Tds Equation and Entropy
I was wondering, with regards to the Tds equation Tds = de + pdv:
1. All of my textbooks state that integrating this equation, although derived for a reversible process, will give the entropy change regardless of the process or whether or not the process is reversible. However, I don't understand this concept because if the Tds equation was re-derived from first law for a general process, I thought that there would be a Tσ (entropy generation term, zero for reversible) in the equation since ds = δQ/T + σ and entropy generation would be path dependent?
2. With regards to the reversible Tds equation, I have trouble seeing how this equation is path independent since, for two fixed states, I always thought there was more than one possible work path or pdv expression in which state 1 can be used to move to state 2 and if this were true it would end up giving different s2-s1 values?
3. I have only found proofs for entropy as a state property for reversible process and argued that by extension it must be a state property for any process but is there a proof that directly show that it is a state property for any process?
Thanks very much