pressure effect

Pressure Effects
The equilibrium temperatures at a specific pressure (1 atm). At other pressures the equilibrium temperatures will differ.
For example Fig.1 shows the effect of pressure on the equilibrium temperaturees for pure iron. Increasing pressure has the effect of depressing the ?/? equilibrium temperature and raising the equilibrium melting temperature. At very high pressures hcp ?-Fe becomes stable. The reason for these changes
derives from this equation: dG=-SdT+VdP

At constant temperature the free energy of a
phase increases with pressure such that
(?G/?P)_T=V

If the two phases in equilibrium have different molar volumes their respective free energies will not increase by the same amount at a given temperature and equilibrium will, therefore, be disturbed by changes in pressure. The only way to maintain equilibrium at different pressures is by varying the temperature.
If the two phases in equilibrium are ? and ?, application of free energy to 1 mol of both gives
?dG?^?=V_m^? dP-S^? dT
?dG?^?=V_m^? dP-S^? dT




If ? and ? are in equilibrium G^?=G^? therefore ?dG?^?=?dG?^? and

(dP/dT)_eq=(S^? ?-S?^?)/(V_m^? ?-V?_m^? ) =?S/?V

This equation gives the change in temperature dT required to maintain equilibrium between ? and ? if pressure is increased by dP. The equation can be simplified as folIows. From
G=H-TS
thus
G^?=H^?-TS^? G^?=H^?-TS^?

Therefore, putting ?G=G^? ?-G?^?, gives

?G = ?H - T?S
But since at equilibrium ?G ^?= G?^?, ?G = 0, and
?H - T?S = 0
Consequently


(dP/dT)_eq=?H/T?V


which is known as the Clausius-Clapeyron