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The last paragraph of page 1352 of IUPAC Technical Report Use of Legendre transforms in chemical thermodynamics" (Alberty, R.A., Pure Appl. Chem., Vol. 73, No. 8, pp. 1349–1380, 2001) says that

The natural variables of a thermodynamic potential are important because when a thermodynamic potential can be determined as a function of its natural variables, all of the thermodynamic properties of the system can be calculated.

And give the example:

All of the thermodynamic properties can be calculated if $U$ is determined as a function of $S, V,$ and $\{n_i\}$, but not if U is determined as a function of $T, P,$ and $\{n_i\}$, or other set of $N+2$ independent variables.

Why not? If I know a thermodynamic potential as a function of any $N+2$ independent variables, can't I calculate all other thermodynamic properties?

(This question may be related to this one.)

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The commonly referred-to as natural variables of a thermodynamic potential are those which appear "naturally" in its total differential. For example, from the first principle of thermodynamics, the internal energy satisfies: $$ dU=TdS-pdV+\mu dN $$ thus $U=U(S, V, N)$ is expressed in such a way that any thermodynamic function of the system is expressed as derivative of the internal energy: $$ T=\left(\frac{\partial U}{\partial S}\right)_{V,N}\qquad p=-\left(\frac{\partial U}{\partial V}\right)_{S,N}\qquad \mu=\left(\frac{\partial U}{\partial N}\right)_{V,S}. $$ Upon action of the Legendre transform $H=U+pV$ enthalpy has:$$ dH=TdS+Vdp+\mu dN \implies H=H(S,p,N). $$ And so on: Helmholtz free energy $F(T,V,N)=U-TS$, Gibbs free energy $\Phi(T,p,N)=H-TS$ and the grandpotential $\Omega(T,V,\mu)=F-\mu N.$

The use of such functions is determined by the experimental setting you can produce for a given system: for example a low-density gas can be easily held at a given volume and given number of particles, at a certain temperature, for it can be enclosed in a box and its volume varied with a piston. Thus it is natural to use Helmholtz free energy, which naturally embeds $T,V,N$ as free variables.

Other materials might be less easy to compress or dilate, so one might choose to vary their pressure, which would lead to using Gibbs free energy in the variables $T,p,N$.

The example you quoted is correct exactly in this sense: if you know the functional dependence of $U$ on its natural variables $U=U(S,V,N)$ then you can use the formulas listed above; but if you have, say, $U=\tilde{U}(T,V,N)$ then how do you calculate the equilibrium entropy $S$?

You have to switch to another thermodynamic potential: $F=F(T,V,N)$ using the Legendre transform: $$ F=F(T,V,N)=U\big(S(T,V,N),V,N\big)-TS(T,V,N) $$ where $S(T,V,N)$ indicates the expression of $S$ obtained by inversion of the relation $$ T=\left(\frac{\partial U}{\partial S}\right)_{V,N}. $$

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  • $\begingroup$ Could you please also comment on the second quote? $\endgroup$ – toliveira Jan 11 '15 at 19:15
  • $\begingroup$ Added an explanation of the example you mentioned $\endgroup$ – Brightsun Jan 11 '15 at 19:34

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