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Jan10-13, 03:24 PM
Sci Advisor
P: 3,633
Quote Quote by Red_CCF View Post
Hi, thanks for the response

So for different polytropic processes to be possible, then the temperature at the end states must not be the same and if it the temperature throughout the process is the same, then the pdv process must satisfy the ideal gas law (assuming an ideal gas is compressed)?

So this would mean that for different reversible processes with the same end states, the sum of the integral of de/T and p/Tdv will always end up working out to be the same value?

I'm also wondering if you can take a look at my first question in the OP.

Thanks very much for your help
Uh, yes, I was assuming an ideal gas here, but only to show that the pressure will be determined by the volume when T is held constant. This will also hold true for other substances.
The integrals of de/T and p/TdV will not always be the same when T is variable, but their sum will be.
Regarding your first question you have to distinguish between a process, which may be irreversible, and a path in the state space of equilibrium states used for calculation of the change of S.
Usually when there is entropy production in an irreversible process, the work is also not equal to -pdV. So there is no contradiction with TdS not being dQ. Think of stirring a viscous liquid which is a form of non-volume work which gives rise to entropy production.