 Now, we will come to exercises on second law. Now, we are in a position to do exercises from the SL and PR property relations. These go from page 8. There are 13 exercises on second law and after that, page 11, page 12 and part of page 13, there are 21 exercises on property relations. Coming to exercises on second law, a few points. In second law, exercise is the questions which we asked. Usually, what is the entropy change of the system? What is the entropy produced during a process? Is this process possible or not? Now, remember that entropy change or delta S is a change in property. So, whatever be the process, this can always be calculated because it depends only on the end states. Set the two states by any hypothetical reversible process. Evaluate dQ by T. This you can do for air and we have for an ideal gas with constant specific heats derived some relations, use them. Using those relations, you can even get relations for sets of isentropic states for an ideal gas with constant specific heats. You will get the famous relation PV raise to gamma is constant just by substitute delta S equals 0 for dS equals 0. You will get PV raise to gamma minus 1 is also constant. And again, you will get a T divided by P raise to gamma minus 1 by gamma that also equal to constant. All these expressions you will get. In case of steam, there is no simple formula. You have no choice but to use the steam tables. Or if you really have a detailed chart available, a large scale multi-page chart, use it. Now, when it comes to question of possibility and impossibility, we have to look at the situation. For example, if it is a 1T cycle, then it cannot be an engine. That means, for a 1T cycle, W net has to be less than 0. If it is a 2T cycle, then naturally we can use, if it is an engine type, you can use eta less than or equal to eta R at the same temperature. Then use the Carnot theorem. For any cycle, we will use the Clausius inequality. If anything of this is violated, then we say the second law is violated. The funny thing happens when we have a given process. Now, for this, you have to check that delta S is greater than or equal to integral dQ by T. The problem arises in sometimes determining T, sometimes determining dQ because we have said that while applying the Clausius inequality, dQ is the small amount of heat absorbed by the system and T is the temperature of the system at the time when this dQ enters the system at the location where the dQ enters the system. Quite often, we have a situation where this dQ is either dQ is not easily determinable or even if it is determinable, we are uncertain about the temperature at the boundary of the system across which this dQ takes place. So, in that case, the possibility or impossibility can be checked by doing the following. What we do is we consider an extended system. For example, if we have our system which has some heat interaction with some neighboring system, executes a process 1 to 2 and we are not sure what this temperature is when we have this dQ interactions. So, in that case, we extend our boundary of the system by including the neighboring system, let me say system S1 surroundings which interacts and provides this dQ. Remember that Q like W is an interaction, so always two systems must be involved. Then, we say that look all that we have to check is whether this extended system during the process which is executed is adiabatic or not. If it is adiabatic, we stop there. If it is not adiabatic, that means one of these is having an interaction with some third system. We include that in our extended system till you reach a stage where the extended system turns out to be an adiabatic. So, we go on extending our system till we reach an extended adiabatic system. This extended adiabatic system in thermodynamics text books is usually called a universe. But remember that this is thermodynamic universe. Why is it called a universe? I do not know, but that seems to be the nomenclature used. And a system is a system which is called a thermodynamic universe or a proper thermodynamic universe will always be an adiabatic system. And our second law says that delta S should be greater than or equal to integral dQ by T. So, that means if our system is adiabatic, dQ will be 0 throughout the process of that system. That means delta S will be greater than or equal to 0. This is for any system and this is for an adiabatic system. And because an adiabatic system which is created by extending our system and including with it appropriate neighboring systems to make it an extended adiabatic system is known as a universe. This we use quite often a symbol of the universe, U here. And this gives us the statement that the entropy of the universe always increases or never decreases. That is okay, but we should not be too ambitious with interpreting the world universe here. This is thermodynamic universe. And thermodynamic universe is nothing but a suitably extended if needed, but adiabatic system. If your system executes an adiabatic process, there is no need to extend it. It is an adiabatic process. So, it by itself forms a thermodynamic universe. So, any process will be executed such that the entropy of that system increases. Now another issue with most of the problems is we can calculate the entropy produced quite often it will be asked what is the entropy change of the universe. And that is essentially the entropy produced in the universe or entropy produced. And if entropy produced is greater than 0, then it is a possible irreversible process. If it is 0, it is a possible, but with a question mark reversible process. Well, thermodynamically it is possible, but we know that in real life it will be virtually impossible to extend it. Less than 0 is an impossible process. And as good teachers and students it is our job to look at the details or look in between the lines and determine if it is impossible process that means what is it that makes it impossible. And if it is an irreversible process it is our duty to try to guess the causes of irreversibility. Now after having said so the recommended exercises are SL1, SL2, SL3, SL4, SL5 actually all of them are important. But I suggest that absolute important are the first six actually I am at a loss to find out what should be dropped. But after that you should definitely do SL10, SL11, the remaining ones you can do later. I will spend some time on SL1 because there we need, there are two ways you can solve it. And in one way you need to determine what is the entropy change of reservoir. So small discussion is needed. Maybe I will show you, see SL1 says that we have a conductor is a system through which heat can flow, connects to reservoirs at temperatures of 1200 k and 500 k. The steady flow of heat from the hot to the cold reservoir is 150 watt. Determine the rate of entropy production by the conductor. So here we have two reservoirs. One is at 1500 k, let us call this T1, another is at, first is at 1200 k and the other one is at 500 k. And we have something called a conductor in between. It could even be free space and the heat transferred by radiation does not matter or it could be a metal block or something. What we know is there is a steady heat flow of 150 watt through the conductor. Now the assumptions here are because it is, let us first look at it from a conductor point of view because we are asked to determine the rate of entropy production from the conductor. So if I just look at the conductor, all that it does is, it cuts through it 150 watt of heat. The, where it enters the conductor, the temperature is 1200 k, where it leaves the conductor, its temperature is 500 k. And we need to assume because it is said that it is a steady heat flow. This means the state of conductor is unchanged or is not changing with time. Hence none of the properties are changing and hence ds by dt of the conductor is 0. Now to determine the entropy production, remember our equation, basic equation is delta s equals integral or sum of dq by t plus sp. We will differentiate this or differentiate the differential part of this with respect to time and we will get dsp by dt equal to. Now this is, this will be rate of heat option divided by appropriate temperature. Let us say this temperature is T1, this temperature is T2. The rate at which heat comes in here is q dot 1, the rate at which it comes here is q dot 2. So this will be q dot 1 by T1 plus q dot 2 by T2 plus s dot p. Now we have assumed that there is a steady state. So there is no change in entropy, this should not be s. ds by dt entropy change of the system and that is 0. So this becomes 0 leading to s dot p being equal to minus q dot 1 by T1 minus q dot 2 plus s dot p. Now we have assumed q dot 2 by T2 and now when you substitute numbers, you will notice that q dot 1 is plus 150 watt, q dot 2 is minus 150 watt. Appropriate temperatures are 1200 and 1500. So q dot 1 is plus 150 watt by 1200 k, q dot 2 would be minus 1500 watt. And as you notice, this will be 150 by 500 minus 150 by 1200. So this turns out to be a positive number and that brings us to the conclusion that this is a possible irreversible process and we can say that the cause of irreversibility is transfer of heat across a finite temperature difference from 1200 k to 500 k without producing an appropriate amount of work. Another way of looking at it is not to bring in the conductor at all. This is not a very strong way of doing things but another way of looking at it is to say that look, let us say that the two reservoirs are such that this heat flow out of the reservoir not is known as q dot 1 and this heat flow into the from the reservoir what we have called q dot 2 are the only thermal interactions they have with the rest of the universe for rest rounding systems. So we can assume that reservoir 1 plus reservoir 2 plus the conductor forms an extended adiabatic system. So this could be our universe. In that case the entropy produced because of this would simply be d s by d t for the reservoir. The conductor plus d s by d t for the first reservoir plus d s by d t for the second reservoir and now d s by d t for the conductor is 0 it is in steady state. For the first reservoir that is at t 1 the heat absorbed by the reservoir is minus q dot 1. So the heat absorbed is minus 150 watt rate of heat absorption. So this will be rate of heat absorption minus 150 watt divided by its temperature 1200 k plus the rate of change of entropy of the second reservoir that is heat absorbed by it which is plus 150 watt divided by its temperature which is 500 k. And you get the same answer as earlier and which is greater than 0. So remember here the idea which we have used here is the rate of entropy change the entropy change of a reservoir is q absorbed by the reservoir divided by temperature of that reservoir. So for a reservoir we will need to know two things one the temperature of the reservoir is fixed at a value specified value t r and if the reservoir absorbs some heat q r or rejects some heat which would be a negative of absorption that entropy of that reservoir will go up and down accordingly to this formula delta s r is q absorbed by that reservoir divided by the temperature of the reservoir. This you may have to use at a few places out here. So a general scheme of thing is look at the problem read through it completely apply first law wherever necessary to obtain a complete quantitative or complete quantification of the interactions or final state and all that. And then you apply second law because generally second law is to be applied and you obtain the result as an inequality. Inequality just tells us something is satisfied or not satisfied something is possible or not possible. Inequality does not give us a final value that final value almost invariably comes out of the first law which is an equality. Now let me take a break for a few minutes.