 So, now if I want to evaluate the entropy change that is the you know the part that we have to be bothered about. How do I go about doing it? So, one thing you must realize is that if a system went from state 1 to state 2 by some method you do not know what that method is. You should not try to find out dq by t for that method. It may have been an irreversible method. If you try to find dq by t for that method that is not going to take you to the entropy of the system. Because dq by t along a reversible path is the only way to find out the entropy of the change in entropy of the system. So, this is something you must remember. So, just because the system executed a process from 1 to 2 using some process please do not use that process unless it is reversible. If it is reversible go ahead you can do it. So, one thing you must remember if I want to evaluate the change in entropy irrespective of what process it as long as I know the end point starting point end point. I should try to find out a reversible path between the two any reversible path. Because I know for sure that along any reversible path dq by t has to give me the same method. So, you should not bother about what reverse along this path will dq if there is no dq then what? You better be sure then along any other reversible path dq will be 0. So, do not consider as long as you know it is a reversible path you know you are on the right track. So, take any two points and try to choose a reversible path and you want to integrate it you will realize that all reversible process by default if I want to integrate they will have to be quasi-static processes. So, this will necessarily be quasi-static is that fine. So, we will now try to figure out how to get these done. So, some things you will realize is that you know one thing you must always realize is that you can always use the equation of state for your convenience as long as it is helpful you should use it. So, for example, for an ideal gas a lot of calculation becomes extremely easy and we will figure out how to you know go ahead with this. So, let me just say now you tell me that if I take an ideal gas or if I take a process where you know I do work, but it is only a one wave work mode. Will that be reversible? Since it is the one wave work mode anyway the second wave was not possible. So, you could not have you know even by let us say dictionary definition of reversibility you could not have reversed this. So, leave alone thermodynamic reversibility is that fine. So, one wave work mode is not going to anything which is dissipative friction etcetera you realize immediately it is going to be a one wave mode. Friction does something you cannot go the reverse way around. So, any kind of dissipation you realize immediately is going to cause an irreversibility. So, as far as let us say simple compressible substance like a gas is concerned the only reversible way is to have a PDV kind of work and not only PDV it better be you know slowly quasi statically increasing with small increases in P. So, that you can actually you know find out what is going happening at every path and it should be reversible. You cannot have certain expansions. If you have certain expansions then that is not going to be a reversible process. So, you will have to have quasi statics low movement of you know PDV kind of work. So, that is the kind of work you are looking for as far as simple compressible substance is. If you have sterrer work your frictional work all this will ensure irreversibility or you will not go along the reversible path is that fine. So, you will realize that if I have let us say I will always begin with the first law. First law for any system is how do I write it q minus w is delta e is that fine or you know q is equal to delta e plus w. Now take I will take the first and simplest case ideal gas. So, an ideal gas is a simple compressible substance. So, let me assume that you know there is no gravitational potential and no movement just to make my life easy. I will just write q is delta u plus w and delta e is delta u using assumption. No movement you are not no change in height you are not bothered about all of this. So, what is delta u directly for an ideal gas m c v d t is that fine and what is this? If it is reversible let me say this is reversible what is this p d v. So, I can write it in two ways q is d u plus p d v or I will write it as m c v is that fine. What is that d q let me write it is that fine. So, now if I divide this by t or rather I will say d q by t is actually d s correct this is reversible. So, q is equal to t d s as long as it is reversible. So, for a reversible. So, do you realize now I have used reversible process for a simple compressible substance work done is only of p d v type d u is d u I mean there is nothing that can be done about it and for a reversible process d q could have been written as t d s. So, but now I have ended up with a relation which is only between properties there is nothing to do with the process suddenly because now this is a property of a system d s is a property u is a property p is a property v is a property and suddenly you know there is no question of the process. So, this is a property relationship. So, it has nothing to do with the process because I am just connecting properties this is a property relationship. So, people often you know think that is this like the first law and t d s is d q that is the wrong way of thinking. So, this by itself is only a property relationship in a reversible process it will turn out that t d s is d q and p d v is w. So, immediately to just ensure you I will just take an example assume a gas in a rigid container container which is adiabatic. So, gas rigid container means d v is 0 and I am going to have a stirrer which is working correct. Now, I can write first law I know what is the first law what is it q is equal to delta u plus w this is adiabatic. So, what is q is this fine no, but sorry I will I will I will change myself first let me write it as delta e then I have made assumption good point. So, let me write it like made assumption that delta k e delta p e all are equal to 0 is that fine. So, yeah it is a good idea that you start like this write the assumption. So, I have written it like this. So, what is this w is equal to minus delta u I mean which is nothing, but our initial you know definition of the energy just that instead of e I am using u. So, I have what on this is this fine, but this is this is a gas which is you know an ideal gas and I have a property relationship T d s is d u plus T d v correct this has to be true it is a property relationship. So, but in this case you did not allow the volume to change in this present process. So, this term is 0. So, that means, T d s is equal to d u. So, T d s in this case is not equal to d q because there was no q correct, but T d s in this equation is equal to d u which is equal to minus w. So, that means, you realize that you know please do not try to start equating T d s with d q it has nothing to do with finally, this this three way this equation is only a property relationship it is not telling you anything about the process it just happened that we derived it using a reversible process. And in the reversible process yes, if there is work done it will be P d v only there will be no stellar work and if there is q it will be T d s. So, only if the process was reversible I could have equate it T d s to d q and P d v to work once the moment the process is you know you are using a one way mode of work you better be very clear that you should not try to equate T d s with d q. So, T d s in this is turning out to be negative of the work correct. So, this is the kind of thing that is going to happen. So, please differentiate this is the this is the property relationship whereas, this is the first law this is irrespective of whether it is reversible irreversible you know it is just the first law. The first law is does not tell you anything about whether it is reversible irreversible. For any process this is always true this is the first law. So, you must use the first law that means often when you are finding out entropy you will realize that very often you must not forget to use the first law because it helps you in finding out what probably happens. Another thing you must not forget is using property relationship if it exists and if there is an equation of state is that fine. So, these are something you must always keep in mind when you are evaluating entropy changes. So, this will help you in trying to get to what you want. So, for example you know if I have an ideal gas you know it is the simplest thing you know equation of state is known property relationship is known you go and go ahead. If you have steam then you know entropy is etcetera well tabulated between any two states more or less you have the steam table you know if you know the initial and final state you just have to look up the steam table. You do not have to do anything else it is a property s is a function of the state of the system you just know the state you find the property you find delta s for the system. Of course, if you want to find delta s for something else that may become a problem. If it is an ideal gas it is again going to be very simple if it is some non ideal gas if it is non then you may run into problem, but you must know if you as long as you can have a very you know an equation of state which you can work out very easily then your problem is solved otherwise things can get slightly tricky, but you know most of our problems we will do is either with an ideal gas or with steam and in both cases I think you should know how to work around. So, choose a reversible path and see what you can do is that fine. So, what do you want to mean by saying derived property we have said that. So, this is a question that you know you are saying I can measure pressure or I can find out pressure using something else. So, again you go back to the state partially you figure out how many independent variables you need to give if you know if you have given the number of independent variables you have known the state of the system. So, that means any other variable that you did not know of can easily be found out because that is dependent only on the state of the system. So, I can always specify s and v of the system and tell go find p and t you should be able to do it because once s and two properties are specified two intensive properties. So, s and specific volume let us say I specify you should be able to go ahead and find the other thing is that fine. So, that way if you are saying derived no I mean it just depends on which two I want to consider as independent any two you can consider as independent is that fine. So, working with ideal gases again t d s is d u plus p d v or d s is d u by t plus p by t d v for an ideal gas I will just write d u is m c v d t and p v is m r t. So, p by t is m r by v. So, d s m c v d t by t plus m r by v. So, that means I have just it is a property relationship I am not going to as long as I have a causal static path where I can integrate I am not going to bother this is only a property it has nothing to do with reversibility. So, I have an equation which is m c v d t by t plus m r d v by v choose the causal static path and integrate you will realize if I want to integrate between one and two I can go in between any way what I will get is s 2 minus s 1 is m let me assume that c v is not a function of temperature or c v can be a function of temperature let me assume c v is not a function of temperature is that fine. So, for an ideal gas this ideal gas with constant this is fine and of course, you know there was that other assumption that delta e is delta u that also went because we use only d u. So, as long as this is this is fine you can get this I can now use again my equation of state if you want to substitute here. So, p 1 v 1 is m r t 1 p 2 v 2 is m r t 2. So, v 2 by v 1 is p 2 by t 1 time p 2 by sorry it will be p 1 by p 2 correct. So, I will just substitute here you will see that I can also get it as m c v l n t 2 by t 1 plus m r l n of t 2 by t 1 minus m r l n of because I will just put everything all the tools together. So, this here you will see it is m l n t 2 by t 1 is common and you will get c v plus r which is this c p. So, I can write it as m c p l n t 2 by t 1 minus m r l n p 2 by p 1. So, this is the same thing I mean it just depend let us say you have a process where t and v changed immediately you can do this if you have t and p have changed immediately you can do this or I can just write t 2 by t 1 it just p 2 v 2 by p 1 v 1. So, I will just write it as m c v l n t 2 by t 1 and then again I will have this one also plus m c v l n v 2 by v 1 plus m r l n v 2 by v 1. So, you get m c v l n p 2 by p 1 plus m c p l n v 2 by v 1. So, I have 3 ways I mean the same thing I have just expressed use the equation of state and come up with 3 derivations the same derivation I mean this is nothing great I have just used the equation of state I could have used this used this use the most convenient one and you will get what is that fine. So, if you are having an ideal gas getting delta s is not going to be a big problem if you can figure out what the t 2 then p 2 is there what the final p n t there. So, maybe sometimes it is a question of getting what the final pressure and temperature is the moment you do this you can actually get this because it was independent of the path I should be able to get delta s. So, if I got it through a property relationship then it should not matter at all is that fine. So, if for an ideal gas this is the very simple way to do it for steam you just look at the steam table absolutely no problem for anything else complicated you know it is up to us. So, I think once this evaluation business is open we can go to. So, see what the first problem is conductor connects 2 reservoirs at a temperature of 1200 Kelvin and 500 Kelvin the steady flow rate of heat from the hot to the cold reservoir is 150 watts determine the rate of entropy production by the conductor. So, now this is a problem where things are happening continuously. So, things are expressed not in joules, but in watts. So, every second some q is being transferred. So, instead of just q it is q dot. So, this is one reservoir 1200 Kelvin this is 4.1 this is your conductor 500 Kelvin and you are having 150 is that fine. This is a situation where 150 watt is flowing from here to here in steady state. So, if something is steady state which means its properties are not with time that is what we mean by steady state. So, if I had written for the conductor dS is dQ by T plus S T I can just you know have dS by dT is equal to d by dT of dQ by T plus dS T by dT correct. This is 0 steady state. So, conductor's entropy is not here. So, this dQ by T is just q dot there are 2 q dot by T. So, I will just write it as q1 by T1 is that fine. So, now the assumption is so this I am taking only the conductor. So, here it is q is coming in what is that dS by dT this one what because it is in steady state. So, in steady state none of the properties of the system are changing. So, whenever I say something is in steady state that means all its. So, it has become let us say it became hot and then now it has remained in its state. So, it is just transmitting some energy from here to here by itself it has not it is now no longer changing that is what we mean by steady state. So, this is I mean I guess I am sure you have seen this regularly in fluid mechanics heat transfer whenever something is steady state the properties of that things are not changing with time. So, this is what I will do this problem in 2 ways. So, you realize what we are doing. So, see what is happening is at this point q is coming in and there is a temperature 1200 is that fine. So, what is dq by q1 by T1 here. So, this is dq by Td by dT of it is q dot I have just put plus sp equal to 0 this is what is happening sorry dsp by dT and these are both q dot that differential is already there. So, q dot because it is coming per second and again as we said this is should be dsp by dT q dot is what here this is the temperature. So, we assume that the temperature here is 1200 here is 500 and as far as this system is concerned only of this conductor that is all we are now looking at. So, q is what 150 1200 why is it that be 0 correct. So, it is getting produced see for this is is it a state property it is not which one sp I will write it as sp dot that is what you can it is a rate of entropy production that is how I must look at it every moment it is getting produced just like q dot it is like sp dot. So, something some entropy is getting produced every moment is how I should look at it it is see it is not a property. So, that is why we should not talk of it like that it is just like q you know if some steady q is coming out you would not say that is 0 it will like q dot only for a property I will say whether it is changing or not. So, if q is changing with time that is something else that is fine. So, you are right let me just you know better for your thing I will just write 150 upon 200 what is it here at the other end it is 150 only correct 500 is equal to 0. So, sp dot is equal to 150 upon 500 minus 150 upon 1200 correct have I put the signs correctly. So, obviously this you know 150 divided by a smaller number is a bigger number. So, entropy dot is coming out a positive quantity which is getting produced which is what we want I mean if it was coming out negative there was something wrong that process cannot happen that it will be. So, this will be some positive quantity you can calculate. So, now your question is this entropy is being produced. So, what is where is it going is some question you can ask. So, you know you should really go back to the first argument we made about isolated system moment you create an isolated system there is no q interaction delta s would be sp dot as long as delta s is greater than or equal to 0 delta s is sp dot. That means, if I had created an isolated system only with this and this I isolated there the entropy of that combined system is really increasing that is where that entropy production whose entropy is increasing you can say. So, the entropy of the universe is increasing what is the universe you create the universe you create an isolated system involving these three that is where the entropy. So, that means, the state of the surrounding system is actually changing that is something you must keep in mind because the entropy is continuously changing is that fine. So, another way you can tackle the same problem is consider this thing you will realize that find out the entropy change of each part of the system consider this isolated system find out entropy change how many parts are there are three parts there is this there is this and there is this is there an entropy change for this it is in steady state there is no entropy change is there an entropy change for this yes. So, what is the definition for entropy change of the reservoir always is t is same some q is coming out it is the q is negative correct. So, it is what minus 150 upon 1200 this is t 1 what is the entropy change of this it is plus 150 upon 500 what and of course these are all watts. So, it is happening joule per second. So, whatever we never sorry I made a mistake I never said what the units of entropy are. So, you should know s was ds was just dq by t what is the units of q joule kilo joule, but we will use joule you are you want to write it like that I mean it is not per kg basis if you are right if I divide everything by the mass it is all specific. So, then this s is also specific. So, if you want if you really have with the expression dq for the entire system no per mass basis. So, the units for s are always joule per Kelvin if you want specific. So, small s then it will be joule per kg Kelvin I mean that is like no difference between s and in this specific entropy and entropy per kg basis. So, if I am going s dot it will be just watt per kg Kelvin or you can write it as joule per kg Kelvin per second however you want to look at it is that fine. So, here you should realize now this is you know this is going to be I should not put the kg here unless I am talking about specific joule per Kelvin. So, here it will totally be and so now if I want to add up the entropy change of each of the parts of this isolated system for the first system it is 150 upon 1200 for the conductor it is 0 and for this it is plus 150 upon 500. So, obviously you realize that this is the same as this one this is the entropy production. So, you are saying that the conductor is producing entropy at this rate, but finally what happened was you went to this isolated system you went to this isolated system you figured out the entropy change of each of its parts you just added up you realize that the entropy has changed for the entire. So, what the conductor was really doing is reducing the entropy of one part and increasing the entropy of another part, but the net entropy change of an isolated system combining the three parts was only improved. So, that is how you must look at the problem. So, entropy change of something can be 0 you should just overall look whether you when you consider an isolated system whether the entropy is equal to 0 or greater than 0 then it is possible for the isolated system if your entropy is turning out to be less than 0 either you are doing something wrong if you thought the process was possible or the process is not possible that is the way you must look at it. What is that? Conductor we said the entropy is not changed it is in steady state. So, it is in steady state its entropy is not changed. So, that is why I said there is this 0 I put here there are three parts to the system one part entropy is reducing one part entropy is not changing and one part the entropy is increasing and the net addition if I do I see entropy is increasing which is what I want is that fine. Steady state means first for this thing the properties are not changing. So, its own entropy is not changing properties any its temperature it is maybe you know maybe it adjusted its length its length is not changing nothing is changing. So, you can consider that as a system by itself. So, its own properties are not changing it is just acting as a medium to transfer something from somewhere, but its own properties are not changing. Let us do problem 4.5 correct. So, actually it is a conductor who is leading to the rate of entropy that is how we are looking at it. If you want it for the isolated system the entropy is getting produced for the conductor itself entropy is not changing it is just you know acting as a medium to increase the entropy of the entire system, but then I have to define which system. Overall. Then I will say if you consider an overall system consisting of the two reservoirs and the conductor yes then I will say that your thing is fine I mean you define what is the system then you will say the entropy production rate in that system is changing. So, the way this line is written is that you know entropy production by the conductor is that it is just like a medium which is helping in increasing entropy. If I agree sometimes the language is not directly you know probably something that you can understand, but this is what it really means that you know it is really leading to an increase in the entropy of some other thing it is just acting as a conduit nothing else its own entropy is not changing. So, I have let us say I can at least do 4.5. So, an insulated chamber. So, moment you see insulated you know we are talking of some probably some adiabatic process. An insulated chamber of volume 2 v 0. So, you know this is my chamber it is insulated is divided by a thin rigid partition into two parts of volume v 0 each. So, there is a thin partition here is v 0 here is v 0. Initially one chamber contains an ideal gas at pressure p naught and t naught. So, here it is p naught, p naught and v naught, v naught I am just copying the other chamber is evacuated. So, the partition is suddenly removed show that when equilibrium is established the temperature is t naught determine the change in entropy. So, I will say partition is removed. So, what happens when the partition is removed? Maybe initially this gas gas will gush through and occupy that other space and it may take some time for equilibrium to establish you know initially here you know slowly the gas is here gas is here then maybe you know something will hit here there you do not know how the pressures will change, but after you wait enough time now pressure everywhere would be the same is that fine. So, you immediately realize this is not a quasi static process. So, I cannot use this process to find any entropy changes or etcetera. So, there is something you must keep in mind there is no way I could have figured out what is happening, but can I use the first law? I can always use the first law. So, that is something again you must know I can always use the first law. So, I will just write for this entire system which consists of both of these here. So, I will draw the system boundary exactly here q is delta E plus W. So, at the end and beginning no change in kinetic energy no change in. So, that I will just substitute this E by U is there any q is there any W? No we have decided free expansion it expanded and. So, what is delta U? Delta U is 0 is it an ideal gas it is written it is an ideal gas. So, I can write delta U is m C v d t. So, if this is 0 g t is 0 is that fine this means t naught final or t final is t naught t final is equal to t naught. So, now, what is the final thing volume is to v naught final. Temperature is t naught and pressure is pressure will be what because it is an ideal gas I will just write initially it as t naught v naught should be equal to t final v final by t final t final and v final are the same correct. So, p naught will be just sorry p final will be just p naught v naught by v final v naught by v final is 1 by 2 correct. So, p final is just p naught by 2 I mean this I am just taking the idea. So, see what I have used I have used the first law and figured out temperature is not changed I have used the equation of state to figure out what is final p because I know what the final v was. So, now, I have final state p naught by 2 p naught and 2 v can I find entropy change? Correct because now I have my equation for ideal gas. So, you choose whichever you want one equation said m c v l n t 2 by t 1 is it correct plus m r l n v 2 by v 1. So, I can choose this relation t 2 by t 1 is 0. So, this is actually m r l n of 2 is this fine I mean you can use any of the other two relations also use the p relationship use the you know both p and t relationship in the p there is a relation where there is a p 2 by p 1 and a v 2 by v 1 you can use that relationship also you will you are going to get this same value back t 2 is equal to t 1. So, t 2 by t 1 is equal to 1. So, l n of 1 is equal to 0 sorry. So, I assume that so this is m r l n 2 is positive correct l n 2 is positive, mass is positive. So, it is the thing I mean is this what you expected. So, entropy of the gas itself has increased which means entropy of the system has increased entropy of the isolated system has increased. So, why has the entropy increased? It is it is really an irreversible process there is no way that on its own the gas is going to come back into one half of the thing and stay there unless you do an external work. So, if you can take a piston and actually push it and you know assuming that whatever is this. So, you take this you have a piston here and start pushing it and you assume that you know this is all leak proof. So, the moment you put there is a evacuation here bring it here then I can create I can always go back to the whole state, but I have to use an external agency for it by itself it is an irreversible process. Delta is how to find out what is the part of entropy generation and what is the part of entropy transfer? Entropy transfer. So, for that I mean see finally, when I am taking an isolated system then the entropy change is equal to you know some part of the system would have you know increased its entropy that is for sure you must keep that in mind. So, if you have two systems interacting separately you should be you should try to find out one person's entropy may be decreasing then the other person's entropy may be increasing, but the sum of the two better be positive because if they are an isolated system the sum of the two better be positive. So, you can find out where it is increasing and where it is decreasing, but some part of an isolated system definitely the entropy is going to increase. So, when I say entropy of the universe increases you know the universe includes you know maybe 2, 3 subsystems one of it the entropy has to increase. So, that is something you have to keep in mind physical significance of entropy. Now that would be slightly. So, you end is saying do not even try to tell it. So, you just say the second law tells us which process is possible and not possible. DSP is a method to figure out whether you know the process was possible or not. If DSP was 0 you say it is a reversible process if DSP is turning out to be greater than 0 you say yes the process is perfectly possible and it is an irreversible process that is the way you can use it. If DSP is turning out to be less than 0 you say you have created a process which was not possible. So, please forget about the process it cannot be done. If you want to this is what I said the irreversible process that has occurred is that now you have occupied a space and you cannot come back to it. So, the whole system has occupied the entire room and the irreversibility is that by itself now you cannot come back to its origin. So, that expansion the free expansion to occupy the entire space which cannot be brought back you say irreversible.