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Question. I have trouble understanding one step in Gibbs adsorption isotherm derivation. I don't understand why we can write $G=U+PV-TS+\sum \mu_i n_i$, how to prove this equation? I saw this equation in Hiemenz's Principles of colloid and surface chemistry, it was given without a proof, so it should be well-known. Some people here claim that this equation is incorrect, is it so? One of my colleges claims that it is correct and it is a definition.

What I already know. I know that $G=U+PV-TS$ is definition of Gibbs energy in case where reagent molar quantities are constant. Also I know that $G=\sum \mu_i n_i$, when p and V are constant, because when $p,T=const$ Gibbs energy is homogeneous function of $n_i$. I know how to prove $dG=Vdp-SdT+\sum_i \mu_i dn_i$. I also know that $d G= \left(\frac{\partial G}{\partial p} \right)_{T,n_i} dp+\left(\frac{\partial G}{\partial T} \right)_{p,n_i} dT+\sum_i \left(\frac{\partial G}{\partial n_i} \right)_{T,p, n_j} dn_i$, where $i\neq j$.

My attempt to solve question.

I am grasping that:

  1. I assume that G has partial derivatives in all path points starting from initial to final point of integration, so there is total differential dG in all path points.

  2. $G_1=U+PV-TS=\int \left( \left(\frac{\partial G}{\partial p} \right)_{T,n_i} dp+\left(\frac{\partial G}{\partial T} \right)_{p,n_i} dT \right)$. I am grasping that it is because $G_1= \int dG , ~if n_i=const$.

  3. And that $G_2=\sum_i \mu_i n_i= \sum_i \int \left(\frac{\partial G}{\partial n_i} \right)_{T,p} dn_i$. It is so because from equation $G_2=\sum \mu_i n_i$ derivation I see that $G_2=\int dG, ~if p,V=const$.

  4. $G=\int \left( \left(\frac{\partial G}{\partial p} \right)_{T,n_i} dp+\left(\frac{\partial G}{\partial T} \right)_{p,n_i} dT+\sum_i \left(\frac{\partial G}{\partial n_i} \right)_{T,p, n_j} dn_i \right)$, where $i \neq j$

  5. So we have: $G=G_1+G_2=U+PV-TS+\sum \mu_i n_i$

Gaps in attempt to solve my question. If my attempt to prove $G=U+PV-TS+\sum \mu_i n_i$ is correct, I have this questions:

a. What are limits of integrals in my proof and why? I am guessing that they are from 0 to $(p,T,n_1...n_i...n_k)$.

b. When I try to plug in my $G_1$ and $G_2$ definitions to $G=U+PV-TS+\sum_i \mu_i n_i$, I don't get $G_1=U+PV-TS$ and $G_2=\sum_i \mu_i n_i$. That is when I try making constant T and p or $n_i$. How to account for this?

c. If there is way to make my proof more rigorous I would like to see it. If there is different proof, I would like to see it. Is there "classical proof", I would like see the source? I am surprised that I wasn't able to find proof of this statement in physical chemistry textbooks.

If whole equation is incorrect, like some people say, where is error in my proof attempt?

P.S.: Frequently, in thermodynamics books authors prove some steps by marking them "obvious" or "self-evident", frequently they add "physical story" that should make this obvious to readers. I am not asking for stories, I am asking for proof. By proof I understand sequence of formulas where each of them is an axiom or a hypothesis or derived from these using inference rules. All axioms and inference rules are listed, and on can't add new ones while proving something. In practice one can use theorems too. I know predicate logic, I prefer Fitch notation. I am a chemist, so things that are obvious to a physicist, might not be obvious to me.

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    $\begingroup$ I may have missed something but in my opinion the correct relationships are : $G=U+PV−TS$ and $G = ∑μini$ $\endgroup$ Commented Nov 13, 2021 at 13:53
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    $\begingroup$ You are having trouble understanding how you can write that equation because it is incorrect. $\endgroup$ Commented Nov 13, 2021 at 14:33
  • $\begingroup$ I saw this equation in Hiemenz's Principles of colloid and surface chemistry: $G=U+PV-TS+\sum_i \mu_i n_i$. It was without proof, so it should be well-known. $\endgroup$
    – Alex Alex
    Commented Nov 13, 2021 at 17:36
  • $\begingroup$ Are you sure it's not a typo : the = replaced by a - ? $\endgroup$ Commented Nov 13, 2021 at 18:30
  • $\begingroup$ I am sure, look for Gibbs isotherm derivation, there is this equation. $\endgroup$
    – Alex Alex
    Commented Nov 13, 2021 at 19:12

1 Answer 1

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According to Bazarovs Thermodynamics textbook in case when state of state of system is determined by generalized forces $A_i$ and conjugated outer parameters $a_i$ then (generalized) Gibbs energy is defined as: $G=U-TS+\sum_i A_i a_i$. So my question is specific case of this equation: $G=U-TS+PV+\sum_i \mu_i n_i$. So its true by definition.

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  • $\begingroup$ "Generalized" should be emphasized here, not parenthetical. The question conflates this generalized Gibbs free energy with the conventional Gibbs free energy $U+PV-TS$; they are two different things. This could easily confuse readers. $\endgroup$ Commented Dec 18, 2021 at 20:13
  • $\begingroup$ Imagine situation where is other than PV work, for example work done against surface tension. Imagine that we are using classical definition of Gibbs energy. In this case Gibbs energy is no longer equal to maximal work, so I guess some thermodynamic relations might fail. $\endgroup$
    – Alex Alex
    Commented Dec 18, 2021 at 20:39

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