 KK Wag, over to you. V and P0 at the reference state, in the morning we have seen the reference state, V0 T0. I want the values and you have said that triple point is having the value 0 for a fluid that is a liquid, but what about gas, at triple point UF is having 0, but UG is having some value, please elaborate on that, thank you. See for any fluid reference state, reference state 0 for U0 and S0, now this has to be decided as either some pressure and temperature or some volume and temperature or something like that. For water, we decide this as saturated liquid at P triple point, T triple point. This is saturated liquid is needed, because at triple point you will have three phases together, F liquid, G vapor and before that S solid. So, at triple point the three phases will have three specific volumes, three energies everything will be three, VS, VF, VG and these are three states in equilibrium with each other. All three states have the same pressure and temperature. So, there will be a US, UF, UG and similarly there will be an SS, SF, SG. Out of these we select this state as the defining reference state. So, that is why UF at the triple point and SF at the triple point are defined to be 0, just as reference values. Actually, because we generally work with the liquid phase and the vapor phase at temperature and pressures higher than the triple point, 0 is ok. We generally end up with positive values, but you go to your steam tables and you will notice that on page 19 you have a table 6, which is the ice and water vapor temperature base. It starts with 0.01 degrees C and goes up to minus 40 degrees C. Nothing special about minus 40, it just stops there and you will notice here that instead of S, they have used the thing I. You see the two values there, the values for vapor and values for ice and the values for ice are all negative, but since we generally do not usually work with ice, these values are ok. Otherwise, if you really work at these low temperatures, you may have used just for convenience a still lower value or define the value of UF instead of reference UF instead of 0 to may be 1000 just for the heck of it. In which case most of these values or may be all these values would have been positive over to you. I have lost your video again. Hello, thank you sir. One more question regarding the properties of fluid, sir. That question is how water is a superheated steam in atmospheric air? Water as a superheated steam in atmospheric air, this question has been discussed in Moodle. The simple idea is this, unless our air is saturated and has drops of dew or something in it, the vapor part is dry air and water vapor in equilibrium. The concentration of water vapor is so small that the partial pressure is of the order of 0.0. A very small fraction of the atmosphere, something like point how much not even about of the order of 1 percent. So, about less than 1 percent may be of the order of 0.1 percent. So, you take atmospheric pressure of 1 bar, this comes to something like point less than 0.01 bar, it is 0.005 bar of that order may be even less. So, the state of vapor out there is temperature of say 30 degree or 40 degree and pressure of a small fraction, very small fraction of a bar and if you look up the steam table that turns out to be superheated steam. It will be saturated dry, you can say it is dry saturated vapor only when the relative humidity is 100 percent. If the relative humidity is less than 100 percent, it will be superheated steam over to you. My question is this previous question that UF and VF, SF all are assumed as a reference state as you have said, but what about UG? I want to know at triple point how they have calculated UG as a value 2375 kilo joule per kilogram. Notice that out of these only two are defined as reference VF and SF, VF is measured out and so is UG measured out. Actually UG is not directly measured, you first measure the latent heat HFG at that pressure and from and then separately you measure VG and from HFG and VG and the pressure being known you determine the UG. Same thing is true of HG over to you. Thank you sir, over and out. Thank you very much. Sivanthrao Samant College, Nagpur over to you. Good afternoon sir. My question is how entropy concept is useful in our day to day life or and one more sir, what are the practical applications of entropy? Over to you sir. Thank you. In our daily life entropy concept need not be involved at all, need not be invoked at all. In our daily life we have a feel for what is happening and what is not happening, that natural feel all of us have. Entropy is only a way of technically formulating it. So that is about day to day life. We do not have to people go through daily life, do excellent work, become prime ministers of our country. I do not think Dr. Manmohan Singh has to worry about entropy at all. He may not even have come across the word entropy in his whole life. There is just no need. Now in the technical word entropy is very useful because that allow us to design and analyze machines like turbines, compressors, nozzles, aircraft propellers, rocket nozzles and all sorts of things. And tomorrow morning when you see open systems actually you will find that we do not have that much of a feel for open systems that we generally have for closed systems. So in open systems the second law and the entropy will really be very active and there is a major parameter for turbines, compressors, etcetera which is isentropic efficiency, not a cycle efficiency, it is a process efficiency. And that will be defined tomorrow morning when we consider the first law and second law for open thermodynamic systems. And at that time you will start appreciating what entropy is, over to you, over and out sir. NIT Calicut over to you. Second law of thermodynamics, the tutorial 4.9, sir how do we calculate the change in the entropy of the universe? 4.9, page number 9, page number 8, how do we calculate the change of entropy of the universe? Tutorial page number 8, question number 4.9, over to you sir, thank you. Your question is about a cell 9, the last question on page 8, 1 kg of an ideal gas at 12 bar 500 k is 1 mixed with 1 kg of the same gas at 5 bar 300 k. The mixing takes place at constant volume, during the process the system rejects some heat to the environment which remains constant at 300 k. Gas properties are specified, determine final state temperature, volume pressure and change in entropy of the universe. Let us formulate the problem, you have a constant volume chamber initially in 2 parts, let us consider 1 has 1 kg at 12 bar 500 k and 1 has 1 kg of the same gas at 5 bar 300 k. So, let us I do not know what exactly the relative volume will be, here the pressure is 2 and half almost 2 and half times higher, but the temperature is slightly less than 2 times higher. So, let us say that it is not worry about sketching it to scale, let us say this is this is part A, this is part B and this is 1 kg, this is also 1 kg. And let us say the initial state here is P 1 T 1 and because it is P 1 T 1, you can determine what is V 1, this is P 2 T 2 and you can determine what is V 2, remember the relative molecular mass that is molecular weight of 32 is given. So, the R for this is the universal gas constant divided by 32, this is in kilo joule per kilogram Kelvin, 32 will be in sorry this is k mole Kelvin, whereas this is kg per k mole and hence you will get R in terms of kilo joule per kg Kelvin. And using that R you can determine because P 1 V 1 is m a R T 1 and P 2 V 2 is m b R T 1 and P 2 V 2 is m b R T 1. So, you can determine V 1 and V 2, you can show the process as follows, if you take P V diagram there will be some initial state 1 of 1 part, some final state initial state 2 of some other part, but finally they will have to come to some uniform pressure, some uniform temperature and some uniform volume. Let us say that state is 3 and since it is going to be a totally non-quasi-aesthetic process, we can just show them by 2 dotted lines, need not be straight, need not be crooked, you will just sketch the way you feel like. You cannot do anything better than this in sketching the process diagram, except perhaps that you can either select P and small v that is specific volume or P and capital V and you can maybe make it more to scale on any type of diagram. So, now here process is W equals 0, why? First is rigid container, constant volume that suppresses any expansion work. Two, we will assume that W other is 0 because there is no mention of a stirrer or an electric heating, but there is some amount of heat absorption and it rejects it to the environment which is at 300 k and that means this Q is minus 150 kilo joules, this Q has been specified. Now, the first law is going to be delta E plus first law will be delta E is Q minus W, W anyway is 0 and we will assume it to be stationary so we will assume delta E to be delta U, this is assumption again. So, delta U is Q, Q is given and delta U will be written down in 2 parts, delta U will be delta U of part A plus delta U of part B and since we are asked to it is not asked, but we will assume air to be an ideal gas with constant C P C V and we have R, we have not air the gas we have R, we have gamma. So, from R and gamma we can determine C P and C V because R is C P minus C V and gamma is C P divided by C V and so we know C P, we know C V in particular we will need C V this is and because it is ideal gas delta U will be M into C V into delta T. So, this will be mass of A C V which is common T 3 minus T 1 that is the temperature change for A plus mass of B C V of B which is just C V T 3 minus T 2. So, this will be equal to Q because delta U is equal to Q that is what we saw in the as the application of first law Q is specified. So, this is specified the only unknown here remains T 3 which now we can obtain. The moment you obtain T 3 you know the final the temperature you know the final volume V 3 is going to be volume of 1 plus volume of 2 M 3 is going to be mass of A plus mass of B. So, since you know T since you know V and since you know M this gives you P 3 the final pressure and also the final pressure and final specific volume and final temperature delta S of the system is going to be delta S of part A plus delta S of part B. We write this as two parts here as well as here, but that is only for convenience in the final part whatever is in state 3 is considered to be essentially the same. So, you can just imagine a partition in between 1 kg on this side and 1 kg on that side that is just for modeling purposes. So, delta S A plus delta S B and since we know the initial and final state delta S A can be calculated delta S B can be calculated. Now, for the reservoir you can determine delta S reservoir S Q reservoir which will be negative of that 150 it will absorb plus 150 kilojoule divided by temperature of the reservoir and then change in entropy of the universe will be change in entropy of the system plus change in entropy of the reservoir over to you. Thank you very much. Sir delta S A is calculated using M C V L N P F by P 1 plus M R L N V F by V 1 am I right? Delta S the question is delta S for an ideal gas consider let us say P V and the initial state 1 final state 2. So, delta S a simple way is to calculate like this. This is V 1, this is V 2, this is P 1, this is P 2 naturally this will be T 1. This will be T 2 and at this point where the pressure is P 1, but the volume is V 2. Let us say the temperature is T 3. So, delta S 1 2 is S 2 minus S 1. Let us take specific entropy which is S 2 minus S 3 plus S 3 minus S 1 and S 2 minus S 2 minus S 2 minus S 3 will be calculated as because it is a constant volume process it will be C V L N T 2 by T 3 and since this is a constant pressure process this will be C P L N T 3 by T 1 and because this is a constant T 2 to T 3 sorry 2 to 3 or 3 to 2 is a constant volume process temperature ratio will be proportional to pressure ratio. So, this will be C V L N P 2 by P 1 because P 3 is P 1 plus C P L N T 3 by T 1 will be V 3 by V 1 which is V 2 by V 1. This is one method by considering a constant pressure and constant volume process either derive it using a constant temperature and constant volume process or constant temperature and a constant pressure process and you will get slightly different formulae. Any one of those three is good enough you can derive it according to your convenience over to you. Thank you very much. PSG Koimtour over to you. Sir, I am from PSG College of Technology. My name is Babu. So, my question is so what is the third law of thermodynamics sir can I explain the third law of thermodynamics and next one is so what is the absolute enthalpy and the absolute entropy what is the absolute temperature. So, these things I need the clarification. So, other one question is so we have the lot of basic assumptions in thermodynamics we have a lot of basic assumptions. Then how will correlate the real systems this is a question over to you. Your first question is what is the third law of thermodynamics in mechanical engineering thermodynamics we do not have to discuss or even talk about the third law of thermodynamics. Third law of thermodynamics is something which is used by physicists and material scientists who work with solid state physics and crystals. They define third law or they state third law as the entropy of a pure crystal perfect crystal. At 0 Kelvin it is 0 that way they are just defining a convenient reference point for entropy. We do not have to worry anything worry about anything like that. So, that is why in this course we are not talking about the third law of thermodynamics at all. The second part was what is absolute enthalpy absolute entropy and absolute temperature there is no first let us come to temperature. There is no such thing as absolute temperature there is only a thermodynamic scale of temperature and the ideal gas scale of temperature and earlier than that there will be many, many there were many, many empirical scales of temperature like Celsius, Fahrenheit, Humor. Anyone of us could define a scale of temperature even now we can define any scale of temperature that we feel like, but others will not accept it unless it there has to be shown we have we show some convenience in using it and directly link it to the Kelvin scale because it is established that Kelvin scale is a as a thermodynamic basis. So, unless everything is anything any other scale is directly related to Kelvin scale there is no point in defining a scale of temperature. So, in the in a crude way you can say the Kelvin scale is an absolute scale of temperature, but there is no nothing absolute about it is only one of many thermodynamic temperature scales, but it is a standardized thermodynamic temperature scale. So, everybody tabulates data using Kelvin scale. Now there is no such thing as an absolute value of enthalpy internal energy and entropy. So, I do not know where you came across this term is absolute enthalpy and absolute entropy etcetera over. Sir, another question in the properties we have intensive property and extensive property for extensive property we are using uppercase letter some exceptional case for mass we are using small lower case. And if you take intensive property we will be using a lower case letter and some exceptional case say for example, temperature we are using capital sorry uppercase. Is there any rule for intensive property and extensive property should be denoted by a lower case letter? Go to user. There is no such rule about extensive and intensive properties. If you look at the extensive properties like mass I can use capital or small m it does not matter, but usually the h u s etcetera are written capital, but I can I know at least one text book in which small letters is extensive and capital letters is extensive sorry small letter is extensive and capital letter is intensive. Now among intensive properties essentially we are looking at only two intensive properties temperature and pressure. The other intensive properties h u s and v are made intensive by creating specific properties. These are specific properties, but for a system in equilibrium they will be intensive properties. The mass divided by mass does not give you anything. So, there is no specific property related to mass, but reciprocal of specific volume that is mass per unit volume is density which also is some sort of a specific property and intensive property. Regarding pressure why pressure is lower case I do not know when we live with the upper case. Temperature is usually written in capital T because small t we will soon have to use tomorrow for time. So, we need time and we need temperature both begin with a t. So, for convenience one is given capital letter one is given lower letter and since school books and mechanics books usually use lower case t for time we continue to use lower case t for time and upper case t for temperature that is about it over to you. Thank you sir another question. In a temperature entropy diagram say for example water as a substance if it is in a liquid state between two constant pressure line work input it will be less, but if the same water if it is in a vapor state for the same constant pressure work input it will be more whether it is correct or wrong sir. I do not understand just by looking at a temperature entropy diagram how can you determine the work input what is the process that you are going to consider over to you. Isentropic process. If you are looking at an isentropic process then the work done in any open system process or in an isentropic process will depend on the see in an open system process it will depend on the enthalpy difference and the enthalpy difference depends not only on the temperature difference, but on other aspects also. For example in the liquid part the specific volume is very low whereas in the vapor part specific volume is pretty large. Since in the liquid specific volume is very low a change in a small change in or a specified change in pressure does not lead to any significant enthalpy difference significant addition to enthalpy difference apart from delta u and also delta u also changes, but that large specific volume for vapor leads to reasonably large enthalpy differences of vapor. That is why larger work needs to be put in to compress a vapor, but similarly large work can be expanded extracted from a vapor over. Thank you sir over and out. NIT Calicut over to you. The request from NIT Calicut is the physical function of physical significance of thermal Helmholtz function A Gibbs function G and physical significance of Maxwell's relations. Let me come to the last part I do not think there is any explainable physical significance of Maxwell's relations except that a combination of the first law and second law of thermodynamics imposes certain restrictions on the properties. We saw the most visual depiction of that is that for a reversible cycle the area reversible cycle executed by a simple compressible system let me be more specific. The area under the area of the cycle on the PV diagram and area of the cycle on the TS diagram will be the same and that automatically results because of this property equation property relations into the Maxwell's relations. So, you can say that this first law that represents areas and second law which also provides the meaning to the area a combination of these is extracted as our four Maxwell's relations maybe that is the nearest I can come to significance. This is the significance of the Gibbs function that it is a potential the decrease in which represents the maximum work other than expansion work that can be obtained in an isobaric or isothermal process. And I was about to tell you that a electrochemical process as that as it takes place in the fuel cell is a process which is an isobaric come isothermal process no change in pressure no change in temperature at least under ideal conditions and hence and it does not do even if it does an expansion work we do not extract it it is just exposed to ambient so that goes in moving the ambient fluid around but you extract electricity out of it or we provide electricity to the fuel cell to charge it. So, that is a situation where the change in Gibbs function will decide how much work we can obtain. My question is regarding the entropy I have read that entropy is a molecular disorder in one book. So, can you tell about this molecular disorder whether we should relate entropy to molecular disorder or it is said that entropy of the universe is continuously increasing does it mean that the molecular disorder is increasing continuously over to you sir. This has also been discussed on the Moodle. So, I recommend that all of you keep track of the discussion on Moodle. We have derived entropy and we will be using entropy without any reference to any molecular disorder or without any reference to molecules. Overs is classical thermodynamics for us all material is a each and every piece of material is a continuum. So, we do not have to talk about molecules and we will not talk about molecules. Many physicists and chemists have to or they have tendency to talk about molecular disorder but finally when it comes to actual computation they do not do any disorderly computation they do computation very similar to ours. But they talk about molecular disorder the disorder directly is not measured. It is only an indirect representation of something else which is happening. And second thing is when you talk of a universe remember our thermodynamic universe is a well defined adiabatic system. If you are talking of the universe as our physical universe or our astronomical universe the first situation that we confront is no astronomer no physicist no one knows what are the boundaries of our physical universe. And if we cannot define the boundaries we cannot define the system and if we cannot define the boundaries we do not know whether our system is adiabatic or not. So, we cannot say whether the entropy of the physical universe is increasing or not. So, there is no question of arguing in that direction at all. Over to you. In equation of state for an ideal gas you have not talked about Wanderwald's equation just I want to know something about Wanderwald's equation how it has been. Thank you. See Wanderwald's equation of state is not for an ideal gas Wanderwald's equation of state is a model of a gas which is not ideal. And the importance of Wanderwald's equation of state is that it is perhaps the simplest equation which shows the existence of a critical point. It is only from that point of view that the Wanderwald's equation is important. Another reason for everybody to fall in love with the Wanderwald's equation of state is that it is not very difficult to manipulate mathematically. It is a cubic in volume, but it is linear in pressure and linear in temperature that is why we can do lots of calculations with it. That is the only reason why Wanderwald's gas equation is popular. That P plus A by V squared and V minus B is claimed to represent A by V squared represents molecular attraction and the B represents molecular volume. That is only a model. These are corrections to the ideal gas equation of state which is P into V equals RT to hoping to be able to take care of molecular attraction and molecular volume. It does not do complete justice to it because we know that it is not an exact fit for almost any gas. Gases like hydrogen and helium which have slight deviations or a small deviation from the ideal gas law. For them it can be used, but that can be used only in small ranges not over a very wide range of state space over to you. Thank you sir. Over and out. VNIT is over to you. If the heat transfer due to factors causing irreversibility is given in addition to the direct heat transfer Q then will we be now able to represent such irreversible process by a definite curve on the TS diagram? One thing I do not understand what do you mean by the causes of Q which causes irreversibility. That is something which is not understandable by me. For Q it is very simple. If the donor system, see Q is an interaction, so there should be a donor system and a recipient system. If the donor system is in equilibrium, recipient system is also in equilibrium in its own right. Then the Q can be considered to be a reversible Q only if the temperature difference between them is 0. If the temperature difference between them is not 0 then naturally that heat transfer process at least that dQ across a finite temperature difference is not, does not lead to a reversible process and remember that a reversible process is not really representable on paper except as a line. But a line represents any quasi-static process. So just because we are able to show a locus does not mean it is reversible. For example, you have a stirring and expansion of a fluid say at constant pressure or stirring and heating of a fluid at constant volume. If the stirring is slow enough, you know a very patient stirring then that is a quasi-static process very well represented on that state space PV diagram, TAS diagram whatever you choose. But it is nowhere near reversible. So you have a representation but the process is not reversible. We should not be under the impression that just because a process is nicely shown on the state space by a continuous curve it is likely to be reversible far from it. A reversible process is a very very special process and there is no special representation of that in the state space. Over. Thank you sir, over to you. That brings us to the end of day number 6.