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

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.

• Please clarify what type of explanation you're looking for. It takes energy (termed latent heat) to break bonds. All of this energy needs to supplied essentially in lumped form to turn the lower-temperature phase into the higher-temperature phase. Is this what you're asking about? Dec 9, 2022 at 21:18
• I'm confused on how the condition that the first derivatives of the Gibbs free energy are discontinuous leads to the result that the internal energy is also discontinuous. Dec 9, 2022 at 21:28

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.