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Jan10-13, 08:48 AM
P: 492
Tds Equation and Entropy

Quote Quote by DrDu View Post
The state variables p, V, E, S etc. are only defined for equilibrium states. In classical thermodynamics, all states proceed from one initial equilibrium state to a final equilibrium state. The entropy difference can be calculated for any path in equilibrium state space to the final state.
Hi, thanks for the response

I understand and have accepted the statement but I have trouble understanding how this is reflected, which is why I posed the questions above.

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. To me this seems to reflect different s2-s1 values despite having different processes from the two states. I have the same confusion regarding the derivation of the Tds equation if entropy generation is added as I see that it would change the form of the Tds equation which many books say is applicable for both reversible or irreversible processes.

Any help would be appreciated.
Thanks very much