 Sir, my doubt is in steam tables, sir in steam table entropy is there, entropy property is also there. You said that entropy, change in entropy only we can, yeah we cannot find entropy at a point and another doubt is internal energy also mentioned. What does it exactly, what is that internal energy, that quantity 2500 or something, Ugr, Uf, what is that quantity indicates, over to you. The question is about the steam tables and what is tabulated in the steam tables, while it is true and while not not while, it is true that energy and entropy and all related functions like enthalpy, Gibbs function, Helmholtz function, they are all defined as differences. So, when you look at your steam tables, you should be wondering how come a single value is written, for a delta U. This is U2 minus U1, so we will need to specify two states and the table will become very complicated. So, what we do is we write this, we define a reference state and U0 at that state. Consequently, what is tabulated, what this will be calculated as U2 minus U0 minus U1 minus U0. So, it is this which is tabulated, this is a fixed state U0. So, what is tabulated is U minus U0 and you notice on page 5 of your steam tables, first line in the, in table 2, you will notice that the specific internal energy of saturated liquid at the triple point is defined as 0 and specific entropy at the saturated liquid state is also defined as 0. I am opening the steam tables, you will see it here, not table 1, first line in table 2. This value, this is defined to be 0 that is saturated liquid at the triple point. So, you can highlight this value which is defined to be 0, you can highlight this value, the specific entropy of saturated liquid at the triple point, it is defined to be 0 and the another value which is defined to be so in this table is the Celsius temperature at triple point, 0.01 degree C. Those are the three exact values in the steam table, over to you. Hello sir, we have one more question. So, regarding the efficiency of the Carnot cycle, so it speaks about the efficiency is 1 minus T2 by T1. So, but it says that this is not bothering about the working, working substance or a gas in it, but if you see in that the other form H1 minus H2 that this amount of heat absorbed and amount of heat delivered that basically depends on the working substance. So, in that way one can tell that efficiency depends on the working substance. Is it correct sir? Over to you. Basic idea of the second law of thermodynamics says that efficiency of any reversible 2 T heat engine is a function only of T at which heat is absorbed and T at which heat is rejected and this is independent of working substance. Now, using this we define thermodynamic temperature scale and using that thermodynamic temperature scale, efficiency of a reversible 2 T heat engine turns out to be 1 minus theta rejection divided by theta absorbed. This is on the thermodynamic temperature scale. Now, up to here no working substance come in, then we said that use air or an ideal gas with constant Cp Cv in a reversible 2 T heat engine. This is the Carnot cycle. Now, because we are using an ideal gas the efficiency of this reversible heat engine will turn out to be in the parameters of ideal gas. It turns out to be let us say it will be function of the molecular weight of the ideal gas Cp Cv of the ideal gas and also the ideal gas temperature because PV etcetera is related to the T absorbed and T rejected on the ideal gas scale. When you do the analysis we find that efficiency of a reversible 2 T heat engine turns out to be 1 minus T rejected by T absorbed where these are temperature on the ideal gas scale and then we say by Carnot theorem the efficiency whether calculated this way whether calculated this way or whether calculated this way. This is not calculated whether defined this way should be the same and that gives us the confidence to write that this expression must equal this expression and from there we show that temperature on the ideal gas scale and the thermo temperature on the thermodynamic temperature scale are the same. I think that should be understood over to you. I have 2 Carnot engine or reversible engine which is functioning the temperature between T 1 and T 2 having the same work W. Now, is there any possibility of superposition of the work W plus W or like that? This is over to you. You have 2 Carnot engines that means 2 reversible 2 T heat engines same T 1 and T 2 and same W net. Now, what happens is let me sketch this is T naught this is sorry this is T 1 this is T 2 and there is 1 engine let me say E 1 again reversible. So, R E 1 it produces a W we have R E 2 if that also produces W. Let me write this as W 1 and W 2 that means W 1 is W 2. Now, since they are reversible heat engines same T 1 and T 2 that means by Carnot theorem corollary you must have eta 1 eta 2 equal to eta 2 and since efficiency is the ratio of work done to heat absorbed. If this is Q 1 and Q 2 that also implies Q 1 equal to Q 2 and by energy balance or by first law sorry this is Q 1 1 and Q 2 Q 1 of 2 and this is Q 2 of 1 and this is Q 2 of 2 and Q 1 of 1 should be Q 1 of 2 and Q 2 of 1 should be equal to Q 2 of 2. So, except for perhaps their internal working externally where it comes to work interaction efficiencies and heat interactions the 2 machines have to be identical, but they may differ in their working internally. One may be based on air another may be based on water vapor or some other material it does not matter. Come in Pune over to you. Good afternoon sir. My question is related to SNTEL in which a system is considered to be absorbing heat from condensing steam and rejecting heat to the atmosphere while undergoing expansion and after applying the first law and second law we could derive we could able to derive the given equation, but we could not understand what do you mean by show that any irreversibility in the process gives rise to a wastage of thermal energy of steam by an amount. So, I want the interpretation of the statement please which gives rise to a wastage of thermal energy of the steam over to you If you look at this is a non-cyclic process a system undergoes a change in state represented by delta E delta S and delta V. It is given to us that the it absorbs heat from steam at T s either you can say T s is here or this is better write this as T s and you say this is Q is the amount of heat absorbed from the T s. Then you have an interaction with the ambient say at temperature T naught and let us say that the heat absorbed from the ambient is Q naught. This Q naught is not mentioned here, but let us leave it as an algebraic thing then this delta V let us say that delta V could be modeled as a cylinder piston stuff of the thing and there is a P naught all that it means because it says here that the and work done in displacing the environment that is the only work output. So, that means there is a W which will be equal to P naught delta V. Now, you should have applied first law to this first law is Q equals delta E plus W and Q is made up of two components Q plus Q naught this is delta E plus W W is P naught delta V. This is first law I think when you apply second law second law says that delta S is greater than or equal to integral d Q by T or summed over Q by T naught as appropriate. In this case this is this gets replaced by delta S and delta S should be greater than or equal to Q by T S plus Q naught by T naught. Now, the question is if you eliminate Q naught between the two you will get an expression which says Q is greater than equal to something some expression and in that expression if you put equal to that you will get Q min and Q minus Q min is the answer. Now, the interpretation is let me say this is equation one this is equation two. Equation two can be converted to an equality equation two can be converted to using S P entropy produced which will always be greater than or equal to 0. So, that way your second equation can be written down as S P plus Q naught by T naught plus Q by T S is delta S and let me say this is three. Now, if you eliminate Q naught from one and three you will get an expression for Q equal to something something and a term containing something into S P. Remember that S P is always greater than or equal to 0. So, Q min will be the same right hand side of this with the condition S P equal to 0. So, the minimum heat will be absorbed when you produce no entropy and Q minus Q min is the wastage of thermal energy of steam. I suppose now you will be able to appreciate the wastage of thermal energy of steam over to you. Thank you sir, over and out. Amruta, Coimture over to you. Your question is can entropy decrease. Remember entropy is a function of state. So, over a process delta S equals S 2 minus S 1 can be positive that means it can increase, can be 0 need not change, can be negative. There is nothing wrong about entropy for any process to be negative. For example, our basic entropy relation is delta S is greater than or equal to integral d Q by t or for a process element d S is greater than or equal to d Q by t all that is it is greater than equal to that is needed. So, it is possible that this is four units, this is three units, it is possible that this is 0 units, this is minus 2 units, it is also possible that this is minus 3 units and this is minus 4 units, nothing wrong in it. This is algebraically greater than or equal to that is it. Now illustration take for example, your let us take steam and you have wise steam you just take let me go to the next page. You have a just a pail of hot water say it is at 1 bar and it is at 80 degrees C and the ambient is at 30 degrees C. So, you wait for some time there is some amount of heat absorption which is heat absorption that means heat absorbed is heat loss. So, heat absorbed is less than 0. So, this temperature may come down from 80 degrees to say 50 degrees C. So, your state 1 is 1 bar 80 degrees C, state 2 is 1 bar 30 degrees C and you will find that S 2 is greater than S 1. So, many natural processes have S 2 less than S 1. So, many natural processes have S 2 less than S 1, there is absolutely nothing special about this. Remember, we wrote delta S is greater than or equal to or in the differential form we wrote d I think I have it somewhere, but I do not have access to it just now. We wrote d S equals d Q by T plus d S P and this we change say is change in entropy, this can be positive, this can be 0, this can be negative. Then d Q by T is related to heat flow, you absorbed heat it will be positive, you do not have any heat transfer adiabatic it will be 0, you reject heat this will be negative. So, all three are possible here whereas, here this is so called entropy produced and actually this is the defining equation for that. The second law says that d S has to be greater than or equal to d Q by T that means this term can never be negative. So, this can be positive, this can be 0, but this can never be negative, it cannot be negative is dictated by second law. When it is 0, it is a reversible process and when it is positive, it is an irreversible this is an impossible process it just cannot take place. These the rows are not linked to each other, the columns are important. I suppose now you should be able to understand over to you. We have one more question sir. Sir this is Krishnan from Amrita. My question is is there any example for quasi static process, any practical example? Another thing if so, but can we consider the natural degradation of a dead body as quasi static process or the natural discharge from a battery as quasi static? Over to you. The quasi static process each state equilibrium in actual practice this means each state in almost equilibrium that means very near equilibrium. How near equilibrium? It is for us to decide and it will depend on what you want to do with it. Now for example just now I took an illustration of a pail of water reducing in temperature from 80 degrees C to 50 degrees C under 1 bar pressure. Now this is a common enough thing we have boiled water or we have heated it water or milk. We can drink it only when it is near 40 or at most 45 degrees C. So, we want to before serving it to people we wanted to bring it down to tea, milk even warm water or something like that. We want to bring it down to somewhere near 50 or 45 degrees C. We let it cool by rejecting heat to the ambient and for rejecting it faster maybe we stir it or do something. Now suppose we stir it move it then you will notice that as you stir it there are pressure variations from one point to another point. Sometimes we dip it in another big vessel of water containing water at say 30 degrees C. If you are a hurry you will put ice cubes in it. When that happens temperature here will be slightly lower than the temperature here and if you really take a you know microscopic measurement or very fine measurement you will say that it is a non quasi static process. But once you are near 80 once you are near 50 you take it out it is almost 50 and you can determine the amount of heat loss by just using your C P delta T formula or whatever. So, for our practical purposes it is quasi static. So, whether to consider a process to be quasi static or not is not told to us by thermodynamics it is for us to decide. And as you said many natural processes of decay at our scale now human beings have a patience of a typically a few minutes or at most a few hours. So, anything which seems to be slower than that for our purpose is really quasi static. But something which you know starts decaying pretty fast goes bad in minutes then naturally that will not be considered quasi static by a human being. So, it is a matter of degree and it is for us to decide. Generally, natural decay processes are can be modeled or can be considered quasi static. But then other processes for example, you have a hot plate and you spray fine drops of water in it. They immediately sputter into vapor and they flash into steam that may not be quasi static if you want to analyze it. It happens in fraction of a second over to you abatic and isothermal at a time. So, is there any process and can we give the example of a thermo flash for that over to you. Processes are defined by either interaction or the path they follow. For example, qualifications like isothermal isobaric constant volume these are states sets of states on the periphery of on the state space. A diabetic is something which happens at the boundary. I will sketch a diagram and show it to you. See you take a system containing a gas or a fluid. It has a pressure, it has a volume, it has a temperature and let us say this is the p v diagram of that and these are the isotherms. When I say that I have an isothermal process that only means that if I start from here, I am staying along this isotherm during the process either one way or the other. That is all it means. Similarly, an isobaric process means if I start at this pressure I must remain at that pressure and a constant volume process means if I start at this volume I must remain in that volume. It does not say anything about how that process is executed particularly what happens at the boundary. So, q is an interaction w expansion and w stirrer are also interactions. Now, if you say this interaction is 0 then it becomes adiabatic that is it. Now, remember that constant volume if v is maintained constant that means because expansion requires a change in volume this you are preventing, but you may still be allowing stirrer work. So, a constant volume does not automatically mean no work it only means no expansion work. So, remember you can have a process which is adiabatic as well as isothermal. You can have a process which is adiabatic as well as isobaric you can have a process which is adiabatic and constant volume. If you want an illustration open your exercise sheet F 1.4 this is adiabatic plus isobaric. Now, F 2.13 I will go to the next page. So, that in fact this is an incomplete problem in the sense interactions are not possible are not determinable. So, you take F 2.13 you have 1 kg of saturated liquid water at 100 degree C evaporates isothermally into dry saturated steam. So, let me draw it on say the T s diagram this is initial state 1 this is final state 2. Process given specified 16 is 1 specified isothermal 2 phases together. Hence, isobaric also. So, as it is the process will have to be isothermal plus isobaric. Now, if you consider this to be a cylinder piston in which it is expanding like this. We are maintaining temperature constant, but it turns out that the pressure also turns out to be constant. Now, the third thing naturally I cannot have volume constant the isobar goes like this. This is the pressure of that process this is the temperature of that process and it remains isothermal and isobaric because the process is from saturated liquid to dry saturated state. Now, how does this process come about? Remember that if you apply first law and assuming that it is stationary that means, no change in energy other than in delta u. So, you will get delta u equals q minus w delta u is fixed initial state final state nothing changes. Now, out of w one component is also fixed that is p delta v plus I may have a stirrer here providing me w stirrer apart from the possibility of any heat transfer. So, p delta v plus w stirrer. So, all that I get is delta u is fixed p delta v is fixed all that I have to do is match q minus w s t. So, that this equation is satisfied I can have no w s t only q or some w s t and some q in which case it will be a non adiabatic process, but can be made adiabatic in which case all that I have to do is say that q will be 0. Hence, my delta u will be equal to minus p delta v minus w stirrer. So, provide enough w stirrer. So, that this equation satisfied in that case I will have a process which is isothermal plus isobaric plus adiabatic there is nothing special about this over to you. Sir regarding any practical example this is a practical example. I ask you regarding the this thing thermo flask can it be taken as an example or any other example in practical or a thermo flask is a perfectly insulated stuff. You do not expect any change of state there all that I can say is assuming the it is perfectly sealed then I can say and if you neglect the thermal expansion or contraction of that then you can say that it is a very slow quasi static process slightly non adiabatic because there is a heat transfer, but it is a very slow heat transfer that is about it, but that thermo flask is not a situation I can say at most it is very good approximation of a constant volume process if it is sealed properly. And if it is like a you know something which is in which you put liquid nitrogen, but keep the spout open you can say that it is a constant pressure process, but apart from that you cannot say much about a thermo flask. K. Pune over to you. My question is regarding the morning topic that is property relation can you explain property relation for Clausius-Cliffron relation. So, that is given in our even the syllabus also I did not get that. The so called Clausius-Cliffron relation there are many ways of deriving the Clausius-Cliffron relation I will derive it by one way purely by remaining within the domain of property relations. In fact, we will use the Maxwell's relations for it the other way of deriving it I will tell you later and leave that to you as an exercise. The Clausius-Cliffron relation starts of or tries to determine the slope of the any saturation line either the liquid vapor or solid liquid or solid vapor. So, the slope of this line will be d p by d t sat. Since, we are moving around the saturation line pressure is a direct function of temperature. So, we get an ordinary derivative. Now, rewrite this in terms of the temperature say temperature volume diagram. This is the saturated liquid line f this is the dry saturated vapor line g and let us say we are at a point somewhere here this is at that particular if you are saying this is the p naught t naught this is our t naught and of course, the constant volume line the constant pressure line p naught will go like this along this whole line the temperature is t naught the pressure is p naught and we are looking at a point in between. Now, consider the variation of temperature at a higher temperature you will have a higher pressure at a lower temperature you will have a lower pressure. Now, I want to know how does the temperature behave with pressure. So, although I have d p by d t sat I can write d p by d t sat, but in the two phase zone I can decide to go along a constant volume line or I can go along say a constant entropy line does not matter whichever way. So, I can take this as partial of p partial with respect to t along the v in the two phase zone say liquid vapor or I can say that this must also equal to the volume of the partial of p with respect to t along a constant entropy line in the two phase zone. Let us look at this this choice also is possible and I will leave it to you to show that whether you take this or this the final result is the same. Now, notice partial of p with respect to t at constant v does it have a candidate in the Maxwell's relation yes it has because if I write this is p v t v and I can replace p v by t s. So, I get t s t v since t is the common factor here I can write this as partial of s with respect to v at constant t. So, my again I am writing down this equation of the previous page d p by d t d p by d t sat which we took as d p by d t at constant volume, but in the two phase zone now turns out to be partial of s with respect to v at constant temperature in the two phase zone. Now, in the two phase zone how does s vary with t how does s vary with sorry the entropy vary with volume remember we are now looking at variation of entropy this with volume and how does it vary remember that we are looking at a constant temperature and constant temperature line in the two phase zone is this line from point f to point g. And the variation of f is linear with x variation of v is also linear with x. So, this becomes s g. So, this can now be written down as s g minus s f divided by v g minus v f and this gives us the first and the last give us the Clausius-Clapeyron relation d p by d t sat equals this is written as s f g divided by v f g. And you can also show that s f g is t minus v f and this gives us sorry is h f g divided by t. So, this can also be written down as h f g divided by t into v f. Now, on the previous page we had selected the first option d t by d p at constant v and not the second option d p by d t at constant s. If you had to select the second option all that would happen is instead of this d s by d v at constant t you will end up with d s by d v at constant p. But in the two phase zone d s by d v at constant t or d s by d v at constant p mean the same thing because the isotherm is also the isobar in the two phase zone. So, you end up at precisely the same result this result is the Clausius-Clapeyron relationship. Here we have derived it using the property relation the other way of deriving it I will only hint at it and leave it to you to complete the derivation. You do the following you take say a let us say you can sketch this on a p v diagram let us say this is the liquid dome and let us say that these are two constant pressure lines and hence with the two phase zone from f point to g point these are also two isotherms although the constant pressure lines will be horizontal the isotherms will go like this does not matter this is the f part this is the g part. Now, let this be differing by a small pressure difference delta p and hence a small temperature difference delta t. Now consider two isentropic within this two distinct isentropic. Now, remember that this looks like a Carnot cycle it works between temperature t and temperature say t plus d t and some you say some s 1 and some s 2 how much is the work done work done will be the area of this curve how much is the heat transferred because it is an isothermal process the heat transfer will be the temperature at which heat is absorbed multiplied by the entropy difference remember d q is t d s for a reversible process and all the processes in a Carnot cycle are expected to be reversible process. So, calculate the area of this as integral d p d v that is the work done calculate the area under this curve it simply the higher temperature t plus d t divided by whatever is the entropy difference that you have assume that is the heat transfer divide work done by heat transfer and that will give you one expression for efficiency, but we know that a Carnot cycle is a reversible two t heat engine cycle. So, the efficiency would be one minus heat rejection temperature divided by heat absorption temperature equate the two and simplify and you will get the Clausius-Clapeyron relationship although I wrote it as t t and t plus d t your algebra will be simpler if you write it as working between t the upper end and t minus d t the lower end the algebra will be that much simpler otherwise it is essentially the same derivation. Kukas Jaipur over to you. Sir good afternoon my question is what is the can you once again explain the procedure for getting the properties of subcooled liquid over to you. The question is properties for subcooled a system containing a subcooled liquid one you should remember is a simple compressible system. Hence, specific volume of subcooled internal energy entropy must be a function of temperature and pressure almost in compressible relation for low p say up to 50 bar may be up to 100 bar approximate computation of properties of subcooled liquid assume subcooled liquid to be in compressible that means I will sketch the p t diagram and I will say this is the triple point and this goes into the critical point and we are somewhere in this zone say this is p naught t naught but we are saying the liquid here is incompressible. So, there is no effect of pressure on the properties no effect which are properties the basic properties of course, we have said incompressible. So, obviously there is no effect of v but incompressible means no two way work mode you cannot compress it you cannot expand it. So, properties must be functions of one variable temperature that means v must be a function of temperature only u must be a function of temperature only s must be a function of temperature. Now, consider this point and along the same temperature this is t naught go along this line till you come to saturated liquid t naught and we say from here to here in this zane in this zone same u s why because same u s must be a function of temperature only. So, this is v t and hence I will go to the next page now and hence we say v at t naught p naught will be v f at t naught u at t naught p naught will be u f at t naught and s at t naught p naught is h f at sorry s f at t naught provided t naught p naught in the sub cool. The exception for this is h because h at t naught p naught is h f at t naught p naught by definition is u at t naught p naught plus the pressure p naught multiplied by v naught which is a function of p naught t naught. So, if you substitute this this becomes u f at t naught plus p naught into v f at t naught. So, h d naught is h f at t naught. So, it depends because p naught is physically present here. I think that should make the idea of our incompressible liquid scheme or model for estimating properties in the sub cool liquid zone very clear over to you. Thank you sir. Thank you over and all. Will I pan well over to you? Sir, there are two questions. First one, can you explain internally reversible and externally reversible process with examples and the second question is that LPG cylinder what type of system it will be simple complex or rudimentary. So, the first question is what is internally reversible process and externally reversible process we do not make any such classification over definition of the reversible process is very clear. Two systems which are interacting with each other and execute a process which takes a from a 1 to a 2 and b from b 1 to b 2 is reversible. If the whole process can be reversed that means b can be brought from b 2 to b 1 a can be brought by a 2 to a 1 using the same forward path which they traverse and by reversing all the interactions in all details that they had with each other. So, it is exactly like a movie in all its detail being able to be played back and still be true and consistent that means the requirement is that if a reversible process is executed from a 1 to a 2 and because it is reversible it is reversed back and systems come back to a 1 and b 1 then no trace of that process having been executed is ever maintained. You cannot even make out whether that process had taken place or not it is a very, very strict requirement and such a process we can only think about and not even think of being able to execute. There is no need at all for us to define internal and external reversibility. There are some books which define it as internal, internally reversible and externally reversible and so on, but I do not want to get into it. The second question was about the LPG cylinder, the cooking gas cylinder. The question is whether it is a simple system or a complex system. By cooking gas cylinder I think what you mean is the gas inside the cylinder and let us assume that the cylinder has a connection and the process is the pressure outside is low, the pressure inside is high and it is liquefied petroleum gas. So, you have some liquid and some vapor and since it is liquid vapor in equilibrium and if the cylinder is not insulated, the pressure inside the cylinder will depend on its temperature which will depend on what is the temperature in the ambient. Now, since we do not have a stirrer and we do not have any other type of work mode, I think it remains a simple system. There is no need to consider it as a complex system. It is an open system, if it remains closed then there is nothing happening to it except pressure changing because of changing temperature, because of changing temperature of the environment. The moment it is an open system and something flows out, flowing out requires some work to be done which is an expansion type of work as we will see tomorrow when we consider the open systems. So, and expansion type of work can be done only by a simple compressible fluid or at least by a compressible fluid simple or otherwise. So, in this particular case, I think the gas in the LPG cylinder or the fluid in the LPG cylinder can be very well modeled as a simple compressible system. But in a way you are right, if this cylinder is closed then it is some sort of a restricted system which is a rudimentary system because in that case we do not have a pressure gauge but if you have a pressure gauge on it then that pressure will indicate the temperature of this and that temperature will depend on the ambient temperature. So, you will have a very crude thermometer, a vapour spring thermometer as they say that is available to you over to you. Thank you sir, over and out. Truba Indore, over to you. So, my question is that is it mandatory that isentropic process should be reversible adiabatic process over to you? The question is is it mandatory for an isentropic process to be a reversible adiabatic process? No. If you look up our yesterday's scheme of things, we looked at the thing first the adjectives which we came across is adiabatic. Adiabatic means d q is 0, then we came across an adjective which was reversible. Reversible just I explained in the previous discussion what reversible meant, but if we define the entropy production then reversible process is d s p is 0 and the third one was isentropic. The isentropic process was defined as one in which the change of entropy is 0. So, these three are three distinct adjectives but they are not all that distinct because d s d q and d s p are all related by this relation. The entropy relation and the second law in terms of entropy production I will cross this indicating this represents change of state whereas, these two represent interactions and because of this is something like a equals b plus c, any one of them can be 0 the other two can have any appropriate value to satisfy this equation, but if two of them are 0 the third one has to be 0. That means of these three adiabatic reversible and isentropic you take any one that is other two need not be applicable because if d s is 0 the restriction here is only that d s p be greater than or equal to 0. This d s p cannot be negative if d s is 0 you could have this positive this negative and you will still have this 0. If d q is 0 that means adiabatic you can still have this positive this positive. So, the other two need not be satisfied if you have d s p equals 0 that only means d s is d q by t either of them can be positive or negative. So, any one does not imply any other two. So, the answer to your first question is an adiabatic process or an adiabatic process should it be reversible isentropic the answer is no or is an isentropic process does it have to be reversible adiabatic the answer is no. Similarly, a reversible process does it have to be adiabatic isentropic the answer is no, but you take any two of these three you make 0 in this equation the third has to be 0. So, you take any two this automatically implies the third. So, if you make a process adiabatic and reversible it should automatically be or it will automatically be isentropic you make a process isentropic and reversible it will automatically be adiabatic and if you make a process adiabatic and isentropic it should automatically be reversible. So, that is the question and at this stage I am saying over and out because I am going for a cup of tea.