 Okay let me start we were discussing the ensemble we are discussing the probability of a state P i we showed this was e power – ? e i by q remember q is a function of n vt we said e bar is equal to sum over e i P i this is first of all here and d e bar is equal to sum over i d e i which we said partial of e i with respect to v d v this is equal to sum over e i P i – p bar d v because this was – p use small p for pressure when we said this corresponds to d e in classical thermodynamics which is T ds – p d v so we said ds was equal to is equal to 1 by beta T no I have to do this e i d P i sorry 1 by T times sum over i of e i d P i we said e i was after all – 1 by beta you take it from here take the log of all this into log of P i plus log q you had ds is equal to – 1 by beta T sum over i of ln P i d e i d P i plus ln q into d P i we said sum over P i is 1 therefore sum over d P i is 0 so this term is 0 and we rewrote this exactly for the same reason as – 1 by beta T sum over i I put the d outside P i log P i because d of log P i d P i is log P i d P i which I have here plus P i times d log P i d log P i is d P i by P i P i will cancel sum over d P i is 0 now if you take two independent systems I am going to give this argument very quickly which are rather important argument if I have two systems A and B at the same temperature and in equilibrium with one another you know that the entropy of the combined system is simply the sum of SA plus SB the corresponding probabilities of two systems are independent you know that the probabilities of independent systems are products of the probabilities so if you take the product term you will find SA plus SB being the same requires that beta T is a constant I will give you references the book that I have the only book I have read cover to cover is Hill Hill has two books one he wrote he was actually at Oregon University he was actually originally at Yale with on Sagar code and all those big people then he moved to Oregon and he wrote a book called statistical mechanics in 1950s when people said it is too difficult to read so they called it big hill they said you have to write a small hill and so he wrote another book which is actually fatter than the first book but simpler to read it is called statistical thermodynamics and both of them are really classics Terrell Hill T something else I forgotten this middle initial and is called introduction I think one is called statistical thermodynamics maybe it is called introduction to statistical thermodynamics the other is called statistical mechanics and please look at this essentially what that will tell you is if the entropy is to be an extensive property which if you have two systems and the entropy is additive and the probabilities are automatically if they are independent systems you know the probability of an event that is the probability of finding your system A in state P I and system B in state PJ is simply PI times PJ PI of A times PJ of B so if that is true then the logarithms will become additive and you will get the same result that SA plus SB is equal to the sum of the two entropies and for that to happen beta T has to be an absolute constant so we will simply say here beta T is equal to an absolute constant and then by proceeding further and deriving the properties of an ideal gas you can verify that this constant surprise is actually the Planck's constant it is surprising that there are so few universal constants this Planck's constant comes up again and again and again so this is this was not necessary for statistical mechanical theory it is exactly like your absolute temperature and the ideal gas temperature being proportional to one another with a proportionality constant of one so all I need in the theory is that this should be constant but measurement by measurement by comparison ideal gas properties you can show that it is actually K now let me get back here so S is therefore equal to minus sum over PI log PI for an isolated system we are to looking at a closed system right now but if you isolate the system then all the probabilities are equal PI is simply equal to 1 by Omega all where Omega is the number of states so if you put PI equal to 1 by Omega we will simply get S is equal to log Omega incidentally this is inscribed on the tome of on the burial stone of Boltzmann Boltzmann Boltzmann's burial stone carries this inscription S is equal to log Omega in some senses it is the most profound result in statistical physics simply tells you that the number of possible states in which an isolated system the logarithm of that gives you the entropy so entropy is about choice and if you have a larger number of such possible states then the entropy increases and the famous statement it has to be said in German to sound very profound the entropy der Welt Zimmer half a button always increases got in the word for increases in German anyway the entropy of the world always increases of the universe always increases that is the famous statement the idea is that you get more and more choices but there is nothing to prevent the system from going back to the original choices but it never does that is what gives you a reversibility that is if you have for example this is the most classical example that is always given I have the same gases say hydrogen gas on either side I have an impermeable partition at t is equal to 0 I break this partition let us say I have hydrogen gas here and vacuum here initially this is gas this is vacuum I cut this membrane at t is equal to 0 there is nothing that tells the hydrogen atoms they cannot stay on this side but they want simply very large number on an average you know with time they will spread out evenly the number of states if these two partitions are equal size the number of states available to the system has doubled because you have got twice the volume if it is doubled then the entropy will change as log 2 times the original entropy and this is this is the entropy and delta s is equal to R log 2 is the classical if you do this for v1 and v2 you will get ds is equal to if you do it at constant temperature ds is merely minus cv dt see let us let us do this okay we will write it in any in cv dt by t plus partial of s with respect to v s with respect to t then s with respect to v is partial of p with respect to t right s increases with v and this is for an ideal gas this is p by t this is r by v so as s will integrate to R log v2 by v1 right the entropy change for an ideal gas at constant temperature is R log v2 by v1 so if you double the volume the entropy increase will be R log 2 and this tells you that the number of states has doubled so the entropy will go as I should not say ln should say k but remember this is molecular entropy if you do it for one mole you will multiply by Avogadro's number so you will get R log ? and this is done per molecule but you have to actually calculate per mole because your practical chemical engineering measure of sizes in terms of moles not in terms of molecules so you simply multiply by Avogadro's number this is the result that you will use because this is by definition for an isolated system and you cannot verify it directly for an isolated system because the minute you make measurements on a system it is no longer isolated so you get into this problem of relating to an isolated system so you can only about isolated systems you can say anything you want and trouble is logic will lead you to conclusions about a closed system for which you can make measurements and you will be exposed at that time and if you are exposed nobody will read your theory afterwards so this is the final result this is the most important result that you have from statistical thermodynamics the other result is this this is an important result to I should write this as if I write this as DEI it may go back say this DE bar is equal to sum over PI DEI plus sum over I EI DPI this says the energy of a system can change only in two ways one is because the energy levels themselves changed the other is because the probabilities of occupation of those energy levels changed right and this turns out to be work where the energy levels change if the volume changes EI changes because quantum mechanics tells you EI is a function of N and V as you change V the energy levels change and as the energy levels change you do work on the system whereas the energy levels remain the same but the occupation changes the number of systems in that state changes then this energy change is because of heat so this has the interpretation of ? Q and this has the interpretation of – ? W depends on the minus sign only comes because of convention E with respect to V is – P then you will get otherwise you can say work done on the system is positive and then become plus in terms of measurable variables of course the equations are always the same but this is formally work this is heat so this is something that falls out of molecular theory naturally but this is what took Joule thousands of experiments to prove to the Royal society long after they were convinced he had to continue convincing them because he was not convinced anyway after so the biggest achievement in thermodynamics is to find that there is a perfect differential which is the algebraic sum of two quantities heat and work and there are only two quantities that you have to worry about heat and work that we have said but in statistical thermodynamics you get expressions for them the way to derive thermodynamic properties finally to make life easier you go back to L and Q you go back to Q this is true of I have done this for the canonical party canonical system that means the closed system Q is simply sum over e power – ? E I so let us look at partial of Q with respect to ? this Q is a function of N V and ? ? is 1 by KT okay it is not I am writing ? instead of T ? remember is 1 by KT it is the more natural variable in statistical mechanics so if I take partial of Q with respect to ? I get – ? over EI e power – ? EI if I divide this by Q divide this by Q this becomes – ? over EI P I which is – E bar the second result it is a function only three variables so let us differentiate with respect to all three variables partial of Q with respect to V is – ? over I am differentiating this I am going to – I am going to get – ? ? EI by ? V e power – ? EI now you know the game divide by Q divide by Q both sides this ? EI by ? V is – P so this is equal to – pressure in the ith state there is a ? of course times P I which is the probability so this is equal to – ? P bar and then ? Q by ? N let me divide by Q in anticipation would be – ? again sum over partial of EI with respect to N e power – ? EI because I have divided both sides by Q is also a Q there at the moment I am going to write this as – ? Mu I will tell you why the minute or actually I will go to an easier proposition then we will get this result let me do this let us look at a is E – Ts so this will be E bar – Ts incidentally we do not put an S bar or an A bar you write only E bar I told you the averages can be calculated only for mechanical quantities quantities that are defined in the individual state which will be pressure energy entropy arises because of the existence of multiple states so it is not an average of a property in a particular state you do not take the average value over several states it exists only because there are multiple states and you have a choice if you had exactly one state if ? was 1 and that is the that is how the 0 law comes up 0th law of thermodynamics tells you that entropy of a perfect crystal at 0 degrees K is exactly 0 and that comes because the number of states in which a crystal a perfect crystal can exist a perfect crystal is simply a monatomic substance in a crystalline form at 0 degrees so it cannot have translational kinetic energy it cannot have vibrational rotational energy if it is a monatomic substance such a crystal we can exist only in one state it has no choice therefore entropy is 0 what I want to show is that this is equal to E bar we have seen this is okay let me write this out this is sum over E I into P I minus T into sum over entropy is K into P I log P I what I want to show I have to do some I have to show a is equal to minus KT log minus K log KT log Q you see if that is easy to show yeah this is sum over that is quite easy E I plus this is 1 by beta I have log P I times P I but you know log P I is is minus beta E I plus log Q so log P I by beta plus E I is log Q by beta log Q times sum over P I by beta so this is simply log Q by beta so you have this result Q is equal to minus KT I am sorry A is equal to minus KT log Q so the whole aim of molecular theory is to calculate Q if you can calculate Q you take the logarithm you get A then you can differentiate with respect to A with respect to V for example at constant and at constant temperature will give you S right you know DA is this is for a closed system DA is minus STT minus PDV so if you want the pressure you differentiate A with respect to V you differentiate A with respect to T same thing you do here set when you differentiate Q you differentiate with respect to beta because beta is a more convenient variable. So in fact I could have started with any system I started with NVT system this is called the canonical system okay I could have done the NVE system which is an isolated system this will this is called micro canonical because it essentially if you take the NVT system and say all systems will be in state EI one particular state it becomes an isolated system so it is called micro canonical you can have an open system you can have sorry VT mu N is not constant but I can have a system in which VT and mu T represents equilibrium with the surroundings mu represents again equilibrium mass equilibrium so this system would be called grand canonical system and so on these are different ensembles there is also another ensemble it is convenient use which is called the NPT ensemble or isobaric ensemble because it holds the pressure constant the point is molecular theory has very different ways of looking at a system but you know the thermodynamic properties of a closed system I mean classical thermodynamic properties are completely fixed if I give you this and if I take a closed system and give you two variables all other variables are fixed in classical thermodynamics in a pure system you can extend the argument to mixtures readily then the question is if you have different ensembles will you get different answers you are not supposed to so you have to show that you do not actually get different answers the flip side of it is simply here that after all I calculated energy as an average quantity is there a fluctuation in the energy so the energy itself is sum over EI times PI what is the fluctuation in the energy by fluctuation I want to know what is E minus E bar the whole squared or I will calculate E minus E bar the whole square essentially what I am going to do is to calculate this average quantity E minus E bar average is 0 but E minus E bar squared average will give me the fluctuation to do this what I do is simply take this EI minus E bar whole squared and PI any quantity if I want to find its average I simply find the quantity in state I am multiplied by so this is sum over I this is sum over EI squared minus E bar squared minus 2 EI E bar into PI sum over I this EI PI is E bar so you will get 2 E this is plus so these two will cancel I will get simply sum over I EI squared minus sorry EI squared PI minus E bar squared so what I do is go back to Q Q is the source of all inspiration now in the God you worship now is a new God it is called the partition function so Q is sum over E bar minus beta EI if I want to bring down EI I differentiate with respect to beta so I get delta log Q by delta beta is sum over EI E bar minus beta EI is that the minus sign sorry delta Q by delta beta so if I want to bring down one more EI I differentiate again with respect to beta squared it is a thing we never grow up near a kid you do the same thing again and again right you do the same thing here except that you do it in a very sophisticated way with symbols that others do not understand therefore you appear very clever so we just differentiate again with respect to beta I have got EI so this is the quantity that appears here because if I now divide by Q I get EI E squared bar but this can be rewritten what is delta log Q by delta beta and take delta squared log Q by delta beta squared this is delta by delta beta of delta log Q by delta beta which is 1 by Q delta Q by delta beta which will give you 1 by Q delta squared Q by delta beta squared this is one term minus 1 by Q squared times delta Q by delta beta the whole squared sorry what I am trying to do is find out what E minus EI squared bar you know that quantity the reflux variation in the energy that I can expect is in terms of thermodynamic properties to do that I have to always go back to Q log Q and its derivatives now log Q this I know because log Q with respect to beta is minus E bar minus delta E bar by delta beta right this is the same as this delta log Q by delta beta is simply minus E bar this is equal to the quantity I want is this it is equal to this quantity here 1 by Q this is this is what I want okay these two are the same this is 1 by Q delta squared Q by delta beta squared minus this is delta log Q by delta beta the whole squared that is simply E bar squared. So I have got both these quantities I wanted EI squared times PI right which is EI squared bar EI squared bar accordingly or E squared bar is therefore equal to this is this is E bar squared this is E squared bar so E squared bar is equal to E with respect to beta what is E with respect to beta minus delta E bar with respect to beta is CV remember this is a canonical ensemble when you differentiate with respect to beta you are holding V and T constant I mean V and then constant so it is for a closed system at constant V so you will get CV if you differentiate with respect to T times dou T by dou of 1 by KT this is minus KT squared times CV it is minus KT squared. So E squared bar is E bar squared plus E bar squared plus dou E bar by dou beta dou E bar by dou beta is this actually it is minus dou E bar by 2 so it is plus KT squared CV in what you are interested in is E minus E bar the whole squared you are interested in E squared bar minus E bar squared E squared bar minus this is KT squared into CV and I want to find what this is with respect to I want to find the relative deviation I want to show you that this is of the order of 1 by n because CV if you are doing it per mole CV is NK this is NKT of the order of NK this is of the order of NKT so if you square this you will get an N squared here you will get only 1 N in the numerator so the ratio will be simply 1 by n so what it tells you is that sigma E is of the order of sigma E by E sigma E is this squared minus this the square root of that is of the order of 1 by square root of n that means any fluctuations that you observe in the calculated energy will be of the order of 1 by square root of n compared to 1 so if you have 10 to the power 23 molecules unlikely that you see any difference at all if you have 100 molecules you have a good chance of seeing 10% deviation so you get the smaller and smaller and smaller numbers your prediction basically your assumption in classical thermodynamics is that you do not worry about the constitution of matter that all properties vary continuously in particular the density varies continuously density never varies continuously because you can only add a molecule and take away a molecule but if you add a molecule to 10 to the power 23 molecules you will not notice that the differences not differential but if you have 10 molecules and you add and take away a molecule you have trouble which is why nano systems are so special in the sense that in nano systems you have number of molecules is so small that if you add or take away a molecule the average property will change significantly that is if you predict an average property for nano systems you could have if you have only 10 molecules in the system then you could have a deviation of 1 by square root of 33% a priori deviation is permitted that is why human behavior is also very predictable because you are talking of a billion people in India it is even more predictable then in a small country but fundamentally you are talking of large numbers all these are laws of large numbers in fact the whole concept of irreversibility and time arrow and all that comes only because of large numbers it is only when you have large numbers because if you had this box it and only two molecules and this set and a vacuum on this set and you open this up these two molecules the high probability that they will stay on the left side and nothing to make them move so there is no it is very difficult to predict that arrow of time you will say I will watch it after a time if molecules are on both sides then I know time has passed all that bunker won't work with two molecules but if you have a million molecules maybe very sure that about 500,000 will turn up here the law of large numbers is so true that people have applied it to animate systems although the fundamentals of these are not applicable to animate system because fundamentally you assume here that every system has a state which is described by variables that are not a function of history right that is one basic assumption I started with internal energy the proof this two laws of thermodynamics simply established that you internal energy and s are functions of state in an animate system they are not functions of state they depend on memory in the system so if you have memory is not applicable but one of the schools of thought it is called social economics and so on is that even in animate systems the law of large numbers is so overwhelming that memory does not matter they will still behave the same way so you can predict social behavior based on that it is also a reasonable assumption in except in traumatic cases if you have a second world war and the Jews have been exterminated for a long time Jewish populations behavior will defend depend very strongly on that persecution so other than such traumatic events normally seems true that in social dynamics also the law of large numbers works to large extent the predictions are valid there is a book by Georges Q Roma Georges Q what is his name as Nicholas Georges Q sorry count Nicholas you got the Nobel price for economics in 60s essentially applying second law of thermodynamics to the equivalent of the second law to economic problems it is a very beautifully written book there is a variation of that by Jeremy Rifkin it is a popular book okay let me get back I will just do thermodynamic properties of an ideal gas very quickly to show you that all this is not just talk I want to calculate Q e power – beta ei there are several ways of doing it I will do it in the classical mode this is summation over I this means you multiply by ?i which is 1 right you take integral counts you say state 1 state 2 state 3 etc I changes by 1 so you multiply by 1 if I want to convert this to an integration I must do it over all the unit cells in phase space so I go to phase space I have P and R I already told you that a cell in units in this space if I have only Px and x for example I know that this cell has to be greater than or equal to H so if it has to be minimum this size that is what quantum mechanics tells me so all I do is divide this phase space into cells of size H then ?i will actually be I will rewrite this e power – beta ei will be P squared if I write it in continuous form I am going to write this as integral P squared by 2m plus potential energy d of here I integrate over dpi and dr I will have to do this right now I do this by H I have to take a product over many of these cells because one there are actually 3n of them so I will write 3n if you like actually what I mean is I have Px P ypz like that for every particle so it is P1x P1 yp1z and so on since these are independent particles actually I know this is going to be Q is going to be of the form Q power n where Q is for individual particles so I can take or I will write Q power 3n e power – integral – beta Px squared by 2m plus the potential energy is 0 so I have dpx and dx I am sorry this is Q I will write small Q as this and capital Q as Q to the power 3n this will give me L times this is simply e power – beta x squared dpx and the momentum in the x direction can go from – infinity to plus infinity the integration over x also has to be done this will be from 0 to the container size let us take a cubical container eventually the properties do not depend on the shape of the container so you will get L times this integral e power – beta Px squared by 2m into dx dpx this is the error integral do you remember what its magnitude is something like pi by the square root of this right e power – ax squared dx do you remember – infinity to plus infinity e power – ax squared dx is equal to square root of pi by a you recall that it is of that form I mean there may be a factor in front put it as proportional to that means if I do this integral I will get Q is of the form L times this is kt 2m kt will come the denominator so we will get square root of pi into 2m kt so Q cube will be L cube will be the volume V then square root of pi 2 pi m kt so Q will be simply Q to the power 3n so it is V cube I am sorry V to the power n 2 pi m kt to the power 3n by 2 Q cube is this to the power cube this is volume this is cube at this stage actually if you go back to quantum mechanics you have to in quantum mechanics you will get the correct answer in classical mechanics you make a small mistake because the particles are indistinguishable you have to divide when you calculate this this way you have not taken into account that the particles are indistinguishable so this is what is called the correct Boltzmann counting and write down in quantum mechanics you do not need this if I had done it quantum mechanically I would have got the correct answer in all this will be the same divided by you have to divide by the number of permutations that are possible of the particles and we would not write n factorial because in quantum mechanics in statistical mechanics n factorial is very ugly so we will write it as n by e to the power n using sterling's approximation so this will be if I write this as n by e I will get e V to the power m 2 pi m kt to the power 3n by 2 divided by n to the power n if you do not put n to the power n you will get you can show that entropy is not an extensive property effectively so this does appear sneaky but it actually comes in automatically if you do quantum mechanics because quantum mechanics recognizes the distinguishable indistinguishable states in classical mechanics you do not make that distinction that is the only difference now once I have Q I take log Q to get the pressure I am going to differentiate with respect to V to get the energy I am going to do differentiation with respect to T so I am only worried about T and V dependence so I will get n log V on top and then T dependences plus 3n by 2 3n k by 2 or 3n by 2 log T I am looking at plus some function of n and if you like mass so what is my energy this is k if I take kt log Q I get the Helmholtz energy a is simply kt log Q think there was a minus sign minus kt log Q so if all I have to do is multiply by the kt n kt log V minus 3n kt by 2 log T plus a function of n and m, m is the molecular weight the mass of a single molecule it is not the molecular weight mass multiplied by Avogadro number will give you molecular weight so if I want for example if I now want the pressure P is minus partial away with respect to P at constant volume so I will get nk by V nkT by V is something the matter kt log Q I get some out terms and not so I am sorry this is with respect to V and what I am writing yeah partial with respect to V so only one term this is RT by V low and behold you have the ideal gas so it looks like you start with obscure assumptions and get to a beautiful result gives you complete mechanistic picture it talks I think really you have to look at these things in historical perspective but the whole result is beautiful you can either classical thermodynamics does not have much scope in terms of derivations because you have to imagine mechanisms and do it in statistical mechanics you can go to the equilibrium part and derive these results you can also discuss things like diffusion yes other things and stop there as far as the course is concerned it is over now formally declared closed.