MSEG 803
Equilibria in Material Systems
11: Phase Transition
Prof. Juejun (JJ) Hu
hujuejun@udel.edu
Ehrenfest classification of phase transitions

First order phase transition



Transitions that exhibit a discontinuity in the first derivative of
TD potential with respect to some thermodynamic variable
Boiling: liquid-vapor phase transition
G
G
 S
V
T
P
Both S and V are discontinuous in liquid-vapor transitions
Second order phase transition


Transitions that are continuous in the first order derivative of
TD potential discontinuous (divergent) in a second derivative
Second order derivatives:
Cp
 2G
 S 
 
 
2
T
T
 T  P
 2G
 M 



  
2
H
 H T
Equilibrium between phases


At constant T and P : 1  2
For a single phase, stable system: dG = - SdT + VdP
G
 S  0
T
Cp
 2G
 S 
 
0
 
2
T
T
 T  P
G
V  0
P
 2G  V 

  V T  0
2
P
 P T
G
G
T
P
Gibbs free energy of phases
G
S v  Sl  S s
Vapor
Gv
Gl
Gs



T
T
T
At Tm, Gl  Gs
 H l  Tm Sl  H s  Tm Ss
  H m  Tm  Sm
Liquid
Solid
Ts
Tm
Tb
T
Calculating phase transition temperature

At 293 K, the enthalpy and entropy changes during the
vapor-liquid phase transition of water is 44.17 kJ/mol and
119.5 J/mol·K-1, respectively. Estimate Tb of water.
  G 
T
  S
G293K  H 293K  T S293K  9.16kJ / mol
At boiling point, G = 0
Gb  G293 K  Tb  293K  
  G 
~ G293 K  Tb  293K   S293 K
G293 K
 Tb ~ 293K 
 370 K
S 293 K
T
0
Deviation from
373 K due to:
 S cP

0
T
T
Phase diagram (CO2)
Critical point
Triple point
Phase diagram (H2O)
Negative
slope
Slope of P-T coexistence lines

a
b
Along the coexistence curve dG = 0, i.e. G1  G1 , G2a  G2b
dG1a 2   S a dT1 2  V a dP1 2 
a
b
a
b
V

V
dP

S

S


 12 
 dT12

b
b
b
dG1 2   S dT12  V dP12 
dP  S
H


dT V T V
Clapeyron equation
P
P
a
a
1
b
T
2
b
T
Ice melting under pressure


A bag of mass M = 8 kg are hung by a rope over a rod of
ice. The contact area between the rope and the ice rod is
S = 20 mm2. The latent heat of fusion of water is 336
J/gram, the density of liquid water is 1 gram/cc and ice
cubes float with ~ 0.8 of their volume submerged.
Calculate the melting point of ice under the rope.
Solution:
336 18  J / mol

dP
H
6




4.9

10
Pa / K
6
3
dT T V 273K  4.5  10 m / mol

Mg
P 
 4  106 Pa
S
Tm  0.8 C

 dP 
 T   P 
  0.8 K
 dT 
Ice melts faster under high pressure
Clapeyron equation

The impact of pressure on phase transition temperature
is most significant for liquid-vapor transition


If you try to cook an egg in boiling water while camping in the
Rocky Mountains at an elevation of 10,000 feet, you will find that
it takes longer for the egg to cook because water boils at only
90oC at this elevation
The coexistence curve between solid-liquid phases are
almost perpendicular to the temperature axis

Typical values:
H s l ~ 10kJ / mol
Vs l ~ 1cc / mol
dP
~ 107 Pa / K ~ 102 atm / C
dT
Clausius-Clapeyron equation and vapor pressure

Vapor-liquid phase transition



Neglect the volume of liquid phase: Vv l  Vv
Assume the gas is ideal: PVv  NRT
Clausius-Clapeyron equation
dP
H
H
P H



dT T V TVv
NRT 2
 d ln P  
H m
1
d  
R
T 
P H m
 If  H m is temperature-independent: ln 1 
P2
R
 If H m  H m T0   cP  T  T0  ,
P
ln  1
 P2
 H m T0   cP  T0

R

 1 1  c
T 
     P  ln  1 
R
 T2 T1 
 T2 
1 1
  
 T2 T1 
Generalized Clapeyron equation

General system
dU  TdS   yi dxi
i

Legendre transform:   U  TS   yi xi
d   SdT   xi dyi
i
i

Generalized Clausius-Clapeyron equation
dyi
S
H


dT
 xi
T  xi
Degrees of freedom in single component systems




Variables in a single component system: T, P, and 
In a single phase system,  can be written as a function
of T and P (Gibbs-Duhem relation d    sdT  vdP ),
thus the independent variables are T and P
In a system containing two phases I and II, chemical
equilibrium condition  I   II reduces the number of
independent variables to 1
In a system containing three phases I, II and III, two
chemical equilibrium conditions  I   II   III leads to a
unique set of T and P values (0 independent variables at
the triple point)
Gibbs Phase rule





Consider a system containing r components and M
phases in equilibrium. The complete set of variables are
T, P, and x1, x2, … xr-1 for each phase, where xi is the
molar fraction of the component i
Only r-1 molar fractions are independent as  xi  1
The total number f of variables are thus 2 + M·(r-1)
Between each two phases I and II, chemical equilibrium
holds for all r components giving r·(M-1) equations:
1I T , P, xi   1II T , P, xi  , 2I  2II ,..., rI  rII
Thus the total number f of independent variables are:
f   2  M  r  1   r  M  1  r  M  2
Binary phase diagrams



Variables: T, P, and x1
Usually plotted for constant P values: f  r  M  1
An excellent phase diagram library:
http://www1.asminternational.org/asmenterprise/apd/

Categories



Solid solution: two components completely miscible
Eutectic
Peritectic
Solid solution
36%
66%
Calculating phase composition

Lever rule





1 and 2 two phases coexist in a x-y two component system
System composition: x component a%, y component (1-a)%
Phase 1 composition: x component a1%, y component (1-a1)%
Phase 2 composition: x component a2%, y component (1-a2)%
The system contains phase 1 b1% and phase 2 b2%
b1  a  a1   b2  a2  a 

b1  b2  100
Example: 0.5 Si - 0.5 Ge alloy at 1200 °C

a% = 50%, a1% = 36%, a2% = 66%
b1  a  a1   b2  a2  a   b1 : b2  8 : 7
 b1  53.3 b2  46.7
Eutectic
liquidus
solidus
Eutectic
Vapor-Liquid-Solid (VLS) growth of silicon
nanowires
Interface where growth
continues
Growth
direction
SiCl4 vapor + H2
Melted gold
nano-particles
saturated with
SiH4
Substrate
Substrate
Heterogeneous nuleation of Si
at the interface
Temperature for Si deposition on a planar substrate
(requires homogeneous nucleation): ~ 800 ˚C
Diameter of nanowire
is determined by the
gold droplet size
Nanowires formed by VLS growth
Gold nanoparticle
Growth
direction
“Germanium Nanowire Growth Below the Eutectic Temperature,”
S. Kodambaka, J. Tersoff, M. C. Reuter, and F. M. Ross, Science 316, 729 (2007).
Why gold?
Reason #1:
Low eutectic
temperature
Question: what would
happen if we were to
grow nanowires at
high temperature?
“The influence of the surface migration of gold on the
growth of silicon nanowires,” Nature 440, 69-71 (2006).
Why gold?
Reason #2:
Small gold
solubility in Si
(Eutectoid
transformation)
Question: why is
such small
solubility
important?
99.8%
Solubility: 0.2%
Albert P. Levitt, Whisker Technology (1975).
Tapering during nanowire growth
“A Systematic Study on the Growth of GaAs Nanowires by MetalOrganic Chemical Vapor Deposition,” Nano Lett. 8, 4275-4282 (2008).
Peritectic