# Why do $C_p$, $\alpha$ and $\kappa_T$ all approach to infinity when two phases coexist at the mean time?

My textbook (writed by Chinese) says that

when two phases of a system coexist at the same time, this system's isobaric thermal capacity $C_p=T(\frac{\partial S}{\partial T})_p$, coefficient of thermal expansion $\alpha=\frac{1}{V}(\frac{\partial V}{\partial T})_p$ and isothermal compressibility $\kappa_T=-\frac{1}{V}(\frac{\partial V}{\partial p})_T$ all approach to infinity.

However, it doesn't give the physical explanation to me. I wanted to know the true mechanism, and I have tried to derive those from Clapeyron-Clausius equation, but I failed. So I hope someone could tell me and analyze that why them approach to infinity under the circumstance.

• Take the heat capacity. If there are two phases coexisting with each other, than any added heat will go towards the enthalpy of the phase transition (shifting the relative amounts of the phases), not raising the temperature of the system. So, energy in, no change in T makes for an infinite heat capacity. – Jon Custer May 3 '16 at 13:17
• @JonCuster Do you mean entropy? – Wang Yun May 4 '16 at 1:14
• @JonCuster Because there are two phases coexisting with each other, the system should arrive at the equilibrium state. And entropy is maximum. Why will entropy rise if the system absorbed heat? – Wang Yun May 4 '16 at 1:20
• No, I really mean enthalpy. Remember that $S = -\partial G / \partial T$. But, in the bigger scheme of things, you are focusing on the minutiae of the equations (and there are many in thermodynamics!) at the expense of the bigger picture. At a first-order phase transition, the temperature does not fully specify the state of the system - you have a mixture, and the relative amounts of the phases depend on how much enthalpy you have put in to the system. Since you don't have fixed quantities of the phases, how can you define any of the quantities you list? – Jon Custer May 4 '16 at 12:56

In single phase system, you can have $$dQ =m C_p dT$$ or $$C_p = \frac {dQ}{m dT}$$ When two phases co-exist, the heat added will not increase system's temperature. The heat is consumed by latent heat for phase changing from for example water to vapor. So if you still use, $$C_p = \frac {dQ}{m dT}$$, you have finite value for dQ but you have zero for dT. This gives you mathematically infinite $C_p$.
• So, how about $\alpha$ and $\kappa_T$ ? – Wang Yun May 5 '16 at 3:51