noc19-ch17-lec24

[Music] [Music] welcome back in today's lecture we are going to talk about equilibrium and stability in the earlier stage of this lecture we looked at the concept of the entropy maximum for defining an equilibrium condition okay so the earlier definition of what we have used was something called entropy maximum principle so I'm going to just write it down this is something which we have used to obtain conditions such as temperature equality pressure quality and chemical potential equality for an equilibrium system so the definition is that the equilibrium value of any unconstrained [Music] internal diameter is such as to maximize the entropy for the given value of the total internal energy so for given internal energy the system tends to maximize the entropy okay in order to achieve the equilibrium condition so that is what we used in the earlier stage of the course and we used the maximization of entropy to obtain various different conditions now what we're saying is this particular concept or principle is equivalent to the energy minimum principle okay which says that the equilibrium value of any unconstrained internal parameters so it says the equilibrium value of any unconstrained internal parameters is such as to minimize the energy for the given value of the total entropy so it says that for fixed entropy the energy is going to get minimized to achieve the equilibrium State ok that is what is saying that so this two statement or two principles are equivalent so in other word what we are saying is the following that for the case of the first u.s. is function of UV n so for a given u we are saying that the first derivative with respect to some dummy variable legs at a constant u should be zero and the second derivative X should be less than zero so this is because this s has to be maximum so this is the definition of the maximum the first and second derivatives should be zero and less than zero respectively now this we are saying that this is equivalent of saying that if we consider u as a function of s B so for a given total s okay so this is equivalent in some sense okay so let us try to get this proof of kind of equivalence so for in order to get that let me just first consider the following so the question is how do you prove this so let us consider the chain rule okay where X could be anything here this must be one okay right so you have su X and that is what this relation we know now this implies that we want to understand that we want to obtain this relation again this is nothing but del u by Del X at constant s that must be minus 1by and del s by Del u X del X by Del s u or del u by Del s X and del s by Del X u okay now by definition the derivative of U with respect to s is nothing but T so this is nothing but t del s by Del X u okay now for maximum okay entropy or in general for equilibrium if you consider this then del s by Del X at constant u should be equal to zero so this at maximum X should be equal to zero which essentially means del u by Del X at X should be also equal to zero okay so this we are trying to get a coolants of these two different relations which we are saying that this is a BOTY : so for a given s you should be minimum or vice versa okay and we are talking about unconstrained at this point internal variables so let me also get the second derivative second derivative we have to prove for minimum you they should be negative how do we do that so let us first start with the expression okay we want this to be evaluated so let me just break this again here del u by Del X okay for the sake of simplicity we are saying this has a M okay this is M then it becomes del M by Del X okay at constant X okay so we need to find out this now M is basically a function of U and X okay from the definition you can clearly see M is a function of U and X hence I can write DM as D del M by Del u constant X dou u plus del X constant u del X okay so this is by simple now if you do take the differentiation now since we all want to del M by Del X so I take this differentiate this with respect to X so they run by the racks are keeping s constant so this will be del M by Del u X del u by Del X that's constant plus okay now this will be one and what about this at extremum so add extremum this should be zero as we have already defined it here so which essentially means that the second derivative del n by Del X s is nothing but del M by Delta X u so this is what we can obtain from this derivation for the case of extremum okay so now we go back and put the variables here so this is nothing but del square u by Del X square yes that is by definition that's delam by Del X was nothing but this right and now I can write it this as follows del by Del X del u by Del X s so if you look at this expression on the left hand side I can simply write del square u del X square s that is by definition of M and now on the right hand side which we have derived it that this is this expression is equivalent to this expression I plug in the definition of M here which is nothing but del u by Del X s but then U is constant here now here this point what we can do okay this is the point we can represent this del u by Del X s which was used earlier as nothing but minus T del s by Del X u okay so this is nothing but minus T it is a constant here and I bring here T del s by Del X you you okay this is what we defined earlier this was you at a constant you so this comes here as you now with this we have now the expression minus T del square as del X square you okay now this we know at extremum if you look at the definition this must be less than zero okay so if this is less than zero that means this is less than zero that means this is greater than zero so essentially this gives you the second condition del square u del X square as is greater than zero and of course the first derivatives del u by Del X at constant s is equal to zero so this is something which we had derived and that means if you look at the original question was that this principle was we used to evaluate the equilibrium conditions by maximizing the entropy for a given internal energy okay this is equivalent to saying that U is minimum for a given and dropping okay so that is an important expect okay so this essentially can be drawn also so can one can come up with this kind of diagrammatic explanation of this that for s as a function of X which is unconstrained variable okay this reaches a maximum for a fixed u VN and this is equivalent to saying that you or yeah some kind of for this u as a function of again X unconstrained variable and this is something which we get a X equilibrium so this is X okay for a fixed espy okay so this is something which is equivalent that is something which we have just derived so using this concept of minimum energy we can actually try to again evaluate the conditions of equilibrium such as thermal equilibrium and so forth so for the sake of illustration I'm going to just do one example okay so this would be just establishment of thermal equilibrium and similarly you can do that for pressure and chemical potential so establishment of thermal equilibrium using energy minimum okay so consider their closed composite system okay with the rigid so we can draw it like this let's say okay so this is rigid impermeable and tighten so essentially the heat can exchange back and forth okay but it's a closed system okay and it's a composite so it's a closed composite system so what is the total energy of this overall system we can write it U as u s1 we and N and so forth right one stand for this this could be two first compartment second compartment okay and this plus U of the second one okay whereas here we'll be sick for the second and similarly for the VN n so what we know is that total volume and number of volts these are constant and known so total energy is fixed okay and in the initial stage s and 1 and s 2 can change because of the heat exchange okay so we can write the D U as d u1 plus D u 2 and since there we are saying that you know the only contribution is through the heat so the change in the internal energy for 1 & 2 are corresponding to the heat and we can write it there considering is an equilibrium process a slow process we can write this directly as this okay so this is something which we can write now this has to be zero because total energy is constant okay now we also can consider there at equilibrium of course at equilibrium what is the conditions of this temperature so at equilibrium of course we are considering that will the total energy would be fixed here and thus we can write that you know s 1 plus s 2 should be constant do you as T 1 minus T 2 D s 1 okay and this equality is indicates your T 1 should be equal to t 2 so just considering that an energy is minimum or considering simply D u we can obtain this thing if you recall that earlier exercises we have just made use of D s equal to 0 as far as the total DX is concerned here so we are not making use of that at this point but of course if we can clearly see here we made use of the total entropy is constant okay that from the different perspective here ok so you can do these similar exercises for the pressure quality and chemical potential equality okay so what we're going to do is we're going to change these conditions now we'll be using the energy minimum principle to find out what is the thermodynamic variable which we should look at if we change the natural variables so as if now the natural variables are as we and for the total for the internal energy but since we have now developed thermodynamic functions for different change different conditions of the natural variables for the case of RTV n it's a harmless free energy for the case of you know TP n it's a gives free energy so considering those changes what is it natural variables or changes in the natural variables or the corresponding thermodynamic functions or considering different constraints what would be the conditions on the thermal area functions if there is a change in the equivalent state total that's a composite system at uvn we still can consider that but our systems or the system which we interested in is at TVM by making sure that you have a die thermal rigid wall but heat heat is being exchanged with a bath which is at temperature T okay so this could be a TV okay so with this but essentially the system temperature will reach at T so we can consider this to be all isolated so total composite could be total energy u VN n okay now with this we can apply minimum energy principle okay and find out what is the appropriate variables which would give the indication of equilibrium condition oh I know under the word what would be the conditions of the thermodynamic functions associated with T V M okay will it be positive will be negative so that is something which we would like to lead to that kind of conclusion so let me start with this a basic definition so essentially u Plus u B should be minimum that is what it says that right at equilibrium okay subject to total entropy is constant as plus as v is equal to constant okay so that is what we are saying so because minimum energy principle says that the energy will be minimum for a given total entropy so for a given total entropy the U Plus u B is it has to be constant so this would give us two conditions one is the D of U Plus u B should be equal to zero and the second derivative d square u Plus u B should be greater than zero so again subject to the D of s plus s B is zero okay oh and the other thing which we know that this is a rigid here okay the only thing is that thermal okay so rigid and die thermal which essentially means that your DN and DV these are going to be 0 okay what is it this expression so this is d u Plus D u B should be 0 and do you this is nothing but DQ plus DW right but there is no work associated with this bath because of the rigid walls so this is nothing but your TB DSP okay now from here we know that D s is nothing but minus D s B so therefore do you minus t d s ok right so the other thing which we can do is that we can also write the D u in for the system and this would be nothing but again TDS minus TB DS so this is the expression we have now now I can write D s in terms of s be also so from here I can obtain t minus TB now with this you have t minus TB d s is equal to 0 so this would mean that T is equal to TB that is the condition of thermal equilibrium essentially where the entity but from this exercise we could get that okay but our intention is to find the appropriate thermodynamic function which would be indicative of the equilibrium condition for such a conditions where you where you have fixed the volume temperature n number of moles okay so in that case let us consider this expression again so that is fine way as far as the thermal condition is concerned two-level consider again okay so this is the condition of this is the minimum energy principle here here I'm going to represent this you as this as simply do u minus DV DT years that is something which we have written here right now this DP is a constant anyway because bath temperature does not change but is supposed to have a large volume where this heat exchange would not change the temperature of the system in that case I can write this as d u- right but TB is nothing but t here so this is we can write u minus TS is equal to 0 and u minus TS is nothing but help us free energy okay so that means one of the condition is of course that DEA should be equal to zero for such a system here okay we have a maintaining the temperature volume and number of moles now let us look at the second derivative that is the disk well u Plus u B okay okay we can write this as d of d u plus d ub and this can be written as simply again back you from here I can get it u minus t s greater than zero okay so right and this is nothing but d square a greater than zero so this indicates that both this condition indicates that for the case of such a system where you maintaining temperature volume n and n your u minus TS must be minimum so you minus TS or a is minimum at equilibrium so something which we have derived okay another thing which we have derived from this is of course the temperature equality okay so that was for this constraint where we put temperature of course you can extend this understanding in analysis for more difficult conditions now what we have done is all this derivation was at an equilibrium so but you can also consider the case where you're a little bit off equilibrium in that case your intra P is not going to be constant is going to be greater than zero total entropy so how do you do that what will be the appropriate variables which you are going to get as far as this almost free energy is concerned so let us look at it okay because in the okay say you can clearly see that da is equal to zero the first derivative and second derivative is greater than 0 that means a is to be minimum for such a condition so now that is little bit extend it so that we can understand the complete picture here so consider for any process that takes place from and off equilibrium to an equilibrium state okay so we are considering a process which takes from a Norfolk libram and going towards an equilibrium space in such case your total entropy should be greater than zero total energy still remains zero so this is d u Plus D u B is equal to D u plus DB TSB right this is still equal to zero okay so we are still considering slow and reversible process so in this case I can write this as write so this is nothing but your d s+ d SB okay this is greater than zero okay remember that so so by definition we can now write this as dou u plus so here as B can be written as because this is d s t can be written as d d s t minus T DS this is equal to zero so because we're considering temperature to be constructive okay so all I can rearrange this thing that TDS T is TDS minus D u so this we have considered to be simply T here so with this I can now plug in here I can consider this as minus T of D DT yes this is d u minus TDS and since T is constant here considered to be cool already the temperature equality is there then u minus TS okay so this anyway is going to be because DTS is greater than zero this has to be negative zero that this means this has to be less than zero which essentially means D of a equal to less than zero so what we what we did see by doing this simple exercise is when you have a process that takes place from an offic Librium to an equilibrium state the thermos free energy it turns to and eventually at equilibrium the DA is going to be zero so it will reduce until the point at equilibrium reaches so you can draw this like this you have a and some X is unconstrained variable so it's going to be like this of your like here then if you're like off process going towards minimum then essentially the DA has to be less than zero okay similarly for the case of whether is off equilibrium here so this is your executor for a given T again okay so you can you can do this exercises for other cases as well but for now I think we can stop and we will take this further in the next class okay so see you in the next class [Music] [Music] [Music]