 Welcome back, we are now going to look at a major theorem known as the Lausius Inequality. It will be derived using the Carnot theorem. Actually you can say that it is an extension of the Carnot theorem to any cycle. After Carnot theorem we looked at the equality part and we are able to derive a thermodynamic basis for temperature scales. Now we are going to look at the inequality part. Before we do that something about symbolism. Whenever we see this symbol less than or equal to in thermodynamics, you should remember Carnot's theorem because this symbol we first came across when we said, Carnot said that the efficiency of a 2T engine will be less than or equal to that of a reversible 2T engine working between the same pair of temperature. So that is this less than or equal to sign and I tend to sketch the less than or equal to as two separate symbols one below the other because here the equal to pertains to the reversible case whereas the less than pertains to the irreversible case. So just to be conscious of these two distinct possibilities, I have a habit of writing less than or equal to using two distinct symbols because that way we are conscious that the equality part pertains to the reversible limit. Now what we are going to do is first state the Clausius inequality, then we will demonstrate it for a 2T heat engine, then we will prove that for a 3T machine and then we will provide a general proof. That way we will be able to appreciate the general proof and we will be able to appreciate the Clausius inequality itself. Now what is Clausius inequality? The Clausius inequality considers a system, a closed system which executes a cycle. I am showing PV diagram that you may consider any two appropriate thermodynamic properties x1, x2. Since it is a cycle, we know that the initial and final state would be the same and I am sketching any general cycle partly quasi-static partly non-quasi-static does not matter. Now the Clausius inequality says that if the system executes during part of a cycle a heat interaction, absorbs some heat DQ and if the temperature of the system or temperature of that surface of the system across which this DQ takes place is T, then the Clausius inequality states that over the cycle the integral DQ by T must be less than or equal to 0. This is the Clausius inequality. Remember that it must be a cyclic process, the integration must be over a complete cycle and that is why we represent it with an integral sign with a circle superposed on it. DQ is the small amount of heat absorbed by the system during a small part of the cycle and T is the temperature of the system at that instant and across that boundary or at that boundary across which this DQ interaction takes place. And of course this less than or equal to is the same sign that we have in the Carnot theorem equal to pertains to the reversible limit less than is the otherwise general case. The next thing we do is to quickly see that this is indeed a consequence of the Carnot theorem by looking at the Carnot's efficiency inequality itself. Let us look at a Carnot engine, a reversible 2T heat engine. And let us look at the general 2T heat engine working between same temperature levels. This is any engine, this is reversible engine. We know that the efficiency of this general engine will be less than or equal to efficiency of the reversible engine. Now, if the heat absorbed by the general engine is Q1, heat rejected is Q2 and the power output is W, the work done is W, then we can write this efficiency is 1 minus Q2 by Q1 which is actually by the definition W by Q1. The efficiency of the reversible heat engine and the definition of the thermodynamic temperature scales tell us that eta r is 1 minus T2 by T1, where T2 and T1 are temperatures on the Kelvin scale. And of course we have this less than or equal to. Now let us proceed from here. First thing we notice is that one is common on either side. So we end up with minus Q2 by Q1 is less than or equal to minus T2 by T1. Then we notice the following. We notice that Q1 is heat absorbed by the engine. So that will be a positive number. So if I multiply this equation by a positive number Q1, the sign of the inequality is not going to change. Similarly, I notice that T2, the thermodynamic temperature is also a positive number. All thermodynamic temperatures are positive numbers. And hence if I divide by T2, again the sign of the inequality will not change. And hence what we do is multiply this equation by Q1 and divide this equation by T2. And we will get minus Q2 by T2 is less than or equal to minus Q1 by T1. Now transposing Q1 by T1 with the negative sign on the left hand side, we will get Q1 by T1 minus Q2 by T2 less than or equal to 0. Now notice here that Q1 is the heat absorbed by the engine from the reservoir at T1. It is a positive number. Whereas our nomenclature is that Q2 is the heat rejected by the engine. If we write this in terms of heat absorbed for both these terms, I can write this as heat absorbed by the reservoir at T1 divided by T1 plus heat absorbed from the reservoir at T2 that will be minus Q2 divided by T2 less than or equal to 0. Now this is also heat absorbed. This is also heat absorbed. It is rejected so it is minus Q2 with this symbolism here. And hence we now see that the Clausius inequality is actually hidden inside the Carnot theorem inequality using thermodynamic temperature scale. This was the first step. In a while, we will derive and demonstrate the Clausius inequality for a 3T machine. Thank you.