# Thermodynamic Stability - Convexity - Concavity of Thermodynamical potential

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".

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?

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.
• All potentials are indeed concave in $P$ and $T$. Is this what you are asking? – Themis Dec 19 '18 at 11:35
• 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? – Themis Dec 19 '18 at 12:00