/sci/ - Science & Math

>>3712314
Are you fucking serious?

>>3712340
yes and please let me know. does helmholtz free energy have natural variables other than V and T?

>>3712314
engineer is dumb

NOTHING NEW TO SEE HERE PEOPLE

>>3712349
It will work with any variables so if you have V(T,P) you have
dA=-SdT-PdV=(-S-P ∂V/∂T) dT - (P ∂V/∂P) dP.
This is legal. The point only is, that A(T,P):=A(V(T,P),T)
does not characterise the system uniquely. If you use "bad" variables like that, then you have more possibilities to describe the same system and you need an extra condition to compute all the real unique values. Like "The value of P(T,V) in configuration XXX is 4.53 bar."

>>3712395
addon:
As an example consider the fact that the natural variables of U are S and V, but coming from the historical termodynamik approach dQ and T are defined earlier, as well as
C=dQ/dT=∂U(T,V)/∂T
So people will write
dU=CdT-PdV
as you surely have seen before.

>>3712423
>>3712395

I think I understand now, thank you very much.

So:

A=U-TS usually becomes dA=-PdV-SdT because U was a function of S and V. But if your process has U as a function of other variables, then you'd get a different dA.

but if U started off as a function of other variables, would it be ok to call the new variables "natural" or does natural imply that U must start as U(S,V)?

>>3712481
To answer your questions actually make sense only in the reverse oder:
>would it be ok to call the new variables "natural" or does natural imply that U must start as U(S,V)?
There is no choice but U(S,V), since
P=-∂U/∂V resp. P=-T(∂S/∂V)
and
T=(∂U/∂S)^{-1} (I set k always 1 here, for simplicity)
are both derived variables.
Notice that the U(T,V) only worked because back then, T was experimentally defined via PV=NT.
From a technical POV there is no temperature before the entropy S, which can be understood best in the context of statistical mechanics. So U(S,V) is the way to go because S=S(U,V,N) is just a computable number and V is a parameter which is obvious from experiment, unlike P and T which are rather abstract, although both can by motivated quite well.

>But if your process has U as a function of other variables, then you'd get a different dA.
You get A(T,V) only from U(S,V). The helmholz free energy A is a quantity defined via Legendre transform of U(S,V) with respect to S.
http://en.wikipedia.org/wiki/Legendre_transform#Thermodynamics
This tranformation is a way on NOT losing the information you would when only setting U(T,V):=U(T(S,V),V).
You have dQ=TdS, a heat flow and if heat flows than the energy U changes, since dU=dQ-dW (or +dW in your convention). The very purpose of defining A is to construct an object which is _free_ from heat variations, so you define A in such a way that
dA=d(U-TS)=dU-SdT-TdS=dQ-PdV-SdT-dQ=-PdV-SdT
The very thing you wanted to achieve is get rid of dQ in dA and that's what happened.

>>3712737
(continued)
The same thing happens when you define Enthalpy. You want something which is resistent to volume change, and what you end up with is the function H, which satisfies
dH=SdT+VdP
Observe that the work dW=PdV is not part of this energy quantity anymore. This is why Engineers/Hydrodynamics people fap over it. It's a measure of more or less heat only (the experimental parameter is usually kept constant, so dP=0, such that dH=dQ)
Then comes the Gibbs free energy, which gets rid of both, S AND V. So
dG naturally turns out to be
dG=-SdT-VdP.
This doesn't depend on pretty much anything (other that T and P, which you control in parameters).
But wait! We have forgotten something. There still is the parameter N, the number of particles <-- this is the origin of that one:
http://en.wikipedia.org/wiki/Gibbs%E2%80%93Duhem_equation

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>>3712314