Re: [Phys-l] thermo differential and extensive/intensive variables

[Stefan Jeglinski <jeglin@4pi.com>]

> 1) Interesting part:  The words talk about energy, which is correct ...
>   so make the symbols agree with the words.  Just get rid of every mention
>   of "Q" and "d'", and then the rest is correct.  Specifically:
> >         "you heat the water, d'Q = CvdT"
>   becomes "you heat the water, dE = Cv dT".
>
>   It's just that simple.  Thermodynamics is not nearly so complicated and
>   ugly as the thermo books make it seem.

OK, at the risk of making it harder than it is, and to satisfy my own 
need for more rigor (this really is the crux of where I'm going with 
this example):

I have some water and I'm going to put E in. Experimentally I know 
that T will change as long as I'm not changing the phase. I suspect, 
but maybe I'm not 100% sure until I study it further (work with me 
here), that S changes also. So I'm inclined to express the math this 
way:

[1] dE = (dE/dS)(const T) dS + (dE/dT)(const S) dT

But those partial derivatives are not particularly simple 
thermodynamic quantities, so I want to recast this in terms that are 
more obvious, namely:

[2] dE = (dE/dV)(const T) dV + (dE/dT)(const V) dT = CvdT (for dV = 0)

But I can just as easily write:

[3] dE = (dE/dS)(const V) dS + (dE/dV)(const S) dV = TdS (for dV = 0)

So for this example, dE = TdS and dE = CvdT, both true for dV = 0.

Fine. If I know dE, I can get dT and then with some simple math, dS. 
Rigorous enough for me, no Q, and no d'. Interestingly, I found 
myself guilty of doing exactly what I was trying to avoid. That is, I 
was trying to avoid casting the problem as splitting dE up into "d'Q" 
and "d'W," but saddled myself by insisting on splitting it up into dS 
and dT. Not in the same league perhaps since there is nothing wrong 
with Eq 1, but it was not convenient for the problem.

Now we turn to the phase change for which dT = 0. For this case, Eq 2 
implies that there can be no phase change without a volume change, 
unless (dE/dV)(const T) is exactly zero. In fact, more than implying 
so, it seems to constitute a proof of sorts (?)

Anyway, for the phase change Eq 1 would seem to be my choice here, 
since I don't a priori know how the volume changes. I know that dE 
will be the latent heat of vaporization for the water-steam phase 
change, or that of fusion for the water-ice phase change. But to use 
Eq 1, I shall need to find out what thermodynamic quantity(s) 
represents (dE/dS)(const T). Or use an even better relationship 
(E(S,P) and enthalpy comes to mind obviously).

My greater concern at this point is the observation about linear 
independence of, say, T and S in Eq 1. How does one avoid this 
pitfall in general? From Eqs 2-3 I know that dS = Cv(dT/T), so Eq 1 
should be "safe" albeit inconvenient. But that was in no way obvious 
until I had "solved" the problem.


Stefan Jeglinski