 Low-temperature Gibbs free-energy expansion

At fixed pressure and material content, $S=-(\partial G/\partial T{)}_p$. If the equilibrium entropy tends to zero, the Gibbs free energy approaches its zero-temperature value with zero slope. When $C_p=a{T}^r+o(T^r)$ with $r>0$ and zero residual entropy, integrating $dS=C_{p}dT/T$ gives $S=(a/r)T^r+o(T^r)$ and $G=G(0,p)-a{T}^{r+1}/[r(r+1)]+o(T^{r+1})$. Nonzero residual entropy $S_0$ instead contributes $-S_{0}T$ to $G$; a horizontal limiting tangent requires the zero-entropy hypothesis.