 So, we begin the afternoon session on day 3. This pertains to exercises set F1 that is the first set of exercises known as shown here as first law 1. We have 6 exercises F1 1 to F1 6 plus additional exercises which you should look up later. I will also be taking some questions. So, this is an exercises session plus discussion. So, we can say the session is exercises F1 plus. Pillai Panvel, any questions from you? A number of independent intensive properties required to define the state of a closed system. So, in that sir if we consider a simple system consisting of piston and cylinder arrangement. So, if we take the PV diagram. So, in PV diagram only pressure is the intensive property. So, this is the minimum criteria or what is the. Yes that is a good question, but remember the qualification, the number of independent intensive properties of a closed system. Cloth system means a fixed mass. So, an extensive property like volume can all if you are more convenient with that an extensive property like volume can be divided by mass and converted into the specific volume which will be an intensive property. So, we need not although the statement says independent intensive properties because it is a closed system and it has a fixed mass that is not a very severe restriction and intensive an extensive property like volume say in meter cube can always be converted into an intensive property specific volume which will be meter cube per kg. The volume of the system is fixed because it is a closed system we are talking about. So, we can overcome this problem over to you. But sir many times we do not convert the volume to specific volume we directly consider the PV diagram in case of the closed systems also over to you sir. See that is not an issue when we realize that pressure and specific volume will define the state of a system the diagram can be drawn in terms of P and capital V and not in terms of P and small v. But if you want to draw it in terms of P and small v it is ok. Now it is only with air that we will air or an ideal gas that we have this problem of the total volume that is V extensive volume and the intensive volume or the specific volume small v. When it comes to steam etcetera all our tabulation will be in terms of all intensive properties or specific properties. So, this issue will not arise over to you over and out. Good afternoon I am G D Jauri from Krishna Institute of Engineering Technology Ghaziabad. Sir my question is whenever we are injecting the medicine through the syringe to a patient is it only the work instruction which takes place so can it be an adiabatic or not? Question here was when you are using a syringe to inject a medicine into a patient is the interaction adiabatic or not. I will answer this question first thing is the syringe and the needle if we take that as our thermodynamic system that is an open system because something is going out of it. Second thing I will notice is because the doctor or the nurse or the attendant is pressing the piston a PDV work is being done. The third thing is your question whether it is adiabatic or not. It all depends on the situation if it is quick enough the injection is given quick enough then we can neglect the amount of heat transfer between the medicine through the cylinder walls to the surroundings. But for example, if it is given very slowly sometimes an injection is say 20 cc or even 50 cc and it is administered very slowly not directly, but through may be a flexible tube almost like a drip. In that case it is possible that the person who is holding the syringe as a warm hand the medicine initially may be at a lower temperature then there will be a process of heat transfer and you cannot model the process as adiabatic. NIT Trichy over to you now. Question related to internal energy whether internal energy is the low grade energy or high grade energy and whether internal energy can be transformed into some other useful form or to you sir. See our U that is thermal internal energy it is energy in one form depending on the process depending on the type of system the internal energy is transformed into other forms of energy other forms of energy are transformed into the thermal internal energy or internal energy or thermal energy. Whether it is a high grade energy or low grade energy thermodynamics does not define anything like that it is later on when we come to a combined application of first and second law. We can define something as a high grade of energy transfer and a low grade of energy transfer energy by itself is not high grade or low grade it is only the transfer either by work or heat and that too at various temperatures can be considered to be energy transfer at a high level or energy transfer at a low level. VNIT Nakpur over to you any questions. Sir in case of a mercury in glass thermometer rise of mercury in capillary against or towards gravity can it be considered as an issue over to you sir. Your question was rise and fall of the mercury column in a mercury in glass thermometer against gravity is it an issue. It is not an issue because it is a change which is internal to the boundary. So maybe it is internal energy changes, state changes, temperature changes. So the internal energy changes but there is no interaction involved in that and this brings me let me clarify. See we have a thermodynamic definition of work and that thermodynamic definition of work is not in perfect alignment with other definitions of work. For example a body simply falls down a projectile falls down in the mechanical domain is the gravity doing work on that projectile. Many mechanics text books you will say that the there is work done by the force of gravity or against the force of gravity and because of that the kinetic energy reduces as the projectile goes up. But as the projectile falls from a thermodynamics point of view there is no work done because there is where is the second system which is absorbing the work. Remember that work and heat are interactions. So two systems must be involved. When a projectile is in a free fall all that we will say is the energy remains constant. So initially it had a high potential energy, low kinetic energy later on becomes low kinetic energy sorry high kinetic energy and low potential energy. It is only a redistribution of the various components of energy and this brings me to an exercise think over this. Now consider the following consider a flight of steps a staircase assume all the steps to be rigid and you consider yourself to be a thermodynamic system is walking up the steps. Initially you are here you neatly walk up the steps and you reach here. Have you done any work? What has happened? What are the interactions? That is the question. Think over and comment on MUL. Any more questions from VNIT Nagpur? Over. Sir one more question. Sir for a process to be reversible or quasi static it is necessary that all the states comprising that process must be in thermodynamic equilibrium. My question is how a reversible process other than as a thermal process follows the condition of thermal equilibrium. Basically for a process to be in reversible it has to follow all the three conditions that all the states should be in thermodynamic equilibrium that is mechanical equilibrium, chemical equilibrium and thermal equilibrium. So in other process mechanical equilibrium and chemical equilibrium it is assured that it can be but for thermal equilibrium temperature should be constant and it can be possible only in as a thermal process. What about other process it can how it can be possible? The question is during a quasi static process at every stage the state should be in thermodynamic equilibrium. Our definition of thermodynamic equilibrium is that it is properly defined as a point in the thermodynamic state space. Thermodynamic equilibrium I mentioned yesterday I did not say that thermodynamic equilibrium is a combination of the other equilibrium. Our thermodynamic equilibrium definition is a state which is represented by unique values of property. During a quasi static process at every stage we want unique values of property that does not mean that we want the same set of properties. And it is not necessary that we should have the same temperature throughout a quasi static process. I should also mention that you use the word reversible. So far in this course I have not used the word reversible whenever there was a possibility of work being done from system A to system B and from system B to system A I called it a two way mode of work. Even there I did not use the word reversible because reversible is a very very special word. So very important very special word in thermodynamics we will define it properly when it comes to the second law of thermodynamics. At this stage we are not considering anything to be reversible or otherwise. We have not used the word reversible and till we come to second law we will not be using it. We are using the word quasi static process and may be a two way mode of work and a one way mode of work. Let us come to the exercises. But before we come to exercises let me again emphasize the following and again let me write the first law. And currently again notice that we are studying only closed systems. We will always write first law as either delta E is Q minus W or Q is delta E plus W. Notice that delta E this is to be computed using initial and initial systems. And final states here may be computed using process details. This will be possible for a quasi static process. If the process is not quasi static this may not be possible. And the third thing to be remembered is Q either specified or computed using this equation. Actually we have written it as delta E equals Q minus W. We have written it as delta E equals Q minus W. But the real equation is Q equals delta E plus W and that is the defining equation for Q. So in this Q must be specified as a number or directly somehow or Q will be computed using this equation. I have said that Q for Q this is the defining relation there is no other relation which relates Q to anything else in the thermodynamics. The second thing about first law before we solve problems is delta E is Q minus W. Here Q and W are interactions. They represent energy is equal to Q. In transit so that means for us to compute Q and W two systems must be involved. You cannot have one system and say what is the work done. Whether you say it or not another system must be involved. There is a small book text book on thermodynamics. I forget what that text book is but the I think the name of the author is boxer but I am not so sure. It gives a very nice analogy says that look delta E is what is contained in the system. Q and W are never contained in a system because they are energy in transit. What is contained in the system is only energy. So it is not proper to say the work in a system or the heat in a system or the heat content of a system. These are incorrect words. So one analogy which that author gives is that of a cloud a lake. Now all of us know that rain is water in transit. What the cloud contains are water droplets may be water vapor. What the lake contains is water and rain is only water in transit. In the same way we have two systems when they interact the energy of system A changes energy of system B changes. There are energies in transit which can be of the W type which can be of the Q type. But once either W or Q takes place and crosses the boundary what the system has is only energy. So one should never use the word saying heat content of a system. We can only say heat transfer from a system or heat transfer to a system work done by a system work done work done on a system. And finally before we come to our exercises again first law sign convention sign convention. Remember there is nothing special about this negative sign. Similarly there is nothing special about the hidden positive sign here and the hidden positive sign here. Our sign convention is work done by a system is positive. This we have used in our operational definition of the thermodynamic definition of work. That is one sign convention. The second sign convention is used when we wrote delta E is minus W a diabetic. So we specifically put a negative sign there indicating that when a system does work or does positive amount of work its energy goes down. So the second sign convention is energy increase of a system is positive. And the third sign convention is the one which we used when we wrote Q equals W minus W a diabetic. We were free to use Q equals W a diabetic minus W no harm in it. However we used out of the two choices we use the choice Q equals W minus W a diabetic. So that gave us a sign convention for Q. The heat transfer of a system is absorbed by or transferred a system is positive. This is our sign convention. This is not the only or the unique sign convention. Most engineers mechanical engineers electrical aerospace engineers follow this sign convention. Physicists follow this sign convention. However chemists chemical engineers and material scientists tend to follow a slightly different sign convention where conventions two and three are the same. But instead of work done by a system is positive they take work done on the system is positive. All that happens is they end up with a plus sign here rather than a negative sign. But although algebraically the equation looks different numerically it remains the same because they flip the sign of all work interactions. So, their value of delta E is the same the values they compute for Q also are the same. Now with these preliminaries we are ready to solve some problems in the exercise sheet under first law one F 1.1 to F 1.6. And just like the example we have done exercises in the work transfer all these six exercises are absolutely essential. Today in the remaining part till 5.30 and later in the evening be sure that you absorb and are able to confidently tackle all the six exercises because these show us the typical situations that we come across. But this is not an exhaustive thing here we have the first law of thermodynamics and the systems contain either something innocuous as in exercise F 1.1 or in exercises F 1.2 to 1.6 we only have an ideal gas or something which can be approximated as an ideal gas and that too simply with constant specific heats. Let me illustrate by considering two exercises I will do exercise F 1.1 first. Again remember the same format system diagram process diagram approximate relations assumptions treating numbers etcetera with respect. So, here let us read F 1.1 a system executes a quasi static process from some initial state to some final state absorbing so much of heat expanding from this volume to this volume against a given constant pressure. The system is then brought back to its initial state by a non quasi static process during which so much amount of heat is rejected. What is the work done during the second process? Since there is a mention of expansion it is a good idea to consider or sketch it as a cylinder piston arrangement. So, this is our system process diagram let us say that the process is the first process initial state 1 to final state 2 and it says that it expands against a constant pressure of 1.5 bar it is also given to be a quasi static process and that means the initial pressure would also be 1.5 bar final pressure would also be 1.5 bar and 1.5 bar would be the pressure throughout the process. Let us say this is the initial state this is the final state. So, the initial volume was 2 final volume was 2.25 meter. So, I am justified in sketching it like this initial state 1 final state 2 and there is no mention of any other kind of work. So, in the process 1 to 2 we are given w we will assume it to be w expansion. This is an assumption and this is to be calculated it is also given let me write this is w12 q12 is also given it absorbs 80 kilo joules of heat plus 80 kilo joules. Now the second process 2 to 1 it is brought back to its initial state 2 to 1 by a non quasi static process. So, there is no path specified you can sketch it any way you feel like dotted line joining 2 to 1 you can show it below you can show it above you can show it anywhere only thing we know it it starts from 2 and it ends at 1 and all that is given is it rejects 100 kilo joules of heat. So, q21 is minus 100 kilo joule. What is the work done during the second process we are asked to determine w21 first we will complete the sub problem which is the computation of w12. We have made an assumption that w12 is only the expansion work as mentioned there was no hint of a stirrer or an electric connection. So, we can calculate w12 as integral 1 to 2 p d v and since pressure is constant I can straight away compute this as p into v2 minus v1 p is 1.5 bar v2 minus v1 is 0.25 meter cube and since everything else is in kilo joules let us convert this into kilo joule and bar is we notice 1 bar is 100 kilo Newton per meter square. So, we multiply this by our conversion factor which is 100 kilo Newton per meter square bar. Now, bar cancels meter square cancels with 2 of these we end up with kilo Newton per kilo Newton meter which is kilo joule. So, we end up with 100 into 0.25 25 into 1.5 which is plus 37.5 kilo joule. Now, let us tackle the problem there are two ways of looking at it one way is to say that 1 to 2 and back to 1 is a cycle. So, one way of looking at it is like this 1 to 2 and 2 to 1 is a cycle and for a cycle the first law says q cycle is w cycle. Now, q cycle the cycle is made up of two processes 1 2 and 2 1. So, q cycle is q 1 2 plus q 2 1 and that will be plus 80 kilo joules minus 100 kilo joules. The left hand side is written as q 1 2 plus q 2 1. Similarly, the right hand side will be written as w 1 2 plus w 2 1. If you substitute the numbers q 1 2 was q 1 2 plus q 2 1 equals w 1 2 plus w 2 1 plus w 2 1 plus w 2 1 plus w 2 1 plus w 2 1 plus w 2 2 1 plus q 1 2 was plus 80 kilo joules q 2 1 was minus 100 kilo joules w 1 2 was plus 37.5 kilo joules plus w 2 1. We have this one equation only equation is only unknown is w 2 1 which you can compute out. That is one option of solving this problem. The other way of solving this problem alternate method is like this. We say that consider the process 1 to 2. In this process we have q 1 2 given to us w 1 2 has already been computed. So, we can apply first law for process 1 2 and calculate delta e 1 2 and q 1 2 is 80 kilo joules w 1 2 is 37.5 kilo joules. So, that gives us delta e 1 2. Then we say that for the process 2 1 what will be delta e 2 1? Delta e 2 1 is by definition e 1 minus e 2 energy of the final state minus energy at the initial state which I can write as minus e 2 minus e 1 which is minus delta e 1 2. So, if I have calculated delta e 1 2 now I know delta e 2 1 which is just the same number with the opposite sign. You can calculate how much does this turn out to be 42.5 kilo joule positive number. So, delta e 2 1 will be minus 42.5 kilo joules. Once you know delta e 2 1 you apply the first law to the second process that will be delta e 2 1 which will be equal to q 2 1 minus w 2 1. We have determined just now delta e 2 1 q 2 1 is given to be minus 100 kilo joules. So, now we can compute the only unknown which is w 2 1. I have not given you the numbers, but I think the numbers you should be able to fill up in no time. So, that was f 1.1 a rather simple question just a warm up exercise. Let us take 1.2 3 kg of air rigid container changes its state from 5 bar 75 degree Celsius to 12 bar while it is stirred. Heat absorbed is 195 kilo joule. Assume air to be an ideal gas with this much sieving. Determine the final temperature change in internal energy and work done. So, we have a rigid container. I do not have to show a piston. This is our system. What does it contain? 3 kg rigid container that only means no expansion work. Then heat absorbed is 195 kilo joule. It is stirred. Notice that we show the work done arrows always going out from the system. That is our positive convention. The heat transfer arrows will always be shown into the system. Heat absorbed is 195 kilo joule. We are told to consider air to be an ideal gas and we are also given its specific heat at constant volume 714 kilo joule per kilogram Kelvin. Initial temperature change is 714 kilo joule per kilogram Kelvin. In this state, state 1 5 bar temperature is 75 degree C. Final state P2 is 12 process diagram. Since it is a gas, at least at this stage it is convenient to sketch the process diagram on the PV coordinates. You can use capital V. You can make small V. It does not matter. But one thing is certain, the process is going to restrict itself on a constant volume line. We are asked to treat air as an ideal gas. So, the isotherms are going to be rectangular hyperbolas. Let us say that the initial isotherm 75 degree C is something like this. So, this is the initial state 1. This is 75 degree C isotherm this is 5 bar. The final state is 12 bar final state 2. We do not have to worry about whether the process is quasi static or not. So, let me just show that the process is going to be from 1 to 2. I will sketch a dotted line, but keep it on the same volume because we know V 1 is going to be equal to V 2. It is a rigid container. So, the at least the gross volume in between is not going to change from whatever the initial volume is. We are asked to determine final temperature change in internal energy and work done. So, we are asked to determine T 2, what is delta E and what is work done. And it is very clear that the work done is asked for the expansion component of the work is 0. There is a specific mention of a stirrer. So, the work done will be the stirrer component of work. There are two relations that we have and one condition. One relation is obviously the first law of thermodynamics. The another relation is air is an ideal gas. So, the equation of state of the ideal gas PV equals RT or PV equals MRT is available to us. Air is also an ideal gas. So, by Joule's law, the change in energy will be change in internal energy will be related to temperature difference through CV. Now, notice that CV delta T is going to give us delta U. What is asked is change in internal energy. So, I should write this as delta U. Now, let us write the equations that we have. We want delta E. We know delta E is Q minus W, but we want delta U. So, we will write this as delta U plus delta E other than U equal to Q minus W. We know nothing about delta E other. So, at this stage, we will make an assumption that this is 0 because there is no mention that our system is being raised in a gravitational field or lowered in a gravitational field or getting accelerated. We will say it is a system at rest. We will assume that and that will mean that delta E other will be 0. No change in potential energy, no change in kinetic energy. W is to be determined, Q is given, but delta U is also to be determined. So, we have one equation, two unknowns. But using Joule's law for an ideal gas, we have this relation. CV is constant. So, delta U will turn out to be mCv delta T, which is mCv T2 minus T1. m is given to us 3 kg, CV is specified. T1 is known, but we do not know what T2 is. So, we must have a handle on T2. There we use the equation of state. So, notice up to here it was first law. Now, this is Joule's law or property relation between delta U and delta T. Now, we use equation of state. We have it being an ideal gas, P1 V1 is mR T1, P2 V2 V2 V3. So, this is the equation of state. So, we know that R is mR T2. V1 and V2 are equal growth system. So, the m here and m is the same. R is also the same. So, let me take the ratio and you will end up with T2 by T1 is P2 by P1. And we know this to be 12 bar by 5 bar. So, we know the ratio of T2 by T1. We know T1, but remember and again this is a common mistake, which I think all of you have done sometime or the other, but now perhaps do not do. In the ideal gas equation of state, the temperature must be on the ideal gas scale, on the Kelvin scale. Consequently, when we write this, this will be T2 divided by T1 by T2 divided by T1 in Kelvin scale, which is 75 plus 273.15. And this will give us, the only unknown is T2. It will give us T2 and that will be in Kelvin. Now, once you know T2, let me say this is equation 1. Let me say this is equation 2 and let me say this is equation 3. Equation 3 has already given us T2. The moment you get T2, substitute that in equation 2. We know M, we know Cv. T2 has now been calculated. Substitute that in equation 2, calculate delta u. Now, in equation 1, q is specified. Heat absorbed is 195 kilo joules. Delta u, you have calculated just now. So, w can be calculated. Only one trap is out here, both in Kelvin, whereas out here T1, T2, either both in, that is because this is a temperature difference. And the temperature difference on the Kelvin scale and temperature difference on the Celsius scale, the values remain the same for the same two different temperatures. That is the definition of the Celsius scale. Whereas here, it is a ratio and this ratio comes out of there being used in the equation of state of an ideal gas and this temperature must be Kelvin. Here, we do not have a choice. One thing you should realize in this problem is that it is a constant volume process, but q is not equal to M Cv delta T. So, our school kid idea that heat transfer is mass into specific heat into temperature difference is not valid anymore. Why? I will leave that question to answer. It is obvious from the way we have solved the problem, but you should notice that q for this constant volume process is not equal to M Cv delta T. Now, I will recommend that you try, exercise 1.3, F 1.3 and F 1.4. In that order, notice that both pertain to an ideal gas. Both execute a constant pressure process, but F 1.3 has a heat transfer involved. It is very clearly specified that the heat addition is 450 kilo joule, but F 1.4 it is very clearly said that a perfectly insulated system and a constant pressure process is involved. So, F 1.4 is an illustration of a process which is a diabetic as well as at constant pressure, but come to that later. First attempt 1.3 and then 1.4. At this stage, I am not interested really in the correct numerical answer. I am interested in the way you solve the problem. Sketch the system diagram, sketch the process diagram, complete the process diagram as you progress. For example, in the previous problem F 1.2, we could not complete the process diagram. We knew this was, we knew the state 2 was here at 12 bar. We did not know what the temperature was. So, we could not sketch the isotherm or even if we sketch the isotherm, we could not have put any label on it. So, the process diagram can be properly completed only when T 2 is calculated and T 2 will be calculated only when you solve the problem after step 3 T 2 is completed. So, it is sometimes it is not possible to sketch the process diagram completely to begin with. Take your time as you continue with the problem complete the process diagram. So, try F 1.3 and then F 1.4.