Given the enthalpy and entropy change of a natural phase transition, derive an expression for the normal boiling point of any substance? Hint: consider that the change in Gibbs' free energy is zero for two phases in equilibrium.

1 Answer
Truong-Son N.
Jul 13, 2017

The hint tells you exactly what to do. Consider the relationship between the Gibbs' free energy, enthalpy, and entropy:

DeltaG = DeltaH - TDeltaS

At equilibrium (more specifically, phase equilibrium at constant temperature and pressure!), DeltaG = 0, so at liquid-vapor phase equilibrium:

DeltaG_(vap) = 0 = DeltaH_(vap) - T_bDeltaS_(vap)

=> color(blue)(T_b = (DeltaH_(vap))/(DeltaS_(vap))),

where T_b is the boiling point.

That is, you are somewhere on the green curved dotted line on a phase diagram:

![https://upload.wikimedia.org/](useruploads.socratic.org)

Once you obtain the temperature in "K", convert to ""^@ "C" by subtracting 273.15 from the "K" temperature.

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