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

[John Denker <jsd@av8n.com>]

On 06/15/2007 03:09 PM, Stefan Jeglinski wrote:

> 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. 

Sure, S changes also.  You can easily evaluate the
change in S by integrating dS, where
  dS = (1/T) dE  (at constant V)             [0]

as discussed at
  http://www.av8n.com/physics/thermo-laws.htm#eq-s-vs-t


> So I'm inclined to express the math this 
> way:
>
> [1] dE = (dE/dS)(const T) dS + (dE/dT)(const S) dT

That's not wrong, but it doesn't get you any closer to the
stated goal.  Wouldn't you rather have S as a dependent
variable, as in [0] above?  The experimental setup is
optimized to have E as the independent variable, with
V=constant and T as the observed dependent variable,
unless I misunderstand the description of the experiment.

> 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.

OK.

> 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.

Those words are hard to parse, but it appears we agree that eq [1]
is inconvenient but not wrong ... whereas [3] is the convenient and
conventional way to solve 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 (?)

No.  The problem is that equations [2] and [3] are not powerful
enough to handle this case.  They are incomplete.  We need a third
term on the RHS of equation [3].  The conventional choice is a term
involving dX, where X is the fraction of ice in the water ... but
choosing S as the "extra" variable is perfectly allowable.

My nonbinding recommendation is to try the problem using V, T, and
X as the independent variables, such that E and S become dependent
variables.  The rationale for this recommendation is that the
previous case (pure liquid) emerges naturally as the special case
where X=constant=0.

> 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. 

You can continue to hold constant the volume of the ice+water system.

> 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? 

I don't know about "in general", but I know that in this case the
phase change gives us an extra degree of freedom.  Gibbs phase
rule and all that.  Things that were linearly dependent in the
one-phase system become linearly independent in the two-phase
system.

Again, to my taste, I find that using X (rather than S) as the
"extra" independent variable is more intuitive ... but as soon
as you have an expression for S in terms of X you can invert it
to find X as a function of S, so at the end of the day all
variables are treated on the same footing.