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Derivation of the thermodynamic potentials using Legendre transformations

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Andromon
#1
Dec20-12, 09:53 AM
P: 3
Hello guys, I'm studying Thermodynamics and I don't totally see how you introduce the potencials using Legendre transformations.

I have seen a non formal explanation showing how you can interpret them, but not a rigorous demonstration of how you get them via the Legendre transformations.

Do you know any site or book that covers it?

Also all the other issues, like the Maxwell transformations and the Euler equations and relations.

Ty.
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Philip Wood
#2
Dec20-12, 11:55 AM
PF Gold
P: 944
I expect I'm not giving you a full picture of what a Legendre transformation is, but here's a systematic way to generate potentials, H, F and G...

Start with dU = TdS- pdV.

From the product rule: dU = TdS- {d(pV) - Vdp}

We can write this as: d{U + pV} = TdS + Vdp

U + pV is usually designated as H. It is the enthalpy potential. Its 'natural variables' are S and P.

We can product-transform TdS instead of pdV, and obtain the Helmholtz function U - TS, with natural variables T and V.

Finally we can transform both TdS and pdV,obtaining the Gibbs function G = U + pV Ė TS, with natural variables p and T.
Andromon
#3
Dec20-12, 12:10 PM
P: 3
Ok, I see it, is an add and subtract trick, but I don't see where there it's used the Legendre transformation, it's not needed at all?

jtbell
#4
Dec20-12, 12:14 PM
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Derivation of the thermodynamic potentials using Legendre transformations

http://en.wikipedia.org/wiki/Legendr...Thermodynamics
Philip Wood
#5
Dec20-12, 12:48 PM
PF Gold
P: 944
I'd be surprised to be told that I wasn't actually doing Legendre transformations in my earlier post, but let wiser heads decide.


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