 begin where we left off. We came to this situation where we have two scales of temperature, the thermodynamic scale and the ideal gas scale. The question is are they equal? Now, to demonstrate that they are equal, we will look at the behavior of a cycle, a celebrated cycle which is known as the Carnot cycle. The Carnot cycle is perhaps a simple cycle, it is a reversible cycle and it can be used T reversible heat engine. The idea is the process will be as follows. We will see what a Carnot cycle is. We will define the Carnot cycle, specify the Carnot cycle. Then we will set up a 2 T reversible heat engine based on the Carnot cycle. We will analyze it and I forgot based on the Carnot cycle and using an ideal gas constant C T and C V. Then we will obtain by analysis efficiency of this Carnot cycle and this will be in terms of if it works between T 1 and T 2, it will be in terms of T 1 and T 2 on the ideal gas scale. Using the corollary of Carnot theorem, we know this as a function of say theta 1, theta 2 where I am using a slightly different nomenclature on the thermodynamic scale. We will compare the two relations for efficiency and then link up the Kelvin scale on the ideal gas thermometric scale and Kelvin temperature on the thermodynamic scale. Let us first define what is we mean by a Carnot cycle. A Carnot cycle has four processes. One process is of heat absorption at some temperature T 1. It is an isothermal process, so that the temperature at which it is absorbed remain fixed at T 1. Two, another process where heat is rejected at a thermodynamically lower temperature T 2. Then there is a third process. The processes may not be in this order is an adiabatic process which reduces the temperature from T 1 to T 2 and the fourth process is an adiabatic process which reduces increases the temperature from T 2 to T 1. It has four processes and all processes are reversible. The heat transfer that is Q absorbed and Q rejected, this also is in processes reversible. Now, let us say that we sketch the Carnot cycle on a T V diagram and let us say that the two temperatures between which it works are T 1, T 2 on the ideal gas scale and theta 1 and theta 2 on the thermodynamic scale. We are using the units Kelvin for both that causes some confusion, but let us live with it because that confusion will soon be over. We have a process of heat addition at temperature T 1, but we will need to have a temperature slightly below T 1, so that heat is absorbed. Let me say that temperature is T A and the temperature difference between T 1 and T A is very small and we can consider in the limit that we can make it 0 by providing large enough heat transfer area. Let us say that we have this process going from state A to state B, temperature is the same, but the volumes will be different. Then we have a process of adiabatic reversible expansion. Now, notice that we need to keep T 1 minus T A approximately equal to 0 to make the process reversible. If we make it large then the process is definitely reversible, even with a small one it is irreversible, but we can make it small at least in theory and see what limit we reach in the reversible limit. Then we execute a reversible adiabatic process and remember that for any process to be reversible the system must remain in equilibrium throughout that means the process should be quasi-static and only two way modes of work should be used. One way mode of work like stirring is out of question because you cannot reverse stirring we have already seen that. Stirrer can only do work on the system, stirrer cannot extract work from the system. So, if you use a fluid you are restricted only to PDV work. Let us say by reversible adiabatic process we bring the temperature down to C then at a temperature slightly above T 2 say at temperature T C such that T C is higher than T 2, but T C minus T 2 is also a very small number. So, that we can expect to reach the adiabatic limit. We extract heat, here heat is absorbed and then although I sketched it the other way this heat rejection is over at point D, D to A is the reversible adiabatic compression to raise the temperature back to T A. So, A B C D is the Carnot cycle and in the limiting case where T A is brought very near T 1 and T C is brought very near T 2 it is a reversible Carnot cycle. So, for a Carnot cycle of this kind using any fluid the efficiency will be 1 minus theta 2 by theta 1. This is by the way the adiabatic compression by the corollary of the Carnot theorem and our definition of the thermodynamic temperature scale where theta 1 and theta 2 are on the thermodynamic temperature scale and T 1 and T 2 are on the ideal gas temperature scale. Now, let us use an ideal gas with Carnot constant C p C v as the working fluid and let us say m is the mass of the fluid in the Carnot cycle. We can execute the four processes like this. We can have a reservoir at T 1 and our gas in a cylinder piston arrangement can be brought in contact with the reservoir at T 1. Initially the state is A and we expand it to bring it to final state C. So, during this process Q absorbed is taken in. There is an expansion work done and we will assume that we are working with a stationary system. So, delta E is delta U. So, delta E is delta U. So, our first law will give us Q absorbed is delta E plus W and delta E is delta U and W will be integral p d v from state A to state B. I think this is state B by mistake I wrote it. Now, since it is an isothermal process delta U is 0. It is an ideal gas. So, U depends only on temperature. So, we have integral p d v and if you do the I will do the calculations here and then erase it. Notice that for an isothermal process integral p d v will become integral p is M R T by v from v A to v B and this will be T is constant at T A. So, I can write it as T A and take it out. So, I will have M R T A logarithm of v B by v A. So, I will end up with Q absorbed is T A logarithm of V B by V A. Now, we will have the second process in which we will remove the thermal reservoir. We will start our process from B state B and expand it adiabatically U is 0 from state B to state C. There will be W equal to W expansion. You will have Q which is 0 equals delta E plus W. If you look at it in a differential way and assume delta E equals delta U. You will get the relation 0 equals d U plus p d v replace d U by M C v d T and p by R T by v M R T by v integrate that relation out and remember exercise 1.6 in a similar way. You are doing only expansion work. You have a quasi-static expansion of an ideal gas with constant specific hits and you will end up with this relation T B v B raise to gamma minus 1 is T C v C raise to gamma minus 1. I will call this equation 1. I will call this equation 2. Now, this way we have tackled two processes A B and process A B and process B C. Now, all that remains is to tackle the remaining two processes C D which is similar to A B in which heat is rejected and D A which is similar to B C except that instead of expansion it is compression. So, I will not do that detail. I leave it as an exercise for you to complete the two. All that I will be able to tell you is the following. Just the way we derived this for Q rejected you will get an expression M R T B l n v C by v d. This is the expression M R T B I call relation 3 and similar to relation 2 you will get T C sorry T D v D raise to gamma minus 1. This is for the process D A. This process D A small mistake here this is not T B this is T C equal to E A B A raise to gamma minus 1. This is equation 4. And remember why are we using T A T B T C T D here because these equations come out of the equation of state of that ideal gas. So, this temperature here in this relation is the ideal gas temperature not the thermodynamic temperature. Whereas in this expression for efficiency which we have written down from the corollary of the Carnot theorem we have not even seen whether it is a Carnot cycle or whether it uses ideal gas. We only have seen that between the thermodynamic temperatures theta 1 and theta 2 we have a reversible 2 T engine working on some cycle. So, the efficiency by the corollary of the Carnot theorem has to be 1 minus theta 2 by theta 1. Now that we have these interactions determined you absorbed and Q rejected we should be able to determine the efficiency using this analysis of this cycle. So, from our analysis the efficiency of the same cycle is 1 minus Q sorry Q rejected by Q absorbed. And if you substitute our relations you will notice that Q rejected divided by Q absorbed M R is a common factor. You will have T C by T A and then you have these ratios of volumes. So, let us write it this will be 1 minus T C by T A and logarithm of V C by V D and logarithm of V B by V A logarithm of V C by V D and logarithm of V B by V A. Now this part seems to be similar what about this part to take care of this part. We look at these two relations we have T V V B T B V B raise to gamma minus 1 is T C V C raise to gamma minus 1 and T D V D raise to gamma minus T A V A raise to gamma minus 1. Now you will notice that A to B is an isothermal process and the temperature is essentially T A or essentially T 1. Similarly, C to D is an isothermal process the temperature essentially T 2 in the limit of reversibility. So, when you do that substitute here substitute for T B and T D T B and T A the temperature T 1 and substitute for T C and T D the temperature T 2 and then you will be able to show that these two ratios are the same. So, this particular factor is 1 and hence this turns out to be 1 minus T C by T A. Now notice that we have two relations for the efficiency of our Carnot cycle the efficiency of the Carnot cycle by our analysis has been shown to be 1 minus. Now I will write this in terms of T 2 and T 1 because we are taking the reversible limit and by the definition of efficiency using the Carnot theorem colliery. We have this to be 1 minus theta 2 by theta 1 where remember this T 2 and T 1 are on the ideal gas scale and this theta 2 and theta 1 are on the thermodynamics scale. And since the efficiency has to be the same now it is obvious that T 2 by T 1 is theta 2 by theta 1. That means the this first implies that the ideal gas Kelvin scale or temperature on the ideal gas Kelvin scale and temperature on the thermodynamic Kelvin scale will be proportionate to each other. But then we realize that there is one point the triple point of water where the temperature on the ideal gas Kelvin scale has been defined as 273.16 and on the Kelvin scale has also been defined as 273.16 Kelvin if you want write say thermodynamics. So, that means the two scales are proportional to each other. So, it is something like saying y is proportional to x. But then you say that when x is 3 y is also 3 the same value has been defined. So, that means they are equal to each other and because of this common reference point the two scales or temperatures of the two scales are equal to each other. And that means what we have been able to show is that temperature on the ideal gas Kelvin scale and temperature on the thermodynamic Kelvin scale and temperature on the thermodynamic Kelvin scale are equal. And hence now we do not distinguish between the scales. What is the advantage of this? There is a great advantage of this. The advantage of this is measurement of temperature on the Kelvin scale thermodynamic. What will it require? It will require of us the ability to construct and operate a reversible 2 T heat engine. Whatever be the working fluid, whatever be the detail. But the word reversible itself is scary. There is absolutely no process in real life which can be reversible or which can even come close to being a reversible process. So, although it is good to define a thermodynamic Kelvin scale and say that now we have the definition of temperature and the temperature scale which does not depend on any property. When it comes to actual operation it will be very difficult to measure in a laboratory or anywhere the temperature directly on the thermodynamic scale. This equivalence of the ideal gas Kelvin scale and thermodynamic scale gives us a way out. And the way out is we measure the temperature on the ideal gas scale and that can be reasonably easily set up. Although we do not have a gas which declares itself to be an ideal gas, but any gas hydrogen and helium in particular particularly helium at reasonably low pressure behaves excellently like an ideal gas with an excellent or a very good approximation. So, use that measure the temperature on the ideal gas Kelvin scale and have no hesitation that is equal to the thermodynamic Kelvin temperature. So, the advantage is no need to set up a 2 T reversible heat engine to measure temperature on the Kelvin scale. So, now we say that T is the same thing as theta and we need not make any differentiation. Now, actually I have skipped a few things which you will find in any text book on thermodynamics. We did not talk about the Clausius statement of second law of thermodynamics, but during our derivation indirectly we have proved the Clausius statement. We proved it here, we demonstrated that if a 2 T engine works like this, then heat transfer is possible only like this. Direct heat transfer from a lower temperature to higher temperature is not possible, but you will be able to show that if you put a cyclic device here and provided it some work to do this transfer from a lower temperature thermodynamically lower temperature to thermodynamically higher temperature will be possible. And then you can prove the so called Carnot theorem for refrigerators and the corollary of Carnot theorem for refrigerators, but those now remain only exercises and you can do them yourself. Let us proceed now. What we have provided is the Kelvin Planck statement a lot of derivations and the Carnot theorem, the Carnot cycle and the equivalence of the two Kelvin scales. So, now we can just talk of one Kelvin scale and we started out by saying that second law will tell us or give us tools to determine whether something is possible or not possible. First come to engines, the direct consequence of Kelvin Planck statement is a 1 T heat engine is not possible. So, whenever an engine is proposed check whether it is a 1 T heat engine, if it claims to be a 1 T heat engine we know it is not possible. If it claims to be a 2 T heat engine then determine its efficiency, measure the two temperatures on the Kelvin scale we know how to measure them. Using the Kelvin temperatures determine the efficiency of a reversible 2 T engine working between the two temperatures. Compare the efficiency of the proposed engine with the reversible engine working between the two temperatures. If the proposed engine has an efficiency not exceeding that of the reversible 2 T engine we say it is possible. But these are very restricted situations. Suppose somebody comes up with a 3 T heat engine or any cyclic process or for that matter any process we should have tools to determine whether that process is possible or not. We now proceed in that direction and the first step we do in that is prove state and prove what is known as the Clausius inequality. This has nothing to do with the Clausius statement of the second law that we consider as a non issue now because it talks of a lower temperature and the higher temperature which cannot be defined by anything before Clausius inequality unless we take the Kelvin Planck statement as the primary statement of the second law of thermodynamics. The Clausius inequality pertains to cycles any type of cycles and here as we applied to various cycles we will also realize certain tricks. It says that take any system and let it execute any cycle in the state space. If you want you write here P v you write T v, T h whatever. Since it is a cycle it will start from some point and may be quasi static non quasi static process it will come back to the same point. This is a cycle what we do is during the cycle there will be some work interactions let there be we are not worried about the work interactions. During the cycle if there is any heat interaction we look at the heat interaction in some detail we take interest in the heat interaction and we say that say during some part of the cycle there is a heat interaction dq and when that heat interaction dq takes place let the temperature at the interface of the system across which this dq takes place. Remember system is defined by its boundary dq is a heat interaction which must cause that boundary. So, when it crosses that boundary measure what is the local temperature there is a minor assumption here saying that although part of the process may be non quasi static including that part where this dq takes place but we will assume at least for the time being for the purpose of argument that there is some sort of a local equilibrium. So, that the temperature can be measured and determined then take the ratio dq by t q is in typically joules t in kelvin. Remember whenever we write temperature by default it is kelvin now integrate this over the cycle the Clausius inequality states that this must be less than or equal to 0 and remember we will soon realize that this less than or equal to 0 has come out of the Carnot theorem where it says eta is less than or equal to eta r. So, the equality will hold for a reversible cycle the inequality will hold for a general cycle this is the Clausius inequality. Now, how do you prove this there is a rigorous mathematical way of proving this but we will we will see the approach in proving it you will find it in books or you can yourself do a general integration of this. Let us start with the Carnot theorem let us take a cycle which is 2 t and which works as an engine let our system be this this is some engine let our system be this the cycle plus any connections that the cyclic device has with t 1 and t 2. So, we know that when heat absorption takes place let me put it as q 1 the temperature of the interface is t 1 and when the heat rejection takes place the temperature is t 2. Now, we know that the efficiency of this engine will be less than or equal to the efficiency of a reversible engine working between the same two temperature levels. What is the efficiency of this engine the efficiency of this engine is 1 minus q 2 by q 1 the efficiency of a reversible engine working between the same two temperatures by definition of the Carnot theorem and the temperature scale is 1 minus t 2 by t 1. Solve this cancel out one transpose terms and you should be able to show that q 1 by t 1 plus minus q 2 by t 2 is less than or equal to 0 the same less than or equal to 0 which occurs here same less than or equal to 0 gets transferred here. And now see what is this this is something similar to our integral take the amount of heat absorbed divide it by the temperature of the surface of the system when it is absorbed and sum it up in this case because I am writing it over a cyclic thing it is an integral. Here there are only two terms one heat absorption here one heat rejection here. So, the cyclic integral is made up of a sum of two terms and here the heat absorbed is q 1. So, I write q 1 by t 1 here the heat absorbed is minus q 2 a negative number. So, I write it absorbed is minus q 2 because our symbolism is the other way divided by t 2 is less than or equal to 0. So, using the Carnot theorem we can show that the Clausius inequality is a direct consequence of the Carnot theorem for a two t heat engine now it is a good exercise to do to show that take a three t device whether heat engine or otherwise you can even do it for a two t device whether heat engine or otherwise. Let us say that we have a cyclic device and it has three heat interactions when the temperature is t 1 the heat absorption is q 1 positive or negative when the temperature is t 2 during some other part of the cycle the heat absorbed is t 2 and when the temperature at the interface is t 3 the heat absorbed is q 3. The Clausius inequality will tell us that we must have q 1 by t 1 plus q 2 by t 2 plus q 3 by t 3 is less than or equal to 0. To prove this we assume that this is not true assume q 1 by t 1 plus q 2 by t 2 plus q 3 by t 3 is greater than 0 just the negation of this then what you do is you consider a some reservoir at t naught and if t naught is different from t 1 run a two t reversible machine between t naught and t 1 such that q 1 is supplied by that machine and may be some work is done positive or negative. Similarly, set up another reversible machine between t naught and q 2 at t 2 and t naught and q 3 at t 3. Notice that if this is a cyclic device is absorbed q 1 q 2 and q 3 it must be doing a work w equal to q 1 plus q 2 plus q 3. Add to this work transfer the work produced by all these individual reversible machines and you will be able to show that if you assume this to be true this whole device our cyclic device plus all these reversible two t machines give you a one t heat engine you can prove this and that means this cannot be true it has to be false and that means this is true and this way you can generalize that means wherever you have a cyclic device wherever there is a t with a d q take a reservoir at t naught provide a reversible engine to provide this d q by working an engine between t naught and t and you will get some work output this work output summation of this work output over all such d q plus the inherent work output of this cycle because of this d q will give you a one t heat engine if you assume that integral d q by t over that cycle is greater than 0 this will also give you a one t heat engine and that means assuming this also leads to a contradiction and hence the Clausius inequality gets proved the advantage of the Clausius inequality is the following we can determine whether cyclic process possible or but remember that this is restricted only to a cyclic process all that we have to do is determine for that cyclic device d q by t check whether it is less than or required to 0 if yes possible if no not possible but remember this is all for cyclic processes we still have to devise tools for processes non-cyclic now we come to the situation where we are a step or two from defining the property entropy we now look at the corollaries of the Clausius inequality corollary one consider a reversible cycle what will the Clausius inequality will say for a reversible cycle integral q by t I will use the word re v to indicate that everything here is reversible will be equal to 0 the less than sign will not work and you can argue this out by saying this is the equality out of the Carnot theorem or you can say that look when the cycle is reversible. Execute it by Clausius inequality you will get integral d q by t to be less than or equal to 0 the cycle is also reversible so reverse it all interaction will get inverted so d q will become minus d q but all states would remain the same they will be executed in the other direction so t will remain the same so now you will have integral of minus d q by t to be less than or equal to 0 or integral of d q by t to be greater than or equal to 0 and both these inequalities will have to be simultaneously satisfied and that will be possible only when the integral is equal to 0 now this immediately tells us something suppose let us take a cycle p v or whatever coordinates you want it is a reversible cycle let us say that the cycle goes from state 1 to state 2 and comes back to state 1 and it is a reversible cycle so integrating this cyclic integral I will apply this to cycle and my cycle will be 1 is to 2 is to 1 and I will say this part is 1 a 2 and this part is 2 b 1 I will say 1 a 2 b 1 and the integral can now be summed up as d q by t reversible for 1 a 2 plus integral d q by t reversible for 2 b 1 equal to 0 and now what will be we will do is we will take this on the right hand side writing it as minus d q by t reversible from 2 b 1 and then we will say it is a reversible cycle so I will reverse it so minus d q by t reversible from 2 to 1 through b will become d q by t reversible 1 to 2 through b so I will end up with integral 1 to 2 through a d q by t reversible is integral 1 to 2 through b d q by t reversible and this is for any cycle any 2 states 1 and 2 and for any 2 reversible processes a and b linking the 2 so what this means is like our when we did our first law there we said that w a diabetic is independent of the path so here also now we say that integral 1 to 2 d q by t for a reversible process is 1 to 2 d q by t reversible process independent of the path and what does this mean if it is independent of the path that means it represents a change in some property just the way we said when we came to first law that the diabetic work between 2 states must be representing a change in some property and we define that property as e and here we define this property as the entropy so definition of s and c s has come out of the integral sign otherwise there is no reason for that s to be there so the property s like the energy e is defined as a difference between because what we do is define the change in some property and this is defined as this integral integral 1 to 2 d q by t for this indicates any reversible this is s 1 2 and this will be s 2 minus s 1 just the way we have defined for energy any reversible process this is the definition of entropy now that was one corollary of the Clausius inequality where we considered a reversible cycle now we are at a place where we can apply the Clausius inequality to a general cycle but rather than any general cycle we will create a cycle like this what we will now do is we will take a system and let us say that let the system execute from state 1 to state 2 any process any general process and I will call it process 1 2 what we will do is we will link up 1 and 2 through any reversible process this process I will call 1 2 r e this is any general process and this is a reversible process there may be more than 1 reversible processes possible linking 1 and 2 does not matter we take any one of them because it is a reversible process I can even reverse it and make it to 1 reversible so this process can go either like this or like this so what I will do is the following we will consider a cycle which is made up of 1 2 through any process and then back to 1 by the reversible means so the cycle has gone ahead like this let me change the color I start the cycle from 1 go along this any process execute that process first and then come back from 2 to 1 along this reversible process thus completing the cycle now let us apply the Clausius inequality to this cycle 1 2 through the reversible process back to 1 I cannot insist that it is equal to 0 because I am not sure what type of process 1 to 2 is it is any process I am not assuring myself that it is a reversible process now we will split it into 2 parts integral 1 2 that is for any process d cube by t plus integral 1 2 that is for any process d cube by 2 1 reversible this way d cube by t it is a reversible process I will write it like this less than or equal to 0 now what we will do is we will transpose this to the right hand side giving us minus d cube by t reversible from 2 to 1 and because it is a reversible 1 the integral will be 1 to 2 reversible instead of minus d cube now we will write d cube and we will end up with the following integral 1 to 2 any process d cube by t is less than or equal to integral 1 to 2 by the reversible process d cube by t is less than or equal to 0 again let me emphasize that this is a reversible process we have used the fact that it is reversible by changing the sign from minus d cube to d cube and changing the limits of the integral flipping the limits of the integral and what is the right hand side the right hand side by definition is delta s 1 2 so we get from the reversible process the second corollary of the Carnot theorem the change in entropy between two states of a system is greater than or equal to the integral of d cube by t for the same process this is known as the entropy relation for any process. Out here out here what we have is entropy definition whereas out here we have entropy relation now what is the advantage of this now we can apply our second law through the entropy definition and entropy relation to any process to check any process determine the definition of integral d cube by t for that process the s 1 2 how is delta s 1 2 determine go back to the entropy definition select any reversible process joining the two states evaluate d cube by t that is delta s 1 2 then after determining delta s 1 2 for the proposed process or candidate process actually determine d cube by t and compare these two if the entropy change is higher algebraically than integral d cube by t the entropy relation is satisfied the second law is not violated entropy relation is violated process is not possible. Now from here we see whether what sense does it say people say the entropy of the entropy always increases but here we do not see entropy increasing delta s 1 2 could be positive it could be negative so let us see when do we see that entropy always increases a special case consider any adiabatic process underline the word adiabatic twice adiabatic process means throughout the process d cube is 0 so if d cube is 0 integral of d cube by t will always be 0 so the entropy relation reduces to delta s 1 2 is greater than or equal to 0 that means entropy in a process will never decrease for an adiabatic process when people say or a student says but sir entropy always increases bring his attention to this fact or bring her attention to this fact that this is true only for an adiabatic process. Now there are a few issues again a play of words and unfortunately the meaning one has to be careful people say that this means the entropy of the universe always increases and there are two problems here by universe I think they mean the physical universe or astronomers universe and I think we have already discussed that we are not sure whether the astronomers universe is define able as a thermodynamic system because we do not know what the boundaries are we cannot define what the boundaries are and if we cannot define what the boundaries are then we cannot define and even measure the entropy change determine the entropy change. So consequently the second law of thermodynamics at least we are not capable of applying the second law of thermodynamics directly to our astronomical universe so it is not proper to make such a statement that the entropy of our universe is also increasing but you have to be careful with the word universe because if you look up your exercise sheets and any textbook on thermodynamics you will find problem saying do this do this do this and finally determine the entropy change of the universe. So what universe is that that is something we have to determine now let us look at there are three words which appear to be confusingly related to each other one is adiabatic we came across it quite early adiabatic means dq is 0 then we have the word reversible what does it mean it means that well things can be reversed and Carnot theorem will become an equality Clausius inequality will become an equality etc. The third word which we will now use is isentropic isentropic means delta s0 or entropy not changing or ds is 0 let me put it as ds0 so just the way isothermal means temperature not changing isentropic means entropy not changing adiabatic means no heat transfer we need a handle for reversible and for that we do the following we go back to this entropy relation and I will write it in a differential form instead of delta s I will write it as ds is greater than or equal to dq by t the entropy relation for a process element can be written down as ds is greater than dq by t look at it as a relation where the left hand side is always higher than the right hand side so if I have to make them equal I will have to add some positive quantity or definitely a non negative quantity on the right hand side that quantity we define as sp for a process or dsp for a process element and the technical term used is entropy produced by definition dsp is defined as ds minus dq by t and if you look at our entropy relation ds minus dq by t will always be greater than or equal to 0 so when you define this the second law tells us that dsp must be greater than 0 or for a full process sp will be defined as delta s minus integral dq by t and if you look at our entropy relation ds minus dq by t will always be greater than or equal to 0 so when you define this the second law tells us that dsp must be greater than 0 or for a full process sp will be defined as delta s minus integral dq by t. And that will also be greater than or equal to 0 again as in the first law notice in first law we had three terms q delta e and w q and w where path functions not changes in properties related to the process delta e what change in property not a path function in a similar way here sp and integral dq by t are process related they are path functions they are not state functions whereas d delta s is a change in property whereas this depends on the process this also depends on the process. And out here ds is an exact differential dq by t and dsp will be in exact differential after having said so let us write one relation I will just rewrite this equation ds equals dq by t plus dsp. Now ds is a change in property I will write three columns the rows do not mean anything so ds can be greater than 0 ds can be equal to 0 ds can be less than 0 it is a change in property depending on the two states initial and final states it can go up or down dq by t if during a process heat is absorbed it can be 0 if no heat is absorbed it can be it can be greater than 0 if dq is absorbed if no heat is absorbed it can be 0 if it is rejected it can be less than 0 dsp on the other hand remember it is always greater than or equal to 0 it can never be less than 0 because that is dictated by the second law dsp is greater than 0 or it can be equal to 0 but it can never be less than 0 this is just not admissible. Now what do you call a process where ds is 0 we will call this an isentropic process by definition there is no name for processes in which entropy increases and entropy decreases the process in which dq is 0 is adiabatic process well physical chemists use processes in which dq is greater than 0 as endothermic processes heat absorbing processes where dq is less than 0 they call it exothermic processes heat releasing processes but we do not have to use it in thermodynamics what about dsp when is dsp equal to 0 dsp is equal to 0 when delta s equals dq by t and that is true only for a reversible process so ds equals 0 means it is a reversible process ds greater than 0 means it is irreversible process look at the meanings of the three words isentropic adiabatic and reversible and now quickly look at it I can have a process which is isentropic but it need not be adiabatic need not be reversible because this can be greater than 0 this can be less than 0 giving you an isentropic process. So an isentropic process need not be adiabatic need not be reversible let us look at an adiabatic process it only means dq is 0 but that does not mean this should be 0 this should be 0 this can be positive this can be positive so an adiabatic process need not be reversible need not be isentropic and finally look at a reversible process reversible process only means dsp is 0 and that means dq by t and ds that means dq by t and ds can both be positive or can both be negative they need not be 0 that means of the three adjectives isentropic adiabatic and reversible you can apply any one to a process then the other two do not automatically follow that mean an isentropic process need not be adiabatic need not be reversible and so for the other two but because of this relationship ds equals dq by t plus dsp make a process which is isentropic and adiabatic it has to be reversible similarly make a process adiabatic and reversible then it has to be isentropic and make a process isentropic and reversible it has to be adiabatic that means although these three adjectives are reasonably independent of each other to the extent that any one of them does not apply any two of any one of the others any two of them you take the third one automatically follows so an isentropic adiabatic process must be reversible adiabatic reversible process must be isentropic and an isentropic reversible process has to be adiabatic