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 1


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}


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