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I've just been reading through Van Ness' Understanding Thermodynamics, and I'm having a little trouble following his argument at one point. He is deriving the useful differential equation:

$$dU=T \ dS -P\ dV$$

By noting that:

$$dU=dQ-dW$$ for all processes, and then also noting that, only for reversible processes,

$$dQ_{rev}=T\ dS$$ and $$dW_{rev}=P\ dV$$

and substituting these last 2 equation into the first equation. As long as we're talking about reversible processes, I'm fine with this. But then he then goes on to say:

Now we derived this equation for a reversible process, but once derived we see that it contains just properties of the system, and so it must not depend on the kind of process considered. What we have really done is derive an equation for a special case, and then conclude that it must be general.

Maybe this is blindingly obvious, but I'm really struggling to follow his logic through. He's derived an equation for one particular case, and then, just because this equation doesn't explictly refer to anything other than properties of the system, he concludes that it must hold for all cases. I mean, for the reversible case, sure, but I'm struggling to see why, logically, $dU=TdS-PdV$ would hold for a given irreversible case.

To give a very rough analogy of my thinking, I feel live I've found an equation like Boyle's law:

$$P\propto\frac{1}{V}$$ and can then conclude that, just because this equation (a) holds for one particular system (e.g. low pressure, high volume gas) and (b) only contains 'properties of the system' $(P, V)$, then it must therefore hold for all situations, which is of course nonsense.

Anyway, I'd appreciate any help, thanks

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2 Answers 2

Some properties of systems are special because they don't depend on the path taken to get there. The are called "functions of state". To take a trivial example, the volume of a system is a state function. A given system in a given state will always have the same volume regardless of how you prepared that system.

The equation for $dU$ only contains properties that are functions of state. That means it doesn't depend on the path taken for a change, and therefore it applies to changes in irreversible systems as well as reversible systems.

For more details (and I must admit I had to look it up to check) see http://en.wikipedia.org/wiki/Fundamental_thermodynamic_relation

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Thanks for your help, I do really appreciate it, but I'm still not 110% satisfied with your logic. Perhaps could you point out what's wrong with this 'counter-example'. There's a class of reversible processes (in particular, adiabatic ones) which, on top of obeying the complex equation above, also obey the simpler $dS=0$. Now we know that $S$ is a state function, so why can't we use the same logic and say that, just because $dS=0$ holds for all adiabatic, reversible processes, it must hold for all adiabatic processes, including irreversible ones (obviously ridiculous)? Thanks –  tom Jan 25 '12 at 2:27

Your book really doesn't word it well. Actually, it should say that the equation holds for any equilibrium state. As such, it can be integrated via any reversible process (because reversible processes are always at an equilibrium state).

Now comes the interesting bit: since those equations do not contain path-functions ($\delta Q$, $\delta W$), only point-functions ($dU$, $dS$ etc), the results of their integration between two points (states) do not depend on the path taken between them.

That means that, any reversible path you take between the same two points, the result will be the same.

That also means that, any irreversible path you take between those points, well... you can't integrate through that path, but it doesn't matter. You can use the results of the reversible path!

Conclusion. If you have a system that underwent an irreversible process between the points A and B, you can pretend it underwent a reversible one (or a series of them: isothermal, then adiabatic, then...) and integrate through that (or those). When you're done integrating, you can use the results in the irreversible process.

Hope I cleared it up.

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I asked this question years ago, and your response is exactly what I'd say to my younger self today :) The only thing I dislike is the use of the word 'reversible' - I prefer quasi-static, and reserve reversible for quasi static isentropic processes, in line with callen's thermodynamics text. –  tom Sep 6 at 23:58
    
Glad to hear that :) Will you accept this answer then? About "reversible", a quasi-static non-isentropic process like same-temperature heating can be reversible as well. Nothing wrong with it I guess. –  André Neves Sep 7 at 0:15

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