# 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.

## 1 Answer

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}