 So we are going to look at property relations and in property relations we will consider relations between properties of a simple compressible system that is a fluid system. So one two way mode of work and that is expansion PDV. We will generally assume a unit mass of the system. So we will be working in terms of specific properties small s, small u, small v, p anyway it is small and t will always be capital. The tools which we are going to use is and if a few of you are uncomfortable with tools we will spend some time on it. The calculus of exact differentials we will come to it when we need to know what it is about. Calculus of partial derivatives related to it and we are going to look at energy functions. Everything is a play on energy functions. By energy functions we mean the energy itself and since we are going to consider actually here I should have said simple compressible fluid system at rest. So by default we are going to assume that delta E or DE will be replaced by DU. Although we know one energy that is the thermal internal energy U which is defined by first law. For convenience we have also defined other energy functions enthalpy which is U plus PV. We call it an energy function because it is a definition using the definition of the internal energy U and there are two more functions which we will define for convenience and all this set of four functions U, H and the two more to be defined will be known collectively as energy functions. And why do we define them because it is helpful to define them just because we found that U plus PV is a common sub expression which we come across. So we give it a short form enthalpy so we will do with two more energy functions. So the first law defines U, second law defines S the entropy and you have already seen for a simple compressible system the basic property relation which relates the two. Remember that this relation TDS is DU plus PV is a that is the basic property relation. It includes first law it includes the definition of entropy. We will also use the first law. Now here the first law is DQ equals DU plus DW where DW includes all modes of work. Whereas in the property relation the left hand side becomes TDS because we are considering a small elemental reversible process linking two neighboring properties and naturally because it is a reversible process DW has to be nothing but PV that is the difference. We will also be using the second law which dictates that TDS is greater than or equal to DQ or TDS is greater than or equal to DU plus DW. Now here if you have not realized already what you can do is the following. Compare this equation property relation with this equation. TDS has to be greater than or equal to DQ. On the right hand side DU is the same what can you say about PVV and DW? If the left hand side TDS is higher than DQ, PVV has to be higher than DW. But DW is PVV plus some other component say stirrer work or electrical work which for a fluid can only be one way work. So that work component has to be what? Always has to be less than 0 and that means for a fluid system work other than PVV work for example stirrer work which will all be one way work modes will always be such that you can do work on the system. System cannot ever do the work on some mode by which one way work is done. So that explains our observation that if you put a stirrer in a fluid, fluid is initially at in a state of equilibrium whatever you do the fluid will not be able to stir the stirrer. The stirrer will be able to stir the fluid. Similarly the fluid will have some electrical resistivity. So you can put two electrodes and pass a current and you can provide electrical work input to that fluid. But you put two resistors and say push a current through it, well it cannot because you know it does not have the capability of getting charged or discharged. It is not an electrolyte. So now look at we will now start playing with our energy functions. The internal energy u related to dq and dw by first law and if I expand dw it will become du plus pdv plus dw other that is what I have said. We have done this yesterday and if I consider a constant volume process for which pdv is 0 then dq at constant volume is du plus dw other. This also we know nothing special here. So let us go away quickly. Similarly in a way enthalpy can be treated h is defined as u plus pv. So expand it dh is du plus pdv plus vdp. Now du plus pdv can be replaced by dq. So we get dq which is du plus pdv plus dw other. This is the same relation which is here. Can now be written down as dh minus vdp plus dw other and if I consider a constant pressure process which makes dp equals 0 then dq at constant p is dh plus dw other. Fair enough? Nothing no surprises here. This is all known. That is because we are using only first law. Yes sir? No significance. It is an algebraic term. We will discuss that offline that earlier when we used to have reciprocating compressors. If you consider one complete stroke or one complete cycle because of the associated flow work involved the work per cycle can be shown to be equivalent to integral of minus vdp. But that is under very very special conditions. But now we define two more energy functions. The first one of them is known as the Helmholtz function. In mechanical engineering we do not routinely use them but when it comes to physical chemistry these are routinely used. These two functions are named after two great contributors to thermodynamics Helmholtz function and Gibbs. The Helmholtz function usual symbol is a but traditionally the symbol used to be f. So you may still find and also the Helmholtz function and Helmholtz Gibbs function. Other names are Helmholtz energy or Gibbs energy and sometimes Helmholtz free energy and Gibbs free energy. It is defined as u minus ts. So s enters in the definition unlike in enthalpy where it is u plus pv and because for u it is defined by first law and pv does not have entropy. So we do not have to use second law when we played with du and dh as on the previous slide. So if you differentiate it you will get da equals du minus ts minus s dt replace du by dq minus dw transpose terms you will get dq minus ts on the right hand side which by the second law is less than or equal to 0. Remember ts is greater than or equal to dq. So dq minus ts would be less than or equal to 0 transpose terms and you will get dw is less than or equal to da minus s dt. I am neglecting this part I am saying this less than or equal to 0 means da plus dw minus s dt is less than or equal to 0. So dw is minus less than or equal to minus da minus s dt and work done in an isothermal process would be less than or equal to minus dA t. And why do we get that inequality here because we have to invoke the second law and what does this mean? This means that A the Helmholtz function is similar to a potential the decrease in which represents the maximum work that can be obtained in an isothermal process. Compare this with water at a height it has a potential for some body m of mass m in gravitational field G is at a height with respect to some datum of h1. We reduce its height to h2. So h1 minus h2 is the decrease in potential energy that is the maximum work we can extract out of it. If there is no friction while it is decreasing that will be the work which will be extracted out of it. In a similar fashion A can now be considered to be a potential. So we can even call it Helmholtz potential. The decrease in which represents the maximum work that can be obtained in an isothermal process. Now the last of the energy functions which we are going to look at is the Gibbs function. In fact this is more important than or more popular and more celebrated than the Helmholtz function. It is defined as u plus pv minus ts everything is put into it. U pv ts all properties are included in it. Of course using the definition of enthalpy it can also be written as h minus ts. Differentiate du plus pdv plus vdp minus tds minus sdt. Use first law so you will get dq minus dw other plus vdp minus tds minus sdt transpose term to get dq minus tds on the one side which by second law is less than or equal to 0. And this means dw other other than pdv work will be less than or equal to minus dg plus vdp minus sdt. And that tells us that if we have a process which takes place at constant pressure and constant temperature then dw other during that process has to be less than or equal to minus dg at pt. So what is the meaning of that? The meaning of that is the Gibbs function can be considered as a potential. The decrease in which represents the maximum work other than expansion that can be obtained in an isobaric or isothermal process. And this is important because many of our chemical reactions start point is one atmosphere 30 degree C. We want end point also to be one atmosphere 30 degree C. So during this reaction what is the maximum energy that we can extract as work? So for fuel cells, electrolytic cells it is the Gibbs function which is of importance. And again the inequality here turns out to be the inequality due to the second law of thermodynamics. Now chemists and chemical engineers would use this idea of Gibbs function very heavily always they cannot live without Gibbs function. But unless we go into combustion and kinetics of combustion we do not really have to worry about Gibbs function. We will use these four functions as a handle to obtain some very important property relations. And for that we have to use some characteristics or some properties of exact differentials. So we will now look at the properties of these four functions but we will be using the specific property version. Whatever I showed you in all these slides was capital properties. So now we will divide everything by the mass of our system. Everything will be written in terms of specific properties. We will use the property relation and also the following formula. And this is one major characteristic which we should remember that if z is some function of x, y and dz can be written down as m dx plus n dy and it is an exact differential. And if z is a proper function what they say analytical function of x and y dz will be an exact differential. We should remember that property is a function of state. So for us it is an analytic function of other properties. So all properties differential of properties will be exact differential that is important for us. Then m and n can be represented in terms of partial derivatives of z with respect to x and y. Now y am I writing this y subscript here and x subscript here. The reason for that is simple. The reason is when it comes to mathematics when you write some f of x y or z of x y partial of with respect to x y is assumed to be constant. There is nothing special about it. But in thermodynamics we have a choice of selecting any two variables as our independent variables. Suppose you select z to be entropy for a fluid system. x and y can be p and v can be p and t can be p and u can be h and t any combination. Particularly if I select x to be t y can be volume pressure enthalpy any of the other functions that I have defined except entropy because entropy is the z part. So I can have variation of entropy with pressure at constant volume variation of entropy with temperature at constant volume at constant pressure. So I have to be specific about what is the pair of properties that I am using. So when I write partial of z with respect to x at constant y that means I am considering z as a function of x and y and then differentiating it with respect to x keeping y constant. So if I write partial of s with respect to t at constant v that means I am considering t and v as independent variables considering the property s as a function of t and v and then differentiating s with respect to t by keeping v constant. Similarly I can have a different derivative partial of s with respect to t at constant p. Here the whole basis is different. I am considering t and p as independent variables to make that clear with subscripts like x and y will always be used in thermodynamics and also the cross derivatives are independent of the order. So the second derivative of x first with respect to x keeping y constant and then with respect to y keeping x constant is equal to the second derivative obtained in the reversed order and because of that the partial derivative of x with respect to y sorry partial derivative of m with respect to x with respect to y with constant x and partial derivative of n with respect to x at constant y because both of them would be second derivative of x second cross partial derivative of x order of differentiation is an important. So these two or three relations are of utmost importance to us. Later on we will look at the properties of Jacobians. We will come to that when we need to use them. So now we really go through our basic exercises in the next four slides. We look at the internal energy u. We know that it is tds minus pdu property relation. Tds is du plus pdu. Now notice on the left hand side we have du on the right hand side we have ds and dv. So we consider that for this equation that equation tells us that for u s and v are very convenient independent variables. So we consider u as a function of s and v and because of that we can straight away write notice the factor here of ds is t. So t must be partial derivative of u with respect to s at constant v. Similarly the coefficient of dv is minus p. So minus p must be partial derivative of u with respect to v at constant s and many people consider these t and p as important. These two relations are considered very important and they are considered to be the thermodynamic definition of t and thermodynamic definition of p. Because t here is related to thermodynamic variables u, s and v. Remember my very first lecture that any system will have at least one property and that is volume because the boundaries are defined. u comes out of first law, s comes out of second law. So this is a quantitative definition of t with respect to u, s and v. And pressure we have defined using fluid mechanics but here we have a thermodynamic definition of pressure which has nothing to do with forces or anything. It is just the partial derivative of internal energy with respect to volume at constant entropy with a negative sign. So these two relations are often considered as thermodynamic definition or thermodynamic basis for temperature, thermodynamic basis for pressure. But what is more important is this is the m and n relation. But now let us look at the cross relation. This means the partial of temperature with respect to volume at constant entropy will be partial of pressure with respect to entropy at constant volume of course with a negative sign. And this relation is an esoteric relation. We never expected it. It relates variation of temperature with volume at constant entropy to variation of pressure with entropy at constant volume. This such we are going to derive three more such relations and these three relations plus this are going to be called Maxwell's relations or Maxwell's equations. This is one such equation based on the internal energy u. Now this is algebra but after going through the four slides of algebra we will come to a basis and we will understand what these four Maxwell's relations really represent. So let us continue. We consider enthalpy h and expanding dh as a differential and then using the property relation we obtain dh is tds plus vdp. All that I have done is du has been expanded as tds minus pdv. Now that means h is related to ds and dp. So we consider h as a function of s and p. And hence t must be partial of h with respect to s at constant p and v must be partial of h with respect to p at constant s. And the cross derivative gives us this second Maxwell's relation. Again notice temperature pressure entropy volume entropy pressure. Some funny combination between thermodynamic variables is dictated by our property relations algebra and calculus. We get two important relations when we work with Helmholtz function and Gibbs function. In this slide as well as in the previous slide you will notice that entropy occurs on either side here as well as here. Similarly on this slide entropy occurs on the left hand side as well as the right hand side. But when we come to Helmholtz function and we differentiate it and use the property relation notice here that du has been expanded as tds minus pdv. You get da equals minus sdt minus pdv. So natural independent variables are t and v. We consider a as a function of t and v. The first partial derivatives give us s and p as partial derivatives of a. And the Maxwell's relation which comes out of cross derivatives is partial of s with respect to v at constant t. Notice on the right hand side we have only pvt no entropy no enthalpy no internal energy. So the third Maxwell's equation this equation which comes out of a is important because it relates the variation of entropy with volume at constant temperature purely to pvt data that is the equation of state and to nothing else. So although entropy is an energy function defined in a wake dq by t reversible. Its variation with volume at constant temperature is related only to pvt to our equation of state and to nothing else. It is not related to cp, cv, nothing because we know pvt data has nothing to do with the heat interaction. It is just an equation of state data. Similarly the fourth Helmholtz function based on the Gibbs relation function which is h minus ts expand dg here I have used dh is tds plus vdp which we have already derived. So dg comes as minus sdt plus vdp. So natural independent variables would be temperature and pressure. So the first derivative give us entropy and volume as partial derivatives of Gibbs function but the Maxwell's relation which comes out of it gives us the variation of entropy with pressure at constant temperature only in terms of pvt information. So the last two Gibbs functions are important. They are so important that many people tend to simply many students would be doing is simply you know learn this by heart or have a small cheat somewhere in their pocket listing this down. So called cocksheet or they have some mnemonic or some shloka which puts all these things in order. These things are important particularly the last two are very important because they relate entropy, variation of entropy with volume and pressure respectively to purely pvt data. pvt data is the simplest data to measure. No cp, cv is required. So it is important to be able to use this and to use this properly we must be able to remember this and we will be able to remember this if we are able to derive them very quickly. And for that we will now look at the thermodynamic basis for these. Before that these relations as well as their reciprocals. For example you take this relation partial of s with respect to v at constant t is partial of p with respect to t at constant v. You can go through the calculus and this becomes equivalent to saying partial of v with respect to s at constant t equals partial of t with respect to p at constant v. We have analytic functions so there is absolutely no harm in saying that. So these relations and their reciprocals are known as Maxwell's relations and they are very useful property relations. And as I have already said the third and fourth ones relate entropy variation to purely pvt data. And they help us reduce the requirement of cp or cv data for mapping the state space. Now how do we remember them? We come to an absolutely basic stuff in thermodynamics. Let me go to this. Let me ask you the following question. Let us say that I have a fluid system and I consider a process some process and I say that I consider a process from 1 to 2 which is a quasi-static process. I have shown it in a continuous line on the pv diagram so this is a quasi-static. What does the area under the curve represent? This represents expansion work done. Now I consider a TS diagram and consider a process maybe the same process quasi-static maybe it gets represented like this. It also is a quasi-static process. What does the area under it represent? Does it? See it represents integral Tds. Just the way this represents integral pdv which comfortably we say uneditingly is work of expansion. But this represents integral Tds which we will be able to evaluate because it is a defined curve. dq is what? dq is see Tds is greater than or equal to dq. So dq is less than or equal to Tds. So the q will have to be less than or equal to this area. When will it represent the area? If it is a reversible process. So a TS equals dq only if process is reversible. So remember the difference. This area is q only for a reversible process. It is integral Tds alright. Integral Tds will be dq only for a reversible process. Whereas integral pdv quasi-static process you are able to evaluate it so it is W expansion. No doubt. So get this idea and now let us go back to our, I considered a process. Now instead of a process I consider a cycle executed by a fluid system and let us say I am considering a reversible cycle. So it is executed in an absolutely reversible way. Otherwise there is no restriction on the processes which could be there in the cycle. It could be a small cycle, dp dv type or it could be a big cycle taking it anywhere it feels like. We represent it on a pv diagram. We also represent it on a TS diagram and we will see something like this. We will see a pv representation and we will see a TS representation. It is a reversible cycle. So the closed area of the pv diagram of the loop in the pv diagram will represent work done per cycle or in that cycle. Will it be total work or will it be pdv work? But I have said it is a reversible cycle. It is a fluid system. So the only work done will be pdv work. So this would be the work done in the cycle which also is the pdv expansion work because it is a reversible cycle no other mode of work is admissible. It is a reversible cycle. So the area on the TS diagram is heat absorbed during the cycle, net heat absorbed during the cycle. Are the two areas equal? Yes they are. What does it mean? We said they are equal because cyclic integral of pdv will be cyclic integral of Tds for a reversible process. That means w cycle is q cycle that is first law. But this is true for any of those cycles. I did not assume any particular cycle. So that means when I transform a loop on the pv plane which represents a reversible cycle to a loop on a TS plane which represents the corresponding reversible cycle. The area does not change. What does it mean for the transformation? It is not only area preserving but it is 1 is to 1. We say that this and that means that the Jacobian of the transformation is 1. And what do you mean by Jacobian? Have you understood what is meant by a Jacobian? See when we have let us say this is pv and this is TS. And I consider a loop here. And I consider the corresponding loop here. And I consider integral dpdv double integral over the cycle. And this area is double integral dTds again over the same cycle corresponding mapped point to point. If it were a 1 dimensional transformation and if I want the length I would have just the scale ratio coming in. Here it is a 2 dimensional transformation. So the scale ratio is known as the Jacobian of TS with respect to pv. We will define this. The scale ratio is defined the area scale ratio. This can be derived using analytical geometry. If you look up books by Thomas or Apostol you will find it there. Area scale ratio is known as the Jacobian j of say TS with respect to pv or pv with respect to TS. The other symbol used for this is partial derivative of TS divided by partial derivative of pv. And the expanded form of this is like this. It is a determinant 2 by 2. And when it is partial of TS partial of pv it means partial of t with respect to p at constant volume. Partial of t with respect to v at constant pressure, partial of s with respect to p at constant volume, partial of s with respect to v at constant pressure. This is the definition. This is the symbolism. And the name is Jacobian. Jacobian of transformation or transformation Jacobian. And it is a determinant. So if we know t as a function of p and v and s as a function of p and v we can evaluate. And because our areas are equal, this means that for us this is equal, this equals 1. If t, p and s, t, p, v and s represent temperature, pressure, volume and entropy respectively. Otherwise if you write x, y, z or p, q, r, s it is just a symbol. But we have shown that because of the first law of thermodynamics and the definition of entropy. If t is temperature, s is entropy, p is pressure and v is volume, the value of this Jacobian is 1. Now we will demonstrate that the four Maxwell's relations are hidden inside this. And for that, see one of the greatest properties of Jacobian is remember that the chain rules for partial difference are very complicated because more than one variable is involved. However, the characteristic of a Jacobian is that a Jacobian can be treated almost exactly like an ordinary derivative. And we will see what the properties of Jacobian are. So this means the transformation of Jacobian of the transformation is 1. So I have just now written this part but even the reciprocal is true. Now we will use the properties of the Jacobian for relating them. Now this is like an ordinary derivative. Jacobian of u, v with respect to x, y multiplied by Jacobian of x, y with respect to u, v is 1. So it is like saying dy by dx, ordinary derivative will be 1 divided by dx by dy. In one case you consider y as a function of x, in another case you consider x as a function of y, perfectly analogous. Not only that, the chain rule of Jacobian is also dx by dz is dx by dy, dy by dz. If x is directly a function of z that is the left hand side and directly a function of y and y directly a function of z, that is on the right hand side. So partial of that is Jacobian of pq with respect to tu is partial of pq with respect to rs, the intermediate pair of variables multiplied by partial of rs with respect to tu. But the importance is if instead of two variables in the numerator and denominator, if there is a common variable in the numerator and denominator then what happens? It turns out that the Jacobian of u, y with respect to x, y will have the same value if you just interchange the order, Jacobian of yu with respect to yx because you know the determinant does not change its value even if you flip pros and columns. But if you interchange one of them yu xy then you will get a negative sign or uy yx you will get a negative sign. But important thing is any one of them is equal to partial of u with respect to x at constant y. So all that we have to do is remember this line and you can remember it as a symbolism that partial of u with respect to x at constant y can be written down as Jacobian of uy with respect to xy. Sometimes while saying we even say partial of uy with respect to xy that means Jacobian if there are more than variable. This chain rule for Jacobians and let us see where we reach using this. Now let us say we want partial of t with respect to v at constant s. Is this a candidate for Maxwell's relation? Let me go back a bit I think I missed something. When see in thermodynamics there are so many variables, so many properties that you will come across partial of p with respect to v at constant h and something like that. We have to decide whether a particular partial derivative is a candidate in the Maxwell's relation or not. For that you notice one thing remember we remember pv diagram and ts diagram. So in our brain pv is a comfortable pair ts is a comfortable pair but pt is not pt vt or ps vs are not. Here you see one pair is t and s take the numerator variable and the constant variable. One pair is t and s another pair is v and s. On the right hand side one pair is pv another pair is something useless sv. Similarly here there is a ts pair here also there is a pv pair in some order there is a ts pair there is a pv pair there is a ts pair there is a pv pair. So when you come across a partial derivative partial of x with respect to y at constant z look at the pairs xz and yz. One of them should be pv or ts in some order then it is a candidate you will find it somewhere in the Maxwell's relation. So based on that I am considering one possible candidate which is partial of t with respect to v at constant s there is a ts so this will be somewhere found in the Maxwell's relation. So we write it first as a Jacobian s is constant so this is Jacobian of ts with respect to vs. We want to transform it so we take it and multiply it by Jacobian of pv with respect to ts. Remember what I told you about conversion factor conversion factor is dimensionless the value of conversion factor is 1. So this is an area conversion factor value is 1 dimension dimensionless so this will not change anything. So when I multiply it by 1 pv ts what happens by chain rule I can simply cancel out the ts from here and cancel out the ts from here look up the chain rule for Jacobians. So this becomes partial of pv with respect to vs and you have one variable common in the numerator and denominator that would be the common variable for the derivative this represents and the uncommon variables are not one below the other they are across a diagonal so there will be a negative sign. So this becomes minus partial of p with respect to s at constant v check out whether this is a valid Maxwell's relation you have the four relations with you check them out which one first one meanwhile you try second or third one whether you get it by this method are you getting it okay. So now we should not be scared of any partial derivative in thermodynamics find out whether it is a candidate for Maxwell's relation how will you find out take the variable in the numerator and the constant variable it could be if it is a pv ts in some order then yes if not look at the denominator variable and the constant variable it should be pv or ts if not what to do that we will see when we come to exercise and if one of them is pv or ts continue represent it in terms of a Jacobian use the Jacobian transformation because you know one of them will be pv or ts so when you use either this or its reciprocal that pv will be get will get replaced by ts and you will get the other part of the Maxwell's relation you should be absolutely comfortable with this and now remember that Maxwell's relation is nothing but mathematical depiction of what we know that if we have a reversible cycle of a fluid the area on the pv diagram which represents the work done during a cycle will equal the area on the ts diagram which represents the heat transfer during that cycle and the heat transfer during that cycle is represented by that area because it is a reversible cycle and because the cycles are reversible the directly the property relations can be used we do not have to use greater than or equal to which comes out of the second law of the model and Maxwell's relations are mathematical ghosts which come out of this so try with this and may be by today you are scared of Maxwell's relation should go away so what next the if you look up item 10 which is property relations we have come up to derivation using area relation 10.6 now I will just flip it around we will do some exercises and then we will go to the Clausius-Clapeyron relation in Clausius-Clapeyron relation we will realize what is the link with not just steam tables but the link with the change of phase and we will also be able to link the fact that remember that school boy knows two facts one the melting point of ice reduces when it is under pressure so you put a wire and hold it under ice with weight the wire goes into the ice right a school kid also knows that ice floats on water so the specific volume of ice is larger than the specific volume of water he also knows that ice requires some heat to be melted to a liquid are all these facts link to each other for example if ice some modification of ice where not to float but to sink in water would the melting point of ice instead of lowering with pressure increase with pressure these are linked and we will link them up using Clausius-Clapeyron relation.