# Partial derivative of the Gibbs free energy with respect to temperature at constant enthalpy

I am trying to find the expression for $$\left(\frac{\partial G}{\partial T}\right)_H$$ as a function of the entropy $$S$$, temperature $$T$$, $$C_p$$ and $$\alpha=\frac1V\left(\frac{\partial V}{\partial T}\right)_p$$.

So far from the definition of enthalpy $$H=U+pV$$ and the differential form of the internal energy equation $$\mathrm dU=T\,\mathrm dS-p\,\mathrm dV$$ when setting $$\mathrm dH=0$$ due to the process being isenthalpic, I got that $$\left(\frac{\partial T}{\partial S}\right)_p=-\left(\frac{\partial V}{\partial p}\right)_S$$ but from this point I got stuck trying to somehow use it in the differential form of the Gibbs free energy equation.

We differentiate the differential form of the Gibbs free energy with respect to the temperature at constant enthalpy \begin{align} \mathrm{d}G &= V\mathrm{d}p - S\mathrm{d}T \\ \left(\frac{\partial G}{\partial T}\right)_H &= V\left(\frac{\partial p}{\partial T}\right)_H - S \\ \left(\frac{\partial G}{\partial T}\right)_H &= \dfrac{V}{\color{blue}{\left(\dfrac{\partial T}{\partial p}\right)_H}} - S \tag{1} \\ \end{align} and we identify in Eq. (1) the Joule-Thomson coefficient. Replacing it by its known expression leads to the final result after some algebra \begin{align} \require{cancel} \left(\frac{\partial G}{\partial T}\right)_H &= \dfrac{V}{\color{blue}{\dfrac{1}{C_p} \left[T\left(\dfrac{\partial V}{\partial T}\right)_p - V\right]}} - S \\ &= \dfrac{V}{\dfrac{1}{C_p} \left[T\dfrac{V}{V}\left(\dfrac{\partial V}{\partial T}\right)_p - V\right]} - S \\ &= \dfrac{V}{\dfrac{1}{C_p}(T\alpha V - V)} - S \\ &= \dfrac{\cancel{V}}{\dfrac{\cancel{V}}{C_p}(T\alpha - 1)} - S \rightarrow \boxed{\left(\frac{\partial G}{\partial T}\right)_H = \dfrac{C_p}{T\alpha - 1} - S} \end{align}