# Isn't this spooky variable a problem?

1. Jun 9, 2010

### nonequilibrium

So in my first four chapters of Thermo we basically said U(S,V,N), same old, same old, all work was purely volumetric.

Now in chapter five we discuss H, F, G and for example, delta(G) basically tells you how much work is still available in the process after deducting the required volumetric work on the environment and the required heat. But... can we say 'there can be more work done'? There aren't any other variables in our system! Some undefined variable out there can just take our work? How do I know my entropy, for example, is not dependent on that variable that suddenly exists? This doesn't seem kosher, does it?

EDIT: for example, dE = TdS - PdV was derived in the assumption there is no other work possible than volumetric work. So you even lose the fundamental identity... Are all these things just ignored? Is my book just bad? Am I overlooking something?...

EDIT2: Well to fix that I suppose you can implicitly assume $$dE = TdS - PdV + XdY$$, but that would also imply S is dependent of Y. But what if we already deduced a formula for S independent of Y, isn't that in contradiction with $$\left( \frac{dS}{dY} \right)_{E,V} = - \frac{X}{T}$$?

Last edited: Jun 9, 2010
2. Jun 10, 2010

### Gerenuk

I haven't analyzed the question deeply, but I think this might help for a start:
You indeed need to add XdY. In fact, it is your job to first identify all required variables X/Y. Once these variables are kept at given values, the allowed states must be unique! That's how you know if you are done finding all variables. The energy of this state is what is called "heat". It can be a set of states transforming into each other, but they may not have separate subgroups for the dynamics.

No, it doesn't. One of the variables depends on all the other independent variables, but it is your choice which ones to take as independent.

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