Transcript Document

PHASE TRANSITIONS AND CRITICAL
PHENOMENON
Phase Transitions
Conditions for phase equilibrium
First order Phase Transition:
Clausius - Clayperon equation
Second order phase transition
The critical indices
The law of corresponding states
21-Jul-15
Presentation by Dr. K.Y. Rajpure
1
Phase Transitions
Solid
Matter
rigid
sharp Bragg reflection ordered arrangement of
atoms or molecules
Liquid
fluid
no long range order
Gases
Phase transition
Transition of a matter from one phase to another, which coexists
with the first.
• Change in mutual arrangement of molecules
• Change in thermodynamical properties of the system
21-Jul-15
Presentation by Dr. K.Y. Rajpure
2
Phase transition points
 Solid - liquid: melting point
 Liquid - Vapour: boiling point
 Solid - Vapour: sublimation point
Three phases in equilibrium - at
definite T and P
Line OA - melting boundary
Line OB - sublimation boundary
Line OC - boiling (saturation)
boundary
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Presentation by Dr. K.Y. Rajpure
3
Melting Curve
Melting
Vaporization
Freezing
Condensation
Sublimation
Saturation Curve
Deposition
Sublimation Curve
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Presentation by Dr. K.Y. Rajpure
4
Phase transition points
Saturation curve: terminates
at C; Critical point .
O; Triple point
Pressure, P
P- V phase diagram of one component system
Solid
C
Gas
Triple point
Volume, V
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Presentation by Dr. K.Y. Rajpure
5
Types of phase transition
First order –
• Discontinuous change of int. energy and sp. Volume
• Heat evolved / absorbed
 e.g. Solid – Liquid, Liquid – Gas
Second order –
• Smooth change of int. energy and sp. Volume
• No heat evolved / absorbed
• e.g.
 Iron (Ferromagnetic) – paramagnetic state at Tc
 Liquid He I – Liquid He II at T
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Presentation by Dr. K.Y. Rajpure
6
Conditions for phase equilibrium
System - two phases 1 and 2 at equilibrium
Phase 1
N1, U1, V1
For isolated system:
U  U1  U 2  Const.
V  V1  V2  Const.
m  m1  m2  Const.
Phase 2
N2, U2, V2
Therefore,
U1  U 2
V1  V2
m1  m2
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Presentation by Dr. K.Y. Rajpure
7
In thermodynamics, S = Constant
S  S1  S2
S1  S1 (U1 ,V1 , m1 )
Since
 S1 
 S1 
 S1 
 U1  

 m1
S1  
V1  
 U1 V1 ,m1
 V1 U1 ,m1
 m1 V1 ,U1
From the thermodynamic considerations, we have
 S1 
1



 U1 V1 ,m1 T1
 S1 
P1



 V1 U1 ,m1 T1
 S1 
1



T1
 m1 V1 ,U1
1 is the chemical potential.
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Presentation by Dr. K.Y. Rajpure
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1
P1
1
S1  U1  V1  m1
T1
T1
T1
Similarly,
P2
2
1
S 2  U 2  V2  m2
T2
T2
T2
S  S1  S2  0
We have,
1
P1
1
1
P2
2
S  U1  V1  m1  U 2  V2  m2  0
T1
T1
T1
T2
T2
T2
1
P1
1
1
P2
2
S  U1  V1  m1  U1  V1  m1  0
T1
T1
T1
T2
T2
T2
U1  U 2 V1  V2 m1  m2
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Presentation by Dr. K.Y. Rajpure
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1 1
 P1 P2 
 1  2 
  U 1    V1   
m1  0
 T1 T2 
 T1 T2 
 T1 T2 
This is possible only when the coefficients
vanish individually.
 1  2 
1
1 
 
  0
    0
 T1 T2 
 T1 T2 
of U1,  V1 and  m1
 P1 P2 
    0
 T1 T2 
Therefore condition for phase equilibrium is
T 1 = T2
P1 = P2
1 = 2
(Thermal Equilibrium)
(Mechanical Equilibrium)
(Chemical Equilibrium)
Thus if two phases are in equilibrium; temperatures, pressures, and
chemical potentials of these phases must be equal.
21-Jul-15
Presentation by Dr. K.Y. Rajpure
10
We know that G (T, P, N ) = N G (T, P )
Where N: is the number of particles and
G (T, P ): Gibb’s free energy per particle
Also G =  N (From GCE theory)
So if 1 = 2  G1 = G2
If the chemical potentials are expressed as the functions of
pressure and temperature then,
1(P, T) = 2(P, T)
Equivalently, the molar Gibb's free energies of the two phases are
equal i.e.
G1(P, T) = G2(P, T)
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Presentation by Dr. K.Y. Rajpure
11
G1(P, T) = G2(P, T)
Phase 1
2
=G
1
G1 < G2
G
Pressure, P
Which defines the curve in the
(P, T) as shown. At a point on
the curve the Gibb's free
energies of two phases are
same. Two phases can't be in
equilibrium with each other at
all pressures and temperatures.
Two phase equilibrium curve (G1 - G2)
separating two phases 1 and 2
G1 > G2
Phase 2
Temperature, T
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Presentation by Dr. K.Y. Rajpure
12
First order Phase Transition: Clausius - Clayperon equation
Consider two phases (1 and 2) in equilibrium.
The molar Gibb’s free energies of the two phases are equal
G1(P, T) = G2(P, T)
(1)
Or g(P, T) = G1(P, T)  G2(P, T) = 0
(2)
Solution of equation (2) –
 phase equilibrium curve of P as function of T
o Melting curve Solid – Liquid transition
o Pressure curve Liquid – Gas transition
Equilibrium of three phases is governed by
G1(P, T) = G2(P, T)
G2(P, T) = G3(P, T)
(3)
Solutions of these equations (Ptr, Ttr); called Triple point
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Presentation by Dr. K.Y. Rajpure
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21-Jul-15
Presentation by Dr. K.Y. Rajpure
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(a)

G
Phase 2
Phase 1
T
In Fig. (a): molar Gibb’s free
T0
energy vs. T at constant P

U
Phase 1
Phase 2
At T0: Two phases in equilibrium
(b)
T < T0: First phase stable
T > T0: Second phase stable
T

S
Phase 1
Phase 2
(c)
T

CP
Phase 1
Phase 2
(d)
T
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Fig. The variation of G, , S
and CP during the first order
transition
Presentation by Dr. K.Y. Rajpure
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 G 
 G 
 1    2 
 T  P  T  P
Change T of phase by dT and P by dP : Two phases in equilibrium in
new state, Gibb’s free energies of new states: g(P + dP, T + dT) = 0
Lets make Taylor series expansion of g:
 g 
 g 
g ( P  dP, T  dT )  g ( P, T )    dP  
 dT  0

P

T
 T

P
At equilibrium g(P + dP, T + dT) = 0 and g(P, T) = 0.
 g 
 g 
  dP  
 dT  0
 P T
 T  P
dP  sdT  0
since
where
 - molar volume &
s - the molar entropy
(s2  s1 )dT  ( 2 1 )dP
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 g 
    and
 P T
 g 

  s
 T  P
Clausius - Clayperon equation
dP ( s2  s1 ) S


dT ( 2  1 ) V
Presentation by Dr. K.Y. Rajpure
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Second order Phase Transition
A phase transition of the second order is continuous in the sense that
the state of the system changes continuously, but symmetry changes
discontinuously at the transition point.
In case of the transition between different crystal modifications, there is
a sudden rearrangement of the crystal lattice and the state of the
system changes discontinuously.
Ba
O
Ti
Cubic lattice of BaTiO3
21-Jul-15
Cubic lattice:
Ba atoms – vertices, O – face centers and Ti – Body center
As T  below Tc,
Ti and O atoms  relative to Ba atoms, || el to the edge of
the cube.
Symmetry changes – becomes tetragonal, instead of cubic.
In this process configuration of the crystal changes
continuously.
Such transition from one crystal modification to another is
called Second - order phase transition.
Presentation by Dr. K.Y. Rajpure
17
 The change in symmetry of the system takes place not only due to displacement
of atoms, but also due to change in the ordering of the crystal.
o e.g. consider BCC lattice of   brass at low temp.
 completely ordered alloy of Cu and Zn
 cubic lattice, [Fig. (a) below]
 Zn atoms at vertices, and Cu atoms at center of cell.
Zn
Cu
Fig. (a)
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Fig. (b)
Presentation by Dr. K.Y. Rajpure
When alloy heated –
disordered.
Cu and Zn – change
their positions (Zn at
center of cell, and Cu at
vertices) [Fig. (b) ]
18
o There are probabilities of finding atoms of either kind at every lattice site
until the probabilities of finding Cu or Zn atoms at the vertices and at the
center of the cells become equal – symmetry unchanged
o When the probabilities of finding Cu or Zn atoms at the vertices and at the
center of the cells become equal – symmetry raised.
o Crystal – B.C.C. Bravais lattice – Disordered.
Ehrenfest – suggested the possibility of Second order phase transition
o where the second derivative of G might be finite
o but different for two phases (1 and 2) in equilibrium.
21-Jul-15
Presentation by Dr. K.Y. Rajpure
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Consider phase transition between two phases (1 and 2) in
equilibrium and let,
g = G1  G2
Thus
g(P, T)
=0
Change T by dT and P by dP so as to be in equilibrium,
g(P+ dP, T+ dT)
=0
Lets make Taylor series expansion of
g:
 g 
 1  2 g

2 g
2 g
 g 
2
g ( P, T )    dP    dT    2 (dP)  2
dPdT  2 (dT ) 2   ....  0
PT
T
 T  P  2  P

 P T
g is always zero at equilibrium
21-Jul-15
Presentation by Dr. K.Y. Rajpure
20
For satisfaction of above equation, all the bracketed terms = 0
 g 
 g 
  dP  
 dT  0

P

T
 T

P
which leads to the Clausis - Clapeyron equation (Ist order phase
transition).
When,
terms.
 g 
 P 
&
 g 
 T 
g  2 g
2 g

dP 
dT  0
P P 2
PT
g
2 g
2 g

dT  2 dT  0
T PT
T
individually zero - consider second order
Adding two
2 g
2 g
2 g
2
2
(
dP
)

2
dPdT

(
dT
)
0
2
2
P
PT
T
Second order phase transition.
21-Jul-15
Presentation by Dr. K.Y. Rajpure
21
This corresponds to a phase transition of second order. If such an
equality holds for an arbitrary dP and dT , the following
conditions must be satisfied:
2 g
2 g
2 g
P
 g
0
T 2
2
 g
0
P 2
2
2
(dP)2  2
PT
dPdT 
T
2
(dT )2  0
2
2 g 2 g  2 g 

 0
2
2
T P  PT 
Thus
 2 (G1  G2 )
0
2
T
 2G1  2G2

2
T
T 2
 2 (G1  G2 )
0
2
P
 2G1  2G2

2
P
P 2
 2G
s
Cp




T 2
T
T
 2G v

2
P
P
21-Jul-15
 2G1  2G2

2
T
T 2
 2G1  2G2

P 2
P 2
 2G
v

PT T
Presentation by Dr. K.Y. Rajpure
and
 g 
 g 

 dP  
 dT  0
 P T
 T  P
 G 

 
 P T
 G 

  s
 T  P
22
 2G
s
Cp




T 2
T
T
 2G v

P 2 P
 2G
v

PT T
g  2 g
2 g
 2 dP 
dT  0
P P
PT
g
2 g
2 g

dP  2 dT  0
T PT
T
v
v
dP 
dT  0
P
T
v
s
dP 
dT  0
T
T
dP v / T

dT
v / P
dP   (v) / T  P

dT   (v) / P T
dP s / T

dT v / P
dP   (s ) / T  P

dT   (v) / P  P
v  v2  v1
s  s2  s1
At a second order phase transition, not only the thermodynamic
potentials are continuous, but their first derivative (G/P)T and
(G/T)P are also continuous; however their second derivatives
undergo a jump.
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Presentation by Dr. K.Y. Rajpure
23
The critical indices
Phase equilibrium curve
terminates at C; Critical point
Beyond C - No difference of
phases
Thus at C - Two phases
becomes identicalAt C,
isotherm has a point of
inflexion satisfying
From thermodynamics

1  P 

 0
Cv  V T
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 P 
   0
  T
 2P 
 2   0
   T
The Critical point is defined as the point
where this equality holds.
Presentation by Dr. K.Y. Rajpure
24
The critical indices………...
• Many thermodynamic properties - anomalous values near C.
• A basic problem - phase transition: to study the anomalous
behaviour of a physical system near C.
• These anomalies - large density fluctuations characteristic of
the critical region.
• These fluctuations give rise to singularities in many
observable quantities.
• The nature of singularities - expressed in terms of a set of
critical indices.
• Near C, property varies - simple power of the temperature
difference from the critical point.
21-Jul-15
Presentation by Dr. K.Y. Rajpure
25
The critical indices………...
The singular behaviour of a physical quantity X is usually
described by a simple power law: X  T  Tcm
m - positive number, critical exponent, which is determined by
 ln X 
m  Lim 

T Tc  lnT  Tc  
The anomalous behaviour of a number of physical properties can
be described by similar power laws.
e.g. consider the isothermal compressibility KT defined by
KT  
21-Jul-15
1  V 


V  P T
becomes infinity at C.
Presentation by Dr. K.Y. Rajpure
26
The critical indices………...
If the divergence is represented as a power law in
| T – Tc |, then for T > Tc the singular behaviour of KT
along the critical isochore ( = c) is represented as
KT  A | T  Tc | ; T  Tc
where  is a positive constant, called a critical exponent.
For T < Tc the divergence of KT can be represented by
KT  A'| T  Tc | ' ; T  Tc
The critical exponent ' is not necessarily equal to .
21-Jul-15
Presentation by Dr. K.Y. Rajpure
27
Fig – critical isotherm (i.e. for T = Tc)
horizontal tangent C (Vc, Pc)
If the position of this line near C is given by P – Pc, then
Pressure, P 
P  Pc  D(T  Tc)
Pc
T > Tc
C
T = Tc
T < Tc
Vc
Volume, V 
Isotherm for a gas – liquid phase transition
21-Jul-15
Presentation by Dr. K.Y. Rajpure
28
The critical indices………...
If the position of this line near C is given by P - Pc, then
P  Pc  D(T  Tc)
• The numerical values of the critical exponents - Identical for all
substances.
• critical exponents - independent of nature of the substance.
• critical exponents below and above Tc - same.
 '
  '
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Presentation by Dr. K.Y. Rajpure
29
The law of corresponding states
We know the Van der Waals equation of state
a 

 P  2 V  b  NkT
V 

The Van der Waals curves give
maxima and minima in the region
represented by a horizontal line in
the experimental curve.
These maxima and minima points
come closer and closer with the
rise in temperature. There exist a
temperature Tc, called Critical
Temperature, at which the kink in
the isotherm (i.e. maxima and
minima) disappears. The point of
inflection is called the Critical
point.
P
This equation has been applied empirically to study the real gases over a
wide range of densities and temperature.
C
Pc
T = Tc
T < Tc
V1
Vc V2
V3
Volume, V 
21-Jul-15
Presentation by Dr. K.Y. Rajpure
30
It is only below T < Tc that the Van der Waals equation of state give a kink
in an isotherm.
For a given P and T < Tc, P has generally three roots in V (e.g. V1, V2 and V3).
The critical point for Van der Waals gas can be determined by the equation
P
NkT
a
 2
V  b V
A
By imposing the conditions
NkT
2a
 P 

0

 
2
3

V
V


V b

T
 P
2 NkT 6a
 2  
 4 0
3

V

T V  b  V
2
Equations A and B on solving gives
Vc  3b
8a
Tc 
27 bNk
a
Pc 
27b 2
21-Jul-15
B
See details
Where Pc, Vc and Tc are
respectively known as critical
pressure, critical volume and
critical temperature
Presentation by Dr. K.Y. Rajpure
31
In terms of Pc, Vc and Tc the Van der Waals equation of state becomes

3 
1
8
 Pr  2 Vr    Tr

3 3
Vr 

Where
Pr 
P
,
Pc
Tr 
T
,
Tc
Vr 
V
Vc
How ?
Here Pr, Vr and Tr are called as reduced pressure, reduced volume
and reduced temperature respectively.
Above equation does not contain any constant characteristics of gas
and hence, it is a universal equation valid for all substances. This is
called the Law of corresponding states.
Possible questions
Statistical Mechanics: Theory and Applications - S.K. Sinha
Chapter No. 10, Page No. 283.
21-Jul-15
Presentation by Dr. K.Y. Rajpure
32
21-Jul-15
Presentation by Dr. K.Y. Rajpure
33
To find Pc, Vc and Tc:
P

NkT
a
 2
V  b V
(a)
NkT
2a

0
2
3
V  b V
(b)
2 NkT 6a
 4 0
3
V  b V
(c)
Multiplying equation (b) by
3NkT
6a

0
2
4
V V  b  V
NkT  2V  3V  3b 
0

2 
V  b  V V  b 
NkT
V  b2
NkT
 V  3b  0
3
V V  b 
 V  3b  0
3
V
Vc  3b
(d)
Putting this value in equation (b)
Subtracting equation (d) from (c)
2 NkT
3NkT

0
3
2
V  b V V  b
NkT  2
3

0
2 

V  b V  b V 
21-Jul-15
  V  3b 
V V  b   0



NkTc
2a

0
3b  b2 3b3

Presentation by Dr. K.Y. Rajpure
NkTc
2a

0
4b 2 27b 3
34

 8a 


27
b
 a
Pc  
2b  9b 2
NkTc
2a

0
2
3
4b
27b
 27bNkT c  8a
0
3
108b
 27bNkTc  8a  0
Tc 
8a
27 bNk
Pc 
8a
a

54b 2 9b 2
Pc 
4a
a

27b 2 9b 2
Pc 
4a  3a
27b 2
Pc 
a
27b 2
Putting values of Vc and Tc in eqn (a)
Pc 
NkTc
a
 2
Vc  b Vc
 8a 
Nk 

27
bNk

 a
Pc 
3b  b 
3b 2
Thus
Vc  3b Tc 
8a
27 bNk
Pc 
a
27b 2
Back
21-Jul-15
Presentation by Dr. K.Y. Rajpure
35
To show

3 
1
8
 Pr  2 Vc    Tr

3 3
Vr 

Consider values of V, T and P
at critical point
Vc  3b Tc 
8a
27 bNk
Pc 
a
27b 2
We know the Van der Waal’s equation of state
a 

V  b  NkT
P


2 
V 

If the values of V, T and P are expressed in terms of their critical values
by relation
Then
P  Pc Pr
V  VcVr
T  TcTr
So that
 a
a 
 8a 




P

3
bV

b

Nk

Tr
r
2 r
2 2 
9b Vr 
 27bNk 
 27b
3ab 
3 
1   8a 


P

V



Tr
r
2  r
2 
27b 
Vr 
3   27b 
Back
21-Jul-15

3 
1
8
 Pr  2 Vc    Tr

3 3
Vr 

 a 
P
P
2  r
27
b


V  3bVr
 8a 
T 
Tr
27
bNk



3 
1   8a   27b 
 Pr  2 Vr    

Tr
V
3
27
b
3
a
 
 

r 

Presentation by Dr. K.Y. Rajpure
36
1.
2.
3.
4.
5.
6.
Explain P-T diagram of one component system.
Distinguish between First and Second order phase transitions.
State and explain conditions for phase equillibrium.
Derive Clausius-Clayperon equation
What do you mean by ‘critical indices’?
What is law of corresponding states.
Back
21-Jul-15
Presentation by Dr. K.Y. Rajpure
37