Why is the internal energy discontinuous in a first-order phase transition?

Question

Using the Ehernfest classification where first order phase transitions are those where the 1st derivative of the free energy has a discontinuity, I can follow why the entropy and volume are discontinuous $S=-\frac{\partial G}{\partial T}$ & ${V}=\frac{\partial {G}}{\partial P}$. However it is not to clear to me what relationship results in the discontinuity for the internal energy.

Answer 1

Once we have a closed thermodynamic system having a discontinuity of the first derivatives of the Gibbs free energy $G$ as a function of temperature and pressure, the discontinuity of the internal energy $U$ as a function of the same variables is a trivial consequence of the relation between $G$ and $U$: $$ U(T,p)=G(T,p)+TS(T,p)-pV(T,p) $$ $G(T,p)$ is a continuous function of the variables $T$ and $p$ (it is a consequence of its concavity). Therefore, the discontinuities of $S$ and $V$ imply the discontinuity of $U(T, p)$.

Notice that the functional dependence on $p$ and $T$ is essential. The internal energy, as a function of $S$ and $V$, is continuous. A rarely stressed side remark on Ehrenfest classification is that it is based on the order of derivatives of a thermodynamic potential as a function of the intensive variables only.