 1195 Bansal College, Mandideep Bhopal, over to you. Hello sir, good afternoon sir. This is Rajeev Divedi from Bansal Engineering College, Mandideep. I would like to a question sir. Sir, so far we have discussed energy, heat and work. Heat and work both are path function and energy we have studied, we have concluded it is a point function or state function. While heat and work both are transient form of energy, why? Could you explain sirs? See, it is not proper to say that heat and work are transient forms of energy. Heat and work are interactions, they are energies in transit. Okay, so energy is something which belongs to either system A or system B. Heat and work are energies during the process or during the interaction when they come out of system A and system B. It is like a transaction. For example, the money which you have in your bank is the storage, that is like the equivalent to energy in the bank. The money which you have in your purse is another system, another storage of energy. If you take out some money from the bank and put it in your purse, there is a transaction. Money has come out of the bank, your property in the bank has reduced, your property in your pocket or in your purse has increased. And because there is a transaction of withdrawal of money from the bank, our Q and W are these types of transactions. Okay, you cannot say that system A has so much of heat or so much of work stored in it. What is stored in it? It is simply energy. Similarly, system B has its own storage of energy. When you take out energy from system A and provide it to system B, there is an energy transaction. Thermodynamics tells us that there are two types of transactions, the work type of transaction and the heat type of transaction. So, energies are energies, heat and work are energies in transit, nothing more. 1-1-5-0, Sushila, Godavath Institution, Kolhapur, over to you. Sir, we are already discussed that. Example number 7. So, can you change this medium like air gun replaced by the thermal gun? So, situation is different, differ or not? I don't understand what you mean by a thermal gun. Sir, question number 7, there will be medium is air gun. So, replace a thermal gun. So, situation will differ or not? Yes, it will be different because what you have been told or what you have been asked to do is only the part which is the sort of a following part. I say that in the chamber, you already have air or a gas at a certain pressure, I think PC and the temperature Tc. This is specified. The actual thing in a, not in an air gun, but in an actual thing in an air gun is instead of air there, there is a mixture of the appropriate gun powder or whatever is the fuel, fuel and oxidant and there will be a trigger by which a reaction will take place and energy will be released. That means the chemical energy will get converted into thermal internal energy and that energy will be used to cause the movement of the bullet. The second part would be similar expansion but the first part or precursor to this would be creation in that chamber of a situation which is at a high pressure PC and may be at a high temperature Tc. What we want really is a high pressure, high temperature is incidentally. Even if it is not high temperature, it doesn't matter but because we want the bullet to accelerate, we want PC to be as high as possible. Over to you. Temperature changes from 50 to 600, right. So it is a situation like that expression of a volume. So we are obtained this expression of volume in air gun. So thermal gun also changes that. Yes, some expressions will change because it will not be air, it will be a complicated mixture of gases. So the properties will be different and because the properties will be different and maybe they are functions of temperature, so our PV raise to gamma type of situation we may not be able to get at. It will be a more complicated expression. But yes, the process of determination remains more or less similar. But because of the variation of properties and mixture present, you may have to integrate it out numerically rather than analytically. Is there another question? Good evening sir. Sir, while explaining about the first law of thermodynamics, you have given the example of the brake and the drum system. Right. I think you are hearing sir. And in that brake and drum system you considered two boundaries. One before the outer surface of the brake shoe and one below the circumference of the drum. Right. And another one exactly at the interface of the brake's shoes outer surface and exactly the circumference of the brake drum. Okay sir. And here the area of contact is only partial of the circumference. Then why we have to consider the whole circle as the boundary sir? And another thing here one system is, sir one system is constant that is brake shoe and the drum is rotating. And like this how we are calculating the thermodynamic calculations in these type of situations sir. See this situation was provided to get your thought process going. I am not providing any computational thing. Okay and I just sketched a part of the boundary but actually you should sketch the complete boundary. I just showed you a brake, a shoe liner or a brake but if you consider a rope dynamometer that rope will come you know surround the drum completely. If you are more comfortable with that situation you can use that situation. That situation I just sketched to demonstrate to you that in the in one case where the boundary of the system is totally within the drum you have one type of interaction. If the drum is rotating and the brake is trying to stop it. If the boundary is totally within the drum you can show that it is definitely and completely a work interaction. If the boundary is outside the drum say inside the rope or inside the brake liner one or more how many whatever is the number then you can definitely demonstrate that it is a purely heat interaction. But if you insist on putting the boundary exactly at the interface it will be very difficult for you to say whether it is purely a work interaction or purely a heat interaction. This is just to impress on you that a small relocation of the boundary of a system redefining a system can change the type of interaction completely from the one type to another type. This is the simplest example and I like this example because we have seen this situation in our laboratories IC engine or fluid power laboratories. But there are many such examples. Okay sir and the practical application of this concept is in the railways where we are using the sand a fine sand between that brick shoe, brick shoe and the drum and what is that concept what is the part of that sand in the thermodynamic processor how we can imply that. That is that sand essentially there is used for increasing the friction okay. However the since the sand comes in and some sand gets thrown out you are adding more complication because then it becomes an open thermodynamic system. The brake liner may be a closed thermodynamic system the brake drum will be a closed thermodynamic system but the sand particles or the collection of sand or even the oil film on it if there is any oil or lubricant or whatever material which you have will be an open thermodynamic system and analysis of that will require not just the principles of thermodynamics but that also of fluid mechanics, materials, sand sand whatever you have okay. Thermodynamics is a subset of physics so there are certain types of problems which thermodynamics does not look at or is not capable of looking at. Yesterday somebody mentioned that electrical resistor if it is not cooled and if its temperature rises the resistance value of the resistance may change yes but then how do we cool it that is not a part of thermodynamics that is a part of heat transfer and if it is liquid cooled fluid mechanics is also involved. What we are learning it here are basic principles of thermodynamics so we should accept the fact that there are certain issues which the science of thermodynamics does not look at leaves it to other sciences like fluid mechanics heat transfer combustion and all that. Okay sir last question sir in future I am planning to proceed with the concept of environmental protection in that case how I can go with the thermodynamic concepts to protect the nature and in what field I can do the research. Very general question unless we take this offline and I will not discuss it here okay. Since that was your last question I will say over and out let me go to some other centre. 1006 NIT Tiruchirappadi over to you. This is related to problem number 1.6 the equation of the process TV power gamma is equal to constant that we have to prove for that we have started with Q is equal to delta E plus W and in differential form we are written as DQ is equal to DU plus DW then DQ is written as C into DT is equal to CV DT into DW then it will come in the form of CV sorry C minus CV into DT is equal to TW then in order to prove that the processes PV power k is equal to constant we are unable to process proceed further. Okay I think you are missing out on using the ideal gas equation of state you are right when you write this is 1.6 you write see our differential form would be DQ is DE plus DW. You consider first DE is DU that is the assumption 1 assumption. Second assumption is DW is DW expansion that is another assumption with these 2 assumptions you will get if you divide throughout by the system you will get DQ equals DU plus because of this DW expansion this will become PDV you are right since it is an ideal gas you can write this as CV DT. So here the ideal gas is used and DQ is given as CDT. So CDT is CV DT plus PDV. Okay now all that you have to do is this is one equation the second equation you use is PV equals RT. So since you want a relation between P and V PV raise to K using this replace DT by something into DP and something into DV then collect terms and you will get PV raise to K is constant. From this you can calculate for example PDV plus VDP if you differentiate this will become RDT. So wherever you see DT here here and here replace it by 1 over R into PDV plus VDP and you will get an expression containing only DP and DV and when you integrate it out you have to say that I am integrating because I am assuming the process to be a quasi static process. So that is another assumption which you need to make 1079 Gita Institute of Management and Technology Kurukshetra. I ask you a question that at absolute zero temperature the volume of a substance become zero. It is possible sir? See absolute zero is something which is a thermodynamic one can call singularity after doing the second law we will realize that absolute zero is a temperature which we can only think about we can never reach and our behavior of systems becomes more and more complex as we go to lower and lower temperatures particularly near zero Kelvin and hence in fact material properties become so important although laws of thermodynamics are still applicable but there is no point in discussing anything at absolute zero because absolute zero is as distant as is infinite it turns out to be zero the way we have defined our scales of temperature. It is a ratio scale so somewhere you will have a zero but we could have defined it the other way round and then what is absolute zero could have become infinitely last temperature. So there is nothing special about nothing to be discussed about absolute zero that is a temperature to be only thought about we will not reach it in practice. We will reach micro Kelvin's, Millie Kelvin's maybe even nano Kelvin's but reaching absolute zero is an impossible task over. Absolute zero you mean to say I think it should be or it is you mean to say sir it is approximately zero that is it is tending to approaching to almost zero temperature is it sir. See as you approach lower and lower temperatures as you go to zero Kelvin the state of any substance depends on its inherent properties for example I doubt whether anything remains in the vapor form or the gaseous form everything liquefies helium liquefies hydrogen liquefies so measurement of temperature everything becomes an extremely difficult task by itself and that you know near zero Kelvin or very low temperature cryogenics is applied by itself the basic principles of thermodynamics are still applicable provided there are no quantum effects. But in a basic engineering thermodynamics course it is not within our scope nor is it possible for us to discuss details of the physics of temperature near zero Kelvin. Over and out let me go somewhere 1, 2, 7, 8 SDM Institute of Technology Ujirey Karnataka over to you. Hello sir this is Lavakumar from SDMIT my question is is there any other interaction in the universe other than eat and work. No I will repeat the question and it is an excellent question the question is is there any other interaction in the universe other than heat and work well a very faithful answer to that is we do not know but the way we have defined heat interaction as something other than the work interaction there is an inherent assumption there that any energy interaction other than the work interaction is a heat interaction that assumption seems to work and that assumption seems to work exceptionally well because not only do we have heat interaction being defined and we have number of predictions based on the first law of thermodynamics. The heat interaction also enters into the picture in the detailed formulation of the second law of thermodynamics and many of the predictions created using the second law of thermodynamics either by itself or in combination with first law have turned out to be very reliable predictions hence the basic assumption that the any interaction is either work or heat seems to be a true assumption apart from that but I agree that it is an assumption it is a premise and we believe in it because we have not yet found anything which is neither work nor heat but something else if we discover something else we will naturally we will have to modify our first law and if that third interaction it is say x interaction then just like second law maybe there will be a third law or a fourth law or a fifth law to provide further details of that x and when that happens perhaps similar to entropy we may even have another property associated over to you okay sir that is it thank you over and out 1016 KJ Somya College Mumbai Good evening sir I am Akash from remote center KJ Somya College of Engineering Mumbai I have two questions my first question is basically related with F problem number F 1.8 in this problem the time is given volume also given and pressure is given in such type of cases how can we justify the value of delta E 1.8 is the question we are talking about what is the issue the time is given volume is given volume is 15 liter and pressure also given one bar is in such type of cases how can we justify the value of delta E. Yeah this is essentially a actually this is not a first law problem it should have gone into perhaps the work interaction we only have to determine the work interactions here it is a two work interaction problem one interaction is the charging of the battery the other interaction is the change in volume which is to be modelled as the displacement of an atmosphere. So in F 1.8 let us say that we will consider modelled it like this you have an electrolyte which is may be partly liquid partly gas but our system is the whole system exposed to the atmospheric pressure and there are electrodes and what we are told is the potential across this is 24 volts and the current drawn is 1 ampere and 15 liter of gas is evolved so the delta V will be 15 liters and the atmospheric pressure is not given or is it given one bar P atmosphere is 1.0 bar sketch the system diagram sketch the appropriate process diagram there will be two process diagrams system diagram is already drawn one process diagram would be P against V the constant pressure is P atmospheric and the V will go from some initial volume to some final volume you can assume under quasi-static assumption I have sketched it and let the initial volume be V 1 the final volume V 2 is V 1 plus 15 liters this is one part of the process diagram the other part of the process diagram can be plotted against time so from 0 to 24 time in hours and out here you have the potential E or V and the current I 0 to 24 hours we are given that the E remains fixed at how many volts 24 volts and I remains fixed at 1 ampere is a simple problem if you take a real cell may be initially the voltage is slightly lower as it comes to full charge the slowly the voltage will balance to its correct value initially the current will be large later on as it gets fully charged the current is likely to drop but it is good you brought this to my notice I think we should move this to the work interaction part rather than the first law part because first law is not involved in this so I will write here only work interactions over to you is basically related with the cyclic process formula cyclic integration of dw is equal to cyclic integration of dq so what is the condition of this formula and where this condition is applied this condition is only applied for close part of adiabatic work transfer or other than that no I think I have discussed this earlier somewhere in the morning the first law for a code closed system is always q equals delta E plus w for any process now if process is a cycle that means initial state is the final state hence for such a process delta E is 0 so for a cycle q equals w but since this is for a cycle we should really write q cycle is w cycle so that the condition that this is true for a cycle is also included in the expression itself okay and of course if the cycle is represented by a set of processes over which you can integrate the interactions that means it is an almost or appropriately quasi-static cycle then you can write this as cyclic integral of dq with cyclic integral of dw this circle over the integration sign automatically indicates that it is a cycle cyclic integral over a closed loop in the state space 1088 K.K.Wag Institute Nashik over to you. Temperature is a label given to the isotherms at which the interactions stop but according to the radiation theory every system about 0 K interact means emit radiation so interaction stop or it is continuous. You bring in the idea of radiation remember that the phenomena of radiation like other phenomena of physics is not inconsistent with the principles of thermodynamics. So zeroth law and radiation they live very well with each other when you have two systems which are isothermal they will not interact by radiation even by radiation if the temperatures are equal or if they are isothermal. The radiation tells us that if they are isothermal well radiation says that anybody emits it radiation only talks of emission the basic laws of radiation only talk of emission but we also have emission from the two systems so the net effect will be 0. In fact this idea is used in radiation theory you would have come across it in the your heat transfer course when you study radiation in any in some detail that this is used in what is called the Kirchhoff's principle or the Kirchhoff's laws for radiation. It is using this idea that at same temperature there is thermal equilibrium net heat transfer has to be 0 so that idea is used to derive relations like emissivity equals absorptivity in thermal equilibrium when the whole thing is in thermal equilibrium. So radiation is perfectly consistent with zeroth law or zeroth law is perfectly consistent with radiation over to you. There is one question one more question sir this is regarding just I could not understand why Kelvin scale has started from minus 273 degree Celsius what is the reason? The Kelvin scale has not started from minus 273 or minus 273.15 Kelvin scale has only one fixed point which is defined and that is the triple point of water 273.16 Kelvin based on that it is obvious that if you go to situations which are purportedly lower and lower temperatures that means you find out systems for your thermometer consisting of your gas goes to lower and lower PV values you will go to lower and lower temperatures on the Kelvin scale. And naturally 0 is something which is very attractive but as I have discussed something that 0 is just a mathematical oddity the importance of Kelvin scale will be clear when we derive a thermodynamic Kelvin scale after the second law and then using the properties of ideal gas and the Carnot theorem we prove that the ideal gas scale and the Kelvin the thermodynamic scale are essentially equivalent and that means the thermodynamic temperature and the ideal gas temperature can be considered to be numerically and quantitatively the same and based on that we will show that it is virtually impossible to reach the 0 Kelvin temperature. So that is side tracking but the fact remains that the Kelvin scale does not start from 0 Kelvin or minus 273.16 Kelvin. The Kelvin scale starts and has only one suffix point and that is plus 273.16 Kelvin. Over and out 1, 2, 3, 5 Dr. Mahalingam College Pallachi Tamil Nadu over to you. Senthil Kumar from Mahalingam College of Engineering and Technology can we say the concentration from one area to other area can be deal with chemical thermodynamics can we say that chemical transfer is as an interaction other than the heat and work interaction. The question asked is in chemical thermodynamics when a reaction takes place one component gets replaced by another component products gets replaced by reactants that is not that happens within a system it only changes the form of energy from the chemical energy to some other form of energy it does not lead to any work interaction. A state changes because now the internal components of energy had changed but there is no chemical work interaction involved out there. This is a very simple analogy of this is you consider a your say a projectile as a system a ball thrown up into the air the ball has may be thermal internal energy but it has also a component known as gravitational potential energy and it has another component known as the kinetic energy. If you assume that the thermal internal energy of the ball does not change a solid ball the thermal internal energy would essentially be a function of temperature as it goes up and down all that will happen is the as it goes up the kinetic energy will reduce the gravitational potential energy will go up and as it comes down the gravitational potential energy will reduce the mechanical kinetic energy will go up there will be some interplay between the two components of energy but there will be no work done there is no work done because there is no system to provide the work interaction or there is no system to receive the work interaction. Remember that for work and heat two systems must be involved if only one system is involved whatever happens within it is only reallocation of energy it is not any work or heat interaction over to you. Thank you sir and one more question sir some books refers to the third law of thermodynamics but I am not sure what is called third law and some books text books refers to the third law of thermodynamics can you explain about the third law of thermodynamics or the third law of thermodynamics. I will not say anything about the third law of thermodynamics because we do not have to study anything like the third law of thermodynamics we have three laws they are called zero first and second and that is it in mechanical engineering at least routine traditional mechanical engineering and basic engineering thermodynamics we do not have to study the third law of thermodynamics we do not currently have any applications which require anything other than the three laws of thermodynamics over 1166 Belay institute of technology Durga over to you. In question number 10 we calculated the change in volume change in internal energy and change in enthalpy but the net work done and the expansion work done the expansion work done is coming out to be zero am I right. Expansion work done is coming out to be zero why coming out to as a volume V1 and V2 why zero equal pressure is half temperature is half so V1 equals V2 that does not mean that the expansion work is zero because the process is suspected to be non quasi static that is a very important information and the important hint given here and in this actually I think F1.10 please talk to your coordinator this F1.10 is the illustrative solution of this including the thought processes is uploaded on the coordinators moodle your coordinator should be able to show that to you it is very clear in that how the expansion work is calculated. Expansion work will be zero only if the volume remains unchanged and the process is quasi static throughout at least from that point of view on the PV diagram the process is quasi static for example if it were to be given a rigid container we are not given in that there is a rigid container or anything of that sort here 1089 KLE college Chikodi over to you sir this is a regarding to that modes of work transfers because we have a different types of modes of work transfers is there any standardized methodologies like coefficients constants and any physical significances to relate these modes of this work transfers well there is no standardized method of doing things pertaining to different modes of work transfer that was your question remember work is actually a primitive but we have to treat it with respect because in thermodynamics we have to define the heat interaction and we define the heat interaction as the interaction other than work and hence we have to be perfect in our understanding of what work interaction is that we did by providing a proper operational definition of work in our scheme of things but the details of work interaction it all depends on other branches of physics tomorrow if physics realizes some other type of work interaction all that we will do it check from our point of view whether it is really a work interaction if it is a work interaction we will assimilate it into our sigma w nothing special about it over to you sir under one question it is related to the black holes in the inverse actually where that science that usually the thermodynamics has been stopped any relations to find that interactions in that black holes we are very much interested to know that one. See black holes are astronomical entities thermodynamics is applied to them with appropriate modifications because out there the gravitational fields are very strong perhaps the quantum effects are also very strong our well thermodynamics is applied but with appropriate safeguards and modifications I am not an expert in cosmology forget that of black holes you should talk to a friend in the physics department who is conversant with general relativity and cosmology some somebody who appreciates the work of such people as Chandrasekhar, Naralikar, Hawking and all that and maybe he should be able to explain it to you in a proper way in an understandable way over 1286 U V Patel College of Engineering Kherwa Gujarat over to you. Sir we are having a problem with the 1.3 we are having problem in finding the temperature T2 we are having one way that Q equals m Cp delta T. So sir can you please throw some light on that? See F 1.3 is actually a straight forward problem we have a closed system it executes a constant pressure process initial state is given heat interaction is given we have to determine the change in temperature change in enthalpy change in internal energy and work done assume the air to be an ideal gas with the properties given. We can sketch the system as a gas in a cylinder piston arrangement. So we can say that we have 2 kg of air and it is going to expand so there is going to be a W expansion there is no mention of a stirrer then no mention of an electrical work the only other interaction mentioned is Q and Q is given to be heat addition of 450 kilo joules. So Q is 450 kilo joules initial condition 2.5 bar undergoes a constant pressure process. So this pressure is 2.5 bar the process diagram would look something like this on the PV plane at a constant pressure of 2.5 bar the process will go from some initial state 1 to a final state 2 this will be V1 this will be V2. We are asked to assume air to be an ideal gas it is given that air is ideal gas molecular weight is given CV is given as a constant. So it is ideal gas with constant specific it is given the molecular weight because m is given and CV is given from which you can calculate R and CP as needed. Now the way to proceed is start with first law anyway we have to determine interaction so first law will be involved. So we start with first law Q equals delta E plus W. First step assume delta E is delta U there is no mention of any change in height or any movement assume W is W expansion. So this becomes delta U plus W expansion. Now delta U it is simply we will leave it at delta U just now we need not expand it but now W expansion is integral PDV for which you will have to assume it to be a quasi static process. But since it is given to be a constant temperature process it follows that at least for the integration of PDV for that purpose it can be considered as quasi static from state 1 to state 2. Then because pressure is constant you can write this as P integral dV or P delta V. So the next step is this becomes delta U plus P delta V. But again since P is constant you can write this as delta PV and then you can write this as delta H. So we have ended our first law formulation as Q equals delta H. Q is given so we know delta H and because it is an ideal gas with constant specific heat this becomes M into Cp into T2 minus T1. M is given Cp can be calculated from here R and Cp have been calculated. T1 is known so we can determine T2. Once you get T2 everything else falls in line. We have temperature once you get T2 well Q itself here turns out to be the change in enthalpy. The change in internal energy will be calculated as m Cv delta T and the work done will be calculated as P delta V or if you want we can go back since you have calculated delta U and if you have calculated Q you can calculate it W as Q minus delta U. You do not have to do P delta. That will be a simpler calculation. Over to you. Yes sir my next question is regarding the quasi-static process. Having known the concept of quasi-static at each and every point the state points are uniquely defined. So with this concept can be said that having known one position can be predict the future position. No quasi-static does not mean predictability. Predictability requires that we know what the interactions are. For example you take this this problem itself F1.3. We are given that it is a constant pressure process. It is not given quasi-static but we can even assume it to be quasi-static. But that does not tell us what the final position is. For final position is we need to know the interaction. If you change Q you change the final position. You change Q you change T2. So that way quasi-static does not mean predictability. Quasi-static makes the process analytically simple because direct functional relationships however complicated are available for the process. A non-quasi-static means we do not even know what exactly the value of pressure, temperature, etc. is. So you cannot do any calculations, detailed calculations. We are not in the domain of calculus with a non-quasi-static process. Over. Then sir how does this concept helps in establishing some process details as if for adiabatic PV raise to power gamma is constant or for some polytropic PV raise to power n equals constant. Because until we cannot establish this quasi-static details then how can we establish that thing? See the quasi-static idea comes from mathematics. See in mathematics you can integrate a function f of x dx only when say from a to b only when given any x between a and b a unique value of f of x is defined. This is what we mean by quasi-static at every location every stage during a process the situation of the system the state in which the system exists is defined. Hence we can do manipulations like integration, differentiation on that process or part of that process. Quasi-static only means that the state is defined or is definable all through the process. 1, 2, 6, 0. It is K. N. Simmerger College of Engineering, Ponderpur, Maharashtra. Over to you. So my question is sir in problem number 6 heat transfer during quasi-static process. So here my question is what will be the effect if process is changed from quasi-static to reversible and irreversible? I will not take that question because we have not yet discussed what we mean by a reversible process or an irreversible process. I will leave it at that. We need to have this process quasi-static otherwise we have to remain at the DP-DV level from DP-DV etc. If you want to integrate it and show it that PV raise to K is constant in that case we need to assume it to be quasi-static. Reversibility is not included here. Reversibility will come in when we start studying the second law of thermodynamics. Over. Okay sir. For next problem, problem number 7. If we will take with one case that one bullet is impacting on material and the target material if we will take one case is wood and another case is it is steel. In one case it is happening that indentation will happen and in second case only deflection will happen. So how can we give the explanation with thermodynamics? Well you have an interaction and along with thermodynamics you will in this case you will have to include the conservation of momentum also. The wood being softer the bullet will get embedded in that wood. So there will be one way the momentum transfer will take place but because it is embedded the final state will have some commonality for example the velocity of the bullet and the velocity of the block of wood will be the same. Whereas if it is a steel block you will have more of an elastic interaction and maybe the bullet will ricochet from that steel block. So the final energy forms will be different in the first case with wood there will be hardly any kinetic energy of the bullet whereas in the second case the bullet will still continue to have some kinetic energy. If you consider the appropriate components of kinetic energy and other forms of energy first law can be very properly applied to these two cases over and out. But just before I go out I want to ask you one question. I think this center your center 1260 has been noted as a center in which there is a possibility of 200 candidates. Is that true? Do you have a big hall for 200 people to sit down? 35 registrations are there but actual candidates which are attending these are 30 candidates. Okay. But what is the maximum number of candidates you can handle in principle? We are having capacity 200. Our classroom is having capacity of 200 sitting capacity and this is our computer center. What is the capacity of this? This is not classroom this is computer center. So computer center is having capacity of 66. Okay. Thank you very much over and out. 1126, Vignan Institute of Technology and Science, Nalgonda over to you. I am asking question 1.7. If I take a chamber as a system there is no need of coming the word atmospheric pressure there because they told that the process is an adiabatic. We know that in adiabatic process the work done is T1V1-P2V2 by gamma the work done equal to force into the distance that has moved by the bullet. If the force is considered as mass into acceleration we know this Mb square will come as a work done during an adiabatic process called Mb square and we can find out velocity by using this one. There is no word come across as an ambient pressure there. That is not true. Okay. See the gas in the chamber as it expands through the barrel does work on the bullet. But remember that the bullet is also affected by the atmospheric pressure on the other side. So if you take the bullet as a system there are in principle three interactions. One with the barrel, one with the gas in the chamber which is expanding and one is the atmosphere which is being pushed back. It is given to us that there is no interaction between the bullet and the barrel. It is given to be frictionless. We make an assumption that there is no thermal interaction also. Okay. So the two interactions remain one is the one with the gas which as you said is the equivalent of the gas expanding adiabatically. Maybe finally you get the formula as P1 V1 – P2 V2 by N – 1. But you cannot neglect the other interaction which is the atmospheric interaction. So the volume of the barrel or volume displaced by the volume of the atmosphere displaced by the bullet multiplied by the atmospheric pressure that will be the work done against the atmosphere by the bullet. And that will have to be included otherwise you will over predict to that extent the muscle velocity of the bullet. Over. My system is a chamber sir. Yeah. So if your system is the chamber what happens? The pressure is not directly interacting with the chamber. See the chamber is not interacting with the atmosphere. I agree. The chamber is interacting only with the bullet. But the bullet in term is interacting with the atmosphere. So if we want the final state of the bullet that is its final velocity we will have to consider all interactions of the bullet. The answer which you will get without looking at the atmosphere will perhaps be the correct answer if the gun is fired outside the atmosphere of the earth in the vacuum of space. There is no atmospheric pressure out there or hardly any atmospheric pressure, ambient pressure out there. Over. Can you explain about the basic difference between a quasi-static process and non-quasi-static process? I think I have explained to that. Again I will do this and perhaps this will be or maybe I will take one more centre. If I execute a process from a initial state 1 to a final state 2 and during the process I observe the system and I find that at any stage during the process whichever way it goes I can determine exactly what the state of the system is. Then this is a quasi-static process. I know what is the value of X1, what is the value of Y1 or PVT whichever three dimension four properties all properties are known. So I can plot this route which the process takes in all detail. But if I am unable to do this I will take an extreme case of process in which case I observe the system to be initially at 1 then the process begins but during the process I am unable to make unique measurements of the properties. It could be you know there is no proper word which we can use. The process is so jerky the measurements are so diffused so uncertain if you try to measure that I dare not plot even one point in between. But then after sometime everything settles down and I can confirm that the system has reached state 2. Because at that stage I can determine all the properties of the system and I can create plot the point 2 of the state in which case all that I would say is that initially I was at 1 finally I was at 2. I do not know anything in between. So but I want to represent it somehow so all that I will do is I will draw a line joining 1 and 2 any line but a dotted line. This is just by convention a dotted line indicates that the position of the line and the meaning of the intermediate point is meaningless. Location of the intermediate points is meaningless. I am just joining it as a connecting link between 1 and 2. It is just a symbolic link it does not have any physical significance except that it starts at 1 and ends at 2. Work done equal to integral PDV. Can we apply this equation for non-cause study? In principle yes but since pressure is not known during the process how will you do the integration? That is the issue that we end up with. In principle integral PDV is still applicable but we cannot really apply it we cannot integrate it out. It is like saying you know an extreme case is a pathologically funny mathematical function which say it is defined between 0 and 1. If I tell you that my function f of x is exactly equal to 1 when x is less than or equal to 0.5 and is exactly equal to 0 for x greater than 0.5 I think you can integrate it yes but now I make it more complicated. I say that look my function is 1 if the intermediate x is a rational number and my function is 0 if the intermediate x is irrational number. Now try to integrate it from 0 to 1 a non-cause static process is something like this. At every x you can say well it is defined but try to plot it on the x y you just cannot plot it because you do not know where to plot 1 you do not know where to plot 0. So there is although you can say that well integration limit of the sum etc or even the area under the curve since the curve cannot be perfectly defined here you just cannot do the integration. I will just take one more center and then we will call it quits. Let us go to the capital of India. So I am trying to latch on to 1, 2, 3, 8 Maharaja Agrasen Institute of Technology in Rohini, Delhi. Over to you. So I am there. So my question is the first question that yesterday you talk about the Zodimanti system. So and you give an example that the glass thermometer that is a Zodimanti system but in case of the glass thermometer it is the mercury column that moves and there is an expansion of that. So how would you explain that as a Zodimanti system? The question asked is yesterday I proposed the mercury in glass thermometer as a rudimentary system. The question is since the mercury column expands there must be some work done how can that be a rudimentary system. Remember that the mercury column and whatever is the free space above the mercury whether vacuum or anything else is all inside the system. For work to be done we have to have two systems involved and where are the two systems we are talking about. If you consider the mercury capillary itself or the mercury part itself as a system then we are not perhaps looking at a rudimentary system. But if we are looking at a thermometer as a whole then the capillary, the mercury is something which is inside the system. All that we can observe is the length of the mercury thread in the capillary. When it expands it does not do any work. If it works even if there is a gas and it compresses it that remains inside. I cannot extract that work and try to raise the weight with that and hence there is no work interaction involved or there is no two way work interaction involved and hence that mercury in glass thermometer is a rudimentary system. I hope you understand. Over. Question. Yes, just second question. So there is a question with the muzzle vibalsuit is there that there is a compressed air gun. Yes, 1.7. Yes sir. So in that case you have considered the bullet as the system and then you solved it. Let me explain. This is pertaining to exercise F 1.7. It seems to be very interesting to many of you. No doubt. I am happy about that. The first system we considered is the gas which was initially inside the chamber not the chamber itself. We have to know what is the content of the system. The content of the system was the gas inside the chamber and as the bullet moves the gas expands through the barrel because the bullet acts as a piston. I think that hint is also given here. Assume that the bullet behaves like a leak proof frictionless piston. So the gas or the air in this expands so it does work on the bullet. So if my system 1 is the gas, system 2 is bullet, there is a work interaction between system 1 and system 2. Now that interaction we can compute till the bullet clears the barrel. The second interaction is the second system which I considered in the later part is the bullet itself because we want to determine the muzzle velocity or the final velocity or the exit velocity of the bullet. So that is a property of the bullet since the initial state of the bullet is known at rest. The final state of the bullet is at velocity v0. We have to determine the change of state of the bullet and that means I have to consider the bullet as a system. Now the bullet as a system as I discussed earlier has a number of interactions. One interaction is the work done by the bullet on the gas. This will be a negative number and will be equal in magnitude to the work done by the gas on the bullet. The work done by the gas on the bullet we have already determined in the first part. The second interaction is the work done by the bullet against the atmosphere, the blue bullet on atmosphere and since it is pushing back the atmosphere that work can also be calculated. It is the pressure of the atmosphere multiplied by the volumetric displacement of the atmosphere. The volumetric displacement of the atmosphere is the swept volume by the bullet which is the volume of the barrel. So these two work interactions are involved. There is a possibility of an interaction between the bullet and the barrel but we are told that the bullet is a frictionless bullet. The interaction is frictionless so there is no work interaction and we make a minor assumption that there is no heat interaction either. So in which case we end up with delta E plus W for the bullet is 0. We are neglecting any heat interaction and delta E is essentially the change in kinetic energy of the bullet. The W turns out to be the net interactions between the bullet and the gas on the chamber side and the gas on the atmospheric side. That is how we solve the problem. Over to you. Sir, in that question, so you have talked about the optimum length. So what is the significance of the optimum length? If you obtain an expression for the velocity of the bullet, the question then arises is if I use different lengths of the barrel that means different values of VB. How does different values of volume of the barrel, capital VB? How does the muzzle velocity of the bullet change? It is possible that as you start increasing the value of the barrel volume or length of the barrel, the velocity of the bullet will start increasing because now more work is done on the bullet by the gas in the chamber. It expands over a larger volume but then it is possible that as the pressure in the chamber after expansion becomes very low, at some stage that may even fall below the atmospheric pressure. If that happens, the bullet will get retarded and you will find that an increase in the barrel volume will actually decrease the muzzle velocity of the bullet. That is what I wanted you to explore. Obtain an expression for V0 that is muzzle velocity of the bullet in terms of the chamber pressure, initial chamber pressure which remains fixed, atmospheric pressure which remains fixed, the velocity of the chamber we see which remains fixed but velocity of the barrel which is changing and then you see whether at you know since it is an analytical expression you can differentiate it with respect to VB and see whether at some value of VB the differential becomes 0. Check whether it is a maximum or a minimum. If it is a minimum, it is a maximum that means there is an optimal velocity of the barrel at which the bullet will reach a maximum velocity. Any increase in the length beyond this particular length is likely to reduce the muzzle velocity of the bullet. We want our guns to fire our bullets at the maximum possible velocity so that number one the range is large enough and number two the effect where the bullet reaches is the appropriate effect, good effect so we want a high muzzle velocity that is why this question is asked. Over to you. There is one more question sir whether every isolated system is a Zodimenti system? Well an isolated system is a system which has no interaction so classification of rudimentary system and simple system are the possibilities and isolated system is kept isolated by design. A system inherently is not isolated for example you take a gas in a cylinder piston arrangement I can have a stirrer I can have that piston moving up and down I will have two work modes. If I remove the stirrer I still have one two way work mode. I freeze the piston I will have a zero two way work mode so if I say freezing the piston then only one property of significance will remain then I have constrained it to be a rudimentary system I do not even have to make it isolated to make it a rudimentary system. An isolated system is created or thought about a system does not by itself remain an isolated system. For example we are perhaps getting into some philosophy here but if you have a system which is essentially isolated that means inherently isolated that means it can have no interaction it can have no work interaction no heat interaction and if a system is incapable of having a heat interaction then we cannot apply zeroth law to it and hence that means we cannot determine its temperature. In fact such questions were initially raised when black holes and the science of physics of black holes were was developed. In fact there are quite a few papers discussing whether black hole should have a temperature or not because if black hole has a temperature then the laws of radiation would say that it would start radiating energy out and if it starts radiating energy out and if it gets absorbed somewhere then it is possible that the state will change and if it has a temperature and an interaction and it will have something like an entropy and all those effects will come out. So an isolated system is something which we only think about because an inherently isolated system will be even worse than a rudimentary system we cannot even define its temperature and an isolated system will have if you say it has no interaction then we will not even be able to detect that isolated system. Why do I say that this bottle exists because I have a visual interaction with it if the bottle were really isolated I would not even have a visual interaction with it how do I know that the bottle exists. So if you start really thinking about isolated not just thermodynamically isolated but really physically isolated systems we are getting into dangerous ones. I think it is past 5.45 pm. Thank you and have a nice evening.