 we will now come to the exercises. So we will do 5.1, we will quickly tell you by similarity we can write 5.2 and then we will do some exercises which are not listed here but which you can perhaps include in the text. By default we are doing property relations for a simple compressible system, so that will not be indicated. So consider S as a function of T and P, show that this is known as the first, there are two important TDS relationships of this format, this is the first one of the lot and this is just the way the first counting starts with 1, 2, 3, 4, this is the first exercise in any property relation. Remember on the right hand side we have a DT and a DP, so that is why it is obvious that we have to consider S as a function of T and P. So that is what we do, we consider S as a function of T and P. So naturally this immediately gives us DS is partial of S with respect to T at constant P dT plus partial of S with respect to P at constant T. We write our property relation as TDS is du plus P dv, I am taking a slight detour, this I will write as du plus P plus vdp minus vdp and what is this? This is dh minus vdp. Now rewrite this as dh is TDS plus vdp and you will get, can I go to the next page? Since dh is TDS plus vdp dh by, see what we want is dh by, I want to get to Cp, how is Cp defined? Cp is defined as dh by dT at constant P. Now dh is TDS plus or minus vdp. Now we consider this, consider S as a function of P and T. So DS becomes T partial of S with respect to T at constant P dT plus partial of S with respect to constant P at constant T dP plus vdp. Compare this, this is equal to TDS by dT at constant P dT plus something into dP. What is this? This is going to be partial of H with respect to T at constant P and that is what we have defined as Cp. So the conclusion from here is this particular expression is nothing but Cp, which one? This step, see I have considered S as a function of T and P. This is okay, this expression is okay, this expression is okay. So from here I expand this DS is partial of S with respect to T at constant P dT, because I have considered it as a function of T and P. So just expand it, first one derivative with respect to T at constant P dT, derivative with respect to P at constant T dP and we already have a vdp here. So this that means the inside here if you want is plus TDS by dP at constant T plus v into dP. So we have defined, we have now derived that Cp is TDS by dT at constant P. Now let us go back here, TDS by dT at constant P will be Cp. So if I multiply this let us say equation 1 by T and use definition of Cp. Then what do I get? I will get TDS equals TDS by dT at constant P dT plus TDS by dP at constant T dP out of which this particular part I have decided is Cp. So this will be Cp dT. Up to here it is okay. Now what about this? See in thermodynamics our aim will be the following. Whenever we measure the properties of a fluid what do we measure? Easiest thing to measure is PVT information, the so called standard equation of state. For an ideal gas it is very simple but for any fluid bring it to a particular temperature, heat it to a, sorry bring it to a particular temperature, pressurize it to a particular pressure, measure the volume. You have the PVT data. Then you can do calorimetric experiment for measuring Cp. It is much easier to measure Cp than Cv. So typically Cp is measured. So the aim is to obtain everything in terms of either PVT data or Cp data or both. So our aim is to reduce everything as a function of PVT and Cp or Cv. Here we have reached Cp. Now what about this? Is it a candidate for Maxwell? Yes. And now I would like you to do the following simple trick. It is a candidate for Maxwell. Why? Because we have an ST. I will write it. I am cordoning this off. This is my local cocksheet. It is ST, sorry, it is STPT. I am not writing all those Jacobian symbols. All that I do is I see a TS. I am effectively replacing it by PV. So what does this mean? This means that this partial of S with respect to P at constant T is now partial of V with respect to T at constant P, but with a negative sign. This implies that what I started off that partial of S with respect to P at constant T. STPT is going to be VPPT. So this will be minus partial of V with respect to T at constant P. You could do this by writing out the complete thing, but all this much symbolism is all that is necessary. Check out that this is one of the Maxwell's relation or its reciprocal. Is that right? You have the printed version with you. So now we replace this by purely PVT data and that is why this will be either third or fourth relation or its reciprocal. So we get the first TDS equation. TDS equals CP DT minus T partial of V with respect to T at constant P DT. That is what we were asked to derive. Now I will quickly go through two and three. We have just now derived that TDS is CP DT minus T DV by DT at constant P DP. This we did by considering T to be a function of T to be a function of T and P. In the second problem, we consider this to be a function of T and V and you will get the second TDS relation which will turn out to be CV DT plus T partial of P with respect to T at constant volume DV. Notice the similarity, notice the differences. Wherever there is a P here, there is a V here. Wherever there is a V here, there is a P here. P becomes V, V P becomes PV, P becomes V and wherever there was a partial derivative involved the sign has changed. Now the third problem is essentially combination of one and two. From this relation which I call two and this relation which I call three. The left hand sides are equal. So you equate the right hand side and you will have a relation between DT, DP and DV. Take any one of them to the left hand side. Take say for example DT to the left hand side. On the right hand side you have DP and DV. That is what will remain. So now consider T as a function of P and V. Obtain the partial derivative with respect to P at constant V with respect to V at constant T and equate with the expression and you will end up with the result given in 5.3. I leave that to you as an exercise. Let us do now something which we have said we know. We have said that an ideal gas has two relations. It obeys Boyle's law and it obeys also Joule's law. U is a function only of T and when T is constant it is not a function of P and V. What does this mean? This means that if you consider U as a function of say T and P then you should get partial of U with respect to T sorry with respect to P at constant T is 0. Similarly if you consider U as a function of T and V you should get partial of U with respect to V as a function of T to be 0. Let us see whether we can show this. What we will do is we will start with 3. So let me rewrite 3 again. We have TDS just now I have not invoked ideal gas. TDX is CV DT plus T DT by DT at constant V. Am I right? Now what is DU property relation? TDS is DU plus PDV. So DU is TDS minus TDV. This also is a derivation from the property relation. This also is this is the basic property relation. Substitute for TDS from here. So you will get CV DT plus T DT by DT at constant V minus PDV. Algebra. Now what does this mean? Consider U as a function of T and V then CV is a coefficient of DT. So CV must equal partial of U with respect to T at constant V. No surprise there. That is how we have defined CV in its consistent. What about this? We now get partial of U with respect to V at constant T should be the coefficient of DV which is T DT by DT at constant V minus V. This is true for any fluid. What happens to an ideal gas? And remember that this partial of U with respect to V at constant T depends only on PVT data. Equation of state. What is the ideal gas equation of state? PV equals RT. Substitute that and see what happens. Do it. Turns out to be 0 with no conditions attached on PV and T except that PV equals RT. So this closes that forward link which we talked about earlier. That ideal gas obeys two laws, Boyle's law and Joule's law and at that time we said that this is thermodynamically consistent and now we notice that it is thermodynamically consistent. That means if a fluid has an equation of state which is of the type PV equals RT, its internal energy should be temperature dependent only. Is this clear? So this turns out to be equal to 0 if PV equals RT. Now let us do 5.4. Derive an expression for the variation of CV with respect to V at constant T. Remember we notice that CP for a fluid can be shown to be equal to partial of S with respect to T at constant T multiplied by T. We have shown it earlier while deriving the TDS, first TDS equation. Similarly, you can show that CV is T into partial of S with respect to T at constant V that you should be able to show. Now what we want is partial of CV with respect to V at constant T. That is what we want to determine. So how do you do this? We differentiate the right hand side. Anything the differentiation is with respect to T. So I can take the T out of the differential. So this becomes T d by dV at constant T of CV out of which T I have taken out. So this is partial of S with respect to T at constant T. Now is this a candidate for Maxwell? I have a pair SV and I have a pair TV. Neither of them is T S or PV in some order. So this is not a candidate for Maxwell. It does not matter. But now we have two partial derivatives. So this is a second partial cross derivative T V V T. So I can change the order of derivative. So this can become T. What was the inner derivative? Partial with respect to T at constant V. So partial with respect to T at constant V. Outer derivative partial with V at constant T. So partial with V at constant T. Is this a candidate for Maxwell? S T V T. What is the corresponding? It will be S T V T. So there is a T S. So I call it PV. It is partial of P with respect to T at constant volume. Am I right? So this becomes partial of P with respect to T at constant volume. No I do not think there is a negative sign. Is there a negative sign? No there is no negative sign. And which Maxwell's relation is this? Third one. So now you notice that after this replacement using Maxwell, the inner derivative is with respect to T at constant V. Outer derivative is with respect to T at constant V. Same differentiation, same variable is constant. So we can confidently write this as T. Second derivative of P with respect to T. Now I will leave it as an exercise to you. All of you know about the Van der Waals equation of state. Slightly more complicated than the ideal gas equation of state. Do you think the internal energy of a Van der Waals gas is a function of temperature only? You can substitute the equation of state. We have this relation, right? Variation of U with respect to V at constant T. Substitute in Van der Waals gas and you will notice that the internal energy of a Van der Waals gas is not a function of T alone. It will be a function of T and V or T and some other variable. But you substitute CV with respect to V at T and you will find that for the Van der Waals gas, CV will be a function of T alone but not U. Do that and check it out. Now let us come to the an important part of property relation and that is Clausius-Clapeyron equation or Clausius-Clapeyron relation. We have always been looking at the phase plot of water, triple point to critical point, triple point to solid liquid line, triple point to solid vapor line. And to us the interest always has been, what are the slopes of these lines? We know that the slope of the liquid vapor line is always positive. Wherever you increase the pressure, the boiling point increases, the saturation temperature increases. But for the ice liquid, that means the solid liquid transformation for water as you increase the pressure, the freezing point decreases. So what is the relation? Let us say this liquid vapor interface Dp by Dt along a say liquid vapor saturation line. Maybe I will include a modified question in which we derive it in a different way. I typically leave it for a quiz or an NSM examination but I will just mention it to you. But here we will derive it using the fact that in the two phase flow pressure and temperature are not independent variables. And then we will link it to the Hfg or Hfg and the temperature and that point. What we will do is we will consider for the liquid vapor saturation a small part. Now instead of Pt, let us consider the Pv diagram and let us say this is the saturated liquid line, this is the dry saturated vapor line. And we are considering one pressure and let us say the isotherm will go like this. So along this line from the saturated liquid point F to the dry saturated vapor point G, it is a constant pressure line, it is also a constant temperature line. Now I consider some point here, I consider some point here, what is it doing? And I consider two lines, one is a volume line, another one perhaps is a constant entropy line. Now I consider points on the constant volume line and on the constant entropy line but at a higher pressure, higher temperature and a lower pressure correspondingly lower temperature. And I am looking at variation of pressure with temperature along the constant volume line and variation of pressure with temperature along the constant entropy line. At that point, will these two partial derivatives be different? My point and the neighboring points are fully within the two phase zone, the liquid vapor zone. In the two phase zone pressure is a function only of temperature. So when I change my state, whether volume changes or entropy changes does not matter. If I am changing pressure by one bar, I am remaining the saturation line, the temperature will change correspondingly, it does not matter whether I am keeping volume constant or I am keeping entropy constant. So in the two phase zone, this must equal this simply because both of them should basically equal dp by dt along the saturation line. dp by dt along the saturation line can be written down either as dp by dt along the constant volume line or dp by dt along the constant entropy line. If that is understood, the next step is using Maxwell. We will use Maxwell both ways. We have said dp by dt along the saturation line is either partial of p with respect to t along a constant volume line in the two phase zone or is also equal to partial of p with respect to t along a constant entropy line also in the saturation zone. If you want, you say during the saturation zone, during the through the saturation zone. Is this a candidate for Maxwell? Yes. What is the replacement? Use Maxwell, partial of s. What is this? This will be partial of s with respect to v at constant pressure also along the saturation. Now go back to our, we want variation of s with respect to v at constant temperature at our point. Go back at this point. What is the variation of s with volume? When what is maintained constant? Temperature is maintained constant. And what is that value? When pressure is maintained constant. How does, when pressure is constant we are moving along this line? And how does entropy vary? And how does specific volume vary? Can I write, I will ask you this question. Can I write this as sg minus sf divided by vg minus vf? And can I write this also to be the same thing? Constant temperature and constant pressure means the same line going from saturated liquid to saturated vapour. If you want any one of these, so whether it is t or whether it is p does not matter. So s is going to be sf plus x into sfg. v is going to be vf plus x into vfg. So write this as partial of s with respect to x divided by partial of v with respect to x both at constant t or p and you will get sfg by vfg. This relation is the Clausius-Clapeyron relation. Now sometimes it is written in a different way. It may be written down as it is reciprocal instead of dp by dt. Now remember, what is our expansion for dh? dh is tds plus vdp. Now consider the process from f to g. It is a constant pressure process. Also a constant temperature process. This is p is constant, t is constant. So dh along this process is going to be tds simply and t is constant. So you are integrating it from f to g. hg minus hf is going to be t into sg minus sf. You can even check this out from your steam tables. Of course t here has to be in Kelvin. So this can also now be written down as hfg divided by t vf. Now come to our ice problem. Latent heat of ice is positive means numerator is positive. We have to provide heat to melt it. What about vfg? When ice melts, is vfg positive or negative? Vfg is negative because ice has lower density, higher specific volume. So out in vfg for ice, g is the specific volume of water. f is the specific volume of ice. So for water steam transition, steam is lighter than water. It rises. Latent heat of steam is positive. So whenever temperature increases, the vapor pressure increases. For when pressure increases, the boiling point increases. But for ice water transition, latent heat is positive. So numerator is positive. But ice floats on water. So the specific volume of ice is higher than that of water. Density is lower than that of water. So vfg is negative. Consequently, this is negative and when pressure increases, t decreases. Yes madam, Maxwell's equations are derived for a simple compressible system. Since it is a wet steam, it can be considered as a simple compressible system. Any part of steam and wet steam is no expression, no exception. Except when we substituted PV equals RT here. Up to this point, even this relation is for any simple compressible system. We said that it is 0 only if PV equals RT. So Maxwell's relation, TDA's relations including this, the relation in 5.3 are for simple compressible system. So do not be under the impression that Maxwell's equations are for an ideal gas or even for a gas. They are for a simple compressible system. They are property relations. So do not bring in a process. No, the idea of a reversible process has been used only to derive a link between ds, du and pdv. See, it is simple. Our basic relation is dq equals de plus dw. We are looking at the properties of a system. So we look at a system which is constrained to do only two-way work and that too in a small way, reversible way, quasi-static reversible way. It is a stationary system. So de becomes du, restraining it to the two-way mode of work and small quasi-static reversible changes reduces dw to pdv. And because it is now reversible, dq can be replaced by Tds. So Tds is du plus pdv is our basic property relation. The idea of a localized reversible process is used only to derive this. Once it is derived as a property relation, it is applicable everywhere because it is a relation between properties of state. And that is why in my first slide I said we will use the property relation which is Tds is du plus pdv. We will also use the first law which is dq equals de but because it is stationary it is du plus dw where dw could be pdv plus anything else which it can do. And there is no guarantee in a real process that dq equals Tds. Now one trick which you should realize is that there are two relations which we have not had the opportunity to use. And those are what are known as cyclic relations. The most celebrated cyclic relation is this. If there are three variables, then partial of x with respect to y at constant z multiplied by partial of y with respect to z at constant x and partial of z with respect to x at constant y. Note the cyclicity x, y, z, x, y, z, x, y, z. And similarly x, y, z, x, y, z, x, y, z. Product of this is minus 1. The second cyclic relation which we will generally not have the opportunity to use but you never know where T is a common variable. You may have to use this. And also the reciprocity relation that partial of x with respect to y at constant z is the reciprocal of partial of y with respect to x at constant z. Yes, yes, yes. Now this is useful when we come to say an exercise like 5.9. Just one minute. I will not go through the whole thing. See it is given that mu is defined. This is the Joule-Thomson coefficient. Partial of T with respect to P at constant h. For Joule-Thomson refrigerator, we expect the working fluid to have this value negative, sorry positive. So that when we reduce pressure, the temperature reduces. Now this is nowhere near our Maxwell. So we are stuck. So we expand this using the fact that this multiplied by partial of P with respect to h at constant what will be h? P h T then partial of h with respect to T at constant P and divide it by whatever we have multiplied it with. Partial of P with respect to h at constant T. Partial of h with respect to T at constant P. I have not changed anything, multiplied and divided by this. But in the numerator now I have the cyclic relation. So this becomes minus 1 divided by what is this partial of h with respect to T at constant P? C P. So the second factor in the denominator is the C P. The first factor in the denominator, I can take it into the numerator and write partial of h with respect to P at constant T. And remember we have derived a partial of u with respect to v at constant T using the expression for du. Similarly, consider h as a function of P and T and you can derive this in terms of PVT data and you will get the answer. Use Maxwell's relation. But here we have used the cyclic relation as a trick to get out of this funny partial of T with respect to P at constant h. The moment you get something like this, it is a hint. Use the cyclic relation because then you will get del h by del T at constant P which will be C P. And the other factor you can take care of. Now that brings us to the end of property relations.