 Now with this we are ready to take on some exercises but since it is 10 minutes to lunch time I would spend this time in telling you something about again the procedure for exercises. It will be a good idea for you to take out your exercises sheet. We are now ready to attempt, analyze, digest and then proceed with the solution of exercises in the set called first law 1, F 1.1, 1.2 up to F 1.11 and of course there are some additional exercises. You can read this and you will find that there are you know there are process specifications for example quasi-static process, process at constant pressure, non-quasi-static process, then rigid container indicating no change in volume, then we have a constant pressure process, then we have an insulated system indicating an adiabatic system, a constant pressure process and all that. The process specification you will find in two different ways and one should realize the differences in the two ways of specifying processes, two perhaps even more ways. One way is to say that the process is quasi-static, the very first exercise system executes a quasi-static process. This only says that look the process is such that all through the process you know exactly what the value of properties of the system are. At any time you know what the temperature is at any stage not time, time is not of a sense here. At any stage what is the value of pressure, temperature, volume, mass, energy, any property and not only that all these properties are related to each other through the equations of state. So not only can we draw the path, not only can we determine the properties, the properties are properly related to each other and we can make use of these relations for our computations and derivations. That is one specification. So whether quasi-static or not is important. There are some exercises where it may be specified that the process is quasi-static, there may be some exercises for example exercise I think 1.10, there is a hint given that the process is suspected to be non-quasi-static, 1.11. The process is not necessarily quasi-static. This is warning to you, for senior students I may not give this warning, they are experienced enough or they are expected to be experienced enough to guess that the data is such that the process is likely to be non-quasi-static. So do not make a default assumption that it is quasi-static. The second type of specification is a specification where the process is specified on the state space. For example in equation 1 it says expands at constant pressure against a constant pressure. So specifications like constant pressure, constant volume given by rigid container or even constant temperature, isothermal process. These are processes which we can sketch on the state space. The constant pressure or isobaric process means during the process the pressure is known to be this fixed value that of the initial state also of the final state. Constant volume means you are in a rigid container whatever you do the volume is not going to change. So we can plot on a PV diagram with constant process line as a line parallel to the V axis or perpendicular to the P axis. Constant volume line can be a line parallel to the P axis or perpendicular to the V axis. An isothermal line would be represented along an isotherm. That means it would be a rectangular hyperbola if the system contains an ideal gas and of course it will be some complicated locus if it is a non-ideal gas. But there is a third type of relation which sometimes is directly given, sometimes is indirectly given. For example I do not think any such thing is in the first thing a rigid container. There will be a specification where the interaction is specified. For example adiabatic. Adiabatic implies W only hence Q is 0. There will be some other specification we will say rigid container. This means dV will be 0 and that means W expansion will be 0. Do not jump to the conclusion that dV is 0 hence W is 0. There could be other components of work. If you are conscious while reading and appreciating the problem about the process specifications and constraints which are provided and take care of them properly, you are unlikely to get into any serious trouble. The two major traps which I find my students quite often fall into and these are not traps of my setting. These are the traps which exist in any thermodynamics problem is by default assuming things to be quasi-static and even for any non-quasi-static process merrily using the equations of state throughout the process to not distinguishing between the interaction specifications and the specifications of the constant P, constant V and constant T type. The third and a very dangerous trap is not starting with Q equals delta E plus W, reading something and directly writing a final form so called final form relating Q to something else. A typical thing is you see constant pressure and you write Q equals MCP delta T. You see constant volume and you write Q equals MCV delta T. Forgetting it is obviously quite often that a constant pressure process can be an adiabatic process. A constant volume process can also be an adiabatic process. A specification of this type can always go with a or quite often can go with a specification of the third type. So do not ever make a mistake of beginning to write or writing your first law or a consequence of first law as anything other than Q equals delta E plus W. If you are on the correct track and proceed it properly the consequential derivation relation between Q and something else will come out automatically but do not jump to that conclusion. I think I have kept track of time so we have finished item 7. We are now ready to tackle the first law 1 this set of exercises going from F 1.1 to F 1.11. I recommend that you attempt all of these preferably in the order but not necessarily they are generally from simpler to more complex and this is one of those exercises in which you must be able to chew digest and assimilate each and every part because what you do later significantly depends on what we do here. I will also be open for discussions after some time but I would tell you something about each of these exercises. So have your notebooks ready and I will just do some thinking allowed for you for a few of these. F 1.1 is more or less straight forward. We have two states of a system given 1 and 2. First quasi-static process is executed from 1 to 2. Amount of heat absorbed is given and the expansion is specified. Initial volume, final volume and the pressure maintained constant at a given value. This means for the first process almost everything is specified of interactions except that only the work interaction is specified. So to complete the analysis and to determine delta E you will have to assume that components of work other than expansion work are 0 or negligible. The second part then says that the system is brought back to its initial state by a non-quasi-static process and one of the two interactions is given that it rejects 100 kilojoules of heat. And what is the work done during the second process? Note that it does not ask for expansion work done. It simply says work done. So for the second process again you will have to use appropriately Q equals delta E plus W. Q is given as minus 100 kilojoules. Delta E will be based on that calculated in the first part. W is the answer to be extracted. In F1.2 we have 3 kg of air in a rigid container. Rigid container immediately should alert you to the fact that W expansion will be 0 but not W. W expansion is 0. It changes its initial state from 5 bar 75 degree C to 12 bar while it is stirred. Assume air to be an ideal gas with the C V value given. If air is not to be given as an ideal gas at this stage you will have to assume it to be an ideal gas. We do not know how to solve it otherwise. Otherwise something else will have to be specified. Determine the final temperature, change in internal energy and work done. Whenever change in internal energy is specified read carefully. Here first law only says Q equals delta E plus W. Q is given to be 195 kilojoules. Delta E can be calculated provided W is determined. Now W is determined from the fact that it is a rigid container so W expansion is 0 but the W stirrer is to be determined. So that remains an unknown. Now we are given the initial state 5 bar 75 degree C. We are given also the final state as 12 bar. Another common thing between the initial and final state is that the volume will be the same because it is a rigid container. So you have P1 V1 by T1 equals P2 V2 by T2. V1 equals V2. P1 and P2 are specified. So P2 is to be determined. Notice that 75 degree C is specified. So before using it in the equation of state convert it into Kelvin. And when you convert it into Kelvin by default use 273.15 as the shifting factor. Do not truncate it to 273. Does not generally matter but in a few cases it could be crucial. And then once the final temperature is determined you will be able to use delta U is integral Cv dt. If you assume ideal gas then you will write delta U is integral Cv dt and Cv is specified at a constant. So it is Cv delta T. So delta U is known under the assumption that delta E is equal to delta U. You can determine the work done. Notice in this problem that because there is stirrer work the default jumping to conclusion formula of Q equals m Cv delta T is not applicable. Be conscious of that fact. Now F1.3 is more or less straightforward. We have given mass of air at a given state pressure temperature specified closed system. Air to be assumed as an ideal gas with molecular weight and Cv given. So you can calculate if need be Cpr whatever is needed. Constant pressure process is executed with heat addition as specified. We have to determine final temperature change in enthalpy, change in internal energy and work done. Although it talks of enthalpy do not jump to a conclusion that it is a constant pressure process. So Q equals delta H. Always start with Q equals delta E plus W. Assume delta E equals delta U if appropriate. Expand W as W expansion plus W other. Check whether there is any hint of W other. If not leave it as W expansion. Constant pressure process will simplify those expressions and maybe you will get everything in terms of the final thing in terms of enthalpy. F1.4. Now here remember you have a perfectly insulated system. So the specification is Q is 0 but it also executes a constant pressure process. Now do not get confused. One indication is that of interaction Q is 0. The other indication is the process as executed on the or in the state space. The two can be together. Do not be under the impression that an adiabatic process is not a constant pressure process. You can have an adiabatic constant pressure process. You can have an adiabatic constant volume process and you can have an adiabatic isothermal process. So here you have an adiabatic constant pressure process. Hydrogen not given but you have to assume it as an ideal gas that is one of the assumptions. Initial temperature is given. Initial pressure is given. Final temperature is given because it is a constant pressure process. Final pressure is also known. So you can determine the final volume if need be. Now write the first law as Q equals data E plus W. You will have to assume data E equals data U. So Q equals data E plus W. Now W is in two parts. There is a W expansion plus W stirrer. The expansion work can be written as P data V if you assume that it is a quasi static process and constant pressure is given. So it will become P data V and data U plus P data V that part will become data H because P is constant. But then you will end up with Q equals data H plus W stirrer. Q is 0. So you will have 0 equals data H plus W stirrer and you can proceed from there. Similarly in 1.5 you have a closed system containing air. The type of gas is not mentioned. So we may assume it to be an ideal gas. It is stirred. So there is stirrer work and it expands. So there is expansion work. During the process the temperature of the system is maintained constant at 150 degrees C. So we know the initial temperature. We know the final temperature. We know the initial pressure. We know the final pressure. So I think the two states are properly specified. The stirrer work is given. So now you will have to determine the expansion work and the heat transfer. For heat transfer use the first law Q equals data E plus W and proceed from there. I would like you to look at exercise F 1.6 very carefully because this is a general case of our PV raise to gamma is constant for a certain type of adiabatic process. When you do this completely you will notice that an adiabatic process leading to PV raise to gamma being constant is a very, very restricted case. A large number of conditions have to be satisfied and the purpose of this exercise is to make you conscious of these conditions and to force you to derive these conditions. F 1.7 is essentially a cylinder piston situation that the gas as it expands, expands through the barrel of a gun, an air gun and the piston here is actually the bullet. So you have a number of choices here. You can consider the gas to be the system, the bullet to be its environment. Of course the barrel is also part of the environment and by making certain assumptions you can neglect the interactions with the barrel that is part of the game. Certain assumptions are listed that the bullet behaves like a leak proof frictionless piston, air expands adiabatically and air behaves like an ideal gas with constant specific heats. Since we have to determine the muscle velocity of the bullet, we will have to consider the bullet as another system in another part of the problem or you may consider the air and bullet to be together a system but then you will have to make certain assumptions about the behavior of air and behavior of bullet. Those assumptions will come into picture anyway one way or the other. Now I think I have given enough of loud thinking about the first seven exercises. Do the same thing about exercise 1.89, 10 and 11. Remember that it is not proper to assume each and every process to be quasi-static. In 10 and 11 there are enough hints that it is suspected to be quasi-static not necessarily quasi-static not expected to be quasi-static and such stuff. So take care of those hints and proceed.