Replacement version, part 1. This part was correct all along.
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]
> 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.