 get started we are discussing thermodynamic properties set this equal to 0 then entropy we have S of T0 P0 plus not minus so it is R by P ? V by ? T there is a minus RLNP that is the correct expression you can check it out there may be a plus RLNP 0 minus RLNP is that okay yeah thanks so these are the two expressions that we use and we started drawing the chart not enthalpy here it is usually one intensive variable on the y axis and clearly the isenthalpic lines are simply vertical lines and the isobaric lines are like this question is what about the other lines you looked at isotherms you look at isotherms write this as TDS plus VDP and DS is CP by TDT minus partial of V with respect to TDP you look at isentropic lines for S is equal to constant you get partial of H with respect to P remember partial of H with respect to P is the inverse of the slope on this diagram and this is clearly equal to T H with respect to P at constant entropy sorry V okay then if you want V is equal to constant you get partial of H with respect to P at constant V several ways of deriving it this is equal to partial of H with respect to P at constant S if you treat H as a function of P and S you get DH is equal to TDS plus VDP so if you are differentiating this with respect to P you get doh H by doh P with respect to doh H with respect to doh P at constant S plus this term plus T to S by doh P at constant V this first term is of course V H with respect to P at constant S is V doing something you know V plus T doh S by doh P that is correct and to get DS S with respect to P you have to write SDT and VDP so S with respect to P would be the second partial of A with respect to T and P and that is equal to the second partial of A with respect to T and P if you get V with respect to T here this is correct something at constant T V with respect to P is always negative and therefore this term is actually greater than doh H by doh P at V we will rewrite that in terms of measurable quantities when I have declared measurable quantities we said the caloric measurements give you Cp Cv and Cs I will discuss this later the other quantities are P, V and T or I will put capital T itself absolute temperature and then it is derivatives these are usually reported as 1 by V doh V by doh T this is the expansion coefficient 1 by V doh V by doh P since volume always decreases as pressure increases minus of it as a positive quantity so with the minus sign this is called the constant T is called the isothermal compressibility so happens that the isothermal the isentropic compressibility or the adiabatic compressibility under reversible conditions this is adiabatic sometimes referred to as adiabatic should actually I should write isentropic this is of experimental importance because I can show you that and this is actually directly related to the velocity of sound measurements in a fluid in velocity of sound relatively easy to measure compared to most the modernic properties so this is an important property this is kai sub S I do not exactly remember what Smith and Van Ness uses usually they use a for this expansion coefficient and this is beta or kai some books it is kai in some books so when I have partial of V with respect to P at constant temperature so this is actually simply minus V times kai of T so this is equal to V into 1 plus kai of T T times kai of T T times kai of T is dimensionless actually it would have been nicer if people had defined that as the coefficient and you do not have to worry about dimensions but in practice kai of T is reported in degree k minus 1 you multiply by T most tables give you degree k so first of all the isentropic lines when V is equal to constant so let me write the slope of these lines apart from these measurable quantities the caloric measurable quantities will include enthalpy changes phase changes enthalpy changes during a phase transition we will come back to that there is some more measurable quantities but these are fundamental ones let me get back here and say the slope here you are looking at doh P by doh at constant volume is simply 1 by V and constant S and doh P by doh H at constant volume is 1 by V into 1 plus T times kai sub T so this slope on the pH diagram is less than doh P by doh so the isentropic lines have a slope that is less than the isochoric lines so on this graph I will draw this in anticipation I am drawing some figures the graphs usually look like this the all thermodynamic diagrams some exceptions like carbon dioxide it come back to it later there is usually a liquid region a liquid plus vapor region vapor region let us say I am looking at the vapor region for now I am looking at the vapor region we have already seen that isotherms would go like this or you should draw the critical isotherm and then one far away okay this is T is equal to TC above this is vapor actually technically this portion is called gas in this portion would be vapor between the gas and below the critical line but we would not make the distinction as far as property representation is concerned gas and vapor the equations are stated the same everything is same but in this region this is the critical isotherm if I am looking at this region I am looking at slope of these curves first of all the slope here is always positive unless the volume gets to be infinity if you like I will write data then equal to 0 and the slope here is less than this slope but it is also positive so it is always a positive slope line so one line could be like this the other line which will be of a higher slope so this is constant s lines this is constant V lines so you get these lines isentropic lines and you got isothermals isentropic and isocores and then you have got the isobars which are horizontal and the isenthalpic lines which are vertical you have essentially all the information you need so if I am looking at work comparisons suppose I have a pure fluid I have an alpha phase and a beta phase if I am if I write the two laws down I have TDS for a closed system TDS-PDV these equations are valid for a homogeneous phase so if I am if I am writing a mix of two phases u is equal to u alpha plus u beta simply the internal energy sum of the alpha phase internal energy plus beta so if I rewrite this I get du alpha or I would like to write it actually for G it is more content write it for G DG will be minus SDT plus VDP similarly for G also this is equal to DG alpha plus DG beta I should write here is less than or equal to so this is also less than or equal to so this is minus s alpha DT alpha minus s beta DT beta DG is equal to this is less than or equal to plus V alpha DP alpha plus V beta DP beta actually written it for I am looking at this I am looking at the closed system which is the sum of the two phases since I am looking at a closed system I should have started with a let me because basically what I have to do is put constraints on the system and ask when it goes to equilibrium in the constraint I can put on a closed system which contains two phases as you do not do any work that means I cannot be a volume change so the total volume should remain a constant if I want to impose a total volume constraint I should consider a is a natural function of T and V I am going to consider an isothermal system with a constraint of V equal to constant total V so if I am considering T and P constant then DG is the natural variable so let me come back and write this VA is less than or equal to minus I can come to the same conclusion here but it is much nicer to do it to the correct variables in you have a choice of variables you may as well choose the natural ones if you are doing adiabatic isothermal systems I am sorry not adiabatic isothermal adiabatic systems with pressure as a variable then you have S and P as the natural variables you choose H it is the equations are simply U is a natural function of S and V H is a natural function of S and P a is a natural function of T and V and G is a natural function of T and P so we will use this as a thumb rule whenever depending on the variables and depending on the constraints we have to impose in the system we will choose the independent variables you can choose any two variables as far as pure systems are concerned incidentally that is purely empirical information we know from experience that given two variables I can reproduce the state of pure substance it is always empirical information at all levels so if I have this then I have VA a plus VA beta is equal to 0 for equilibrium with VA plus V beta equal to constant this comes from you lay the condition that no work is done on the system is done on or by the question itself is asked under those conditions if you keep on doing work on a system it is not going to be at equilibrium so first is first you stop interfering with it and then you ask when does it reach equilibrium this is a closed system but if you look at a for a phase alone this can exchange molecules with the beta phase so not only is the temperature and the volume of the a phase not only are the temperature and the volume of the a phase variable the extent of the a phase is also variable so I should first write a in an open system of constant composition a is a function of net not just T and V it is also a function of the number of moles if I keep the number of moles constant a this is a this is specific Helmholtz energy it is a function of T and small v specific volume but if you write it for the whole system which is what we are writing here I should have written this this implies VA is equal to partial of A with respect to T which would be minus S because that partial is taken holding V and N constant if you hold it and hold N constant you are talking of a system of fixed number of molecules so this is still minus S DT there is still a minus P DV when you differentiate with respect to V you get minus P plus a with respect to N DN now if I write the same thing for G I get DG is equal to minus S DT plus V DP plus G with respect to N DN notice this a with respect to N is holding T and V constant the G with respect to N is holding T and P constant and write that out explicitly TV and this is T, P and by comparing these two you know G is equal to a plus TS so if you a plus PV so if you differentiate DG plus D of PV you get minus S DT minus P DV exactly so by this comparison you know these two are the same partial of A with respect to N at constant T and V is equal to partial of G with respect to N at constant T and P the variables that you hold constant are different in fact this is also equal to partial of U with respect to N holding SNV constant is also equal to partial of H with respect to N holding SNP constant this is because of the redundancy that we introduced U is the only independent quantity you introduced H A and G U and S were the independent quantities U the others were introduced by combination this quantity comes up again and again and for reasons that gives intuted it was called the chemical potential I will come back to why it is called the chemical potential so when I write three lines it is a definition so up to here I am right I need to add a term here from equation for a you get minus S alpha DT alpha minus P alpha DV alpha a is minus S DT minus PV plus mu DN so I will have to add here plus mu alpha DN alpha and here I have to add plus mu beta DN beta similarly here also I have to add the same thing I will just put the arrow there so that term will have to be added to both those equations I have to redraw this I do not know when we are getting the full board we have one more constraint because V alpha plus V beta is constant and because the closed systems just like this is constant because no work is done the system as a whole two phases together do not exchange mass with the surroundings we are looking at a closed system therefore N alpha plus N beta has to be equal so in these equations DV alpha is equal to minus DV beta DN alpha is equal to minus DN beta so I have DV alpha is equal to minus DV beta DN alpha is equal to minus DN beta so I have DA is equal to for the whole system is minus S alpha DT alpha minus S beta DT beta then minus P alpha minus P beta into DV alpha minus mu alpha minus mu beta plus DN alpha this should be equal to 0 this is the criteria of equilibrium the two laws combined will tell you DA is less than or equal to on the right hand side this is the criteria this is equal to 0 at constant T and P at T and V at constant T that is all for a closed system it is at constant temperature because I have already held the volume constant I put V alpha plus V beta equal to so at constant temperature I know A has to be a minimum therefore DA has to be equal to 0 now out of these I have constant T which means then I have to specify further conditions when I say equilibrium I ask for three kinds of equilibrium I ask for thermal equilibrium which means T alpha is equal to T beta is equal to T say we just label it as T the two temperatures have to be equal I ask for mechanical equilibrium which means P alpha is equal to P beta is equal to P some given pressure P so first I have thermal equilibrium which means temperatures have to be equal I know if temperatures are not equal heat will be transferred from one to the other that is my 0th law so as long as changes occur I cannot have equilibrium so this term is 0 because of thermal equilibrium P is equal to constant and thermal equilibrium thermal equilibrium demands that the two phases are the same temperature and I have put T is equal to constant mechanical equilibrium demands that P alpha is equal to P beta therefore this term is 0 I am not setting P equal to constant all I am saying is the two pressures have to be equal the criterion of equilibrium says DA should be 0 at constant temperature for a closed system I am saying if the pressures are unequal then one phase will do work on the other phase it will expand till the pressures are equal and so I have to have P alpha equal to P beta whatever the value of P if this has to happen this implies this criterion of equilibrium says we call this equation 1 equation 1 then implies mu alpha should be equal to mu beta so a is the natural variable to consider because I can build in the work function correctly I could have done it with the G equation as well notice that the law really says DA is less than or equal to 0 and what you get is DA at constant temperature and conditions of mechanical equilibrium is equal to mu alpha – mu beta from D this is thermodynamics the two laws at constant temperature constant no work this is calculus from calculus alone you would not draw any conclusions in thermodynamics you have to always use the law at some point calculus is just going to get equal give you equalities so you got this result and from thermodynamics this should be less than or equal to 0 and I am saying at equilibrium it has to be equal to 0 because to reach the minimum and if mu alpha is not equal to mu beta let us say example mu alpha is greater than mu beta then DN alpha can be of any sign if I choose DN alpha to be positive I change increase the number of moles in the alpha phase if DN alpha is negative I decrease the number of moles in the alpha phase in this I have a choice and by definition as a thermodynamics I will be as perverse as possible and you still have to show that the law is valid so if you give mu alpha is greater than mu beta this is positive I will simply choose DN alpha to be positive and so the energy will increase I mean I will choose sorry DN alpha to be negative so I can decrease the free energy further if I can decrease the free energy further I am not already in a state of equilibrium if I am currently in a state of equilibrium it should be possible for anybody to reduce the free energy of the system now I let us say I am now in a state of equilibrium then there must be no possible way for me to decrease the free energy if this was greater than this I would simply take away moles from alpha phase and decrease the free energy I should not be able to do that conversely if mu alpha was less than mu beta I will choose DN alpha to be positive I will add moles to the alpha phase and decrease the free energy again and thermodynamics as you can never do it if you are already at equilibrium therefore the conclusion that is then DN alpha less than 0 would lead to decrease in a if mu alpha is less than mu beta DN alpha greater than 0 would lead to decrease in a neither is possible if you are already at equilibrium neither is acceptable if the system is already at equilibrium so this implies mu alpha is equal to mu beta and Gibbs called it the chemical potential because if T alpha is greater than T beta then you have a temperature gradient and heat flows in the direction of the decreasing thermal gradient similarly if mu alpha is greater than mu beta then heat and moles would be transferred in the direction of decreasing mu alpha so that the free energy decreases so you call this the chemical potential probably the most important variable in chemical thermodynamics you know if you have read this is your reduction I had observed improve when you sort of assume let us assume mu alpha is greater than mu beta then by transferring moles away from the alpha phase I can decrease the free energy which is absurd because I am already at equilibrium so you assume first of all that you cannot decrease the free energy and then you show that if this is true then you can decrease the free energy then you could not have been at equilibrium or this is not true so what we have is we proved that mu alpha is equal to mu beta then let me also integrate the basic equations and again this is very typical of thermodynamics in particular I will take this equation this is the most easy convenient equation for most purposes you can do this with any of the equations that you have in thermodynamics so I am writing the basic equation for an open system basic equation for open systems now I want to integrate this I am going to do something this is system at t is equal to 0 t time equal to 0 system at time t now at time what I am going to do is I am going to connect it to a reservoir this contains fluid at some temperature and pressure p in the reservoir I have the same fluid at the same temperature and same pressure my process is the following the whole thing is a thought experiment I simply take this box I add more fluid to the reservoir and make sure its volume changes so that the t and p remain the same all the only difference between these two stages I have n moles here I have k times n moles here I keep the extensive variables the same so that the specific state of the system remains the same if I take one mole of the substance here one mole of the substance here the state is the same because t and p are the same first argument is if t t and p are the same in this process for this process if I took take a calculation of dg this is 0 this is 0 mu by definition is partial of g with respect to n at constant t and p I claim this is an intensive variable because it is the property it is the partial molal property so I am really calculating g per mole of addition g is extensive n is extensive I am asserting that partial of g with respect to n is an intensive if it is an intensive variable it can be a function only of t and p might say this is in a fundamental way an assumption in classical thermodynamics but let me come back right now if mu is an intensive variable it must remain the same here as here so for this process g final minus g initial is simply equal to mu times n final minus n initial this is k minus 1 times g initial which is simply g mu times k minus 1 times n ni and this is same as g use the notation in the initial state initial state has n moles and it is free energy gives free energy is g and k need not be 1 therefore I can strike this out this is valid for all k so g is simply equal to n times mu so although we introduced a variable mu as the partial of g with respect to n in the case of constant composition systems or in the case of pure substances that we are now dealing with mu simply happens to be small g so this is g by n which is by definition small g is also equal to me so the specific Gibbs free energy is exactly the chemical potential you can see the relationship with other variables because you have introduced this redundancy h is g plus ts so if you are doing partial of h with respect to with respect to n you will get partial of g with respect to n which is mu plus t partial of s with respect to n and you can do the same thing with a as g plus p v so h with respect to n is mu plus a quantity the only partial that is exactly equal to the specific property is the Gibbs free energy specific Gibbs free energy is the chemical potential okay having said that no let me go back and look at our equations for equilibrium so the equilibrium equation said mu alpha is equal to mu beta if you have two phases you can derive the Clausius-Clapeyron equation from this because d mu alpha is equal to dg alpha you have done that before but this gives you h alpha minus ts alpha is equal to h beta minus ts beta h alpha minus h beta let us say is delta h is equal to t delta s in delta h is a measurable quantity you know that there is a latent heat you can measure you can do this in experiment so there is a positive difference in enthalpy between two phases so we have done this part this is your critical isotherm in this region we have seen two sets of curves I think this is constant s this is constant v right now if you come to if you go below the critical region the lines here are vertical as we have pointed out because it is ideal gas at low pressures you can exactly calculate this ideal gas behavior is vertical lines so at constant temperature we are saying thermal equilibrium means you are looking at an isotherm mechanical equilibrium means you are looking at the pressures being the same we are saying the enthalpy of one phase is greater than the enthalpy of the other phase so there will be a shift here in the enthalpy this quantity is your delta h for phase change this in case can be the alpha phase this can be the beta phase the enthalpy is larger by convention for vapor so you get vapor here and you get liquid here this is liquid plus vapor in between this delta h is a function of temperature if you go to lower temperatures you find the width experimentally you can show the width actually increases so delta h will increase as you go down in temperature we can do the calculation for that in a minute so I have done this portion the isotherms here you can calculate the isotherm slopes you again have DH is equal to CP DT plus V minus T doh V by doh T to DP so at isotherms this is along isotherms this is 0 so partial of P with respect to H is simply 1 by V into 1 minus alpha this is the expansion coefficient 1 by V doh V by doh T is a measurable quantity minus alpha T sorry for an ideal gas alpha T becomes 1 and therefore you get infinite slope those lines here also the slope is reasonably steep because 1 minus alpha T becomes much small here actually 1 minus alpha T is non-zero but V itself is small in the liquid region so you got high pressures you get steep slopes so the isotherms are always somewhat steep so what happens is I have got these isotherms I have got the difference in the enthalpy so the isotherm will have to come like this there will be a difference in the enthalpy then there is the other half so this side is the vapor region this side is the liquid region what we can do is actually calculate this whole thing and like you to compute it for a van der Waals fluid because in order to give you information I have to give you CP and so I have to give you CP the specific heat and PVT information PVT information for example you can take the van der Waals equation you said P plus A by V square into V minus B is equal to RT the van der Waals fluid is of great importance for two reasons first it was the first historical equation of state that took into account attractive forces and secondly it has no qualitative contradictions you have to show that the quadratic form is positive if you show that you can show what are called stability criteria you can show we always has to decrease as pressure increases things like this which you take for granted or you take for granted that some of these coefficients are positive they all come from this requirement of stability stability is not usually discussed in undergraduate classes so I will not discuss it further it is only simple calculus it is putting down criteria of calculus on the second derivatives in the equations in order to show that A is actually a minimum and not a maximum at equilibrium so this equation has no qualitative contradictions it satisfies all the thermodynamics stability criteria qualitatively it does not agree with experiment but qualitatively it is one of the few equations of state that is completely free of faults therefore it is of importance.