Phase boundaries
Vapor pressure
from
At phase boundary:
Dynamic equilibrium
between two phases
∆G = 0
Nils Walter: Chem 260
Phase boundaries: Where are they?
Phase 1: dGm(1) = Vm(1)dp - Sm(1)dT
Phase 2: dGm(2) = Vm(2)dp - Sm(2)dT } in equilibrium
Along the phase boundary, the molar Gibbs energies stay equal
⇒ the changes in their molar Gibbs energies must be equal
Vm(1)dp - Sm(1)dT = Vm(2)dp - Sm(2)dT
⇒ [Sm(2) - Sm(1)]dT = [Vm(2) - Vm(1)]dp
∆trsS ∆trsV
dp ∆ trs S
⇒ =
dT ∆ trsV
Clapeyron equation
Nils Walter: Chem 260
Special case: The liquid-vapor boundary
dp ∆ vap S ∆ vap H ∆ vap H ∆ vap H p∆ vap H
= = = = =
dT ∆ vapV T∆ vapV TVm ( g ) T ( RT / p ) RT 2
trs → vap ∆vapS ∆vapV perfect gas
= ∆vapH/T ≈ Vm(g); approximation
Vm(l) small
p' T'
∆ vap H
d ln p ∆ vap H
= 2
∫ d ln p = ∫
p T
RT 2
dT
dT RT
Clausius-Clapeyron p' ∆ vap H 1 1
equation ⇒ ln = −
p R T T'
Nils Walter: Chem 260
Characteristic points
Same as melting point;
“normal” = at 1 atm Highest T for liquid
Closed vessel:
Pressure increases
until critical point is
Lowest T for reached (Tc, pc);
liquid phase boundary is lost
Open vessel:
Only set of
Vapor pressure equals
conditions
external pressure
where all three
phases coexist ⇒ vapor drives back
atmosphere: Boiling
(water: 273.16 K,
Nils Walter: Chem 260
611 Pa)
How many phases can coexist in equilibrium?
Four phases: Gm(1) = Gm(2); Gm(2) = Gm(3); Gm(3) = Gm(4)
BUT: Only two unknown parameters (p, T) in a phase diagram
⇒ Four phases cannot coexist
in equilibrium!
line
point
Phase rule: F = C - P + 2
F = Number of degrees of freedom
C = number of components
P = number of phases
area
Nils Walter: Chem 260
Phase diagrams: Water
dp ∆ trs S
=
dT ∆ trsV
Liquid water
has a higher
density than
water ice
⇓
ice skating!
frozen ponds
sustain fish!
On Mt. Everest “boiling” eggs
becomes easier
Nils Walter: Chem 260
Phase diagrams: CO2 and Helium
CO2: quite typical He: solid and gas are
never in equilibrium;
He-II is superfluid
Sublimes at
1 atm; exists
as a liquid
only under
pressure
Nils Walter: Chem 260