Show that, for a transition between 2 incompressible solid phases, that deta G is indepentent of the pressure.

Answer 1

Fundamental Thermodynami Relation:

dU = T dS - P dV (1)

Gibbs energy is:

G = U + P V - T S

dG = dU + P dV + V dP - T dS - S dT =

(insert (1) ) =

V dP - S dT

The two solid phases have different Gibbs functions, let's call then G1 and G2. Both functions satisfy the above relation:

dG1 = V1 dP - S1 dT

dG2 = V2 dP - S2 dT

The change in the difference is thus given by:

d(G2 - G1) = (V2 - V1)dP - (S2 - S1)dT

And it follows that the partial derivative of G2 - G1 w.r.t. P at constant temperature is V2 - V1. If the solid is incompressible, then the volume will stay constant during the change and V1 will be the same as V2, so Delta G will then be independent of P.

Answer 2

Good question