Friday, November 2, 2007

Problem 2 on the midterm practice problems

I was told there may be some confusion about part 2b. Here is how to approach that problem.

For any phase, we can write the entropy as a function of the pressure and temperature:

S = S(T,P)

The total derivative is:

dS = (dS/dT) dT + (dS/dP) dP

= (Cp/T) dT - (alpha V) dP

Where Cp is the constant pressure heat capacity and alpha is the thermal expansion coefficient. In the second line, the (dS/dP) term was converted with a Maxwell relation to get -(dV/dT).

If we have a line along which the entropies are equal between two phases:

S1 = S2 (along Kauzmann locus)

The full differential must also be equal:

dS1 = dS2 (along Kauzmann locus)

Substituting the above relation:

(Cp1/T) dT - (alpha1 V1) dP = (Cp2/T) dT - (alpha2 V1) dP
(along Kauzmann locus)

Now simplifying:

(dP/dT) = Delta(Cp) / [T Delta(alpha V)] (along Kauzmann locus)

In general, the Kauzmann locus will extend into regions of (T,P) space where one is comparing a metastable state to a stable one. In the helium example given, the Kauzmann locus occurs in some parts of the phase diagram where the liquid is supercooled and the crystal is stable. It also occurs in places where the crystal is superheated and the liquid is stable. Where the Kauzmann locus intersects the phase boundary between the liquid and crystal, there is a Kauzmann point.

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