can use the weight fraction and if you make a plot you get a remarkable straight line the two data are from completely different sources. This one is the market price and the other is the log of the weight fraction as it occurs in nature and this approximately calculates the work required. So you can see that the thermodynamic work required correlates very well with the price the slope also depends and the slope of this will depend on several factors including the economic conditions the wage wages and so on. Now you look at it is a free market so everything falls in place look at some data for India that we plotted this is 2001 data it is old data but it still gives you a flavour you can see that petrol is much higher that is artificial because you have an excise duty on petrol and you can see that pre urea pre subsidy is right bang on the line the actual price of urea is much lower in order to get the vote of the farmers and to make it available for the farmers the government makes it available at much lower prices sugar again is a political decision to subsidize sugar and so you get sugar at much lower price than the work warrants. So criteria of equilibrium I would not say a few words about the other aspects of thermodynamics at constant T and P thermodynamics will tell you that G should be a minimum and G is n i mu i you can write it in terms of activity coefficients and it is either the activity coefficient is gamma i x i it is a liquid mixture it is equal to P y i times V i if it is a gaseous mixture different ways of expressing the non-ideality and by asking the delta G be equal to 0 for equilibrium you actually come to the conclusion that RTL K K is the equilibrium constant is equal to minus delta G reference the reference state should be chosen judiciously and finally for face equilibrium you can show that the chemical potentials in two phases the same substance should be equal at equilibrium. And let us for example let us take a very simple example almost all the overs appear as sulphides or oxides in the case of zinc it is zinc sulphide you roast it in oxygen what do you expect to get you can show that you expect essentially zinc oxide the reaction possible reactions are listed there zinc sulphide plus oxygen will give you zinc oxide plus sulphur dioxide and the delta G 0 the free energy change in standard state is minus 280 look at all the others sulphur dioxide going to sulphur trioxide zinc plus sulphur trioxide giving a zinc sulphate and zinc oxide giving a zinc plus oxygen all of these are positive and you can say only the first reaction needs to be considered the product will be zinc oxide. So thermodynamic can tell you right away what to expect in terms of product you can of course quantify quantitatively calculate them then you take couple of reaction for example these reactions where carbon dioxide plus hydrogen gives you carbon monoxide plus water and then CO plus hydrogen can give you carbon deposition now you ask if you do not this carbon can deposit on the catalyst and choke and make the catalyst ineffective and so you want to know what should be the pressure at which you should operate or what should be the range of pressures over which you should operate this example that T is equal to 1000 degrees normally the temperature is chosen on the base of kinetic concentrations not on the base of thermodynamic concentration and some data is available you can do this calculation quite quickly and show that for pressures below 2.5 atmospheres you will have no carbon deposition these are very powerful conclusions from where we started. So let me summarize results the thermodynamic theory is so beautiful and so complete that it sort of gives you results of the same form in many contexts if chi is a characteristic intensive property at equilibrium right through thermodynamics will tell you how to calculate chi from one of the criteria that we already discussed and also tell you that the change of log chi with respect to T is given by an equation that says minus characteristic enthalpy change by RT squared similarly the change of log chi with respect to P will give you a characteristic volume change by RT and the maximum minimum work will be change in the appropriate energy function these are the summary of results that you get in thermodynamics they are always the same form and let me now pass and give you a molecular picture for example in the case of one example as I said classical thermodynamics does not depend upon the molecular structure of matter if you look at the picture from a molecular structure however it is quite insightful if you look at the dissolution of salt in water in the salt molecules are added to water the because of the high dielectric constant the ionic forces are much weaker in water and the salt ions can separate. So effectively the salt has more states in which it can exist and therefore its entropy actually increases molecular theory tells you that the entropy depends on the number of possible states in which system can exist and this entropy of the salt increases while the entropy of water decreases because the presence of salt molecules. Now the increase of entropy of the salt is much greater normally than the decrease of entropy of water and dissolution is therefore spontaneous because the resulting entropy change is a driving force the energy absorbed to break the bonds in the crystal in order to do this at a molecular level you have to break the bonds in the crystal and this energy we will call this E, A and energy released when ion water bonds are formed let us call it E, R now the comparison of E, A and E, R will determine whether the dissolution is going to be accompanied by a temperature change for example in the case of NaCl E, A is approximately equal to E, R and the dissolution is more or less isothermal if you take ammonium nitrate E, A is so much greater than E, R that the resulting decrease in temperature is so significant it is often used as an instant cold pack. So what a molecular theory does for you is give you a mechanistic handle and therefore most of the theories in thermodynamics for calculating property changes come from molecular theory. The classical thermodynamic theories are very few as I said I mentioned Van Laar and Dohlser Lake for calculating delta G and they both came up with classical theories we will discuss them in the course. So in summary the whole purpose of this course is to help you calculate thermodynamic properties and therefore calculate changes in thermodynamic properties. It is these changes that are going to be very important as far as the chemical industry is concerned and because you are prediction of both equilibrium and the maximum work that you can get from a system as well as the minimum work you need to put into a system are both determined by these changes. So I think that is the summary of the course you can see that there is a relevance of thermodynamics directly to the industry that plot that I showed you of specially the agreement between the correlation between the price in the market place and the mole fraction or the weight fraction of the substance as it occurs in nature is a particularly revealing. So this is sort of a motivation for studying the subject you must realize that when the master set up the subject it was not at all clear that there were going to be such far reaching applications. It is a remarkable fact that by pursuing the laws and pushing the frontiers you can get results of great significance for the modern chemical industry as well.