Thread: Tds Equation and Entropy View Single Post

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 Quote by Red_CCF Hi, thanks for the response For instance, I imagine that there is more than one polytropic (or any) process from two equilibrium states for which work can be applied, but from Tds = de + pdv, the integral of the p/Tdv term (taking T constant) would result in different values while e2-e1 doesn't change if the initial and final states are the same.
True, but take in mind that to obtain Delta S you have to integrate de/T, not de. Both p and T will be different in different polytropic processes.