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

• From: John Denker <jsd@av8n.com>
• Date: Fri, 15 Jun 2007 16:56:13 -0400

Replacement version, part 2.

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

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

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

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

....

> 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; see below.

The experiment is set up so that
-- V is an independent variable (or independent non-variable if you like)
-- E is an independent variable
-- T is an easily-observed dependent variable
-- X (i.e. the fraction of ice in the water) can be considered
an observable dependent variable.
-- My recommendation is to treat S as a dependent variable.

Experimentally, there is (roughly speaking) a distinction between
independent and dependent variables. Conceptually, though, it
usually pays to treat all variables on an equal footing.

> 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.
This is the least of our worries.

> 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).

It is a pain in the neck to observe S operationally. There is
no handy entropyometer. Therefore it is more practical to measure
something else, and calculate S at a later stage ... but
conceptually, S is a perfectly well-behaved variable, as
illustrated at
http://www.av8n.com/physics/img48/s-t-phase-change.png

> 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?

Assuming dT and dS etc. are not simply multiples of one another,
which is almost always OK (though you have to check), then the
key to having a linearly independent basis set is to not have
*too many* vectors in the set.

Again, it really helps to imagine the "system point" moving in
thermodynamic state space. Here's how I visualize it:
http://www.av8n.com/physics/img48/s-t-phase-change.png

That is, in the calorimeter, in the pure liquid phase, as we
change the energy the point moves along a path of changing E,
changing T, unchanging V, and unchanging X=0. Then when we
hit the phase transition, there is a *corner* in the path.
During the transition, the point moves along a path of
changing E, unchanging T, unchanging V, and changing X.

To repeat, at the corner one variable (X) becomes "alive" and
another (T) becomes "dead". The contours of constant S cross
at an angle to the path, such that S continues to be a smooth
function of E on both sides of the corner, i.e. in both the
one-phase and two-phase regions.

This is tricky, because T and X are "natural" variables at
experiment-time ... but they are badly behaved at analysis-time.

I know I've been talking a lot about linear independence, but
I think in this case the problem is elsewhere.
-- Equation [1] and equation [2] have a problem with the dE/dT
term in the two-phase region. You can't evaluate dE/dT when
there is no possible dT.
++ Equation [3] dodges this bullet, and is the conventional way
to attack this problem.