Subsections

• Heat of Phase Transition
• Clapeyron-Clausius Formula
  • General Law
  • Consequences
  • Example: Liquid-Gas Phase Transition
• Order of Phase Transition

Phase Transitions

Heat of Phase Transition

Take a sample of phase I and slowly transform it into phase II. Some heat Q will be absorbed.

• Since this is reversible,  

  $$Q = T \Delta S$$ (1)

• Since in equilibrium P₁=P₂, the process occurs at constant pressure
• Since T₁=T₂, it occurs at constant temperature

We have

$$\Delta E = P\Delta V + T \Delta S$$

or

$$Q = T\Delta S = \Delta (E - PV) = \Delta H$$

The heat of phase transition is equal to the change of enthalpy.

Clapeyron-Clausius Formula

General Law

On the P-T diagram phase transition occur along a line P(T). Let's calculate the tangent to this line dP/dT.

Chemical potentials are equal:

$$\mu_1\bigl(P(T),T\bigr) = \mu_2\bigl(P(T),T\bigr)$$

Differentiating:  

$$\begin{gathered} \left(\frac{\partial \mu_1}{\partial T}\r... ...l \mu_2}{\partial P}\right)_{T,N}\frac{dP}{dT} \end{gathered}$$ (2)

We know from previous lectures that

$$G = \mu N$$

and therefore

$$\begin{gathered} \left(\frac{\partial \mu}{\partial T}\righ... ... G}{\partial P}\right)_{T,N} = -\frac{V}{N}=-v \end{gathered}$$

with s and v being quantities per molecule

From (2):

$$s_1 -v_1 \frac{dP}{dT} = s_2 - v_2 \frac{dP}{dT}$$

or

$$\frac{dP}{dT} = \frac{s_2 - s_1}{v_2-v_1}$$

Since

$$s_2 - s_1 = \frac{1}{N}(S_2-S_1) = \frac{\Delta Q}{TN} = \frac{q}{T}$$

we obtained  

$$\frac{dP}{dT} = \frac{q}{T(v_2-v_1)}$$ (3)

This is Clapeyron-Clausius equation

Consequences

• For liquid-gas phase transitions q>0. Since v₂>v₁, dT/dP= (dP/dT)^-1>0. Boiling point temperature increases with pressure
• For solid-liquid phase transitions q>0 (exception: $\mathrm{He^3}$). Usually v₂>v₁, and melting point increases with pressure. For water v₂<v₁, and it decreases--that is why skating & skiing is possible!

Example: Liquid-Gas Phase Transition

Assumptions:

1.

Phase transition heat q does not depend on T and P:

$$q = \mathit{const}$$

2.

Gas is ideal (PV=NkT), and its volume is much grater than of liquid:

$$v_1\approx0,\quad v_2 = \frac{V}{N} =\frac{kT}{P}$$

We obtain:

$$\frac{dP}{dT} = \frac{qP}{kT^2}$$

or

$$P = P_0\exp(-q/kT)$$

Order of Phase Transition

If we move from phase I to phase II, thermodynamic potential G is continuous (why?), but its first derivative S jumps. Another possibility: S is continuous, but $\partial S/\partial T$ jumps.

Definition:

if the n-th derivative of free energy

$$\left(\frac{\partial^n G}{\partial T^n}\right)_{N,P}$$

is discontinuous, and all lower derivatives are not, the transition is of n-th order.

Phase equilibrium:

first order phase transition!

Second order phase transitions:

Curie point, some liquid crystal phase transitions...

Practical guide:

if q=0, the transition is of the second order (or higher!).

First-order-close-to-second-order transitions:

q small (liquid crystals)