Week 2 Monday January 14, 2013 page 1
solid-liquid equilibrium
Clausius-Clapeyron doesn’t apply
Clapeyron Equation:
since ∆Sfus =
∆Hfus
Tmp
∆Sfus = Slid - Ssol
P2
T2
∫ dP = ∫
P1
T1
dP
∆Hfus
=
dT T∆Vfus
∆Vfus = Vliq - Vsol
so
dP ∆Sfus
=
dT ∆Vfus
T2
∆Hfus 1
∆Sfus
dT = ∫
dT
∆Vfus T
T1 ∆Vfus
Assume ΔHfus and ΔSfus don’t change with T:
P2 -P1 ≈
∆Hfus
∆Sfus
(T2 -T1 ) ≈
(T -T )
∆Vfus Tmp
∆Vfus 2 1
That’s a good equation for really high pressures, like underneath glaciers.
sold-solid phase transition
polymorphism: when a compound forms several different crystal structures
example: ZnS has two forms:
1. zinc blende, diamond-like structure, tetrahedral network
2. wurzite, hexagonal
polymorphism in elements is called allotropy
example: carbon has 3 forms: 1. graphite 2. diamond 3. Buckyball
Sulfur has several allotropes: 1. rhombic = orthorhombic 2. monoclinic
Sulfur has 3 triple points:
1. rhombic, monoclinic, gas in equilibrium at 95°C
2. monoclinic, liquid, gas in equilibrium at 119°C
3. rhombic, monoclinic, liquid in equilibrium at 151°C
metastable: phase α is said to be metastable with respect to β if Gmα > Gmβ or µα > µβ
two boxes: one down flat and one up on end. The one up on end is metastable.
examples of metastable: diamond, superheated H2O(liq), supercooled liquid, supersaturated solution,
superheated water coming out of undersea geysers
higher order transitions
- everything we’ve done so far is first-order
liquid-gas or solid-liquid: Δ = qp ≠ 0
ΔV ≠ 0 1st order
∂H
CP = ( )
∂T P
CP at mp (melting point) is not defined
higher order transitions
For second order, CP increases by a finite amount.
CP = (
∂H
) does not apply to higher order transition
∂T P
ΔV = 0 ΔH = qP = 0
Example of second order transition: normal conductivity to superconductivity
For Hg: Ttran = 4.2K at 1atm
Clausius-Clapeyron is meaningless, since ΔV = 0
For lambda (λ) transitions, CP approaches ∞ from both sides
examples of lambda transitions: paramagnetism to ferromagnetism
liquid He(I) to liquid He(II)