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The thermodynamic stability principle requires convexity of the internal energy upon all of its independent variables.

When we pass through the Legendre transforms to build all the other thermodynamic potential, the thermodynamic stability principle is stated as "Thermodynamical potentials $(H, F, G)$ must be concave on their intensive variables and convex on their extensive ones".

I've got one question about this argument.

If i take the Helmholtz Free Energy $F(T,V,N)$, following the last statement one would say: F must be concave on T and convex on V [m^3] and N [mol].

It is reasonable to infer that if i take F(T,v,N) with v [m^3/kg] now F must be concave on v?

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The free energy $F=F(T,V,n)$ is convex in $V$ and $n$ and also homogeneous with degree 1. By homogeneity we have

$$ F(T,V,n) = n F(T,V/n,1) $$ where $F(T,V/n,1)\equiv f(T,v)$ is the intensive free energy (J/mol) and $v=V/n$ is the intensive volume (m$^3$/mol). It follows that since $F$ is convex in $V$, then the intensive free energy $f$ is convex in $v$.

Whether we express the intensive free energy per mole, as I did, or per mass, as you did, it makes no difference. However, the intensive free energy is a function of two intensive properties ($T$ and $v$) not three, as you wrote.

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  • $\begingroup$ Thank you for your answer. I have to apologize for the imprecision, I really meant concavity for intensive variables and convexity for the extensive ones. In any case, following your reasoning, the concavity of thermodynamical potentials must be assured just for temperature and pressure, am I right? $\endgroup$
    – iterrate
    Commented Dec 18, 2018 at 14:12
  • $\begingroup$ All potentials are indeed concave in $P$ and $T$. Is this what you are asking? $\endgroup$
    – Themis
    Commented Dec 19, 2018 at 11:35
  • $\begingroup$ My question would be more: are there other variables for which a potential should be concave? As far as I know, entropy must be concave in internal energy, for example. $\endgroup$
    – iterrate
    Commented Dec 19, 2018 at 11:41
  • $\begingroup$ Here is how to think about this: The standard thermodynamic potentials are $U(S,V,n)$, $H(S,P,n)$, $F(T,V,n)$, and $G(T,P,n)$. These are convex with respect to all extensive arguments, and concave with respect to all intensive arguments. This applies only to the arguments that appear in the potentials as written above. For example, we cannot say anything about the the curvature of $U(T,P,n)$ with respect to $T$ and $P$, because these are not "proper" variables for $U$. Makes sense? $\endgroup$
    – Themis
    Commented Dec 19, 2018 at 12:00
  • $\begingroup$ Thank you again for the really clarifying explanation. Let's say we have a potential that is not expressed as a function of the "proper" variables, is there a way to establish any criteria to ensure thermodynamic stability based on the curvature of the potential? $\endgroup$
    – iterrate
    Commented Dec 19, 2018 at 12:58

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