If I understand correctly (which I clearly don't) the condition of stable equilibrium of a closed system at constant $(P,T)$ is that $G$ is minimum.

That implies that $dG=0$ and $d^2G≥0$. Since $dG=VdP-SdT$, if the second differential is positive then $\left(\frac{\partial V}{\partial P}\right)_T≥0$, which is false. What am I doing wrong?

EDIT: I would like the answer to be as mathematically rigorous as possible, and with no hand-waving about differentials. I would also appreciate if you could tell me a textbook/article where stability is discussed with some level of rigour.

  • $\begingroup$ How can the 2nd differential be positive if dT = 0? Also, you left out the summation of $\mu dN$. $\endgroup$ – Chet Miller Mar 7 '17 at 18:31
  • $\begingroup$ @ChesterMiller I said that my system is closed, therefore the number of particles is constant and I don't need that term. $\endgroup$ – mlainz Mar 7 '17 at 18:33
  • $\begingroup$ @ChesterMiller The second differential being positive means that the quadratic form of the second derivatives is positive definite, which is a condition for having a local minimum. Is the multivariable generalization of having positive second derivative. Correct me if I am mistaken $\endgroup$ – mlainz Mar 7 '17 at 18:53
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    $\begingroup$ You left out chemical potential out. With constant P and T, dG is only a function of derivative of the potential! Though it is closed, chemical reaction or mixing will get the composition to such level that its Gibbs free energy is the minimum. $\endgroup$ – user115350 Mar 7 '17 at 21:18
  • $\begingroup$ What are you using for your expression of the second differential? It must include several terms, which means not all of them are necessarily positive. $\endgroup$ – compmatsci Mar 25 '17 at 19:17

... the condition of stable equilibrium of a closed system at constant $(P,T)$ is that $G$ is minimum.

This is true, but it is not supposed to mean that equilibrium happens only for special values of $T,P$ for which some $G(T,P)$ has local minimum. Temperature and pressure are understood as given external conditions, the system is not supposed to change their values. Stable equilibrium ordinarily can exist for almost any values of these parameters.

What the statement means is that when we can express Gibbs energy of a thermodynamic system as a function of $T,P$ and some independent variable $x$ describing state of the system, equilibrium state has to have $x$ with value that minimizes $G(T,P,x)$.

For example, thermodynamic state of a puddle made of mixture of liquid water and kitchen salt can be described, in addition to $T,P$, by the amount of salt in dissolved state $n_d$ (assuming salt conservation). The principle of stable equilibrium then implies that in equilibrium, $n_d$ has value for which

$$ \frac{\partial G}{\partial n_d}(T,P,n_d) = 0, $$

$$ \frac{\partial^2 G}{\partial n_d^2}(T,P,n_d) > 0. $$

  • $\begingroup$ That makes​ sense. Do you know a reference where this is discussed and proved? $\endgroup$ – mlainz Mar 26 '17 at 16:17
  • $\begingroup$ @mlainz You can find the derivation from the entropy principle in the books Pippard: Elements of classical thermodynamics, chapter 7. The thermodynamic inequalities and Zemansky: Heat and thermodynamics, section 15.8 Conditions for chemical equilibrium. $\endgroup$ – Ján Lalinský Mar 27 '17 at 0:57

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