∫dQ/T≤∫dQ(rev)/T * , where both integrals are evaluated between the same thermodynamic coordinates- A and B , say. - I am having trouble interpreting this inequality. -( I understand the derivation in my textbook via the Clausius diagram(considering a reversible and an ireversible process between the two points A and B, which form a closed loop, so together they from the closed loop in clausius inequality and * follows.) - I thought the change in entropy has to be greater for an irreversible process. So I also know that, we derive and introduce the state variable entropy by looking at when equality holds for the Clausius equality, and consequently we can not evaluate the entropy change of an irreversible process directly from dQ/T. - So I now interpret that the extra term required for the entropy of an irreversible process (let this be k) due to irreversibilities is large enough , such that, despite *, the entropy change in a irreversible process is greater than that of a reversible process. (as for a reversible process I have ds(rev)=dQ(rev)/T and for a irreversible process ds(irrev)=dQ/T+k) -My last question, I thought that entropy was a state function s.t dQ/T is the same regardless of the path ...oh wait, this is because entropy has only been defined as equivalent to dQ/T for a reversible process? So does this mean it is correct to say: entropy is path dependent for irreversible processes but is path independent for irreversible processes? One more question along these lines, ΔS system=0 seems to be used quite commonly in some text book examples in order to deduce ΔS surroundings, but surely ΔS system =0 is only correct if all processes which the working fluid undergoes are reversible - this doesn't hold for irreversible processes does it? Thanks alot for anyone who sheds some light on this !