 Now it is time for us to consider the Helmholtz function. Definition in specific u minus d s differentiate d A equals d u minus d d s minus s d t. Use the property relation written like d u equals d d s minus p d v. So, d u minus d d s will become minus p d v. So, this becomes minus p d v minus s d t. So, that tells us that for Helmholtz function v and t are the most natural independent variables. Notice that for energy it was entropy and volume, for enthalpy it was entropy and pressure, for Helmholtz function now these are volume and temperature for temperature and volume the order is immaterial. Now let us continue with our calculus. Compare this with our m d x plus n d y format and you will get pressure equals minus partial derivative of A with respect to v at constant t entropy as with a negative sign partial derivative of A with respect to t at constant v. In the cross derivatives because there are two negative signs are involved they will cancel out. We will get partial of p with respect to t at constant v equal to partial of s with respect to v. Let me call this now unlike the first two m relations they are also derived in the same way and they are also important, but there is something more important regarding m 3. Notice that the left hand side of m 3 contains only pressure volume and temperature p v t data whereas, here we never had a p v t combination either here or in m 1 it was t v s p s v here t p s v s p, but we never had a p v t combination. So here this is we have a relation with change in entropy which is an energy function with respect to volume at constant temperature is obtained only in terms of pressure volume temperature that is equilibrium data. Let us finish this with our last function that is the Gibbs function in specific properties g is defined as u plus p v minus t s or which turns out to be h minus t s. So, we have a d h is we have already defined or derived that d h is t d s plus v d p notice this. So, when we differentiate we will have d g is d h minus t d s minus s d t and d h will be t d s plus v d p minus t d s minus s d t. So, this gives me my d g equal to this t d s t d s goes away and I get for my this thing minus or I can straight away write v d p v d p minus s d t notice p and t. So, that means when you have the Gibbs function the most convenient pair of variables will be pressure and temperature. Mind you when we come to this most convenient pair a v t h s p and u s v and for g it is p t this is only the most convenient pair that does not mean that say Gibbs function has to be considered a function of pressure and temperature. The choice with us, but the basic property relation when written down for that function turns out to be in terms of those two variables. So, these are the most convenient independent variables for us to consider. Now, applying our calculus of exact differential and partial derivatives to this we will get specific volume is partial of Gibbs with respect to pressure and constant temperature and entropy is with a negative sign partial of Gibbs with temperature and the cross derivative partial of v with respect to temperature constant pressure equals negative partial of s with respect to pressure constant temperature I call this m. So, what we have done so far is we have taken in turn the four energy functions internal energy enthalpy, Helmholtz function and Gibbs function applied the basic property relation along with their definitions determine the most convenient pair of variables for each and then related some thermodynamic properties to their partial derivatives of certain kind and using cross derivatives we have determined relations which I have tabulated as m 1 m 2 m 3 m 4 which sort of do like a knight on the chess board relate something to something sort of unexpected something which is not obvious from the the basic idea of thermodynamics which comes out of thermodynamics these definitions and differential calculus why have I called again remember that like m 3 where the variation of entropy with volume is related to purely pressure volume temperature data m 4 also relates variation of entropy with pressure at constant temperature to pressure volume temperature data and that is something to be noted because this pressure volume temperature data being equilibrium data only one state is involved state of equilibrium is involved and hence it is the most convenient and perhaps the most accurate measurement. So, anything which is related directly to only PVT data we are happy because then the process will be convenient and not very difficult these four relations which we have come across in fact I will show them I have already tabulated these four relations coming out of u coming out of h coming out of a and coming out of g here from there to here I might have transposed the right hand side and the left hand side and may be put the negative side on the other side that does not matter, but they are algebraically equivalent to these four relations and they are reciprocals because the science of partial differential tells us that partial of t with respect to v at constant s is equal to the reciprocal of partial of v with respect to t at constant s. So, you take the reciprocal of the left hand side reciprocal of the right hand side and you will get equivalent relationship these four relations are known as Maxwell's relations and those perhaps when it comes to property relations are the most important relations in thermodynamics. So, important that students tend to or students will have to and we will have to reach a stage where we must know these by heart, but now knowing something by heart or cribbing or something about it is not the way to do there is no understanding involved when you create a poem or a limerick or some mnemonic to consider to remember these, but is not there a fundamental thermodynamic principle which relates these and it turns out that yes there is a thermodynamic principle from which we are directly able to relate this using some absolutely simple ideas when you look at it it looks simple from analytic geometry and that is what we look at first we will look at how to derive this using basic thermodynamic ideas and some simple mathematics. The first thing you should notice is these are not any derivatives let us look at what type of candidates are these you know we have any combination of thermodynamic properties and suppose I take a differential of pressure with respect to volume at constant entropy will that appear in the Maxwell's relation. Now, you will notice that pressure differential of pressure with respect to volume or differential of volume with respect to pressure at constant entropy does not appear here at all we have two derivatives here at constant entropy one is temperature with respect to volume and temperature with respect to pressure. So, what is it that distinguishes these equations you do the following take the variable in the numerator and the variable which is constant then you take the numerator and the variable and the variable which is constant you will notice here that the numerator variable and the constant variable forms a t s pair here the numerator variable and the constant variable forms the p v pair similarly here you will notice numerator constant is t s numerator constant is v p, but p v in some order similarly here you will notice there is a t s pair and a p v pair and here also there is a t s pair and a p v pair. So, remember t s and p v is the idea. So, first thing we will notice is now. So, the question is what is the question partial of x with respect to y at constant z is this involved in the Maxwell's relations for this look at the pair x z and the pair y z one of them must be either p v in some order or t s in some order if it is. So, then we have a Maxwell's relation then the question is which Maxwell's relation and how do we come across the fact that a pair should be either p v or t s gives us a hint and that brings us back to the question which was even discussed earlier today in our discussions. We discussed that the question was that the area under a quasi static process on the p v diagram represents the expansion work done. The area under a quasi static process on the t s diagram does not represent the heat transfer, but if the area and if I have a process which is reversible the area under that curve t s diagram will be the heat transfer because for a reversible process d q equals t d s because for a reversible process d s equals d q by t. Now, come back to the p v diagram and we will notice that if I have a reversible process say I have a process here which is reversible much more than quasi static. So, if the process is reversible then this area a t s is q. Similarly, if the process here is reversible then for the reversible process of a simple compressible system this area a p v will represent the work done. The total work done not just the expansion work done because we have said it is reversible in the moment we say it is reversible for a simple compressible system. The only work mode at municipal is that of compression and expansion there will be no stirrer work there will be no any other type of work and hence a p v will be the work done. So, as we saw earlier under a quasi static process area under the p v diagram is expansion work done and area under the t s diagram will be higher than the heat transfer. But for a reversible process area under the p v diagram will be the total work done and area under the curve reversible process curve under the t s on the t s diagram will equal the heat transfer. This equality comes out of the fact that it is reversible and this no qualification here it is the total work done because also it is reversible. Now, if we have appreciated this then we do the next thought experiment and this is where we will take help from analytical geometry. Let us consider a simple compressible system specific volume specific entropy mean 1 kg mass minor. Let us execute a reversible cycle in some direction and represent the p v diagram like this and the depiction of the same cycle on the t s diagram let it be this. Now, let me say the area of this cycle is a p v mind you what is shown here and here is the same reversible cycle the same cycle and the same cycle. Let the area here be a p v the area here be a t s the area here represents work done in that reversible cycle. This area would represent the heat transferred in that reversible cycle total work done because it is reversible heat transferred and equal to because it is reversible. Now, are these two areas equal notice now that it is a cycle. So, the work done in a cycle must equal the heat transferred to the system in the cycle and that means the two areas must be equal for any reversible cycle. Now, go back to again Thomas and Finney brush up your analytical geometry when you have a transformation of coordinates notice that being a simple compressible system temperature is a function or can be considered as a function of pressure and volume entropy can also be considered a function of pressure and volume and I can transform this cycle on the p v diagram to the cycle on the t s diagram or I can sketch the cycle on the t s diagram first considered p as a function of t and s v as a function of t and s and transform that curve by coordinate transformation to the curve on the p v diagram. And when I find that for any such reversible cycle the areas are equal that means the scale ratio for this coordinate transformation the area scale ratio for the coordinate transformation must be 1. And what is that function which represents the area scale ratio that function is known as the Jacobian and the for going from I just I cannot shift it. So, I will write this means that the Jacobian of the transformation must equal 1 that is the conclusion and the first symbolism and then understanding that means way area varies on the t s diagram with the corresponding area on the p v diagram is 1. And also the area way it varies with on the p v diagram the way it varies on the t s diagram is also equal to 1. Now, what is the use of this to consider the use of this we have to again brush up from our calculus particularly analytical geometry the properties of the Jacobian. The Jacobian of u v now I am looking at not u is not internal energy and v is not volume is defined as a 2 by 2 determinant partial of u with respect to x at constant y partial of u with respect to y at constant x that is the first row the second row is partial of v first row has derivatives of u second row has derivatives of v with respect to x constant y partial of v with respect to y at constant x. Notice that the first row are derivatives of u second row derivatives of v first column derivatives with respect to x at constant y and second column derivative with respect to y at constant x this is the definition. And the advantage of the Jacobian or the properties of Jacobian that it behaves almost like ordinary derivative for example, there are some properties of the Jacobian which we must note you know the reciprocal of the Jacobian is just interchange the numerator and denominator. So, first thing is partial of u v with respect to x y multiplied by partial of x y with respect to u v will be 1. Then even the chain rule can be used for example, partial of u v with respect to x y multiplied by partial of x y multiplied by partial of x y with respect to p q will turn out to be partial of u v with respect to p q just like an ordinary derivative. This is like partial of this is say ordinary derivative of u with respect to x multiplied by ordinary derivative of x with respect to p is ordinary derivative of u with respect to p. But for a pure partial derivative that is not true, but when it comes to this combination of partial derivatives known as the Jacobian the Jacobian behaves like an ordinary derivative note this or this. And now we look at one property of partial derivatives and Jacobian which is not usually mentioned in x books, but you can very easily derive it and that is this Jacobian of u y with respect to x y is equal to notice that because it is a determinant if you change the one flip one row or one column you will end up with a negative sign. So, this becomes partial of y u partial of y x I am flipping both numerator and denominator, but if you flip any one of them there will be a negative sign. So, this will be u y this will be y x this will be minus of y u x y, but what is important is this remember here you have one variable common in the numerator and denominator either just below it or a cross variable all of these are equal to the simple partial derivative partial of u with respect to x and we will just take an illustration of this and then proceed with exercises, but before exercises some illustrations for example, an illustration of the use of this and the Jacobian etcetera would be this let us take one of the Maxwell's relation. Let us take this we have partial of v with respect to t at constant p and suppose we come across this constant v partial of v with respect to t at constant p. So, let us see how we proceed let me try let me go to partial of v with respect to t at constant p that was the ratio. Let me go back yes partial of v with respect to t at constant p we write this as Jacobian of v p Jacobian of t p. Now, you notice that one of the pairs is p v. So, what we do is I will write this first as minus by interchanging this partial of p v partial of t p then I multiply this by 1 and what is the value like a conversion factor 1 this either this or this that is partial of t s with respect to p v or partial of p v with respect to t s and since I want to get rid of p v I will put partial of p v in the denominator and partial of t s in the numerator the value of this is 1 that is the thermodynamic derivation which we are using. Now, we use the property of Jacobian that it is an ordinary derivative. So, in something like I can cancel this part with this part and what you get is negative sign remains partial of t s partial of t p and which by properties of Jacobian is partial of s with respect to p and so with this we have obtained the relation the Maxwell's relation. Let us check whether that was what we started off to derive left hand side is what it was this is partial of s with respect to p at constant t with a negative sign partial of s with respect to p at constant t with a negative sign and that is something which we had derived in a long winded way. But remember now here all that we have used is the thermodynamics is now involved here everything else is mathematics for calculus and remember this is absolutely basic thermodynamics which comes out of this idea that if you have a reversible cycle represented both on the p v diagram as well as the t s diagram the areas of the two cycles should be the same and that is basic thermodynamics first law and of course, second law because only the second law dictates that the area of a closed loop on the t s diagram will be the heat transferred and the area of the closed loop on a p v diagram will be the total work done not just the p v work done that is where the word reversible comes across. Now many of you and your students may not be very familiar and comfortable with the idea of Jacobian. So, we will convert this into a simple partial sorry Maxwell's relations for Dumbos or Maxwell's relations for the common man the idea is like this I will take the same this thing partial of v with respect to t at constant p. So, you have partial of v partial of v with respect to t at constant p proceed as follows write down numerator denominator v t variables write down the variable which is constant next to it or even before it make one column after or one column before. Now, notice that in the numerator or in the denominator there should be a p v or a t s in some order here you notice that in the numerator there is a pair p v replace this with t s in the same order replace p with t v with s. So, here you have p v from right to left. So, replace this with t s from right to left notice that p must be replaced by t v must be replaced by s or vice versa if you were to see s t here replace it with v p if you were to say t s here replace it with p v put the other part as it is. So, now you end up with s t t p now you notice that there is one common t here, but the common is along the diagonal. So, the moment there is a diagonal either like this or like this in this case it is diagonal like this does not matter. So, long as it is a diagonal put a negative sign and say that this is equal to minus and whatever is the common variable put it as the constant variable whatever is in the denominator here put it in the denominator what is in the what is in the numerator put it in the numerator what is in the denominator put it in the denominator put the partial signs and there you are by just this this is the mechanical way of doing things, but I would like you to remember this and remember the thermodynamics which goes into this as the important thing once you do this later on you can do this mechanically there is no harm. Now, what is the use of all this the use of all this is the following. Now, we will obtain obtain relationship between variations and let us see how we proceed and we will also show remonstrate some stuff already known take for example, we know that the so called ideal gas is governed by two laws the p v t relationship is governed by the Boyle's law equation of state the p v t relationship is this is Boyle's law, but then we have u is a function only of t that means if you consider u as a function of t and v then it does not depend on v that is joule the question is there a link for example, if I have a gas which obeys this law can I say that this is consistent with that or if the equation of state is p v equals r t or is p v product is proportional to the thermodynamic temperature t can I say that the internal energy has to be a function of t or is it something absolutely independent let us try out. Let us now play with our internal energy u let us say we have u and let us consider d u d u we know by the basic property relation t d s minus p d v now let us say we want to know whether let us consider u as a function of t and v and see what happens to demonstrate this we will consider u as a function of t and v and then we will show that partial of u with respect to v at constant t is 0 throughout the state space because if we show that that means in spite of considering u as a function of temperature and volume then if we show that when temperature is held constant u does not vary with volume the derivative with respect to volume when temperature is constant in 0 that means u does not depend on volume it depends only on t then our proof is over. So, now we want u to be considered a function of t and v. So, we will consider remember this is d s and this is d v now here we will have d t and d v d v is there. So, we do not have to worry about it, but now s also we will consider as a function of t and v. So, let us write it we will write d u where u is considered a function of t and v is t d s where s is considered as a function of t minus p t v no problem. Now, let us expand this this becomes let me take a slight shortcut let us expand on the right hand side we have the right hand side equal to t expand this this will be partial of s with respect to t at constant v d t plus partial of s with respect to v at constant t d v minus p d v. Now, I will not write this t and v, but I get d u equal to now I will take d t together and I will take d v together when I take d t together I will have just one part t partial of s with respect to t at constant v d t and the second part is plus t partial of s with respect to v at constant t and from here minus p. Now, look at it this is d u equals m d t plus n d v and when you have that go back to our basic thing when d phi equals m d x plus n d y m is the partial derivative of phi with respect to x at constant y and so on. So, that means this thing partial of u with respect to v with respect to t constant v is t partial of s with respect to t at constant v that is one part the second part is partial of u with respect to v at constant t is t partial of s with respect to v at constant t minus p. Now, let us look at this part what do we say from it see the left hand side is defined as c v the specific heat at constant volume. So, the conclusion from this part is c v is equal to t partial of s with respect to t at constant v. So, we now can write c v straight away in terms of s t and v what do we do with this let us see what we can do with it what is that relation partial of u with respect to v at constant t is t partial of s respect to v at constant t minus p. We have partial of u with respect to v at constant t equals t partial of s with respect to v at constant t minus p. Now, the moment we have something like entropy here we are not really comfortable we want to have it in simpler form. Let us see whether we can reduce it into p v t form and what do we get is this a candidate for Maxwell. Yes, if I write it write it here this is my you know sandbox or rough work for this. So, I write it as s t v t s and v are the variables numerator denominator t is constant I have it as v t s. So, I replace it by p v in that order denominator is v t. So, I end up with partial of this is t partial of p with respect to t at constant v. So, this partial of s with respect to v at constant t is partial of p with respect to t at constant v. So, putting that back here t partial of p with respect to t at constant v minus. Now, this is the relation for any fluid. Now, if ideal gas then remember the right hand side of this equation only depends on p v t data nothing more than that. So, it depends only on the equation of state and you will end up with now p v equals r t. So, p equals r t by v. So, substituting this partial of u now for an ideal gas partial of u with respect to v at constant t is equal to v at constant t equals t. Now, what is partial of p with respect to t at constant v v is constant. So, r is anyway constant. So, this is something like constant into t. So, the derivative will be r by v and by the equation of state r t by v is p itself. So, this is p minus p which is 0 and that means equation of state means if equation of state is 0. So, if p v equals r t then u is only a function of t because we said earlier that to show this we will have to assume u as a function of t and v and show that partial of u with respect to v at constant t is 0 and that is what we showed here. Partial of u with respect to t at constant v is 0. So, our conclusion is that if the equation of state is p v equals r t then u must only be a function of t. There is some more part. Now, let us not worry about the ideal gas. Let us go back and now notice that we have obtained a relation for partial of u with respect to v at constant t in terms of p v t data and before that anyway we have partial of u with respect to t at constant v which is shown to be equal to this, but which is actually c v. So, going back to our d u this is the basic property relation and from our basic idea we have obtained d u equals partial of u with respect to t constant v d t plus partial of u with respect to v at constant t d v. This is defined as c v and this is d t and this we have shown is equal to t d p by d t at constant v minus p d v. Now, notice this is for any fluid and notice that the change in energy between two states will depend on one part which depends on c v that is the temperature dependence. The volume dependence depends only on p v t data. So, here we need c v that is calorimetric data or energy data and this depends only on the p v t data that is equation of state data. Not only that, when you do this particular part the second contribution will be 0 if it were an ideal gas. If the deviation from ideal behavior is small this will be pretty small compared to this and quite often this is known as the energy deviation and since most of our energy behavior fluids or gases behave like an ideal gas the following trick is done. You have the state space of a gas let us say this is t and this is v when will a gas behave like an ideal gas when v is pretty large. So, let us say that you decide on some v naught a very large value where the pressure is very low and you decide on a t naught reference state and then you say that this is going to be your reference state and let us say here that v naught is large enough the fluid behave like ideal and then you decide on a t naught reference state and you define u equals u naught or may be 0 at that state and then for any other state lower or higher volume or lower or higher temperature this is u and out here you have the reference u naught u minus u naught is integral of d u. From the reference point to the point under consideration if this is the point under consideration that is the point under consideration the question is which path we must say that any path because this is an exact differential being a change in property. So, and the most convenient path to take is a path like this at the reference volume go up to the required temperature and then at the given volume at the given temperature come back to the required this u will be decided by the specified volume v and the specified temperature t whereas, u naught is at v naught and t naught and the consequence of this is the following notice that this o to a let me call this point a o to a is a constant t path and a to our final state is a sorry constant v path and a to our final state is a constant t path constant v path means here t varies and here v varies. So, our integral we will have to integrate this equation from the reference state to the given state, but since we have two parts let me write down the equation u minus u naught we will now become integral c v d t over the full path plus integral what was the second part t d p by d t at constant v minus p d v. Now, each path of this should go from first 0 to a and a to the required state, but now look at it when you go from 0 to a we are going along a constant volume line and since we are going along a constant volume line this contribution will not be there only this contribution will be there. So, this contribution will definitely be there from 0 to a and since a is at the required t and 0 is at t naught. So, this is I will write this as t naught to t and what is the volume? Volume is v naught. So, this c v is actually c v at t with which it may vary and v naught. So, that takes care of this part and in this part only this integral exist this integral does not exist because volume is constant along 0 to a. Now, take the second part a to the required state temperature does not vary. So, in this part this term does not contribute. So, this remains where it is now t naught to t, but here volume changes from v naught to required v and this at required volume required temperature. Temperature does not vary, but if you are looking at different states to determine we will have to have this at different. So, let me write this in short as or again as t naught to t c v t v naught plus integral to t. I will say v naught t to v t the required state of this integral t partial of p with respect to t at constant v minus p and now that means our from the reference state energy at any state can be determined by summing up two integral one along a constant volume line and one along an isotope and how much is the data required. Let us take different points this was one point let us take another point here then let us take a third point here every time you will notice that the first integral will have to be only along this line point a will get shifted, but that integral will only be along v naught whereas this integral will be at different volumes and different temperature. Similarly, for the third point the first integral will contribute from o to this point may be up to b up to c, but in any case only at v naught, but this part of the integral here or here or here will be at different t's and different v's as we go anyway it is a integral with respect to v. So, we will have to be vary, but data at different t's will have to be used that means for this we require c v data as a function of t only at one specific volume v by v naught and here we will require the equation of state data for various that is what I said earlier that this way we can show that the change in any energy function I have shown it for u, but you can show it for h and you can show it for s also can be shown to be a sum of two integrals one along either here we have taken v and t, but more convenient to take is p and t this is along a constant volume line at various t and this requires equation of state data all over this was just a demonstration.