Legendre transform from internal energy to Gibbs free energy

Let the internal energy be written in the energy representation $U=U(S,V,N)$ (see internal energy ). The conjugate variables are temperature $T=(\partial U/\partial S)_{V,N}$ and pressure $p=-(\partial U/\partial V)_{S,N}$.

Definition (Gibbs free energy as a double Legendre transform).
The Gibbs free energy $G(T,p,N)$ is obtained by trading $(S,V)$ for $(T,p)$ via a two-variable Legendre transform :

At the minimizer $(S^\star,V^\star)$, the stationarity conditions enforce

Equivalently, one can perform the transform sequentially: first form the enthalpy $H(S,p,N)$ by transforming in $V$, then transform in $S$ to get $G=H-T S$; or start from the Helmholtz free energy and transform in $V$ to get $G=F+pV$ at the volume where $p=-(\partial F/\partial V)_{T,N}$.

Differential and physical meaning.
The Gibbs free energy obeys

so at fixed $(T,p,N)$ the equilibrium macrostate minimizes $G$. This is why $G$ (the Gibbs free energy ) governs phase coexistence and chemical equilibrium under laboratory conditions where $T$ and $p$ are controlled.

Statistical-mechanical representation.
In the isothermal–isobaric ensemble , $G$ is obtained from the NpT partition function $\Delta(T,p,N)$ by the same rule as other equilibrium potentials:

Derivatives of $G$ recover response variables; for example, the chemical potential satisfies $\mu=(\partial G/\partial N)_{T,p}$, connecting this construction to chemical potential from entropy and the thermodynamic chemical potential .