 Welcome back, it is time for us to look at an example, an example in which we will be using the first law, but we will also have to look at the process detail. We will have to look at evaluation of the work interaction. We will have to look at the contents of the system because we will have to look at the equation of state, both the PVT part as well as the energy part. And finally, we will have to keep track of the information provided and the assumptions that we have to make. Let us say that the situation that we are going to analyze is this. We have a system which is closed, which is the first piece of information that we have. Then we are told that it contains an ideal gas, that the second piece of information we have. The third piece of information is that it executes a quasi-static process which is also an adiabatic process. So it executes a quasi-static adiabatic process. So here itself we have four pieces of information which are given to us. This may not be the complete specification. We may need to assume additional pieces of information so as to reach our goal or maybe to simplify the process. The requirement is to show that under these conditions with appropriate assumptions, the process is representable by this, PV equals constant, PV raised to gamma equals a constant. Show this. And what additional pieces of information we need? Do we have to make any assumptions? If so, which ones? And we will also be conscious of the steps where we will be using the closed system, the ideal gas, quasi-static process and adiabatic process, these four items of information. The first thing we do is sketch our system. It is a closed system, contains an ideal gas, it is a fluid system. So our system depiction will be something like this. In general, it can expand or contract. It could have other modes of work, stirring, electrical, it may absorb some heat. This is the general depiction. And we have used the first piece of information that it is a closed system here. Now it is given that it is an adiabatic process. Adiabatic process means dq will be 0 throughout the process. That is another piece of information, information 4 in our list. And that means this q is going to be 0, there will be no heat transfer. Now let us apply our first law. As good students of thermodynamics, the first law for a closed system should always be written down as q equals delta E plus W. If you do this, you will make very few mistakes. You are on safe grounds. Do not ever write as a first law expression, an expression which is other than this, like q equal to mcp dt or mcv dt, its integration. Those will all be consequences of this basic form of the first law. Now next step is to expand this. For example delta E, we will expand into its components and W, we will expand in its components. So we will get q equals, the major component for us would be thermal energy delta U plus let us write the other components that occur in delta E as delta E other. And W we will split into the expansion work and other type of work we can. Now let us convert this in differential form. So we will write this as dq equals du plus de other plus dw expansion plus dw other. Now we are given that dq is 0. This is piece of information 4 already with us. Let us leave du as it is, but let us assume because nothing else is mentioned. This is something we have to assume. So this is the fifth piece of information, but it is an assumption. Similarly let us leave just the way we have left du undisturbed. Let us keep the expansion work in place. And let us also assume that this is 0 and this will be an assumption, our second assumption and the sixth piece of information. So we have reached a stage where we have simplified our first law to 0 equals du plus dw expansion. Now we use the fact that it is an ideal gas. That is one of the pieces of information given to us. So for an ideal gas du will always be m Cv dt. This comes out of item number 2, it is an ideal gas, dw expansion will always be p dv. Now what we do is transpose the 0 on the right hand side and divide throughout by m. There is an m here and this dv when divided by m will become d of small v, the specific volume. So we will end up with Cv dt plus p dv, so here all that we have done, we divide by m and transpose sides. Now we notice that it is given to be an ideal gas and hence we should have pv equals RT, the equation of state. Take the differential, we will get p dv plus v dp equals RT, we are writing so many equations that we say this is my equation A. Now remember the final form we want is in terms of p and v. So this dt in equation A which would lead to the presence of t in the final form, it is something which we do not need. So let us use this dt here to eliminate this dt from equation A. So if we solve this equation for dt, we will get dt equals 1 over r p dv plus v dp and substitute this equation which is equation B, the dt part into equation A where also we have a dt part, we will get the following. Cv by r into p dv plus v dp plus p dv plus 0. Taking p dv from here and p dv from here together, we will end up with, let me put v dp first, Cv by r into v dp plus Cv by r plus 1 into p dv into 0. Now we divide throughout by the product pv and you will get Cv by r into dp by p plus Cv plus r by r into dv by v. Now we do two things, we multiply throughout by r, so essentially that cancels out this common denominator from the two terms and then we notice that this Cv by r equals Cp that is one of the properties of an ideal gas. So we use this Cv plus r with Cp property of an ideal gas, so item 2 is again used here. So we end up with Cv dp by p plus Cp dv by v equals 0 and then we divide this equation throughout by Cv and note that the ratio of specific heat Cp by Cv is defined to be gamma. So we end up with dp by p plus gamma dv by v equals 0. Let us say this is equation C. Now to obtain a relation between p and v, we will have to integrate this equation and now here we will have to use the following fact that it is a quasi-static process, hence integration is possible. However, we must have gamma to be constant for a straight forward integration, so we will make an additional assumption that Cp and Cv are constants, hence gamma is also a constant, this will be our assumption 7. And once you do the integration, after integration you will get logarithm of p plus gamma equal to some constant or taking the exponential, you will get p into v raised to gamma. Now remember that we have finally come to this derivation but do not jump to the conclusion that an adiabatic process means pv raised to gamma is constant. Remember that the conditions we have is that we have a closed system, second thing we had an ideal gas, the third specification was that it is a quasi-static process, it was also given to be an adiabatic process. And then we made an assumption, our first assumption was dE other is 0, that was piece of information 5, then we assume dW other than expansion work was also 0, that was assumption 6 and finally we assume that Cp, Cv were constants that was assumption 7. In any problem you solve in thermodynamics, any exercise you do, you will have to be conscious of the information provided but you will also have to make consciously some assumption and list them. Thank you.