1. Classical description of a single component system
Lecture 10
Classical description of a single component system
Gibbs free energy and equilibrium phases as a function of temperature
Phase diagram
Molar properties and Clapeyron equation
Problem 11.4

2. Gibbs free energy
Typically we control pressure and temperature - therefore for a single component system to find equilibrium phase we need to consider Gibbs free energy, G (P, T). In general for each phase, :
From chapter 4 we know that

3. G as a function of T for a given pressure
Entropy of vapor > entropy of liquid > entropy of solid
vapor
liquid
solid
At the melting point Gsolid = Gliquid
At the boiling point Gliquid = Gvapor

4. Enthalpy Since G = H-TS also Latent heat - heat of fusion
vapor
liquid H
solid
Latent heat - heat of fusion
Heat of vaporization

5. Phase diagram
solid
liquid
vapor

6. Molar properties - converting extensive into intensive variables
Mole fraction
Molar entropy etc.

7. Gibbs-Duhem equation The original equation for each phase Dividing by
Gives for two phases,  and 

8. Clapeyron equation Along the coexistence line
Also for a single component in  and  phases
Therefore
From which
Since in equilibrium coexistence

9. Coexistence with vapor
slope

10. Problem 11.4
Calculate the vapor pressure at 300 K of an element for which melting point is 1000 K, the normal boiling point is 2500 K, the heat of vaporization is 240, 000 J/mol and the heat of fusion is 12,000 J/mol.