 Now it is time for me to go over relations between properties. Now what I am going to do is I have a set of slides which I am going to use. We have said earlier that second law of thermodynamics provides us three things, first possibility that we have seen. In limits on processes we have seen efficiency is less than or equal to something, delta s less than or equal to something. If you say less than delta s is greater than or equal to dq by t that means that for a given delta s there is going to be a limit on the amount of q that you can have. In fact when we combine first and second law we will be able to quantify this limit in some way the so called exergy analysis. The third thing the second law allows us to do is to derive property relations. And the first property relation for a simple compressible system or a fluid system at rest is we have already seen T s equals du plus pd and that is what we are going to do. In property relations we are going to consider relations between properties of a simple compressible or fluid system at rest and we will be assuming a unit mass of the system. So we work with specific properties and the mass of the system does not have to be explicitly included in the proceedings. The tools which we are going to use are calculus of exact differentials. We have used exact differential but only one property of that exact differential. Now here we will be using some more properties. We will be using calculus of partial derivatives along with calculus of exact differentials. Before we go to the property relations proper we define some energy functions. Energy functions are properties derived properties using the internal energy of a system. The internal energy of the system is the primary function, the energy itself and it is u the thermal internal energy. The second law defines s which is the second more important energy function. Then we have already seen the basic property relation between the two. Tds is du plus pdv. Now apart from basic property relation we will be also using the first law which is dq equals du plus dw. It is important for us to see that this is a property relation. Here pdv is only pdv nothing more. Everything here is property or a change in property. No interaction is involved in the basic property relation. Although you can say pdv represents the expansion work done. Let it represent but here the significance of every term is simply a property or a change in some property. But we will also be using the first law in which dq and dw are interactions. And here dw is the differential of the work including all its components. So do not jump to the conclusion that dw is pdv. It could be pdv plus some other components of work. And we will also be using the second law which is with transposition of T turns out to be Tds is greater than dq or Tds is greater than du plus dw. Here I have replaced dq by the right hand side of the first law expression. Now let us do some exercises. We first have the internal energy u and let us expand this the expression for first law. dq is du plus dw which is du plus pdv plus dw other than pdv. So this just tells us that dq at constant volume is du plus dw other where dw other means dw other than pdv. We are also familiar with the second energy function enthalpy. It is a derived function and we have defined enthalpy equal to u plus pv. Differentiating this we get dh equals du plus pdv plus vdp. If you use the first law dq is du plus pdv plus dw other substitute for du plus pdv from this expression and you will get dq is dh minus vdp plus dw other. From which we see that dq the heat transfer absorbed at constant pressure is dh plus dw other. So these two expressions one should be familiar with because common mistake is assumed that for a constant volume process dq equals du that is not true. It will be true only if dw other is 0. Same thing for a constant pressure process dqp will be dh only if dw other is 0. Now we are going to define two spatial energy functions. The first of them is the Helmholtz function given the symbol A. Kemi's quite often call this the Helmholtz free energy or simply free energy and use the symbol F. So you should be familiar with both the symbols A as well as F. A the Helmholtz function is defined as u minus ts. Notice that ts has the dimensions of energy. So we can combine u and ts. It is defined as u minus ts. If you take a differential you will get da equals du minus tds minus sdt and if you apply first law du is dq minus dw and if you now apply the second law you will get da plus dw plus sdt equal to dq minus tds which according to second law is less than or equal to 0. Keeping dw on the left hand side it tells us that dw will be less than or equal to minus da minus sdt. And now if you consider an isothermal process it tells us that work done by a system in an isothermal process will always be less than or equal to the decrease in its Helmholtz function. Now remember that the work done equals our traditional mechanical way of looking at things is work done is decrease in some potential. The potential energy goes down, the amount by which it goes down is the amount of work you can do. If everything is ideal, if there is friction or some other dissipative process the amount of work which you can extract will be less than the decrease in the potential energy. So using that idea we can consider the Helmholtz function a to be some sort of a potential and hence sometimes it is called the Helmholtz potential. A decrease in which represents the maximum work that can be obtained in an isothermal process. The next important energy function the final one which we derive define it g known as the Gibbs function. The Gibbs function is defined as u plus pv minus ts you can rewrite it if you want as h minus ts or you can rewrite it if you want as a plus pv because they are all related to each other. Differentiating it you will get an expansion in terms of du, dv, dp, ds and dt. Now apply first law du plus pdv plus dw other will be dq so du plus pdv will be dq minus dw other and then keep dq and tds on one side and you will get dg plus dw other minus vdp plus sdt which is equal to dq minus ts less than or equal to 0. This less than or equal to 0 is a consequence of the second law. Keeping dw other on one side now you will get dw other other means other than expansion work less than or equal to minus dg plus vdp minus sdt and hence if I consider a process which is constant pressure, constant temperature. Now remember that if you have a chemical reaction taking place electrochemical or even an ordinary reaction it is possible for us to change the state while keeping the pressure and temperature constant because the composition will change and if the state changes you can always do some work and hence we can show that dw the work done other than expansion work in a constant pressure, constant temperature or isobaric cum isothermal process will be less than or equal to the decrease in the Gibbs function in that process during that process. Compare this with the expression for the Helmholtz function and we can now say that the Gibbs function is a function similar to the Helmholtz function in that it can be considered as a potential the decrease in which represents the maximum work other than expansion. Notice this dw other here other than expansion that can be produced that can be obtained in an isobaric cum isothermal process and since most of our chemical reactions are isobaric cum isothermal processes they take place at ambient pressure, ambient temperature or can be made to take place at ambient pressure, ambient temperature. The Gibbs function is a is an important idea and that is the basis function on which much of physical chemistry and chemical engineering is based. Chemical engineers are very very comfortable with the Gibbs function just the way we mechanical thermal engineers are comfortable with Enthalpy. Now this was the introduction to the four functions u, h, a and g. Now we look at the properties of these four energy functions and we will use the specific property version everything will be considered per unit mass of the system and we will use the property relation and we will be using the following identities from calculus of partial differentials. We will say that if z is a smooth function of x and y smooth means the first and second derivatives are defined and if the differential of dz m dx plus n dy is an exact differential if z is a function then dz will be an exact differential if it is expressible in form if it is expressed in the form m into dx plus n into dy then m must be partial of z with respect to x at constant y and n should be partial of z with respect to y at constant x. Not only that equating the cross derivatives the second partial derivative of z first with respect to x and then with respect to y will equal the second partial derivative of z first with respect to y and then with respect to x. And hence the partial of m with respect to y at constant x will equal the partial of of n with respect to x at constant y. These are the two relations which we are going to use innumerable times henceforth for our derivations. Now the next four slides would apply these to the four functions u, h, a and g. First let us look at the internal energy u. We have the property relation a transpose property relation. The basic property relation is Tds minus pdv. So sorry basic property relation is Tds equals du plus pdv. So here we have du equals Tds minus pdv. So notice that because we have a ds and dv on the right hand side it is natural for us to consider u to be a function of s and v. We can say that s or v are the most natural independent variables to consider u to be a function of. Now when you do that from our laws of partial derivatives we know that T must be partial of u with respect to s at constant v and p must be partial of u with respect to v at constant s. In fundamental thermodynamics these two relations are known as the thermodynamic definition of temperature and thermodynamic definition of pressure or thermodynamic temperature and thermodynamic pressure. For the simple reason that consider this triad of properties u, s and v. We realize that the moment we define any thermodynamic system we are able to define its volume. First law and second law of thermodynamics are able to help us define two properties of state. First law the internal energy u and second law the entropy s. So u, s and v are considered the basic thermodynamic triad of properties for any systems. Based on this now we know that temperature has to be the variation of u with entropy at constant volume and pressure has to be the variation of u with volume at constant entropy. And using this relation particularly the pressure relation we can define the thermodynamic pressure for a system which is not a fluid at all. For example we can define it for a system purely containing radiation. If we know what is the energy of that radiation field is what its volume is and what its entropy is then we can determine its pressure simply by obtaining this partial derivative of u with respect to v at constant s with a negative sign. And now look at the next thing. If we use the cross derivative relation we get how temperature varies with specific volume at constant s and how pressure varies with specific entropy at constant volume. An esoteric relation which we could not have imagined being able to derive using the first principles without going through this algebra of partial derivatives. And we will see more such interesting relations. These are very useful relations in thermodynamics. Note this relation which is in a box here. Now let us come to enthalpy. It is defined as u plus pv. So if you take its differential and expand it and use the property relation for Tds you will get dh equals Tds plus vdp. So now we consider enthalpy to be a function of entropy and pressure because we have ds and dp here. And we immediately get temperature to be partial derivative of enthalpy with entropy at constant pressure and volume to be partial derivative of enthalpy with pressure at constant entropy. Look at the funny relations that we have. And the cross derivative relation gives us partial of temperature with pressure at constant entropy and partial of volume with respect to entropy at constant pressure. Note this expression in the box. Now let us come to Helmholtz function. It is defined as u minus Ts. So expand it, expand du as Tds minus pdv. Combine terms and you will get dA equals minus sdt minus pdv. Consider A as a function of T and v because there is a dt here and a dv here. And you will get entropy to be negative of partial of A with respect to T at constant v and pressure is negative of partial of A with respect to volume at constant T. And look at what the cross derivative tells us. It tells us that variation of entropy with volume at constant T equals the variation of pressure with temperature at constant v. Now this expression is very important because look at what it tells us. On the right hand side we have just the data which we get from the equation of state p, v and T relationship between p, v and T. But the left hand side tells us how does entropy vary with volume at constant T. So this is a very important relation which leads the variation of entropy with respect to volume to purely pvT data. Finally we come to the Gibbs function which is defined as h minus Ts. Take the differential of dg, use the expansion of dh and use the basic property relation and you will get dg equals minus sdt plus vdp. And that means it is proper for us to consider dg as a function of temperature sorry g as a function of temperature and pressure because these are the two differentials remaining on the right hand side. When you do that we realize that entropy is the negative of variation of g with temperature at constant p and volume equals the variation of g with respect to pressure at constant T. And the cross derivative gives us a relation partial of entropy with pressure at constant T is negative of the partial of volume with temperature at constant p. And again this is important because the variation of entropy with pressure at constant temperature is provided to us in terms of only pvT information which is available simply from the equation of state. Now these four blocked relations from the four slides I have put together here and these are known as Maxwell's relations. And even their reciprocal for example take this relation partial of T with respect to p at constant s is partial of v with respect to s at constant p. And if you take its reciprocal we will get partial of p with respect to T at constant s equal to partial of s with respect to v at constant p. So these four relations and their reciprocals are known as the Maxwell's relations and these are perhaps the most celebrated relations in thermodynamics. So these are very useful property relations. The third and fourth ones relate entropy variation to purely pvT or equation of state data. They help us reduce the requirement of Cp and Cv data for mapping the state space entropy variation is available at least in part using only pvT data. And one of the important tasks in the study and application of thermodynamics and one of the hardest job for a student of thermodynamics any student of thermodynamics including us is how to remember them without having any funny material inside your trouser pockets. And we now find out a trick to do that a physical realization and then some mathematical tools. We consider a reversible cycle. Remember the area under a pv diagram area under a quasi static process and area under the same process on the T s diagram that we have realized we have discussed represent a reversible cycle on a pv diagram and also on a T s diagram you will see two pictures like this. On a pv diagram there is a reversible cycle the same reversible cycle sketched on a T s diagram looks slightly different but it is another closed loop. Let the area of the cycle in the pv plane be a pv. Let the area of the cycle in the T s plane be a T s are the two areas equal. If all of you were to have a clicker or the all of you were to have the akash with a clicker application at this stage I would have rested for a few minutes and asked you to answer this question yes or no. It is difficult for me to obtain such a feedback but I suggest all of you write down in your notebook yes or no because soon I am going to give you the answer to this question and proceed from there. I hope you have written that now I will relieve that the answer is yes the areas are equal and why are they equal because we have a reversible cycle. So area under a curve represents heat transfer on the T s diagram area under the corresponding curve on the pv diagram represents the area represents the work done pdv work done and hence for a reversible cycle the work done would only be pv the heat transfer will then be equal to the work done pv diagram and hence the two areas are equal. Mathematically the area of the loop in the pv diagram is the loop integral of dp dv and area in the loop on the T s diagram is the area integral dT ds and this should be true for any reversible cycle of a simple compressible system at rest and now go to our coordinate geometry this is where the coordinate geometry is needed. Remember that if we transform xy coordinate to some uv coordinates then the area element dx dy gets transformed to area element du dv by a local scale which is known as the Jacobian and if the areas are always equal that means the Jacobian of the transformation is 1. I am not going to spend any detail on discussing the properties of Jacobian and how it is defined. I recommend that you look up your elementary calculus and coordinate geometry book but the symbol for a Jacobian in terms of the Jacobian symbol this means that the Jacobian of T s with respect to pv is 1 and so is the Jacobian of pv with respect to T s. Now we will use the properties of Jacobian to derive many of our relations. These properties of Jacobian are important Jacobian is a combination of partial derivatives which behaves like an ordinary derivative that is the special thing about it. So the first relation is Jacobian of uv with respect to xy when you multiply it by the Jacobian of xy with respect to uv that should be 1 this comes from the area transformation if you transform it one way and back transform it you should get back the same area. The product rule of ordinary derivative is also applicable you have derivative of a with respect to b multiplied by derivative of b with respect to c is derivative of a with respect to c. So Jacobian of pq with respect to rs multiplied by Jacobian of rs with respect to Tu should be Jacobian just cancel this partial of rs and partial of rs and you get Jacobian of pq with Jacobian of Tu with respect to Tu. But the real important relations for us are these you know the partial derivative of u with respect to x at constant y can always be written down in this form it is shown it can be shown to be equal to partial of u or Jacobian of u and y with respect to x and y. Notice that what is the left hand column here u and x are the numerator and denominator functions in the derivative and what is the right hand column the common variable y is the variable which is maintained constant. You will find this in your calculus book and you can check this out for yourself using the definition of Jacobian. U and x need not be in the first column the common variable the constant variable y needs to be common in some column either second column or the first column. So you can also write it like this but if you have the common variable on a diagonal either the principal diagonal or the secondary diagonal then you have to have a negative sign. So this is an important relation that partial of u with respect to x at constant y can be written down in terms of Jacobian in four different ways select one which you find convenient and we will use these and other properties of the Jacobian to manage the Maxwell's relations. I will give you an example here first suppose we want to find out the equivalent of partial of t with respect to v at constant s. First we have to determine whether this is a candidate for Maxwell's relations going back we will notice that the four Maxwell's relations have something in common if you go to this slide you will notice what is common if you spend a few minutes with this you will find that the pair T s and the pair p v occurs in a combination where either the numerator and the constant function or the denominator and the constant function is either T s or p v notice there is a T s here there is a p v here similarly there is a T s here and there is a p v here in some order similarly there is a T s here and there is a p v here similarly here there is a T s and a p v. So this tells us whether a given partial derivative is a candidate for Maxwell's relation or not. Now given partial of temperature with respect to volume we notice that there is a T s pair the constant function with either the numerator or denominator should form the pair T s or p v. Now it is T s so we can go ahead with our scheme this will be a candidate in the Maxwell's relations. So you first write this as a Jacobian partial of T s with respect to v s just put S as the common variable in the numerator and denominator. Then because T s is in the numerator multiply this by 1 which is partial of p v multiplied by T s by thermodynamic this Jacobian is 1. Now we use the product characteristics of product property of Jacobian there is a partial of T s here there is a partial of T s here cancel out we end up with partial of p v with respect to v s. And now we notice that a diagonal is common v so this must equal negative of the partial derivative of p with respect to s at constant volume. I recommend that we spend some time in deriving these because you do not have to really go through this detail when we do exercises I will show you a quick symbolic short form by which you can do this. You need not write all these things down although for the first few attempts you should do this. Now after this what we are going to do is derive I will leave it to you as an exercise to derive all these things all the four partial derivatives all the eight partial derivatives involved in Maxwell's relations and the reciprocal and the next thing is going to be exercises which we will do after the lunch break. There are still some seven minutes to lunch break so I have a choice I am just going to go to 1, 2, 4, 1 NRI institute Bhopal Madhya Pradesh. Over to you sir. Sir one question that arises in my brain is we have centigrade scale of temperature we have ideal gas scale and we have thermodynamic scale of temperature. All three scales have independent different origin centigrade scale is based on melting point of and boiling point of H2O whereas ideal gas scale is based on the fact that at constant volume temperature is a function of pressure whereas our thermodynamic temperature scale is based on reversible is based on reversible Carnot cycle. Now the question that arises in my brain is though their origin is independent why or how this coincidence has occurred that the unit the magnitude of unit of temperature is same on temperature scale same on centigrade scale as well as Kelvin scale as well as thermodynamic scale. Over to you sir. Okay good question I would go the other way round I will put the Celsius scale at the end purely from a thermodynamic point of view the only scale which we define is the thermodynamic Kelvin scale. Actually just because the traditional way of thermodynamics is teaching about ideal gas first and then coming to second. If I were to teach thermodynamics in a very pure way I would not talk about any temperature scale till I come to the end of the second law and define the thermodynamic Kelvin scale of temperature. Then I will ask ourselves the question a reversible engine cannot be implemented. So what is the best we can do to implement and measure the Kelvin scale of temperature the thermodynamic Kelvin scale of temperature. Then we would discover an ideal gas and we would show that the efficiency of the ideal gas Carnot cycle reversibly implemented is given by 1 minus T2 by T1 on the thermodynamic Kelvin scale by definition and using the ideal gas equation of state we will show that the efficiency of a ideal gas Carnot cycle will be 1 minus PV product at the lower temperature divided by PV product at the higher temperature and that way I would be able to show that the thermodynamic Kelvin scale can be implemented in practice using an ideal gas by implementing this demonstrated relationship that T2 by T1 or T1 by T, let us say T2 by T1 on the thermodynamic Kelvin scale it shown to be equal to P2 V2 for an ideal gas working a Carnot cycle where 2 is the state representing the lower temperature divided by P1 V1 where 1 is the state of the higher temperature. In this case I do not even have to define the ideal gas Kelvin scale. The ideal gas Kelvin scale had to be defined because we had to work some exercises simple exercises using first law before coming to second law. So this explains that although we have for our convenience defined them to be separately in the end we have shown that the thermodynamic Kelvin scale and the ideal gas Kelvin scale are the same. That is the first question I think that answers the second part of your question that is ideal gas Kelvin and thermodynamic Kelvin. Now when the first part of the question I agree that the Celsius scale or sometimes called the centigrade scale was defined using the ice point and steam point and till the early part of the 20th century that was so. But later on when you know there is an international union on pure and applied physics IUPAP and they decided that look we should not have these two different scales Celsius and Kelvin. So at that time it was decided that the Celsius scale should be got rid of and today Celsius scale is defined as temperature on the Celsius scale is temperature on the Kelvin scale minus 273.15. So today there is no separate definition of the Celsius scale. The Celsius scale is today defined in terms of the Kelvin scale. Now the question arises what is that funny number 273.16. The funny number 273.16 has been derived it is extracted using experimental data excellent experimental data the PV product of an ideal gas and the ratios of these between the ice point the steam point and the triple point of water. It was discovered that the ratio of the PV product of an ideal gas at the normal boiling point of water to the PV product of the same ideal gas at the ice point turns out to be almost exactly 373.15 divided by 273.15 and based on this it was decided to use the standard value of 273.16 Kelvin for the triple point of water because if you define it like that the ice point turns out to be almost exactly 0 degrees C but not exactly 0 degrees C. The steam point turns out to be almost exactly 100 degrees C but not exactly 100 degrees C just to keep the numerical difference in the temperatures of water at its ice point and at its steam point to be 100 units which was the so in the earlier Celsius scale. We wanted to keep that on the Kelvin scale so that all the earlier data based on Celsius scale can be converted to the Kelvin scale just by adding the number 273.15. It was just a matter of convenience. It was open to people to keep the define the Kelvin temperature of the triple point of water to be 1.0 exactly or 100.0 or 1000.0 or any other arbitrary number. All that would have happened was the distinction between Celsius or the relation between Celsius and Kelvin on the temperature scale would not have been as simple as Kelvin minus 273.15 each Celsius. That is all. The idea that Celsius is defined in terms of ice point and steam point is now history. That is the historical definition of the Celsius scale. Today's definition of Celsius scale is in terms of the Kelvin scale. I hope that satisfies you. Over to you. One more question sir. One day one of my students gave an explanation to me that I can create a reversible Carnot engine which can violate the second law and the simple thing that he told that I will make T lower equal to absolute 0 and definitely when T lower becomes equal to absolute 0 then all the heat that the engine takes from the high temperature reservoir is getting converted into water. So what answer I should give to my students over to you sir. There is an inherent assumption in the argument of the student and that assumption is that he will create a system whose temperature is 0 Kelvin. And now we will say that look you are assuming here that you are creating a system with 0 Kelvin and you can simply use the argument that you cannot have such a system or you cannot create such a system because if you create such a system you can work an engine reversible to T engine between say the ambient temperature and 0 Kelvin the efficiency of that will be 1. And the Kelvin Planck statement says that you cannot have a 2 T heat engine the efficiency of which is 1. Actually this argument is an argument to the fact that you cannot create or you cannot have a system whose temperature is 0 Kelvin. Use his argument against him because he is assuming that the Kelvin Planck statement is violated and hence it says second law can be violated. Thank you over and out.