Why is entropy a state function, in general, even for path dependent materials

Question

I have used the fact that entropy is a state function for years, and it is very useful. But when I think about it, I am not sure why it is true. I get the usual argument for a Carnot cycle and an ideal gas, i.e. that

$\oint \frac{dQ}{T}=0$

and hence that

$\oint ds=0$

where we defined: $ds=\frac{dQ_{rev}}{T}$

I get that we can use this statement to show that s(V,T) will be independent of the path selected to get to the state (V,T). But what about something like a viscous fluid, or a viscoelastic solid? There is no path that you can take that will give you a reversible path integral. You might argue that you could take the path very slowly. But then what about a path dependent material, like a metal, that will undergo hysteresis? Even a slowly deforming path will exhibit hysteresis and be irreversible. So I don't see that you can use a "lets take a reversible path" argument for these types of materials, as you can for an ideal gas and the Carnot cycle. So I am not sure I can a theoretical argument proving that s(V,T) is independent of the path for such materials.

Where am I going wrong? Is there an argument for the path independence of entropy for path dependent materials such as metals and viscous fluids? Or is it simply a reliable postulate?

Answer 1

For a system which shows hysteresis even for very slow processes, you can usually divide the system up into smaller parts which are still large enough to be macroscopic, but small enough that the hysteresis goes away on reasonable timescales. That is, one can then find reversible behaviour without having to wait for years in order to be slow enough compared to relaxation times. The magnetic domains in a ferromagnetic material are a good example. You can model a ferromagnet as a collection of such domains. The thermodynamic arguments apply to each domain. I guess that similar statements would apply to most viscous materials, though there are difficulties with a glass.

By the way, the method to use to deduce entropy from a Kelvin statement of the 2nd law is via Clausius' theorem. A good argument here does not need or use an ideal gas, and this is important because you want an argument that can apply at all temperatures.