 start with the same equations start with closed systems where I have taken delta w is equal to p dv I told you in some of the newer books this will be delta q plus delta w but in that case delta w will be minus p dv it is a minor change but I have taken p v t systems closed systems we say we call these p v t systems which are of greatest interest in chemical engineering. So if you write this down so you get automatically at constant s and v du is less than or equal to 0 that means you can only decrease so the equilibrium criteria is therefore u is equal to minimum but this constraint is important at constant s and v the internal energy has to be a minimum if it reaches a minimum then you have equilibrium the reason I did not start off with equilibrium is notice that there are only two variables in a pure component system that you can hold constant so if s and v are constant u is automatically fixed. So this is not a meaningful criterion when it comes to pure substances when it comes to mixtures it is a very important criteria so it is only in multi-component systems that this criterion will be actually used but formally I will complete this you can see similarly if I now subtract if I rewrite this as da a is u minus ts so I will have d of u minus ts is less than or equal to this minus d of ts so you will get da is less than or equal to minus s dt minus p dv and consequently you get the result that a is equal to minimum at constant t and v which is obviously a much more useful criteria of equilibrium because most of the chemistry experiments you will be working at constant temperature in the laboratory and at constant volume and for close systems again similarly h is a minimum remember h is u minus pv so I subtract of d of I add u plus pv is h so add d of pv which means I will get tds plus vdp therefore h is a minimum at constant s and p and simply say similarly and then g is a minimum at constant t and p so these two are the most important criteria of equilibrium in fact gb you get the same result these are all redundant expressions because you know u and s are independent the rest are just combinations of variables but it is just convenient in calculus to everything at constant t and p but also relates directly to industrial processes which are carried out at constant t and p the whole process does not have to be at t and constant t and p as far as I am concerned the industry is a black box something goes in at t and p comes out of the same t and p so if the envelope is completely dark you know nothing about it what goes in is at t and p and the other is at the same t and p then I have a process at constant t and p for which I can calculate when equilibrium is reached so if the product happens to be the product streams happen to be in equilibrium according to my criteria I know I cannot do better than that in the process we will do that in great detail after we do mixtures to begin with I will discuss properties of pure substances we will discuss work calculations from properties in most systems that do work mechanical work particularly you are not dealing with mixtures then we will go on to mixtures go back to work and calculate separation work and also calculate equilibrium criteria. So we will start off with properties of pure substances now so really there are only two sets of conclusions that you draw from thermodynamics one is about work the other is about equilibrium exactly two sets of conclusions and I will show you very general forms of equilibrium expressions which you can use in any situation in fact after a while you will find if you go to the literature they will simply write down the equilibrium criteria write down an expression in terms of partial differential equations which after about a month or two of familiarity you will also write down everything will always be of the same all the expressions will be of the same form except there will be a characteristic enthalpy change there will be a characteristic volume change which you will be able to guess after a while simply a matter of almost a matter of familiarity but you can derive these rigorously. You have du is equal to Tds-pdv you might say this equation is valid for a reversible process but I am now looking at changes of state from one equilibrium state to the other state so I do not care what the process is I only want to know what the changes in you how the changes in us changes in s and changes in v are related now I have a choice here let me write the equations down all that you need sorry Tds-vdp and I told you I have a choice of p and t or v and t as variables and I want to choose the variables I look at this equation so I have ds I have to write an equation for s is going to be partial of s with respect to pdp plus partial of s with respect to tdt it is also equal to partial of s with respect to vdv I can use either of these criteria either of these equations for ds then substitute it back here if you look at the enthalpy since p is already chosen as a variable the expression for enthalpy becomes a change in enthalpy is simply t partial of s with respect to tdt plus then we have to introduce some variables you are already familiar with I will just go through it and say recall that dh is equal to ?q when p is equal to constant so when you hold one of the variables constant here in a closed system then you know the original equation is du is equal to ?q- ?w so ?q becomes equal to dh and therefore this is no longer independent this is no longer dependent on the path it is independent of the path so I can define rates I can add heat to a system measure its temperature change and I can ask what is ?h by ?t ?q by ?t this we define as Cp at constant p one of the earliest people who measured a large number of properties very accurately actually got a Nobel Prize not for measuring properties but this camera link on us I do not know if you know the name 1850s and in early 19 live quite I think 1920 or 30 he died but the camera link on us laboratory in Europe in Netherlands was a place in the 50s and 60s if an American thermodynamicist or a transport guy who is measuring transport properties and visited camera link on us laboratory and he was not considered a good scientist and it is sort of place of pilgrimage it was the Mecca for property measurements you have to turn up there it was called the coldest laboratory on earth or something because camera link on us himself was given the Nobel Prize for liquefying helium he got one of the earliest Nobel Prize winners he worked with Kitschoff became a prof very early we will get back here so the specific heat is defined this way notice this is measurable this is measurable and therefore this is measurable so the important thing about the specific heat says they are measurable everything that is measurable has a special value in thermodynamics and if it appears on the right hand side do not try to simplify the equations further actually similarly the others I can write down partial of you with respect to T at constant V is called CV this is do Q normally you would not be able to divide the heat exchange by temperature differential and get a property because heat itself is dependent on the path but if you constrain the process suitably you can get functions of state changes in functions of state and therefore you get properties another property that is important and I would not define it here is called C sub sigma which is measured along the saturation line if you have liquid and vapor in equilibrium and you change the temperature by dt keeping the vapor and liquid in equilibrium the heat that is added is measurable and then you get what is called C sigma it is a very easy measurement and therefore it is very attractive it is related to Cp Cv we will relate it little later but right now we will use the Cp Cv as measurable quantities so let us take this alone right now sort of the whole thing is so redundant I can do it in 150 ways there is no point in doing it in those ways unless you need it so if this is true then notice at constant you have this equation at constant P you also have 2h by ?t is equal to T 2s by ?t at constant P this term vanishes so this is equal to Tds in first order differentials work like this when I differentiate the left hand side with respect to T if there is a differential here I can simply divide it by you can do algebra of first order differentials second order you cannot do the same thing but anyway so this is equal to Cp so do s by 2s by ?t is simply equal to Cp so this part this is Cp by T sorry do s by 2t Cp by then I need do s by ?p so you know how these things are done I have to go to s with respect to P I should get I have Dg you just have to juggle these around and if you look at second partial if G is a continuous function of T and P this second partial of G with respect to T and P is the same as the second partial of P and P this is true only if G is a smooth function means all the derivatives must be defined the first order derivative must be continuous function should be differentiable and so on these are generally true actually these some of these relationships break down at the critical point there is a singularity at the critical point but the singularity is so close to the critical point mostly chemical engineers work by interpolation they will get to 0.1 degree from this side of the critical point get to 0.1 degree on the other side and then do everything by empirical interpolation the answers turn out to be right but the critical point itself is a very very big point of theoretical interest fact I think there is a physics of fluids this is a very prestigious journal 1963 or something edition it was called the critical edition it discuss the whole it is this thicker volume only discussing the critical point if some of you are interested you should read is a paper by Fisher which is very famous all about this singularity that singularity gives you a lot of information about structure but it will go off into a lot of physics but if you are interested so if I do this differentiation with respect to P first I get V so this is simply V if I do the second differentiation I get partial of V with respect to T and here you get partial of S with respect to P so these are your five famous Maxwell equations you do not have to write all of them down you have got eight variables three choice of variables that is you can differentiate respect to any two in new order so you what you will have you have to choose this G T and P so three out of eight variables so eight C three possibilities and if you want to be very perverse you can do eight P three because you can turn them upside down and so on and you will get any number of Maxwell relations all of which is meaningless you should only derive those that you need for example S with respect to P is not a measurable quantity but V with respect to T is therefore this becomes indirectly measurable my only aim is to get measurable quantities on the right hand side so I have S with respect to P I will now replace it by minus of V with respect to T and notice when I do the differentiation with T P is constant so automatically P here is constant notice if you got V with respect to T at constant S that is not necessarily measurable but this is measurable okay P V T are considered measurable so all their derivatives are considered measurable so I have two equations one is in one is an equation for S which says DS is equal to CP by T DT plus S with respect to P which is the same as V with respect to P D P I would not bother to write P constant here it is sort of from the context it is understood or you can go back and put it then DH is equal to CP by T I am sorry CP DT plus V minus T partial of V with respect to T so these are the two equations that I have to integrate this I am interested in calculating H any two variables U and S if P and T are the variables is more convenient to calculate H and S so this is my equation one this is my equation I want to integrate these two equations now I have further choices simply because of the redundancy if let us say this is P question P and T I have to integrate from point one to point two I have to integrate from P1 T1 to P2 T2 let us consider some reference point shall be P0 T0 reference point I can choose my reference point as I please because I will integrate from P0 T0 to PT and tabulate it and you are not supposed to I can assign any values at P0 T0 but you do not worry about that value because you will always take differences you never take absolute values in classical so I start from here I want to get some point PT which is of interest so since civilized integration requires that you do not go through some path like this you will get the same answer but if you are perverse you can choose a path like that and then see how to integrate over it but since we have enough difficulties without being perverse you choose straight line parts on this diagram either this way or this way even this it is better to decide it once and decide once and for all how you are going to do the integrations because they are entirely arbitrary you must choose one that is based on convenience. So let us call this path a this is path B along we have to look at what information you need in order to do the integrations if you look at let us take a look at the enthalpy equation we could arguments are very similar for entropy anyway enthalpy equation requires that you have CP information at constant P so if I follow this path I need CP at P0 in order to get to this point intermediate point because I am doing that integration at constant pressure at constant pressure if you look at this term will vanish I need only CP okay then after I turn it is constant temperature so this term will disappear I need PVT information here I will call it equation of state at T by equation of state I mean measurements of PV and T and automatically all their derivatives this means equation of state or PVP all our equations state are practically empirical you can derive them also from molecular theory but you will have to have an empirical input in terms of what the intermolecular potential is given the potential I can calculate so this is what I need along this path along this path you need equation of state at T0 and then here I need CP information CP at P now this is where my choice of reference state comes in if I choose my P0 to be low enough I know all substances behave ideally so if I choose this to be low enough what I need a CP at P0 is actually CP for an ideal gas if P0 is chosen sufficiently low in particular and simply choose P0 equal to 0 very often you choose P0 equal to 1 because up to 1 atmosphere gases behave ideally but really where is earlier old systems use to use 1 PSIA 1 pound per square inch absolute so it does not really matter but nowadays the normal charts are prepared with choosing P0 equal to 0 very often and then T0 can be chosen at as any temperature as long as you go to sufficiently low pressure you will get the gas phase and you will get ideal behavior so the advantage here is that if you go back here you look at this information this is 0 for an ideal gas and you know this already if this is 0 then H is a function only of T CP becomes a function only of T if CP is a function only of T we often denote this CP as CP0 for an ideal gas it is not only only a function of T you can actually calculate CP0 from you can get it from two sources one is spectroscopic data you can measure for low pressure gases you can measure the specific heat from spectroscopic information you can get the values the other thing is you can actually calculated from molecular theory so fundamentally you have two sources for CP0 other than direct experimental measurement of CP whereas on this side this CP is no longer a function of T alone it depends on the conditions of the gas so here I need more information than here the experimental information is the hardest to get so you simply choose the path where less experimental information is required so we will always choose path A simply a matter of convenience all textbooks that report data all tables are prepared by doing the integrations along this path because however complicated a molecule I can get CP0 information for that molecule as an ideal gas so I can always go up to the temperature of interest then I can do this integration of course this temperature will vary so you really need data at all temperatures that is a different matter here again that is also true of these lines while you can go on here you will have to vary this at all pressures because pressure of interest is not fixed so what we will do is so we will settle for integration along a path A with P reference often written as P0 as 0 in some books they will say one but if the pressure is low enough if it is one atmosphere or one PSA anything less than one atmosphere it is completely ideal but you have to be careful when you use two tables if one fellow uses one reference point another fellow uses another reference point you have to make the conversion it is trivial it is simply bookkeeping so there is no point telling you a thousand times but if you make mistakes you will get minus one for any every number that you copy wrongly from the chart but anyway this is what we do if we do this let us look at integral of H so I am doing this calculation H from I will write we will do equation to first integrate equation to from T0 P0 to Tp we agreed we would first do it at P0 so you will get H of T P0 I am integrating from T0 to first I will do this as T P0 then from T P0 to Tp two steps so if I am integrating this equation at constant P0 P equal to P0 this term will vanish I am doing this integration this is simply integral Cp0 dt from P0 to T no no sorry I am doing it at constant P0 correct Cp0 is not a constant it is a function of temperature that is known it is assumed to be known want I will write Cp0 of T dt then I have H of T, P – H of T, P0 now I am integrating from P0 to P of this term so what I will do here is replace P0 by 0 because P0 is chosen sufficiently low so the gas is ideal so V – T doh V by doh T is 0 from 0 to P0 so the integrand is 0 below that so I will simply replace this by 0 and then I am going to choose this to be 0 choose or set to 0 that is why you have to be careful when you go from table to table in one table T0 P0 will be different from T0 P0 from another table so if that number is different then you have to add that number on you want to calculate the if you want to compare data from two tables luckily by now people have accepted many results of this kind usually for example T0 is as far as possible chosen as the triple point of the substance pure substance as a triple point which is invariant so usually you will choose that data so these are there is some conventions but they are not always followed because the correction is trivial you just have to check the tables so finally you get one result the integral of this can be now written like this simply H is equal to because it is simply integral 0 to T Cp0 of T dt plus integral this is from T0 to T 0 to P of V – T doh V by dt so you will find any table or chart on top it will say H the T0 P0 values will be given at 0, 80 degrees K is equal to 0 so right on top of the table that one line will tell you what T0 P0 is it will write they will write their H at 80, 0 is equal to 0 those numbers represent T0 P0 in the chart also they will tell you exactly at H is equal to 0 at this condition that is your reference condition now this is a little trickier to integrate I will tell you that the trivial thing in this is simply that V with respect to be T at low pressure will go as R by P you will have to integrate from P is equal to 0 to P is equal to P R by P dp will give you a log P log P goes to 0 as goes to – infinity as P goes to 0 so there is 1 – infinity there so you have to work around it the way you work around it is simply in thermodynamics you simply add R by P and – subtract R by P so you will get Cp by T dt – doh V by doh T – R by P dp – R by P dp. Now we do the same thing as we did there you will get the following result S is equal to when you integrate Cp by T you simply get Cp0 by T dt from T0 to T and when you integrate this you will still get the same result – P0 to P actually P0 will be made 0 now doh V by doh T – R by P dp then you get this here which is – R log P plus R log P0 plus S at T0 P0 so you are integrating from T0 P0 so you will get this is S at T and P so S at T and P is equal to start from S or T0 P0 at P0 you integrate from T0 to T we will simply get Cp by T dt the remaining you have to integrate from P0 to P as far as integrating this is concerned I replace P0 by 0 because 0 to P0 the gas is ideal doh V by doh T – R by P is 0 so that part of the integral is 0 so I have simply added a 0 here then I integrate this I get R log P – R log P0 right now I know there is a problem here because I am going to set P0 equal to 0 but notice that this quantity in parenthesis refers to the entropy contribution at the reference condition that entire thing is at the reference condition so instead of setting S at T0 P0 to 0 I will set this to 0 and it does not matter because I calculate S difference between in entropy between two points I will subtract of R ln P0 that same identical quantity from itself so it will cancel if I take the limit of P0 going to 0 will go to –infinity so I avoid taking that limit and simply set this component equal to 0 this clear so on top of every chart there will be a chart that gives you thermodynamic properties of tables it will say H at 0 at 80 0 equal to 0 S at 80 0 plus R log P0 equal to 0 so they would not write S at is equal to 0 and does not matter because it is the same –infinity that you are subtracting you have to be careful there are different kinds of infinities so you cannot cancel infinities unless they are identical infinities. In fact according to George Gamow there are only three types of infinities in fact he says we have not progressed much as far as infinities are concerned we are the same condition that the hotend tarts where they could count apparently only up to three this is we can also count only three infinities we do not know how to we do not know how to account for more infinities so that is it this is the basis of all charts you have to be careful if the P0 value that you take is different in two charts then the different that R log P difference you have to account for if it 0 everywhere is no problem but if there is a number there and the units are different then you have to be careful about your choice of pieces normally the convention is to choose P in bars take P0 equal to 1 in which case this will become 1 but 1 to 0 to 1 is 0 anyway so this can still remains in 0 to P then make any difference but this number you have to be careful about when you are comparing two different charts it is the same chart you can blindly subtract 99% of the time you lose the same chart same table so all this all that I am saying is of no consequence but if you are taking data from two different tables please be careful so these are the two integrated equations H of T and T is this so this is the basis of all your if you like I will write here plus H of T0 P0 and set it and say this is 0 so let us look at the chart of a substance normally the charts that are used and you are used to PVT course diagrams but the more useful thing is the pressure enthalpy diagram or the temperature entropy diagram the TS diagram is used very extensively by mechanical engineers the TH diagram is in many ways more useful for chemical if you look at low pressures you are talking of really the ideal gas you are putting the enthalpy usually it is a choice the choice is between one intensive variable and one extensive variable because what you would like to do is to show how the phases differ in properties so if you have a liquid phase and a gas phase one of the properties will change the pressure will remain the same for both but the enthalpy will be different and you can read the latent heat by simply taking the enthalpy difference between the liquid and paper now these are prepared experimentally what I will do is draw the curves as they appear normally for most substances you get a liquid plus vapor region I will trace these regions in a minute what you do is trace isotherms isotherms go like this this is a typical I have not gone the wrong place this is temperature T1 less than the critical temperature there is a critical temperature of course this T is equal to TC and then above the critical temperature complete the normal qualitative diagram the one bar line may be somewhere here usually indicated this is one atmosphere or one bar in almost the charts come with the bar looking at this you can say what the lines look like on this diagram on this diagram they also have plots of constant S constant V a forgotten which is steeper and we will find out in a minute you can give qualitative arguments and find out which of them is you may switch that V and S just non qualitative curves for example if I am looking at H and P you have I will go back to this differential equation such that H is equal to CP DT you are looking at pH lines so I am always looking for doh P by doh H to find out what the slope of the line is like it is easier to get doh H by doh P I will do all this for unit mass so you will find in the charts they use small H in order to indicate specific enthalpy so if I do this I get partial of H with respect to P at constant T would be simply V-T doh V by doh T. So if you look at the isotherms at very high pressure they will have low value but the interesting thing is at low pressure doh H by doh P is 0 or doh P by doh H is infinity we are looking at the inverse of the slope here so these lines have to be vertical when they come to sufficiently low pressure here all these lines are vertical isotherms are vertical here at low pressure you are simply saying they are ideal gases and therefore the enthalpy does not change with pressure it is function only of temperature now look at the isentropic and constant volume lines what I am interested in is doh H we will make this H this is small Cp V doh V by doh T I am just writing small Cp is specific per unit mass small V is per unit volume I want doh H by doh P at constant S for that it is actually easier to go here the earlier one you get is equal to Tds plus Vdp so the constant H with respect to P at constant S will simply give me V so the isentropic lines the constant S lines have a slope equal to 1 by V in calculating doh H by doh P the slope on this is doh P by doh so it is 1 by there