 Now it is time to prove the Clausius inequality in its general form. I did not do this derivation directly because this derivation is in the domain of differential calculus. Let us state what we want to prove. Let us say that we have a system and it executes a cycle, cyclic process. Let us say that during some part of the cycle, the heat interaction from some neighboring system is dQ and the temperature here where that heat interaction takes place is T and the work done through the cycle is W. Since work done is for the whole cycle to apply the first law, we will have to say the work done must be integral of dQ and what we want to prove is the Clausius inequality that integral of dQ by T over the cycle is less than or equal to 0. And to prove this, again we will use the trick of reductio ad absurdum and we will say that let us assume that this integral of dQ by T is greater than 0. And now we proceed as earlier. Let us have a reservoir T0 which would run to provide this dQ a reversible machine R which would extract from this reservoir dQ0, provide this dQ to the system and the work would be dW0. So, this is the scheme. Now remember that this is a reversible 2-T machine. So, we have dW0 is dQ0 minus dQ by first law. Then algebra write this as dQ into dQ0 by dQ minus 1 and which because this pertains to a reversible machine, this will be dQ into the temperature at which heat is supplied to this reversible machine that is T0 divided by the other temperature T1 minus 1. Now, let us consider this as a system. What is this? This is a 1-T machine in which it is cycling. Why cycling? Because the system is part of our machine, it is cycling and all these reversible machines as required to provide this dQ are also cycling. So, our extended system shown by dotted line here, the dotted loop here is a 1-T machine which executes cyclic processes. And what is the work output of this 1-T machine? The work output is W plus integral of all these dW0s which is W plus integral of, we write it as T0 minus. Now, this can be written down as W plus T0 cyclic integral of dQ by T minus cyclic integral of dQ. And from this equation, we know that W equals this cyclic integral, so that cancels out. So, this becomes T0 into integral of dQ by T, which turns out to be greater than 0 with our assumption. But this violates the second law of thermodynamics as represented by the Kelvin-Clang statement. And the only way we have violated by assuming this to be true. And that means this assumption is false. So, this is a violation of the second law. And hence, this assumption is false and hence our Clausius inequality is proved. So, this was the general proof of the Clausius inequality. Thank you.