# Generalisation of Reversible Equation to Non-Reversible Situations Because it Only Contains 'Properties of the System'

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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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.
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