 Welcome back after tea and let us see I am now taking questions not only on open systems but on anything that has happened so far. First, I am going to MES Pillai at Panvel, Pillai Panvel over to you. Hello sir, it is regarding question number OS14 in that we have to determine the final temperature of the air in the bottle. So we have tried and we have considered the heat, work, kinetic energy and potential energy zero. So the only thing remaining is change in enthalpy, is our assumption right and the second question is regarding OS17 in that we have to determine the entropy production rate. So how to proceed for the entropy production rate, please explain sir, thank you. I have one question based on OS14, Pillai made an assumption that there is no heat transfer, there is no WS that is right. They also neglected kinetic energy and potential energy of the air as it flowed in that is at the entry to the bottle that also is right but then they said that the initial enthalpy turns out to be the final enthalpy which is not right that means the application is not right. Consider it as a transient process integrated over time, so if this is the bottle, the bottle is rigid that is given that is also insulated it has some volume V naught and it is perfectly evacuated. So the initial state is vacuum, let us say this is our control volume, the control volume is inner surface of the bottle and the opening of the bottle through which air goes in. The process will be shown as initially vacuum and finally contains some thing at P1 T1. So you can say initial state is 0, final state is 1 or let me say the initial state is I because naught is used and final state is F of the system which is open and air goes in at P naught T naught. Now as we proceed I will mention the assumptions and that should be noted. Notice that Q equals 0, Ws equals 0 and there is one inlet and there is no exit. First conservation of mass, this is going to be trivial but let us go through it, conservation of mass is dmcv by dt is m dot i minus m dot e, there is no exit, so this term is gets dropped out it is not that we are putting it to 0 that term just does not exist. Now integrate this over time of filling that means from the time of opening of the bottle up to the time where you close the bottle stopper is replaced when the flow stops. Now when the flow stops remember the outer pressure will have to be equal to the inner pressure we do not know what the temperature would be but when you integrate this, this will give you mass of the final state in the bottle minus mass at the initial state of the bottle equals the mass which has entered the bottle I am writing in it just to not to confuse it with mi but this initial mass is 0 because it is vacuum. So we get the final mass mass in the bottle equal to the mass which has entered that is what I said it is trivial because we know if initially it contains nothing, nothing goes out of it then what goes in would be what is there in the bottle. Now let us apply first law let me write it in reasonably full form d e C v by d t is q dot minus w dot s plus m dot i into h i plus v i square by 2 plus g z i I am not writing those terms minus m dot e in similar h etcetera but this we do not have to bother because nothing goes out. Now this is h i plus v i square by 2 plus g z i now before integrating let us simplify it is a rigid bottle. So all work done would be the flow work which would be included in the enthalpy this work is 0 because it cannot expand contract there is no mention of a stirrer or an electric worker or anything like that it is insulated. So this will also be 0. So the assumptions rigid and insulator are invoked here. Now let me go to another page now if you integrate this you will get e control volume final minus e control volume initial equal to now this is integral of m dot i h i plus whatever v i square by 2 plus g z i d t over the time when you open the bottle and then you close the bottle. Now we do not have to worry about how to do this integral let us assume that at the inlet state the state is dictated by p naught t naught and hence the inlet enthalpy will be the ambient enthalpy of air that is one thing and second thing let us neglect v inlet square by 2 and g z inlet compared to the center of mass of the bottle because this z inlet will not be the absolute z inlet it will be the relative z inlet. On the left hand side the initial energy is 0 it is only the final energy. So what you end up here is the final energy is the final mass which is m f. Now here this will be u f plus other components of energy just the way we have neglected this component of the coming air let us neglect this component of the final state potential energy and kinetic energy. Now on the right hand side if we neglect this and say this remains h naught the right hand side simply becomes the mass which has come in into h naught the mass which has come in is m i into h naught. So the final form of the equation energy equation is this from the conservation of mass we have m f equal to m i that cancel out. So you simply end up with u f equals h naught and all that I will do is write this as the right hand side h is u plus p v p naught v naught. So we end up with this relation u f minus u naught is p naught v naught and now we use the assumption or make the assumption that I do not think it is given here that we will assume that air is an ideal gas with constant c p c v and in that case we will end up with u f minus u naught with c v into t f minus t naught equal to p naught v naught is r t naught and you take t naught on one side you will get t naught into c v plus r which equals c p and finally you will get t f is c p by c v into t naught and this is what we know as gamma. So this will be gamma into t naught this is O s 14 continued I hope this explains it in reasonable detail over to you. Sir question number O s 1 7 in that entropy production rate is to be determined how to proceed for some thank you. Yes the second question is about one of the heat exchanger questions now although we do it manage heat exchangers in heat transfer the thermodynamic way of managing heat exchangers is as follows. Let us say I what I am writing will be good also for O s 18. So this is the heat exchanger type of situation now from a thermodynamic point of view a heat exchanger a typical heat exchanger will have a hot stream going in and coming out another stream known as the cold stream going in and coming out and the two streams do not mix that is the typical transfer type of heat exchanger it could be tube in tube shell and tube compact parallel flow cross flow from thermodynamic point of you let us say there is fluid 1 which goes in and goes out and fluid 2 which also goes in and goes out and let us say this is 1 inlet and this is 1 exit and this is 2 inlet and 2 exit I am not making any assumption that it is parallel flow or counter flow I am just schematically showing an inlet and exit you can you are free to show it anywhere and our main control volume is the full heat exchanger and as in the case of the standard heat exchanger theory we assume that the full heat exchanger is insulated. We also assume that there is no stirrer or anything or extracting of power. So W dot s is also 0 we will assume that a steady state exists now this becomes a 2 stream or 2 inlet 2 exit no mixing type of situation this is absolutely horrible I think I should shift there my G is just whereas the G which I drew is like this showing it like this so with this our first law steady state Q dot minus W dot s is and we will also assume that delta E k delta E p are made negligible for either stream. So this becomes m dot i h i sorry this is steady state so m dot E h E summed over the 2 streams minus m dot i h i summed over the 2 streams since this is 0 and this is 0 this can be written down as by expanding this m dot 1 h 1 E minus h 1 i plus m dot 2 h 2 E minus h 2 i plus m dot 2 h 2 E minus h 2 i plus m dot 2 h 2 E minus h 2 E equal to 0. Now since there are 2 terms summing up to 0 this the heat transfer people will call energy balance or heat balance of the heat exchanger we call it the first law of thermodynamics for the open system to inlet 2 exit no mixing as applied to this heat exchanger. Now remember that this is made up of 2 terms and the sum is 0 so naturally one of these 2 terms will be positive the other term will be negative with the same magnitude there is no choice unless both of them are 0 and if both of them are 0 that means h 1 i h 2 E is h 2 i the state is not changing or the mass flow rates are 0 that means it is not a heat exchanger of interest to us. So let us say that 1 is a 1 term is positive and 1 term is negative the term which is for which h 2 E minus h 2 i is positive will be called the cold side its enthalpy increases as it flows through the heat exchanger this will be called the hot side the enthalpy of the hot side decreases as it goes through the heat exchanger. Now I have just assumed that the side 2 is the cold side and side 1 is the hot side but if I just call them 1 and 2 anything can be hot and anything can be cold. So let me say that this is my first equation which is the overall energy balance heat transfer people will write it as m dot i heat transfer people will write it as m dot i h 1 i minus h 1 E equal to m dot 2 h 2 E minus h 2 i but it is essentially this equation which they are writing. Now you have to determine heat transferred between the streams now for heat transfer between the streams it is necessary for us to consider the heat exchanger split into 2 control volumes 1 is the upper control volume let me call it C v 1 and second one is the lower control volume C v 2 upper and lower is in this figure but the actual requirement is that the control volume 1 and 2 should totally create our full control volume for which we wrote this. There should not be any gaps or overlaps and control volume 1 would contain will not contain anything of stream 2 control volume 2 will not contain anything of stream 1 and since we are assuming a steady state and because of that the thermodynamic properties of the structure of the heat exchanger do not come into operation every part of the non fluid structure and non fluid components of the heat exchanger in steady state would be maintaining their state maintaining their temperature. So, there is no question of any energy transferred to or from them. So, now when you look at these 2 control volumes we can now write an energy balance for any one of them or for each one of them. Let us say that the energy transfer in the form of heat from the hot fluid to the cold fluid is q dot of that heat exchanger and now I apply first law to say C v 1. So, first law C v 1 will be the heat out flowing if you look at the control volume the out flowing heat is q dot h x what goes in and out is m dot 1 inlet enthalpy is h 1 i outlet enthalpy is h 1 e. So, you end up with heat absorbed is minus q dot heat exchanger equals m dot 1 into h 1 e minus h 1 i and if you apply this to control volume 2 you will get heat absorbs q dot h x I am not writing it here, but you will get if you want you write plus q dot x h x is m dot 2 into h 2 e minus h 1 i and either using this equation 2 or equation 3 any one of them can be used to determine q dot h x that is the heat transferred between the stream and this heat transferred between the stream or rate of heat transferred between the stream is what the heat exchanger people called the duty of the heat exchanger. And now the entropy production rate for entropy production rate you consider the full h x as the control volume that is what we saw on the earlier page the full control volume which contains both the streams and this is an adiabatic control volume. So, the second law applied to a steady open system an adiabatic control volume would give you s dot p equals sigma m dot e s e minus sigma m dot i s i. Expanding this will become for us m dot 1 s 1 e minus s 1 i plus m dot 2 s 2 e minus s 2 this is equation 4 unlike equation 1 remember for equation 1 it is a similar equation in terms of enthalpy, but the other side is 0 here the entropy production equation if you look at it the right hand side of 4 is very similar to the left hand side of 1 enthalpy is replaced by entropy that is about it, but the other side is not 0 it is s dot p and we must have s dot p greater than or equal to 0. Otherwise the heat exchanger as specified one will not be able to work in practice it is not a possible situation in either 17 or 18 enough information is available for us to determine the end states may be one of the end states is not given or end enthalpy is not given use our overall equation of energy balance or first law for determining either some m dot 1 or m dot 2 may be missing or one of the 4 enthalpy may be missing that unknown can be determined from this equation. And once you determine on an enthalpy either the pressures must be given or the you can neglect pressure drop if no information is provided make that as an additional assumption and that means you know the inlet and exit states of each stream and calculate the entropy difference between inlet and exit for stream 2 inlet and exit for stream 1 substitute in this equation equation 4 then will be solved for the unknown s dot p I hope that satisfies you over to you. Sir only temperatures are given in this question and we are told to neglect the loss of pressure. So, only with the temperature how to calculate the entropy. In 17 it is given that air is heated and from 30 to 80 degrees C and for the second air it is 150 degrees C. So, assume air to be an ideal gas with constant specific heats in which case you can use terms like both streams are air. Yes in OS 17 both streams are air. So, you can write h 1 e minus h 1 i equals C p into t 1 e minus t 1 i and similarly for h 2 e and h 2 i because if you assume ideal gas then enthalpy does not depend on pressure. So, you do not have to neglect pressure for this, but s 1 e minus s 1 i this will become C p into t 1 e minus t 1 i. So, l n t 1 e by t 1 i there would be a minus r l n p 1 e minus p 1 i, but this requires neglect delta p between 1 i and 1 e and similar equations you can write for the second stream and that way you can now the all the details of the problems including the property relations are clear. Thank you so much sir over and out. Amritha Coimbatore over to you. Good morning sir, I am Ravindran from Amritha. Today morning I ask you one question regarding our test 2 about adiabatic process. The question paper itself state that all four statements are correct. Can you explain sir? Yes, I am taking over this was the one question from test 2. The question was that an adiabatic process and the choices where can it be p equals constant that was I think one choice another choice is can it be v equal to constant? Yes, third choice was can it be p v equals constant and the fourth choice was can it be p v raise to gamma equals constant and the I think the standard answer given was all four are possible. Am I right in my understanding of the question over? All four statements are correct. Now I will demonstrate how that is possible. First thing let us get rid of let me say this is a, this is b, this is c and this is d. I think most of you have answered p v raise to gamma is constant. Now that is because the our first derivation of an adiabatic process is for a quasi static adiabatic process for an ideal gas with constant specific heat in which only p dv work is done. Refer to question f 1, f 1.6 here if c is 0 you will end up with p v raise to gamma is constant. So, d everyone knows is possible, but under what condition you have an ideal gas constant c p c v of course adiabatic process quasi static process otherwise you cannot integrate that equation and not only that only p dv work. No stirrer work or anything like that in under all these condition if you are if you implement all these conditions really you are really executing a adiabatic reversible process and then that is isentropic process for which p v raise to gamma is constant. Now let us consider a for before that let me consider b which is easier to show adiabatic process in which volume is constant consider a rigid insulated box containing some gas some fluid whatever gas vapor any fluid will do it is rigid. So, v equals constant if it is well insulated then it is adiabatic and how do I execute the process adiabatic means work transfer only v equals c prevents pressure p dv type of work. So, I have stirrer work negative, but w stirrer will be shown in this direction this is my system which is adiabatic which is constant volume and that is a proper adiabatic constant volume process. So, I have shown that d is possible and b is possible the next thing we will tackle is a for a I will just modify the earlier one say that put it in a cylinder piston arrangement and you apply a constant force equal to the required pressure p naught into area where a is the area of the piston cross sectional area of the cylinder and assume that the piston is insulated the cylinder is also insulated and this is our system. So, it will remain at p naught because that is the constant on it. So, it is a p equals constant process it is well insulated. So, it is an adiabatic process and it can be executed by putting in a stirrer as in the earlier case of course along with stirrer work there will be a p dv work done also, but let any mode of work take place that does not change its characteristic from adiabatic. So, with this way we have taken care of I think choices three of the four choices the only choice which remains with choice c p v equals constant this looks like an isothermal process of an ideal gas, but then all that I do is the following I take a similar thing, but I modify it now. So, instead of f equals p naught a f is adjustable the in fact that is the only modification otherwise everything else remains the same and here hence because f is adjustable p will be changing and if I allow it to expand there will be w expansion which is greater than 0 I provide the w stirrer which is less than 0 and w expansion would be p dv or f dx I adjust my f remember that if I reduce my f slightly my pressure will reduce as it expands because there has to be a force balance on this and all I need to do in this case is adjust I will simplify this by putting this to be the fluid to be an ideal gas with constant c p c v and if I adjust my w s t plus w expansion to be 0 it is insulated. So, it is adiabatic so q equals 0 there is no other work so my total work is 0 my total q is 0 and hence my change in energy as the process takes place is 0 and since it is stationary under slope quasi static process I can neglect there will be no change in potential and kinetic energy. So, delta e equals 0 will imply delta equals 0 by first law delta u equals 0 because I know there will be no change in kinetic and potential energy and delta u equals 0 along with ideal gas that fluid being an ideal gas with constant specific heats will give me delta t is 0 or change in temperature to be 0. So, by adjusting this f in such a way that w stirrer plus w expansion is 0 in this otherwise adiabatic system I am executing an isothermal process which for an ideal gas would mean p v equals constant I think that should satisfy your curiosity n i t calicut over to you. So, there is a superior shift based on the 10 minus 0.8 how do I evaluate the tan star is equal to 1 is equal to minus 0.8 how do I evaluate the tan star is equal to 1 is equal to 0.8 how do I evaluate from this table this is about question p r 8 where you have 5 data points given at 3 different temperatures and 3 different pressures. So, you have values at pressure of 6.57 and 7.5 bar and temperatures of 490, 500 and 510, but the data is available only at these 5 points the properties are h v and s apart from p and t. So, that means you can determine something with respect to temperature at constant pressure this can be calculated similarly something at constant pressure at varying with pressure at constant temperature can be calculated. Now, when it comes to things like c v you need a constant volume line similarly when you need alpha you need a constant entropy line. So, for alpha convert partial of v into p at constant s to derivatives at constant pressure at constant t or constant t what I recommend is there are 2 chain rules in partial differentiation and you should remember those they come in handy one is the simpler one where the variable t is not 1 of x y and z and the second one which is known as the cyclic rule that comes in handy quite often is delta x partial of x with respect to y at constant z into partial of y with respect to z at constant x into partial of y with respect to z at constant partial of z with respect to x at constant y equals minus 1 remember this minus sign. You may use this plus Maxwell's relations as needed to convert partial of v with respect to p at constant s you should be straight away able to convert into partial of v with respect to p at constant t and I think you get it in terms of this and c p and c v it is possible to get this and it is not a very complicated formula it is some in the three or four steps of a page of algebra calculus you should be able to get this once you do that you need only temperature derivatives and you are able to solve the problem obtain the numerical over to you try it out K. K. Wag Nashik over to you. My question is on OS 9 yes sir the first part of the problem is okay but only the second part if we define the ideal pump as the one which does the pumping isothermally what is the efficiency of the pump that I am not getting over to you sir. In OS 9 we have a water pump which for which the inlet state is given exit pressure is given power consumption is given and the mass flow rate volume flow rate is given at inlet condition so essentially indirectly the mass flow rate is given. So if you say this is our pump open control volume it will have one inlet the inlet is at one bar and the inlet is at 25 degree C the exit is as 180 bar and it consumes 75 kilowatt of power so W dot s is minus 75 kilowatt and calculate the m dot m dot is determine the density at one bar 25 degree C multiply that by the volume flow rate which is 12000 liter per liter per hour you will get it in liters per second and then kg per second. If you look at the P T diagram you are going at one bar 25 degree C at one bar the saturation temperature is almost 100 degree C so one bar 25 degree C is the inlet state I. Apply first law assume steady state neglect heat transfer assume it to be a reasonably insulated pump neglect data e k data e p and you will simply end up with if this is the exit state this will be the process and this will be the pressure exit pressure which is given as 180 bar so use first law and you will end up with q dot which is 0 minus W dot s which is given to be 75 kilowatt minus 75 kilowatt so this will be plus 75 kilowatt this will be m dot which we have calculated multiplied by h e minus h i and h i is known because the inlet state is known you will get h e and h e and p e these two parameters now fix the exit state and then you are able to determine the exit state and the second part of the question is the ideal pump is defined as one which does the pumping isotherm if we define that means now consider a case where the exit state is say let me instead of e let me call it f this would be the ideal pump the exit state would now be 180 bar and 25 degree C so assume this to be the exit state determine how much is the power required and the ratio of that ideal power and the actual power would be the efficiency of the pump I hope that explains it over to you my one more question is there on OS 13 sir I want to know some thing about the throttling kilometer in throttling kilometer we are taking enthalpy before throttling is equal to enthalpy after throttling but there is no consideration of the velocity of the steam sir for a throttling kilometer over to you sir when it comes to throttling calorimeter we will have to explain to the students what is actually meant by throttling calorimeter if we have a boiler and the throttling calorimeter to go with it you should show the students that I typically draw a figure like this when there is a flow of wet steam the measurement of pressure say P i or say P naught P naught pressure and temperature which are the two easily measurable parameters are not good enough to fix the state because we know that if P naught and P naught fix the state on the saturation line then it could be anywhere from saturated liquid to dry saturated vapor so in this case we use the simplest thing to use is a throttling calorimeter and for that what you do is you put a sampling probe with some holes in it so that a representative sample is taken then it goes through a throttling valve and then it goes through a large expansion chamber and at the exit of the expansion chamber we measure the temperature and we measure the pressure which will be the ambient pressure because this is usually exhausted to the ambient but for this it is expected that P naught is reasonably higher than the ambient to be able to give us some flow if P naught is near the ambient or is at ambient pressure then we will have to use a vacuum pump here to get a low enough P naught for the flow. Our control volume here is this throttling calorimeter including the valve up to the exit condition so the inlet condition is what is sampled a small sample is taken out so m dot is small this is the exit condition where P exit and T exit is measured and the whole thing is well insulated it is a rigid structure so no work other than flow work which would be taken care of by enthalpy so our W dot S is 0 and Q dot S is also made negligibly small. Now it is a small contraption so delta E P is essentially 0 and the m dot is small and here we provide a large area and hence we can well assume that delta E k is also pretty small here the density will be larger so even if there is a smaller area does not matter but here the density will be smaller so we provide a large area to keep our velocities within check and with this if you assume steady state then our first law for open system reduces to simply h i equals h e that is what you write down I have not written but you write down the first law fully second for steady state and put Q dot equals 0 W dot S equal to 0 neglect delta E k delta E P and you will be able to reduce that to this form and then you might as well show them the sketch of our h s diagram you show this is one excellent opportunity to make direct use of the but at least for visual purposes of the Mollier diagram h s and this is the g line and if let us say this is the ambient pressure P e and this is the pressure at which the sample is taken and let us say we have wet steam and with throttling you have do not draw it as a continuous line you say this is the inlet state the exit state is at the same enthalpy showing it a continuous line will give an impression that it is a quasi-static process which it may not be because as it comes out of the throttling there will be a lot of flashing a lot of churning that large chamber is needed to quieten it down so that we can make a proper measurement of temperature and pressure and we know pressure at exit we measure now the temperature at exit and E is hopefully in the superheated zone and hence P e and T e give us h e and from our first law we get h i and then P i and h i give us the dryness fraction at the inlet state at this stage it is necessary or it will be nice to point out if no student ask you that if the at this pressure P i which is higher than ambient if the dryness fraction is pretty low then it is possible that even after throttling it will remain wet steam and in that case we will not be able to determine the dryness fraction at the inlet and the answer is yes there will limit of dryness fraction or moisture that a throttling calorimeter exposed to atmosphere can be measured that depends on the inlet pressure and if there is too much of a moisture if your throttling calorimeter is not useful some other tricks will have to be used for measuring moisture over to you thank you very much sir over and out I am talking a lot with Hyderabad over to you Hyderabad Dr. Brahmara any questions from your side over to you yes sir will you please explain about Joel Thompson experiment and inversion curve sir over to you your question pertains to exercise PR 9 this is just an analysis but the Joel Kelvin experiment or Joel Thompson experiment is the experiment which is known as the throttling experiment or in basic books it is known as the porous plug experiment. Joel wanted to characterize all sorts of gases and he noticed by doing such free expansion and porous plug experiments that the internal energy of an ideal gas and also the enthalpy of an ideal gas is independent of pressure. So and since the internal energy is depending only on temperature if you reduce the pressure in a throttling process as we saw in the throttling calorimeter it does not reduce the temperature then they work with various gases in their state space and they notice that real gases do not have this Joel Thompson coefficient mu or Joel Kelvin coefficient mu as given in PR 9 to be 0 once you derive this equation and if you substitute the ideal gas equation within the square brackets on the right hand side you will find that it turns out to be 0 and but they experimented with various gases and found out that for real gases there is a large region of state space where the Joel Thompson coefficient is negative or nearly 0 or slightly negative but there is a small zone where the Joel Thompson coefficient or Joel Kelvin coefficient is positive that zone where the Joel Kelvin coefficient is positive is of useful is of use because the carbon mu is defined as partial of temperature with respect to pressure at constant h. So if you have a gas at higher pressure and you throttle it through a porous plug or a throttle valve or a capillary that means execute a change of state at constant h reducing the pressure the temperature also reduces. So if this is greater than 0 this implies that h i equals h e that means throttling process reduction in p leads to reduction in t and this can be used for cooling things and in that part of the in the state space whichever you write pressure temperature you will find one zone in which mu is greater than 0 and another zone where mu is less than 0 the boundary line is for some reason known as the inversion curve or inversion line that is all about inversion line and the Joel Thompson Joel can be known Joel Thompson coefficient that we need to know over to you. Sir in question number OS4, Senegal air compressor discharge is given and a specific volume at inlet and R e1 so at inlet mass flow rate and outlet mass flow rate become different over to you sir. This is a problem in which only the flow work and change in velocity and mass flow rates are to be computed if you get inlet and exit mass flow rates different that only means that it is not a steady state but flow rate will be flow work will be calculated as m dot e p e v e minus m dot i p i v i if m dot i and m dot e are equal substitute them to be equal if m dot i and m dot e are different substitute them to be different as the actual value. This is a exercise in which this is a minor exercise in which you do not have to apply the first law. This is only calculation of mass flow rate as rho a v change in velocity once you calculate the velocity and the flow work given pressure and specific volume that is it over to you K. K. Wag Nashik over to you. A well insulated cylinder has perfect vacuum in one half and an ideal gas at pressure p 1 in other half. If the partition is broken and the gas occupy the entire volume then how the pressure p 2 is equal to p 1 by 2. It is regarding the test 2, test 2. It is online test question is there sir. Yeah, yeah, yeah. I am taking over. I think this exactly is C exercise SL 0.5. I think that question was exactly this. I will show you how to proceed so that you are clear about it. You have an insulated chamber of volume 2 v 0 divided by a thin rigid partition into two parts of volume v 0 each. Initially one chamber contains an ideal gas at constant pressure p naught and temperature t naught. The other chamber is evacuated. The partition is suddenly removed and show that when thermal equilibrium is reestablished the temperature is t naught. Determine the change in entropy etcetera. Here we are not talking about pressure but just one step to pressure. So, let us sketch the situation. This is a insulated chamber of volume 2 v 0. Initially it contains an ideal gas in the one half which is at v naught. Another half at v naught is evacuated to begin with and this is let us say it is at p naught and t naught. This partition is broken and the final state would be I will not show the insulation again but you sketch it. The final thing is 2 v naught and some p 1 and some t 1. Now, first thing it is a closed system. So, delta e is q minus w but it is well insulated. So, q is 0. Our system boundary is the inside of the chamber. It is rigid nothing goes across it. So, w is also 0. So, that gives us delta e to be 0. Since there is no mention of any movement of this delta u will be equal to delta e. The other components will not change at all. So, this will be 0. Since it is an ideal gas delta u equals 0 implies delta t equals 0. That means the final temperature t 1 will be the initial temperature t naught. Since the final volume v 1 is 2 v naught, your final pressure p 1 will be p naught by 2. Now, since you know the initial state fully t naught v naught p naught and you know the final state fully, you can determine the change in entropy over to u. Sir, one more question. This is regarding again online test. The triple point of water is chosen to be 273.16 degree Kelvin because so, your answer is it is as per your convenience. So, how it is sir? Over to you. The question is t triple point of water is defined to be 273.16 Kelvin. This is defined and why is it defined? The answer was our convenience. Now, what is our convenience? Our convenience means remember that humans are very reluctant to change. So, if we are forced to change from one scheme to another scheme, we will accept it with reluctance and the acceptance of the new scheme will be easier. If the change from as you change over from the old scheme to the new scheme, you have to do something very simple. Right? Numerically, you must be able to do a simple thing like just divide by 2, multiply by 10 or may be just add a number and also some things which we are happy with or we are comfortable with should remain. So, what I was saying was if you give some arbitrary values say x to the Kelvin temperature of triple point of water, it will turn out that t Celsius is a into t k plus b where a and b are some complicate not so complicated, but some functions of this triple point of water. Now, they selected x in such a way that select x such that a is 1 that means you do not have to multiply anything you also you just have to add something and when x is selected such that a is 1, the value of b turns out to be 273.16. Approximately there was something here and then they said that look if we have to define something we must be precise at that time it was very well established that we can do measurements up to 0.01 degrees of temperature different very precisely. So, it was decided that rather than do this let us define this to be 273.16 k exactly. So, that is how the current definition of Kelvin scale came about the current Kelvin scale definition now means last few decades is that t triple point of water is 273.16 k exactly. Now, when you do that remember on the Kelvin scale when you define it to be 273.16 Kelvin exactly then you are not getting an ice point of 0 degrees C and a steam point of 100 degrees C or correspondingly on the Kelvin scale. So, because of that to have not to have two definitions now the C scale is different defined as the other way round C scale is now defined in terms of Kelvin scale. So, temperature on the C scale is now defined as temperature on the Kelvin scale minus 273.16 k this is definition and the consequence of this definition is now ice point is T ice point is now the 0 degrees C which is equal to 273.15 k will say almost, but not exactly. Similarly, T steam point which is 100 degree C is 373.15 k again almost, but not exactly the difference is in may be fourth or fifth place of decimal and we do not have to worry about it even physicists do not worry about it may be only some standard labs may be looking at it if at all, but because of this definition on the Celsius scale there is only one temperature which is exact and that is T triple point of water now turns out to be 0.01 degrees C exactly we can say it is defined to be 0.0. This is 273 this is sorry correct this is 15 because this definition of triple point is 16 everywhere else it is 15 I suppose that should explain the go to Google and search for one Kelvin temperature and search for second the international practical temperature scale. These definitions of scales are ok, but finally for in the laboratories and industry you require a practical definition that means the how do we measure. So, there are even standard methods of measurement of temperatures in various ranges that is specified in what is known as the IPTS international practical temperature scale there was one in 68 may be there is one in 79 or 82 I am not sure, but IPTS 68 is reasonably well established just look it up and you should be able to get a lot of information about thermometer over to you. Thank you very much sir, sir last question from us this is again on the online topic, online test a gas in a rigid container is stirred slowly it is completely insulated. So, how d u is equal to T d s, if you look up our definition of d s d s is defined as d q by T for any reversible process. Now, we also showed that this equals d u plus p d v by T this is definition. This is the application of definition to any fluid that means simple compressible system. So, a to a gas also this is applicable. Now, you have gas plus rigid vessel now rigid vessel means d v is 0. So, that means d s is d u plus p d v by T. So, d u plus 0 by T and which implies d u equals T d s and over and out. Thank you and that is it for today.