Talk:Gibbs–Duhem equation

the proof here is nice to help understanding but its not a real proof.

Can someone who applications in a binary system with x1 and x2 being the two compositions?

Performed a reasonably extensive rewrite of this page, including trying to make variables of consistant case and tracking whether they are intensive or extensive. Added references to Moran and Shapiro and Poling et al. Changed the term "Proof" to what I think is the more appropriate "Derivation" and added a more appropriate form of it. Tried to expand the "Applications" section to include the form of Gibbs-Duhem used in activity coefficient derivation (Margules, van Laar, NTRL, UNIQUAC). Also included basic explanation and link to Gibbs' phase rule, which seemed appropriate.

Thermodude 16:28, 29 June 2007 (UTC)[reply]

Re: updates made by ChrisChiasson -- The chemical potential is defined with reference to the Gibbs Free Energy (or whatever you chose to call it). The expression for the internal energy for the system, while understandable from a physical perspective, is actually derived mathematically from the definition I put back in. Thanks to ChrisChiasson for the other work done on the page; it's always good to see people putting time on on these pages. Thermodude 21:15, 2 November 2007 (UTC)[reply]

Interesting, I always thought that the expression for the (differential of) internal energy came from the first law of thermo, not the defintion of gibbs free energy. I can see what you are saying though. It doesn't really matter, because the equations are all consistent. ChrisChiasson 16:32, 3 November 2007 (UTC)[reply]

---

The derivation of the Gibbs-Duhem relation in this article relies on a crucial fact about the Gibbs free energy, and instead of proving it here, there is a link to the article on the Gibbs free energy. In that article, the derivation of the same fact uses the Gibbs-Duhem relation and links to this article....

I'd suggest rewriting the derivation section. The lecture notes listed as a reference for both articles are also not very enlightening on this point. After some searching it appears Euler's homogeneous function theorem is really necessary for a derivation of the Gibbs-Duhem relation. — Preceding unsigned comment added by 131.212.213.195 (talk) 18:33, 5 August 2013 (UTC)[reply]

"... for a system in thermodynamical equilibrium (and, by definition, in reversible condition) the infinitesimal change in G must be zero ..."

I think this is extraneous and either wrong or unclear. It's true that a system will tend to the state with minimum G given the freedom to do so (e.g. minimize G by exchanging liquid/vapor at fixed T,p), but it can't be true that any infinitesimal change from an equilibrium state has dG = 0, or else G would be flat everywhere, right? Setting the dG equal to each other is sufficient to derive the Gibbs-Duhem equation.

Kjhuston (talk) 22:35, 14 July 2014 (UTC)[reply]

Not defined in link

It shows that in thermodynamics intensive properties are not independent but related, making it a mathematical statement of the state postulate.

The link entitled "state postulate" does not explain what the state postulate is, only thermodynamic variables and state functions. 84.227.253.226 (talk) 17:30, 11 September 2014 (UTC)[reply]

Derivation is cyclical in nature

The current derivation of the Gibbs-Duhem equation is cyclical in nature. It states that, for reasons covered in the Gibbs free energy article,

"...the gibbs free energy of a system can be calculated by collecting moles together carefully at a specified T, P and at a constant molar ratio composition (so that the chemical potential doesn't change as the moles are added together)..."

The problem here is that the Gibbs free energy page invokes the Gibbs-Duhem equation to show this result. The derivations are, therefore, cyclical.

Does anyone have a suggestion/preference on an alternative derivation of the Gibbs-Duhem relationship? The method that I am most familiar with follows Alberty by invoking the complete Legendre transform of the internal energy and plugging in the integrated form of the total differential of the internal energy assuming constant T, P, and n. All that being said, I don't want to have to open up that can of worms on this article as it is, in my own opinion, far too detailed for this particular article.

Any thoughts? JCMPC (talk) 20:39, 19 September 2016 (UTC)[reply]

I just arrived at this page after a very frustrating class today :-) I immediately noticed this circularity (and it was already mentioned above, 5 August 2013). However, I fixed it by editing the Gibbs Free Energy article. The citation of Gibbs Duhem was extraneous; the Gibbs Free Energy article already contained all the information needed to derive the

G=\sum _{i}\mu _{i}N_{i}

result. Nolty (talk) 01:06, 5 October 2016 (UTC)[reply]

Much thanks for your work on this Nolty. It is much easier to follow now and doesn't involve a cyclical argument. Moving the Euler integral further down in the article cleared things up a lot. JCMPC (talk) 18:31, 6 October 2016 (UTC)[reply]

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Expressions equivalence and the place of differential operator

The following formula needs adjustement in LaTex about the place of differential operator:

$({\frac {\partial }{{1-x_{2}} \over \partial x_{2}}})G_{\frac {x_{1}}{x_{3}}}={\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}$ .--82.137.9.177 (talk) 14:23, 22 October 2017 (UTC)[reply]

The following version, done with LaTEX \mathfrak, is found in source: $({\mathfrak {d}}{\frac {G}{\frac {1-x_{2}}{{\mathfrak {d}}x_{2}}}})_{\frac {x_{1}}{x_{3}}}={\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}$

The question is: are they equivalent to each other? Or with the first multifraction expressions from article?--82.137.9.177 (talk) 14:34, 22 October 2017 (UTC)[reply]

Formatting eq test: $({\frac {\partial {\frac {G}{1-x_{2}}}}{\partial x_{2}}})_{\frac {x_{1}}{x_{3}}}={\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}$ --82.137.13.56 (talk) 10:32, 31 October 2017 (UTC)[reply]

The last equality can be integrated from $x_{2}=1$ to $x_{2}$ gives:

$G-(1-x_{2})\lim _{x_{2}\rightarrow 1}{\frac {G}{1-x_{2}}}=(1-x_{2})\int \limits _{1}^{x_{2}}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}\,dx_{2}$ .--82.137.12.134 (talk) 13:38, 23 October 2017 (UTC)[reply]

Applying LHopital's rule gives:

$\lim _{x_{2}\rightarrow 1}{\frac {G}{1-x_{2}}}=\lim _{x_{2}\rightarrow 1}\left({\frac {\partial G}{\partial x_{2}}}\right)_{\frac {x_{1}}{x_{3}}}$ .--82.137.11.166 (talk) 14:44, 23 October 2017 (UTC)[reply]

This becomes further:

$\lim _{x_{2}\rightarrow 1}{\frac {G}{1-x_{2}}}=-\lim _{x_{2}\rightarrow 1}{\frac {{\bar {G_{2}}}-G}{1-x_{2}}}$ .--82.137.10.186 (talk) 14:56, 23 October 2017 (UTC)[reply]

(Some additional properties of the quantities appearing in the above expressions could be mentioned, in case they are somehow relevant:

G=\sum _{i=1}^{m}n_{i}{\bar {G_{i}}},

where ${\bar {G_{i}}}$ is the partial molar $\mathbf {G}$ of component $i$ defined as:

{\bar {G_{i}}}=\left({\frac {\partial \mathbf {G} }{\partial n_{i}}}\right)_{T,P,n_{j\neq i}}.

The first additional relation can be expressed using x_i instead of n_i where x_i is obtained by diving each weight (called mole fractions) by the sum of all n_i in the system. Thus the sum of x_i equals 1.

G=\sum _{i=1}^{m}x_{i}{\bar {G_{i}}},

The weights x_i can be represented geometrically/graphically in case of ternary and quaternary systems using ternary plot and quaternary plot in equilateral triangle and tetrahedron, etc.)--82.137.12.21 (talk) 16:49, 22 October 2017 (UTC)[reply]

The above parenthetical comment represents the content almost lost due to an edit conflict at WP:RD/MA.--82.137.9.55 (talk) 17:19, 22 October 2017 (UTC)[reply]

Dear IP user,

The issues I alluded to on the ref desk were these:

the broken partial derivatives and fractions,
the unclear meaning of the subscript,
the meaning of the bar on the right side, and
the question of why the subscripts appeared only in some places but not in others.

The first bullet seems to have been resolved, the second maybe, the third seems like it is ok in the context of the article, and the fourth I haven't thought about since there were proposals made for the meaning of the second. I am afraid, however, that I do not have the time nor inclination to provide any further thoughts. --JBL (talk) 01:56, 26 October 2017 (UTC)[reply]

Thanks for the enumeration of aspects to be clarified and thus improved. The enumeration and present thoughts are good enough, I'd say.--82.137.9.223 (talk) 12:59, 26 October 2017 (UTC)[reply]

Multiple reversions, insufficiently explained, by user Deacon Vorbis

I notice that user:Deacon Vorbis has recently made reversions to newly added content, without sufficiently explaining them here on this talk page. His edits will be reverted.--82.137.13.123 (talk) 19:00, 24 October 2017 (UTC)[reply]

The notation you're using is unintelligible, as others have pointed out at WP:RD/MA. The added explanation also doesn't make any sense. I'm not especially knowledgeable in chemistry, but you could ask for further assistance at WT:CHEM. (And even though I'm not a chem expert, the math notation is still totally wonky). --Deacon Vorbis (talk) 19:12, 24 October 2017 (UTC)[reply]

Yes, I've just taken a look at WP:RD/MA and seen several comments beside user:JBL first reply which has been very useful. The last reply by Slawomir Bialy is very illustrative about the nature of derivative involved, indeed it's about a derivative taken with respect to x₂ at constant ratio of the other mole fractions x₁/x₃. Is this somewhat unusual constant variable as ratio of two simple variables instead of simple variable what confuses you?--82.137.11.4 (talk) 21:42, 24 October 2017 (UTC)[reply]

Perhaps these nonstandard aspects should be included on the page partial derivative where it be example of use of partial derivatives in the context of math-based sciences.--82.137.15.34 (talk) 11:24, 25 October 2017 (UTC)[reply]

Deacon Vorbis, you still haven't give a concrete example of a particular piece of formula from those you have reverted it may seem nonsensical/wonky to you. If you remain only at the level of vague allegations you'd better stay out of nonsensical objections like those you have done until now. I still wait for you to point to a concrete example of part of formula to be improved as notation.--82.137.11.191 (talk) 20:12, 25 October 2017 (UTC)[reply]

Deacon Vorbis, you could have enumerate a list of aspects for clarification like those mentioned by user:JBL! If you have an additional aspect not covered by JBL, now it's time to hear it from you. If not, you should refrain from making annoying, unhelpful reversions based on vague allegations.--82.137.9.223 (talk) 13:15, 26 October 2017 (UTC)[reply]

I'm still waiting that you say something constructive as aspect that needs improvement, other than those mentioned before by another editor, instead of doubtful edits here about a supposed altering of older comments.--82.137.9.233 (talk) 15:04, 26 October 2017 (UTC)[reply]