 So welcome back. And now I am open for discussion. I have first started 1197 GH Raishwani Vagoli Pune. Over to you. H e is greater than H e star. Right. Okay sir and the difference is again given that but why it is H e is equal to H e star is given there. I am not stating that. However you see this greater than or equal to signs or less than or equal to signs. These are inherited from the eta less than or equal to eta r which comes out of the Carnot theorem. Equal to here means the reversible limit. So in the limiting case it will be equal to. And I put I have said in the morning lecture that I have the habit of putting less than or equal to as two separate signs to be conscious of the fact that equal to is a very special case the reversible limit. I think right from the Carnot theorem I have been using this less than or equal to as two signs stacked one above the other or greater than or equal to as two signs. And I have explained to you and I will again say that that is my habit. I like that habit because it gives the special status to that equal to sign. But you are free to use the mathematically greater than or equal to or less than or equal to sign. The standard which you find in any book. Over to you. The entropy of a system increases as we approach absolute zero. Is it a statement related to some law or a generalized observation in thermodynamics? It is not the statement of any law of thermodynamics because for our purpose we have only three laws of thermodynamics zero first and second. Apart from that we have certain premises which we said about non quantization continuous variation approach to equilibrium and things like that. What the statement which you have made is neither a law nor a general observation because depending on the material the in the state space the entropy will change. It is not necessary that as you approach lower and lower temperature the entropy of any material will go on increasing and increasing or decreasing and decreasing. It all depends on the material at hand. Over to you. Okay thank you. DMS College of Engineering 1, 2, 2, 5 Bengaluru. Over to you. You have told that entropy increases only with adiabatic condition. But there is no such adiabatic condition which we can have it in practically when we do it. But how to consider that it is not that is adiabatic or anything. In micro channels we are working. See let me restrict only to thermodynamics. In thermodynamics everything is about limits. What is possible what is not possible and some constraints. First law puts a constraint of equality. Second law puts a constraint of inequality and that inequality means there is a limit and that limit is the reversible limit. Through that it turns out that the definition of delta S requires dQ by T for a reversible process. So it is a limiting definition of delta S. Delta S or dS in practice is never calculated by doing a reversible experiment and then measuring dQ. We have determined property relations. So we measure one property set and with respect to that you determine the other property set. For example the whole state space of a fluid within reasonable range of pressure and temperature. Say for example water. The whole state space of water all the properties are based on only the following set of information. Number one at various pressures and temperature you determine what is the density of water. So measurement of density all over the state space. Number two from triple point to the critical point the pressure, temperature, saturation, state measurement. Only the pressure and temperature measurement density measurements anyway are done on the liquid end as well as on the vapor end. And third the value of specific heat of water or specific heat of in particular water vapor at very low pressures as a function of temperature. Only these three pieces of information are needed for us to determine all properties enthalpy, entropy, internal energy everything else throughout the state space of water. It can be shown that this is done and in advanced thermodynamics or second course in thermodynamics I provide enough details of how this is done and which is the proper way to proceed. But in this course we do not have enough time to explain all that. But the germ of all that the germs of the heart of all that is all there in our topic 10 and the associated exercise sheet on PR property relation over to you. M K S S S Pune 1101 over to you. Another question that is regarding open system and control volume. Can you please explain the difference between open system and control volume? There is no difference between an open system and a control volume they are one and the same. Two names for the same stuff. Just the way we have a closed system sometimes we simply call a system sometimes we call it a control mass. So wherever you see open system you can replace it by control volume and wherever you see control volume you can replace it by an open system. Both nomenclatures are used in textbooks of thermodynamics all over the place. Second question, in case of compressors we consider ideal compression as isothermal compression. Then for the agglombraces why we define the isentropic efficiency as the ideal efficiency? What is ideal is to be determined by the user? If you are for example in a gas turbine or in a Brayton cycle power plant the ideal compressor will not be considered an isothermal compressor it will be considered a diabetic compressor. For the simple reason that after compressor you have to further heat the gas in the combustion chamber. But if you are considering some industrial process where you need some gas you have some gas at room temperature and room pressure, ambient pressure you want it at a high pressure but again at ambient temperature then perhaps your compressor ideal compressor would be an isothermal compressor. Because your inlet state and exit state are isothermal so the interweaving process if also it is an isothermal that will be a good idea. So in that application you may consider an isothermal compressor is an ideal compressor but in a power plant where the compressor and the combustion chamber together do the job of providing enthalpy rise adiabatic compressor will definitely be an ideal compressor. Thermodynamically there is no such thing as an ideal compressor it is for us to define what is ideal for us. Thermodynamically there is only one ideal and that is a reversible process of any kind over to you. Now F2.15 let us read it aloud we have a rigid metallic container okay rigid means no change in volume obviously expansion work will be 0. Metallic means it is unlikely to be insulated and we should be alert to that fact. It separated into two equal parts by a thin partition okay so we have this metallic container and we have a thin partition in between. Equal part okay equal volume so if this is V this also is V. One part contains 1 kg of saturated liquid water at 100 degrees C so we have here 1 kg water I should not write H2O the chemists will catch me there 100 degrees C saturated liquid. The other part is evacuated this is the initial state the partition is broken and the equilibrium re-established after some time okay during the process the temperature of the system is maintained by immersing it in an oil bath maintained at 100 degrees C. So there is some oil bath can be modeled as a thermal reservoir there will be some heat exchange is required so that the final temperature is also 100 degrees C. Determine the final state of steam work done heat transfer okay. Now let us look at how the process is going to look like I am writing specific volume you are free to consider the actual volume that capital V it will look slightly different. Now let us say that this is the 100 degrees C isotherm initially initial state was 100 degrees C that corresponding pressure which will be 1.01325 bar this is the initial state 1. Now what is common between the initial state and the final state the final state has a volume twice the initial state we know that V2 equals twice V1 and that automatically means because it is a closed system your specific volume 2 will be twice the specific volume 1. So if this happens to be V1 the specific volume V2 will be twice V1. Finally the temperature is also 100 degrees C so the final state will be state 2. I have not shown the process the line which you see is the isotherm representing 100 degrees C. If you want to show the process you will say that look just like that other problem in second law which we saw yesterday this happens to be a irreversible process it is also a non quasi static process. So I can just show it by means of a dotted line from 1 to 2. If you draw a big enough diagram a big version of this this will be state 1 this will be the state 2 but what you see here this line is the isotherm at 100 degrees C. The actual process can only be shown by means of a dotted line. So the final state is given by state 2 is given by V2 which is 2 V1 and T2 is 100 degrees C. So this gives you the final state and then you apply first law to obtain everything first you notice that work done work done is expansion work plus some other type of work if any there cannot be any expansion work because there is no system to receive that work it is purely an expansion into a vacuum. So expansion work is 0 there is no mention of a stirrer no mention of any electrical connection so other type of work can be assumed to be 0. The net result is sub problem B work done is 0 final state of system we have determined in terms of T2 V2 so you can determine pressure will also it will be at some drainage fraction so the pressure will be the saturation pressure at 100 degrees C and you can read out all the other properties as required. And now the heat transfer comes out of the first law remember heat transfer comes always out of Q equals delta E plus W we have shown that W is 0 so delta E will be delta U plus delta E other delta U you can compute out from the initial state 1 and the final state 2 and delta E other you can assume to be 0 because there is no mention of any change in position or change in velocity. You can set up a more complicated version of this problem for example you can say the initially the system is at a height of may be 1000 meter in the final state that it falls at ground level and finds itself in a big pool of water which is at 100 degrees C or some other temperature which you feel like having. In that case the initial state has a large gravitational potential energy final state has much lower gravitational potential energy so that term will have to be included in the heat transfer. I think somewhere pertaining to second law or later there is a problem which is similar to this I think it is combined laws CL6 will come to that on Wednesday on page 19 they have always been put here over to you 1182 MK Orchid College Solapur over to you sir my question is as the pressure increases why does latent heat of evaporation decreases as pressure increases why does the latent heat of evaporation decrease there is no thermodynamic reason for this thermodynamics only says that properties are related to each other change in temperature with pressure along the saturation line the so called latent heat that is Hfg the temperature and the specific volume of liquid and vapor thermodynamic does not say why thermodynamics will only say that if it decreases something else will become like this that is all there is no thermodynamic reason for that. One more question sir in F2.3 it is a constant pressure process X1 is supposed to be a drainage fraction 0.5 and T2 is given as 400 degree centigrade. Right so the initial state is P1 Y bar X1.5 so it is wet steam at a drainage fraction of 0.5 at 5 bar and the final state is 400 degree C and the process which we consider is a constant pressure process. So the final state is P2 is 5 bar T2 is 400 degree C. So you can determine you know locate that state on the state space and you can read off all the properties. More than saturation temperature. Yeah it will be a super heated steam but this is something which happens in a boiler for example in a boiler at some stage you will have a drainage fraction 0.5 as further heat is added if we neglect the pressure drop in the drum and the pipelines then you will first at that pressure it will become saturated at 5 bar you can read off the saturation temperature so the initial temperature will be 5 bar saturation so 151.9 Celsius it will become dry saturated because it is a constant pressure process so it will become X equal to 1 temperature 151.9 pressure will remain 5 bar beyond that it will go to super heat and then if you come to super heat tables page 12 it goes up to 400 degree C so slowly it will go to super heat of super heat temperature of 200, 250, 300, 350, 400 even the tabulation at 400 is tabulated 5 bar 400 degree C that is the final state. It is a process in which the evaporation gets partly completed from 0.5 to 1 and after that super heating takes place over 3 days. College of engineering today which thermodynamic processes available on sun, moon and earth. The moon and earth are connected mainly by gravitation there are hardly any thermodynamic links between the moon and the earth there are of course I mean for example energy gets transferred by radiation from the sun to the earth directly also indirectly by reflection from the moon. The moon has its own temperature so it radiates some energy part of it is intercepted by earth so there are links but the main earth linkage is gravitational if it were not there moon will just go fly away tangentially from its orbit. But of course there are there will be issues when the earth moon link may be considered up in some planetary science or atmospheric science it will have to be considered and if you set up a colony on the moon or some equipment on the moon what I understand is as the moon goes through its various phases we see only one face of the moon but at any point the temperature may be as high as 150 or 200 Celsius if it is on the sunny side and it may drop to something like minus 100 or minus 150 Celsius when it is in the middle of the non-sunny size. So it will go through at the period of typically one month which is the moon cyclic period it will go through significant thermal oscillations. So for equipment and people and things on the moon that will be something which will have to be considered. So it will go through a cyclic thermodynamic process once every 28 or 29 days as needed over to you. Okay sir another question sir so many years we are studying geoslaw, Carnot cycle, like Stupend Boltzmann law so many laws but after that so many paleo-headers done lot of research but this is included in the syllabus or not? See the purpose of PHD is not to produce material which gets included in the syllabus I think that must have stopped happening long ago when the basics were developed. Nowadays I do not think anything the PHD work has become so specialized that you tend to learn a lot about a very small thing and such specialized small things do not generally end up with textbooks particularly do not end up with undergraduate textbooks at all. I think if undergraduate textbooks differ they differ because of newer topics are included not because esoteric high level research work is included over to you. Thank you sir. 1219 prestige institute indoor over to you. I would like to know something about unsteady flow process please. There is nothing special about unsteady flow process if you want unsteady flow process in open systems I have here start solving the exercises for example I will tell you the typical unsteady flow problems we have is OS19 OS20 OS23 these are simple unsteady flow processes. I want to know something about energy balance sir. Energy balance is nothing but first law of thermodynamics if first law of thermodynamics is satisfied you say energy is balanced and the fact remains that energy flows are always balanced because they always have to satisfy the first law of thermodynamics. So when somebody says that the energy balance for my plant is satisfied that only means you have done the measurements and the calculations correctly. So it means mass multiplied by specific energy that has to be considered. Yes you said it right you have done a specific energy. Thank you. 1161 Anurag group of institutions Rangaretti over to you. Up to pressure 30 bar the enthalpy is increasing. Yes enthalpy of dry saturated vapor is increasing. After that it is reducing. Right. Why is it so sir? No this is just thermodynamic characteristic nothing special about it thermodynamics does not dictate it should be so. It just turns out to be so. How many properties need for internal energy when the system is closed? If you have a closed system and if you consider only one work mode one two way work mode then two properties will decide the state of a system and hence two properties will decide the internal energy of the system. If the number of work modes is higher than two the appropriately higher number will be needed. And why it is one for rudimentary system? Can you explain that one? Because for a rudimentary system the typical example which I took is the mercury in glass thermometer for such a system there is no two way work mode which is available. Hence the number of properties which decide the state of the system is simply one and that is why we consider them as important systems on their own because they are very useful in measurement and calibration of temperature over to you. 1195 one cell collate Mandi Deep Bhopal over to you. My question is this is there any example of the process which falls under the isentropic process but it should not be adiabatic process. Is there any example? There are enough examples of this for example you do the following the simplest default example you take a vessel you stir it what happens the vessel contains a liquid like say water its temperature rises you start stirring it at a particular rate temperature will initially rise but then there will be some heat lost to the surroundings stir a liquid lost to surroundings. Now if you keep on stirring the liquid initially the temperature will rise but then you will have a balance in which the rate at which energy is made available by the stirrer is energy which is lost to the surroundings. In thermodynamic this is q dot this is w dot q dot will turn out to be equal to w dot and both will be less than 0 work will be done on the system and heat will be lost from the system whereas the state of system is steady. So this is a situation when it is steady state your ds is 0 but there is a non-zero thermal interaction so this is a situation where you have an isentropic process but a non-adiabatic process over to you. On heating water the generate vapor and on condensing convert back into water is it reversible process if we have water the heat vapor dissipated in atmosphere and if you know how because we are regaining water completely. See you are looking at just one aspect that water is getting evaporated that evaporated steam evaporated vapor is becoming water again but the requirement for reversible is it should be inverted everything should trace back to its original position. You should ask yourself the question what is it that is causing the water to evaporate for example if you are keeping the water in a saucer out in the sun it is absorbing the sun rays and evaporating in the condensation the reverse should require that the vapor condense into the saucer the saucer should re-radiate or back-radiate the radiation from itself towards the sun or if you are evaporating the water on a stove by burning gas well there should be a reverse flame which absorbs heat from the vessel converts it into unburned gas and pushes it back into the cylinder all this thing is required for it to be a reversible process see we should not be under the impression that a real life process can somehow be considered reversible get rid of that idea absolutely no process in reverse life in real life is reversible reversible is only an idea it is a limiting process we have an idea of what it should be but at the same time we realize that it is impossible to create a reversible process that is what I said yesterday that it is a process about which we can think in any detail there is no restriction on thoughts so it is a thought process it is not a real life process at all just the way you have a continuous function and in the limit something happens in mathematics this is simply such a trick it is a very useful trick because it allows us to define certain properties which can be then computed out using its relations with other properties I think earlier in the day I said that although the definition of change in entropy is ds is dq by t for a reversible process entropy is never determined in an experiment in which you execute a reversible process and measure dq by t you always measure PVT data may be some other related PT data and either the value of Cp at one temperature one pressure all through the various temperatures or something else and from that the entropy values are derived over to you and then we are going beyond the critical point there is yesterday you said that there is a super heated fluids it is nothing it is like any other liquid or vapour the only thing is because you are at super critical pressure just increasing its temperature will make it look like a vapour but you will not have a process of boiling in which bubbles form and liquid gets converted into vapour in a discrete fashion that will not discontinuous fashion that will not happen so you know we have those old bookish ideas because the idea that something is liquid something is vapour these were developed when we did not even realize that there was something like a critical point when critical point was experimentally determined and people realize that there are such states as critical states then they realize that because of the existence of a critical point the differentiation between the liquid and vapour is not so clear cut as it was earlier that is why it is better to use a word like fluid whenever you mean either liquid or vapour or some flowing medium as appropriate that is why a fluid is a better word but even then we know when to call a liquid a liquid and when to call a vapour a vapour when you are at subcritical pressures it is perfectly safe to do this it is only when you go to supercritical pressures that you can convert from liquid to vapour without process of so called boiling taking place over to you. Sir my question is can we use the laws of thermodynamics at the atomic level if has what modifications can be done and while defining the laws of thermodynamics we do not consider the size of the system it is at the micro level micro molecular level a micro atomic level or they cannot be used for earth which is a very complex system so we should limit or should not we limit the laws of thermodynamics to a particular class of systems where we can consider these as a macro systems. Yes what you can say that is partially right for example when you come to atomic systems our assumption that our systems form a continuum there are no quantum effects and they are scale independent these things break down so we will have to use appropriately formulated or different laws of thermodynamics but when it comes to earth I think earth is a more or less continuous system into which the laws of thermodynamics are applied however earth in all it is such a complex system that applying the law of thermodynamics will be proper but quantitatively it will be very difficult to do but in spite of that earth scientist and atmospheric scientist keep on doing it and they will keep on doing it and come up with better and better methods of applying our own laws of thermodynamics to it. In our typical earth science or atmospheric science we do not have to worry about quantum effects and discrete non continuum effects the only thing is the atmosphere and the earth remember is never in a steady state it is never in equilibrium so if you want to study you will have to chop it up into local systems which you can consider approximately in equilibrium that is what we always do whenever you have a large system large enough that you will find that it is never in thermodynamic equilibrium but yes for earth our traditional laws of thermodynamics are applicable no doubt about over to you. My question is system and surrounding both are equilibrium then system entropy constant or change? Equilibrium is just a characteristic of state the moment you say that one your system is in a state of equilibrium that means all properties are uniquely known you have unique values and so is its entropy. Surroundings is another system if that surrounding system is also in equilibrium it will have its own unique set of properties including entropy since these are two different systems there is no link between the entropy of the two whenever we talk of a change in entropy we talk of a change in entropy of a given system we do not talk of a difference in entropy between system one and system two in thermodynamics there is only one property which we compare between system thermodynamic basic thermodynamic property and that is temperature and that is brought to us by zeroth law and linked also to the second law but we never compare for example energy of one system with energy of the other system entropy of one system with entropy of the other system that is something which is never done over to you. Chemical equilibrium system and surrounding both are in chemical equilibrium but within a system chemical equilibrium or not? Sir a system will be in total thermodynamic equilibrium only when locally there is chemical equilibrium and that means not only the composition but even the chemical potential is uniform within the system it has a unique value over to you. Suppose we are defining our process one to two and it is a quasi-static process so it will be possible always that it will be reversible process? No not necessarily a quasi-static process is a process for which the path is completely defined but it may not be reversible process I will take an example the example is here in front of you when you stir a liquid it is possible that the liquid is stirred very vigorously in which case as it is being stirred we do not know where the liquid level is and what is the distribution of temperature and pressure in the liquid. So it is not a quasi-static process but if I stir it very slowly without disturbing the quiescentness of the liquid then definitely it is a quasi-static process but we know that stirring of a liquid is never a reversible process. So here we have a situation where it is a quasi-static process but it is not a reversible process and why stirring of a liquid you take for example you pour hot coffee in your vacuum flask the traditionally called thermos flask and seal it it is an almost adiabatic system but we know heat is lost very slowly from the inside to the outside. So the coffee temperature may be initially 70 degree C but as you wait long enough it will slowly reduce and finally reach room temperature but you are not stirring the coffee the temperature is changing so slowly that any time you try to measure the state of that coffee pressure, temperature etc you will find that it is uniquely defined. So this is a quasi-static process but cooling of coffee when it is kept in a vessel is not a reversible process. So there are enough examples of quasi-static processes which are not reversible. In fact as I said a few minutes ago a reversible process is something which we only think about whereas quasi-static processes can be imagined to be good approximation of real life processes. So you can demonstrate an almost quasi-static process but you cannot demonstrate a reversible process. Over to you 1, 2, 6, 3, Techno India, Salt Lake, Kolkata over to you. Sir, actually my question is that regarding phase rule that you have told that when saturated water and dry saturated water and liquid when are in equilibrium that means in the PT diagram it is in the liquid vapor saturated line. At that point you have applied phase rule and shown that degree of freedom was 1 right. That means that 1 degree of freedom means one property sufficient to define that state. This is a very common confusion the state postulate says that since for a fluid the number of two way work modes is 1, two properties are needed to define the state of a system. This is true whatever be the number of phases, whatever be the phase whether it is liquid whether it is a mixture or whether it is a vapor, two properties are always needed. The phase rule tells us particularly for a single component system like water and that is where we are using it. The phase rule tells us about the degrees of freedom and degrees of freedom is of the properties pressure and temperature how many you can take together and whether pressure and temperature are independent or not when you have just single phase it means yes the number of degrees of freedom is 2 and you can use if you want you can use pressure and temperature as two independent variables, but it does not say that you have to use pressure and temperature. It says that you can use any two property when the number of phases is 2 when you have liquid and vapor together the phase rule tells us that the degree of freedom is 1 that only means that of the pressure and temperature you can select only one not the other. The property requirement it is still 2, but the phase rule requirement says if you want to select you cannot select pressure and temperature together you can select one of them the second one has to be something else, but of course the choice is that we can select we need select neither pressure nor temperature. So in two phase for example we can select enthalpy and entropy as the two properties and fix the state, but the only restriction which is put by phase rule in the two phase in the two phase zone is that pressure and temperature together you cannot choose if you want you can select just one of them over to you. Considering only pressure and temperature why we are not considering pressure and specific volume these two are also intensive properties why we are not considering that phase rule talks only about pressure and temperature that is it it does not talk about specific volume over to you. Sir one question is there that is regarding the property pressure that when it is in equilibrium but at different position within a closed system pressure may be different because suppose in case of a bottle of water actually pressure is different at different point then how can we call this as intensive property from this point of view. Okay, calling it as intensive property is the second question I expected question to be that if in a bottle of water you have the pressure variation because of gravity from top to bottom then why do you consider it in the first place to be a state of equilibrium. Now the answer to that remember I said in the initial that property is a relevant characteristic of our system and there were people have argued with me why should that relevant be there and there are some people who argued saying I said relevant property I said no property is a property property means a relevant characteristic and although the characteristics come with our systems what is relevant and what is not is left for us to decide based on our experience. Similarly we know that in real life no system is in equilibrium no property will have a unique value why the pressure in this bottle even the temperature of water in this bottle if I really make precise measurements will not be uniform because I am a body at 37 Celsius so there will be some radiation coming from a 37 Celsius body on it it is exposed to radiation from the lamps nearby including the lamp on this document projector and of course the this is an air conditioned hall so the surrounding temperature is may be 24 or 23 degree C so there are lots of heat transfers involved and I am sure there is a nice temperature field in inside but then the question that arises is if I want to study it do I really at this stage bother about that small variation in temperature and pressure that brings me to the question the same question rephrase yes I agree that there will be non uniqueness in some properties like temperature and pressure but is that variation local variation in temperature and pressure relevant to what I am going to do and what I am going to study if the answer is no it is not relevant then I say yes I can neglect it and go ahead but if it is relevant then I will say I cannot neglect it I cannot consider this whole system to be a system which is in a state of equilibrium and then depending on the variation I will have to convert it into smaller subsystems and may be hopefully each one of them can be considered to be a local equilibrium okay in the study of fluid dynamics in the study of heat transfer we consider small control volumes or small system dx dy dz why do we do that because we know overall the system will be at a non uniform pressure non uniform temperature or something like that and we want to study that non uniformity but we have to apply our laws of thermodynamics so we take small pockets of it and we will say locally let it be in equilibrium so that locally we are confident of defining a temperature and pressure there will be situations and I will bring your attention to the exercise 1 wi 0.4 this exercise wi 0.4 where you have a piston which acts like a dam and which has water on one side this is a situation where it should be clear to you that for the water pressure by itself will not be uniform there will be significant differences from top to bottom okay so here to solve the problem we will have to take care of those pressure variations and that is the first part of that problem but when I play with the water in the bottle I do not have to worry about the pressure difference but if I convert tomorrow this into a fluid mechanics problem and ask you a question that look if I make a small hole here at what rate will the water flow out and if I make a similar hole here at what rate will the water flow out then I notice that look if I make a small hole just at the level hardly any water will flow out but if I make a hole at the bottom a large amount of water will flow out out there I have to consider the vertical variation of pressure so whether these variations of properties are to be considered or not it is for us to study and decide based on the situation at hand over to you sir yes sir I have one more question regarding the problem of second law that you have solved in the two day actually on the third day that my heat transfer is taking place between two reservoirs to a conductor if you remember a simple problem that you have solved for the entropy produced was to be calculated heat transfer was taking place between two reservoirs to a conductor yeah in that problem you have considered three systems two reservoirs and the conductors and you have calculated the entropy production in that case for calculating the entropy transfer of the two reservoirs you have used the formula heat transfer by a temperature and why you have considered these heat transfers are reversible why you are considered this reservoirs heat transfers are reversible okay this is my question because the entropy produced was the due to the of the conduct see the entropy produced is not because of just the conductor the entropy produced is because of the whole process of one reservoir giving up heat and another reservoir absorbing heat and the two reservoirs are at two different temperatures okay now a reservoir is a again like a reversible process a modeled system it is a system with a much larger very large energy capacity or very large capacity to absorb and reject heat without a significant change in temperature okay and although it is difficult to give a direct explanation as to why the entropy change of a reservoir is heat absorbed by temperature if you model a reservoir at say a system with large heat capacity working either at constant volume or constant temperature and if you take the limit of that as the capacity becomes large the mass becomes large then in the limit you can show that delta s is q absorbed divided by temperature where the temperature does not change in the limit this can be even shown mathematically over to you sir I have one more question regarding the PS diagram I actually want to know what is the nature of PS diagram in the solid liquid region can we simply extend that from the liquid vapour region in the if you please show then well I do not have access to a PS diagram for solid liquid region but I think if you go to the old Keenan and Keyes steam tables properties of water and steam I think they have in it in the appendix the data of the solid liquid region I do not know whether they have the PS diagram plotted but since because you have the data for the solid liquid region you can plot if you want it yourself I do not know whether our steam table has any such information we have yes we have here solid water vapour temperature we do not have the solid liquid but we have the sub 0 ice and vapour where steam the sublimation region solid vapour equilibrium it is on phase 19 if you are interested you can plot it and if you go to old thermodynamics book by example I think engineering thermodynamics by an old book called Lee and Sears you will find three dimensional plots of the PVT diagrams and other things out there TS diagrams may or may not be there but the three dimensional PVT surfaces are plotted you can take a look at them okay thank you sir thank you very much well 05 Mahatma Gandhi Mishra College Noida Uttar Pradesh over to you yeah my question is the heat transfer that you are considering it is reversible isothermal because you did not consider the entropy generation am I right sir no no no see entropy generation is defined as delta S minus that dq by T that is our definition see this during the process let it occupy a small amount of heat and temperature at the boundary where the heat transfer takes place is T then the second law dictates that ds must be greater than or equal to dq by T and we define dsp the small amount of entropy produced is defined as ds minus dq by T because of that this equation the second law equation using this definition can be written down as ds equal to dq by T plus dsp this is nothing but this definition turned around ds equals dq by T plus dsp and the second law then becomes dsp is greater than or equal to 0 a statement like I have not taken the entropy production into account etc is not a proper statement to do because entropy production is to be calculated from this entropy production is never I mean it is not proper to say that entropy production is taken into account over to you. Sir if we write this expression entropy increase of the universe let us write the system. So suppose you have a system and you have another system which we call the surroundings the let the entropy of the system change by delta s let the entropy of the surrounding change by delta s surroundings. Now the so called universe is system plus surroundings this is called universe only if actually if and only if it is adiabatic that means across this boundary there is no q but in between them they can do anything they feel like. So long as they together form an adiabatic system we call it a universe and since the universe has been defined as an adiabatic system we must have delta su greater than or equal to 0. In fact this is actually second law says that delta s for any adiabatic system must be greater than or equal to 0 this is second law. So since universe is adiabatic applying second law to the universe we get delta su to be greater than or equal to 0 and since in this case the universe is made up of our system and our surroundings delta s of the universe is delta s of the system plus delta s of the surroundings. So this should be greater than or equal to 0 over to you. So for delta s if we write the expression for delta s for the system and delta s for the surrounding right suppose there is a heat transfer from the system to the surrounding. Now let us concentrate over the heat transfer that is taking place and so I want to introduce the concept here the heat generation that entropy generation that you gave the table very nice table you know three columns three rows. Now in this case the first thing you will notice that this heat transfer is internal to this large system called the universe. So since it is internal we do not have to worry about it we have to worry about it worry about the fact that the universe totally is adiabatic and then we write this. Now if we apply this thing delta s of the universe will be any this sigma q by t for the universe plus sp entropy produced of the universe. Now by definition this will be 0 because universe is an adiabatic system. So it simply means that the change in entropy of the universe means the adiabatic system is nothing but the entropy produced in that universe in that adiabatic system this is always true. If all the processes inside the universe are reversible. All the processes inside the universe are reversible then if that universe by definition is adiabatic then the universe will not produce any entropy delta s of the universe will be 0. Thank you. Over to you. Over and out.