Download CHAPTER 4: PHASE TRANSITIONS

CHAPTER 4: PHASE TRANSITIONS
4.1 INTRODUCTION
A region within which all properties are uniform consists of a distinct phase. For
example, solid ice, liquid water, and gaseous water vapor are separate phases of
the same chemical species ( H 2O ) . Each phase can be distinguished with the
density ρ of the constituent. For example, a portion of the Arctic Ocean in
vicinity of the North Pole is frozen and consists of ice in a top layer and liquid
water beneath it. The atmosphere above the ice contains some water vapor. The
density in these three layers are different, since water exists in these layers
separately in some combination of three (solid, liquid and gaseous) phases.
Similarly. A vessel containing immiscible oil and water contains only liquid. We
have two distinct phases in this case as ρoil ≠ ρ water . Similarly, in metallurgical
applications, various phases may exist within the solid state, since the density
may differ over a solid region that is at a uniform temperature and pressure.
In liquid mixtures that are miscible at a molecular level (such as those of alcohol
and water for which molecule of one species are uniformly intermixed with those
of the other), even though the mixture may contain several chemical
components, a single phase exists, since the system properties are
macroscopically uniform throughout a given volume. Air, for instance, consists of
two major components (molecules of oxygen and nitrogen) that are chemically
distinct, but consists of a single phase, since they are well mixed.
The distinguishing characteristic of a phase transition is an abrupt sudden
change in one or more physical properties, in particular the heat capacity, with a
small change in a thermodynamic variable such as the temperature. Examples of
phase transitions are:
•
•
•
•
•
The transitions between the solid, liquid, and gaseous phases (boiling,
melting, sublimation, etc.)
The transition between the ferromagnetic and paramagnetic phases of
magnetic materials at the Curie point.
The emergence of superconductivity in certain metals when cooled below
a critical temperature.
Quantum condensation of bosonic fluids, such as Bose-Einstein
condensation and the superfluid transition in liquid helium.
The breaking of symmetries in the laws of physics during the early history
of the universe as its temperature cooled.
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The first-order phase transitions are those that involve a latent heat. During
such a transition, a system either absorbs or releases a fixed (and typically large)
amount of energy. Because energy cannot be instantaneously transferred
between the system and its environment, first-order transitions are associated
with "mixed-phase regimes" in which some parts of the system have completed
the transition and others have not. This phenomenon is familiar to anyone who
has boiled a pot of water: the water does not instantly turn into gas, but forms a
turbulent mixture of water and water vapor bubbles ( Fig. 4.1). Mixed-phase
systems are difficult to study, because their dynamics are violent and hard to
control. However, many important phase transitions fall in this category, including
the solid/liquid/gas transitions.
Fig. 4.1 Phase change of water
The second class of phase transitions are the continuous phase transitions,
also called second-order phase transitions. These have no associated latent
heat. Examples of second-order phase transitions are the ferromagnetic
transition, the superfluid transition, and Bose-Einstein condensation.
Several transitions are known as the infinite-order phase transitions. They are
continuous but break no symmetries. The most famous example is the
Berezinsky-Kosterlitz-Thouless transition in the two-dimensional XY model. Many
quantum phase transitions in two-dimensional electron gases belong to this
class. Apartfrom this, there is another phase transition known as Lambda phase
transition.
4.2 FIRST ORDER PHASE TRANSITION
The first order phase transition occurs between the triple point and critical point
(excluding the critical point). A first order phase transformation should satisfy the
following two requirements:
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1. There are changes in entropy and volume
2. The first order derivatives of Gibbs function change discontinuously.
Consider a system consisting of two phases of a pure substance at equilibrium.
Under these conditions, the two phases are at the same pressure and
temperature. Consider the change of state associated with a transfer of dn
moles from phase 1 to phase 2 ( p, T remaining constant). That is,
It is the phase transition between the triple point and the critical point in a T − s
diagram ( Fig. 4.2).
Critical point
p+dp
y (p+dp ,T+dp)
p
X (p, T)
p
S
T.P
T
Fig. 4.2 First order phase transition
dg = − sdT + vdp
(4.1)
For a reversible isothermal isobaric phase transition
dg = 0
(4.2)
g = constant=molar Gibbs function ( kJ/kgmol)
Hence at state X ( p, T ) in Fig. 4.2,
g(i) = g ( f )
(4.3)
g ( i ) + dg ( i ) = g ( f ) + dg ( f )
(4.4)
At state Y ( p + dp, T + dT ) ,
62
Subtracting Eq. (4.3) from Eq. (4.4) we get,
dg ( i ) = dg ( f )
(4.5)
⇒ − s (i ) + v ( i ) dp = − s ( f ) + v( f ) dp
⇒
( v( ) − v( ) ) dp = s( )dT − s( )dT
i
f
i
f
(4.6)
dp s ( ) − s ( )
⇒
=
dT v ( f ) − v (i )
f
From the relation
i
Tds = dq
(4.7)
dp
l
=
f)
(
dT T ⎡v − v( i ) ⎤
⎣
⎦
(4.8)
We get
Eq. (4.8) is known as Clausius- Clapeyron’s equation and can be used to
estimate the latent heat if p, v and T are known. The vapor pressure curve can
be represented by the following correlation,
(4.9)
B
ln p = A + + C ln T + DT
T
1 dp
B C
⇒
=− 2 + +D
p dT
T
T
The three different phase transition processes for water are discussed in the
following subsections (Fig. 4.3).
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Fusion curve
p
For
water
For any other substance
(positive slope)
(Negative
slop)
Critical point
Liquid
Solid
Vaporization curve
p
dp
1
dT
Sublimation
curve
Vapour
Triple point
T
T1
Fig. 4.3 Phase diagram for a pure substance on p − T coordinates
4.2.1 FUSION It is the solid to liquid phase transition.
l fu
dp
=
dT T [ v "− v ']
(4.10)
where l fu is the latent heat of fusion
' (Prime) indicates the saturated solid state
" ( Double prime) indicates the saturated liquid state
dp
→ −ve
dT
dp
→ + ve
For most other substances v" > v ' ( expansion) ⇒
dT
For water, v" < v ' (Indicating contraction) ⇒
4.2.2 VAPORIZATION
It is the liquid to vapor phase transformation.
lvap
dp
=
dT T [ v "'− v "]
64
(4.11)
Usually, v "' > v "
Assuming vapor to behalf as an ideal gas
(4.12)
_
RT
v "' ≅
p
∴
lvap
p lvap
dp
p
=
=
= _ lvap
_
_
dT
2
2
RT
RT
RT
T
p
(4.13)
Clausius- Clapeyron’s equation can also be used to estimate approximately the
vapor pressure of a liquid at any arbitrary temperature in conjunction with a
relation for latent heat of a substance, known as Trouton’s rule.
4.2.2.1 TROUTON’S RULE
(4.14)
_
h fg
TB
≅ 88 kJ / kgmol − K
_
where h fg → latent heat of vaporization, kJ/kgmol
and TB → boiling point at 1.013 bar (N.B.P.)
From Eq. (4.10) and (4.14),
(4.15)
dp
p
= _ 88 TB
dT
RT 2
p
T
dp 88TB dT
⇒ ∫
= _ ∫ 2
p
101.325
R TB T
⇒ ln
88T
p
=− _ B
101.325
R
⎛1 1 ⎞
⎜ − ⎟
⎝ T TB ⎠
⎡ 88 ⎛ T ⎞ ⎤
⇒ p = 101.325 exp ⎢ _ ⎜1 − B ⎟ ⎥
T ⎠⎥
⎢⎣ R ⎝
⎦
where p is the vapor pressure in kPa for any temperature T .
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(4.16)
4.2.3 SUBLIMATION It is the solid to vapor phase transformation.
lsub
dp
=
dT T ⎡⎣v "'− v ' ⎤⎦
(4.17)
where lsub → latent heat of sublimation.
∵ v "' >> v ' and vapor pressure is low,
_
RT
v"' =
p
dp lsub p
=
dT T 2 R_
_ d log p
(
)
= −2.303 R
⎛1⎞
d⎜ ⎟
⎝T ⎠
∴
⇒ lsub
(4.18)
(4.19)
(4.20)
At the triple point:
lsub = lvap + l fus
(4.21)
p l
dp
= tp _ vap
dT vap
RTtp2
(4.22)
p l
dp
= tp _ sub
dT sub
RTtp2
(4.23)
Now,
and
Since, lsub > lvap
at the triple point
dp
dp
>
dT sub dT vap
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(4.24)
4.3 SECOND ORDER PHASE TRANSITION
The second order phase transition occurs at the critical point where saturated
liquid flashes into vapor. There is no distinction of phases at triple point.
However, there is distinct variation of the value of specific heat C p . Following are
the assumptions for second order phase transition
1. There are no changes of entropy and specific volume
2. Molar Gibbs function g is continuous
3. First order derivative of g is continuous
∂g f
∂T
∂g g
= −s f ;
= vf
∂p
∂g g
= vg
∂p
∂g
∂g f ∂g g
= g ;
=
; g f = gg
∂T
∂T
∂p
∂p
∂T
∂g f
∂g f
∂T
∂g g
∂T
∂g f
∂T
= − sg ;
∂g f
= −s f ;
= − sg ;
=
∂g g
∂T
∂g f
∂p
∂g g
;
(4.25)
= vf
(4.26)
= vg
∂p
∂g f
∂p
=
∂g g
∂p
; g f = gg
4. Second order derivatives of Gibbs function g changes discontinuously.
∂ 2 g ⎛ ∂v ⎞
⎡ ∂p
⎤
= ⎜ ⎟ >> 0 ⎢∵ = 0 at critical point ⎥
2
∂p
⎣ ∂v
⎦
⎝ ∂p ⎠T
(4.27)
At critical point the latent heat of vaporization, l is zero.
(Remember: Fluid containing one component exhibits a single critical point.)
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4.3.1 ANALYSIS OF SECOND ORDER PHASE TRANSITION
⎛ ∂s ⎞
Cp = T ⎜
⎟
⎝ ∂T ⎠ p
(4.28)
⎡ ⎛ ∂g ⎞ ⎤
⎛ ∂2 g ⎞
=
−
⎢− ⎜
⎥
⎜ 2⎟
⎟
⎢⎣ ⎝ ∂T ⎠ p ⎥⎦ p
⎝ ∂T ⎠ p
(4.29)
or,
Cp
T
=
∂
∂T
From the definition of isothermal compressibility,
κT v = −
∂ ⎡⎛ ∂g ⎞ ⎤ ⎛ ∂ 2 p ⎞
⎢⎜ ⎟ ⎥ − ⎜
⎟
∂p ⎣⎢⎝ ∂p ⎠T ⎦⎥T ⎝ ∂p 2 ⎠T
(4.30)
Again, following the definition of volume expansivity,
βv =
∂
∂T
⎡⎛ ∂g ⎞ ⎤
∂2 g
=
⎢⎜ ⎟ ⎥
⎢⎣⎝ ∂p ⎠T ⎥⎦ p ∂T ∂p
(4.31)
For reversible isothermal isobaric process
s(i ) = s( f )
⇒ s ( i ) + ds (i ) = s ( f ) + ds ( f )
[at p, T ]
⎡⎣ at ( p + dp, T + dT ) ⎤⎦
(4.32)
(4.33)
⇒ Tds ( i ) = Tds ( f )
(4.34)
⎛ ∂v ⎞
∵ Tds = C p dT − T ⎜
⎟ dp = C p dT − T β vdp
⎝ ∂T ⎠ p
(4.35)
[ 2nd T-dS relation]
or,
∴ C (pi ) dT − T β ( i ) vdp = C (p f ) dT − T β ( f ) vdp
(4.36)
C (p f ) − C (pi )
dp
=
dT Tv β ( f ) − β ( i )
(4.37)
(
As there is no change in volume
v(i ) = v( f )
)
⎡⎣ at ( p, T ) ⎤⎦
Further, at point [ p + dp, T + dT ] ,
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(4.38)
v ( i ) + dv ( i ) = v ( f ) + dv( f )
⇒ dv ( i ) = dv ( f )
⇒ β (i ) vdT − κ T( i ) vdp = β ( f ) vdT − κ T( f ) vdp
(4.39)
⇒ κ T( f ) vdp − κ T( i ) vdp = β ( f ) vdT − β ( i ) vdT
dp β ( ) − β ( )
⇒
=
dT κ T( f ) − κ T(i )
f
i
(4.40)
4.3.2 APPLICATION OF SECOND ORDER PHASE TRANSITION
1. Transition of a super conductor from super conducting to the normal state
in zero magnetic field (It is a true second order transition).
2. Ferromagnetic to paramagnetic transition in a simple model.
3. Order-disorder transformation.
4.4 LAMBDA TRANSITION
It is the third type of phase transition between the two liquid phases of He 4 ,
ordinary liquid helium l, and super fluid helium He ll***. This transition can occur at
any point along the line separating these two liquid phases in Fig.4.4 .
Fig. 4.4 Phase diagram of helium
69
From this figure, it is evident that He l and He ll can coexist in equilibrium over a
range of temperature and pressures, and He l can be converted to He ll either by
lowering the temperature, provided pressure is not too great, or by reducing the
pressure, provided that the temperature is below 2.18 K. A graph of C p versus T
for the two phases has the general shape shown in Fig.4.5, and the transition
takes its name from the resemblance of the curve to the shape of the Greek letter
λ . The value of C p does not change discontinuously, but its variation with
temperature is different in the two phases.
3
c v (J kilomole-1k -1)
50 x 10
40
30
20
10
0
1.5
2.0
T (K)
2.5
3.0
Fig. 4.5 Lambda transition
4.5 PROPERTIES OF SUPERFLUID
Superfluids have the unique quality that their atoms are in the same quantum
state. This means they all have the same momentum, and if one moves, they all
move. This allows superfluids to move without friction through the tiniest of
cracks, and superfluid helium will even flow up the sides of a jar and over the top.
This apparant defiance of gravity comes from a special type of surface wave
present in superfluid helium, which in effect pushes this extremely thin film up the
sides of the container. It was discovered in 1962 by Tisza, who named the
phenomenon third sound. Another unusual result of third sound is the fountain
effect, where superfluid excited by photons will form a fountain vertically upward
off of its surface.
Superfluids also have an amazingly high thermal conductivity. When heat is
introduced to a normal system, it diffuses through the system slowly. In a
superfluid, heat is transmitted so fast that thermal waves become possible. This
70
fourth kind of wave found in superfluids is called second sound, quite improperly
becuase they involve no pressure variations. Following are some of the
remarkable properties of superfluid He II:
1. Ability to flow through microscopic passages with no apparent friction
2. The quantization of vortices
3. The ability to support four wave modes:
(a) Sound which is analogous to sound in ordinary gases and liquids,
(b) Sound which carries temperature and entropy perturbations with virtually no
pressure variations
(c) Sound which are waves on thin films
(d) Sound which are acoustic-like waves in the "superfluid" component of He II.
As in the case of both classical and high temperature superconductivity,
superfluidity is a manifestation of quantum mechanical effects at the macroscopic
level.
*** Helium He 4 vapor compressed isothermally between 5.25 K to 2.18 K,
condenses to liquid helium He l. When vapor is compressed below 2.18 K, a
liquid called super fluid helium He ll results.
4.6 MIXTURE OF VARIABLE COMPOSITION
Let us consider a system containing a mixture of substances 1,2, 3,…….k. If
some quantities of a substance are added to the system, the energy of the
system will increase. Thus for a system with variable composition, the internal
energy depends not only on S and V , but also on the number of moles (or
mass) of various constituents of the system.
U = U ( S , V , n1 , n2 ,.........., nk )
(4.41)
where, n1 , n2 , n3 ,........, nk are the number of moles of substances 1,2,3,……k.
The composition may change not only due to addition or subtraction, bur also
due to chemical reaction and inter phase mass transfer.
For a small change in U , assuming the function to be continuous,
⎛ ∂U ⎞
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜
dS + ⎜
dV + ⎜
dn1
⎟
⎟
⎟
⎝ ∂S ⎠V ,n1 ,n2 ,.....,nk
⎝ ∂V ⎠ S ,n1 ,n2 ,.....,nk
⎝ ∂n1 ⎠V , S ,n2 ,.....,nk
⎛ ∂U ⎞
⎛ ∂U ⎞
dn2 + ............ + ⎜
dnk
+⎜
⎟
⎟
⎝ ∂n2 ⎠V , S ,n1 ,.....,nk
⎝ ∂nk ⎠V ,S ,n1 ,.....,nk −1
71
(4.42)
k
⎛ ∂U ⎞
⎛ ∂x ⎞
⎛ ∂x ⎞
dU = ⎜ ⎟ dS + ⎜ ⎟ dV + ∑ ⎜
dni
⎟
i =1 ⎝ ∂ni ⎠ S ,V , n
⎝ ∂y ⎠V ,ni
⎝ ∂y ⎠ S ,ni
j
(4.43)
where the subscript i = any substance
j = other substance except the one whose number
of moles is changing.
If composition does not change,
dU = TdS − pdV
(4.44)
⎛ ∂U ⎞
∴ ⎜
⎟ =T
⎝ ∂S ⎠V ,ni
(4.45)
and
⎛ ∂U ⎞
⎜
⎟ = −p
⎝ ∂V ⎠ S ,ni
(4.46)
k
⎛ ∂U ⎞
∴ dU = TdS − pdV + ∑ ⎜
dni
⎟
i =1 ⎝ ∂ni ⎠ S ,V , n
j
(4.47)
Eq. (4.47) can be written as
k
dU = TdS − pdV + ∑ µi dni
(4.48)
i =1
⎛ ∂U ⎞
is the molar chemical potential. It signifies the change in
where, µi = ⎜
⎟
n
∂
i
⎝
⎠ S ,V ,n j
internal energy per unit mole of component i when S ,V and number of moles of
all other components are constant. The chemical potential drives mass (or
species) similar to the thermal potential that drives heat transfer from higher to
lower temperature.
Ref. to Eq. (4.41), we can write in a similar manner,
G = G ( p, T , n1 , n2 ,........, nk )
or,
72
(4.49)
k
⎛ ∂G ⎞
⎛ ∂G ⎞
⎛ ∂G ⎞
dG = ⎜
dp
dT
dni
+
+
⎜
⎟
∑
⎟
⎜
⎟
⎝ ∂T ⎠ p ,ni
i =1 ⎝ ∂ni ⎠T , p , n
⎝ ∂p ⎠T ,ni
j
⎛ ∂G ⎞
= Vdp − SdT + ∑ ⎜
dnI
⎟
i =1 ⎝ ∂ni ⎠T , p , n
j
k
Since,
G = U + pV − TS
k
⎛ ∂x ⎞
d (U + pV − TS ) = Vdp − SdT + ∑ ⎜ ⎟
dni
i =1 ⎝ ∂y ⎠T , p , n
j
(4.50)
(4.51)
(4.52)
(4.53a)
or,
k
⎛ ∂G ⎞
dU + pdV + Vdp − TdS − SdT = Vdp − SdT + ∑ ⎜
dni
⎟
i =1 ⎝ ∂ni ⎠T , p , n
j
k
⎛ ∂G ⎞
dU = TdS − pdV + ∑ ⎜
dni
⎟
i =1 ⎝ ∂ni ⎠T , p , n
j
(4.53b)
Comparing Eq. (4.47) and Eq.(4.51), we get,
k
⎛ ∂U ⎞
⎛ ∂G ⎞
dn
dni
=
⎜
⎟
⎜
⎟
∑
∑
i
i =1 ⎝ ∂ni ⎠ S ,V , n
i =1 ⎝ ∂ni ⎠ S ,V , n
j
j
(4.54)
⎛ ∂U ⎞
⎛ ∂G ⎞
∴ ⎜
=⎜
= µi
⎟
⎟
⎝ ∂ni ⎠ S ,V ,n j ⎝ ∂ni ⎠ S ,V ,n j
(4.55)
k
Eq. (4.51) can be written as
k
dG = Vdp − SdT + ∑ µi dni
(4.56)
i =1
Hence,
k
dU = TdS − pdV + ∑ µi dni
(4.57)
i =1
dG = Vdp − SdT + ∑ µi dni
(4.58)
dH = TdS + Vdp + ∑ µi dni
(4.59)
k
i =1
k
i =1
73
(4.60)
k
dF = − SdT − pdV + ∑ µi dni
i =1
Now,
⎛ ∂U ⎞
⎛ ∂G ⎞
⎛ ∂H ⎞
⎛ ∂F ⎞
=⎜
=⎜
=⎜
⎟
⎟
⎟
⎟
⎝ ∂ni ⎠ S ,V ,n j ⎝ ∂ni ⎠ S ,V ,n j ⎝ ∂ni ⎠ S ,V ,n j ⎝ ∂ni ⎠ S ,V ,n j
µ =⎜
(4.61)
4.6.1 GIBBS-DUHEM EQUATION
Let us consider a homogeneous phase of a multi-component system for which
k
(4.62)
dU = TdS − pdV + ∑ µi ∆ni
i =1
If the phase is enlarged in size, U , S and V will increase in size, while T , p and
µ will remain the same. Thus,
k
(4.63)
∆U = T ∆S − p∆V + ∑ µi ∆ni
i =1
Let the system be enlarged to k times the original size.
∴ ∆U = kU − U = ( k − 1) U
(4.64a)
∆S = kS − S = ( k − 1) S
(4.64b)
∆V = kV − V = ( k − 1) V
(4.64c)
∆ni = kni = ni = ( k − 1) ni
(4.64d)
Substituting Eqs. (4.64a-d) in Eq. (4.63), we get
k
( k − 1)U = T ( k − 1) S − p ( k − 1)V + ∑ µi ni
(4.65)
i =1
∴ U = TS − pV + ∑ µi ni
(4.66)
⇒ U + pV − TS = ∑ µi ni
(4.67)
k
i =1
k
i =1
k
GT , p = ∑ µi ni
i =1
Differentiating Eq. (4.68), we get
74
(4.68)
k
k
i =1
i =1
(4.69)
dGT , p = ∑ ni d µi + ∑ µi dni
(at constant temperature and pressure)
When temperature and pressure changes,
(4.70)
k
dG = − SdT + Vdp + ∑ µi dni
i =1
Comparing Eq. (4.69) and Eq. (4.70), we get,
k
k
k
i =1
i =1
i =1
∑ ni d µi + ∑ µi dni = − SdT + Vdp + ∑ µi dni
(4.71)
or,
k
− SdT + Vdp − ∑ ni d µi = 0
(4.72)
i =1
Eq. (4.72) is known as Gibbs-Duhem Equation. It represents simultaneous
changes of T , p and µ .
Now,
k
(4.73)
GT , p = ∑ µi ni = µ1n1 + µ 2 n2 + ........ + µ k nk
i =1
For a phase consisting of single constituent,
G = µn
G
∴ µ= = g
n
(4.74)
Hence, chemical potential is the molar Gibbs function and is a function of T and
p . For a single phase, µi is a function of T , p and mole fraction xi .
4.7 CONDITIONS OF EQUILIBRIUM OF A HETEROGENEOUS SYSTEM
Let us consider a heterogeneous system of volume V with several
homogeneous phases (φ = a, b, c,......r ) existing in equilibrium. Let us further
suppose that each phase consists of i ( i = 1, 2,3,......., c ) constituents. Further,
number of constituents in any phase is different from others.
75
With each phase, a change in internal energy is accompanied by a change in
entropy S , volume V and composition.
c
(4.75)
dUφ = Tφ dSφ − pφ dVφ + ∑ ( µi dni )φ
i =1
A change in the internal energy of the entire system can therefore be expressed
as,
r
r
r
r
(4.76)
c
∑ dUφ = φ∑ Tφ dSφ − φ∑ pφ dVφ + φ∑∑ ( µ dn )φ
φ
=a
=a
=a
= a i =1
i
i
Now, change in internal energy of the entire system involves changes in the
internal energy of the constituent phases,
(4.77)
r
dU = dU a + dU b + .......... + dU r = ∑ dUφ
φ =a
Similarly, changes in volume, entropy or chemical composition can be written as
r
(4.78)
dV = dVa + dVb + ........ + dVr = ∑ dVφ
φ =a
(4.79)
r
dS = dS a + dSb + ....... + dS r = ∑ dSφ
φ =a
(4.80)
r
dn = dna + dnb + ........ + dnr = ∑ dnφ
φ =a
For a close system in equilibrium,
dU = dV = dS = dn = 0
(4.81)
r
Eq.(4.77) ⇒ dU a = − ( dU b + dU c + ....... + dU r ) = −∑ dU j
(4.82)
j =b
r
Eq.(4.78) ⇒ dVa = − ( dVb + dVc + ........ + dVr ) = − ∑ dV j
(4.83)
j =b
r
Eq.(4.79) ⇒ dSa = − ( dSb + dSc + ...... + dSr ) = −∑ dS j
j =b
76
(4.84)
(4.85)
r
Eq.(4.80) ⇒ dna = − ( dnb + dnc + ....... + dnr ) = −∑ dn j
j =b
Energy Equation (Eq. (4.76)) for the heterogeneous system is,
r
r
r
r
(4.86)
c
∑ dUφ = φ∑ Tφ dSφ − φ∑ pφ dVφ + φ∑∑ ( µ dn )φ = 0
φ
=a
=a
=a
r
r
j =b
j =b
= a i =1
r
i
i
c
(4.87)
c
Ta dS a + ∑ T j dS j − pa dVa − ∑ p j dV j + ∑∑ ( µi dni ) j + ∑ ( µi dni )a = 0
j = b i =1
i =1
⇒ − Ta ∑ dS j + ∑ T j dS j + pa ∑ dV j − p j ∑ dV j − ∑ ( µ j )
r
j =b
r
r
j =b
r
j =b
r
j =b
j =b
∑ ( dn j ) + ∑ ( µ j )
r
j
j =b
r
j
j =b
⇒ ∑ (T j − Ta )dS j − ∑ ( p j − pa )dV j + ∑∑ ( µij − µia ) ( dni ) j = 0
j
j
j
Since, dS j , dV j and
( dni ) j
r
j
∑ ( dn )
j =b
i
j
=0
(4.89)
i
are independent and the expression is zero (Eq.
(4.89)), their coefficients must be zero individually.
∴ T j = Ta
(4.90)
⇒ Thermal Equilibrium
p j = pa
(4.91)
⇒ Mechanical Equilibrium
µij = µia
(4.92)
⇒ Chemical Equilibrium
4.8 GIBBS PHASE RULE
Let us consider a heterogeneous system in which c constituents ( i = 1, 2,3,....., c )
are present in all the phases. Initially we consider no chemical reaction to avoid
the complexity. Let us assume that there are φ phases. In each phase, all the
constituents are present, i.e.,
(4.93)
φ = 1, constituents are n1 , n2 ,....., nc
φ = 2, constituents are n1 , n2 ,....., nc
...........................................
φ = 5, constituents are n1 , n2 ,......, nc
77
(4.88)
For ease of description we will identify the constituents by subscript and phase
by superscript.
The Gibbs function of the whole heterogeneous system is
GT , p = ∑ ni(1) µi(1) + ∑ ni( 2) µi( 2) + ∑ ni( 3) µi( 3) + ....... + ∑ ni(φ ) µi(φ )
(4.94)
G = G (T , p, ni )
(4.95)
c
c
c
c
i =1
i =1
i =1
i =1
Since there is no chemical reaction, only way in which n ' s may change is by the
transport of the constituents from one phase to another. In this case, the total
number of moles of each constituent will remain constant.
n1(1) + n1( 2) + n1( 3) + ............ + n1(φ ) = constant
(4.96)
n2(1) + n2( 2) + n2( 3) + ............ + n2(φ ) = constant
...........................................
nc(1) + nc( 2) + nc( 3) + ............ + nc(φ ) = constant
At chemical equilibrium, G will be rendered a minimum at constant T & p
subject to the aforementioned equations of constraints.
∴ dGT , p = 0
(4.97)
Therefore, from the condition of equilibrium of a heterogeneous system (chemical
equilibrium),
(4.98)
µij = µia
µ1(1) = µ1( 2) = µ1( 3) = ............. = µ1(φ )
(4.99)
µ 2(1) = µ 2( 2) = µ 2( 3) = ............. = µ 2(φ )
µ3(1) = µ3( 2) = µ3( 3) = ............. = µ3(φ )
..................................................
µ c(1) = µc( 2) = µc( 3) = ............. = µc(φ )
Eqs.(4.99) , which specify the condition of phase equilibrium are called the
equations of phase equilibrium. There are k (φ − 1) numbers of such equations.
Now the composition of each phase containing k constituents is fixed if k − 1
constituents are known, since sum of the mole fractions of each constituent in the
phase must equal unity. Therefore, for φ phases, there are a total of
78
φ ( k − 1) variables, in addition to temperature and pressure, which must be
specified. There are, then, φ ( k − 1) + 2 variables altogether.
If the number of variables is equal to the number of equations, then whether or
not we can actually solve the equations, the temperature, pressure, and
composition of each phase is determined. The system is then called nonovariant
and said to have zero variance.
If the number of variables is one greater than the number of equations, an
arbitrary value can be assigned to one of the variables and the remainder are
completely determined. The system is then called monovariant and said to have
a variance of 1 .
In general, the variance f is defined as the excess of the number of variables
over the numbers of equations.
(4.100)
f = ⎡⎣φ ( k − 1) + 2 ⎤⎦ − ⎡⎣ k (φ − 1)⎤⎦ = k − φ + 2
This equation is called the Gibbs phase rule.
For chemical reactions, Gibbs phase rule is modified as
f = (c − r ) − φ + 2
(4.100)
where, r = number of independent reversible chemical reaction
φ = number of phases
k = constituent
f = variance or degree of freedom
Consider the T − s diagram for water. At any point on the saturated vapor line,
c = 1, φ = 1 . Hence, f = 2 . This indicates that 2 (two) independent thermodynamic
properties are needed to fix up the state of the system at equilibrium. Similarly, at
a location within the saturated liquid and vapor line, c = 1, φ = 2 ⇒ f = 1 . Hence,
only one thermodynamic property is sufficient to fix up the state. At the triple
point, c = 1, φ =3 ⇒ f =0 . It is a unique state where all the three phases exist in
equilibrium.
Exercise 4.1:
To make baking soda ( NaHCO3 ) , a concentrated aqueous solution of Na2CO3 is
saturated with CO2 . The reaction is given as
2 Na + + CO3− + H 2O + CO2 ⇔ 2 NaHCO3
79
Thus Na + ions and CO3− ions, H 2O, CO2 and NaHCO2 are present in arbitrary
amounts, except that all the Na + and CO3− are from Na2CO3 . Find the number of
degree of freedom of this system.
Exercise 4.2:
Determine the number of degree of freedom for the system at each lettered point
and state the variables for a cadmium-bismuth system (Fig. 4.6)
400
Liquid solution Cd+Bi
321 C
A
300
271 C
200
B
Liquid solution and
solid Cd
Liquid
solution and
solid Bi
1 44 C
C
D
E
100
Solid Cd + solid Bi
20
60
40
80
Weight % Cd
100
Fig. 4.6 Phase diagram of cadmium-bismuth system
Note: We can have several separate solid phases in a system, and the same is
true for (immiscible) liquids. On the other hand there can be no more than one
gas phase, because all gases freely intermingle.
4.9 THIRD LAW OF THERMODYNAMICS
Marching towards absolute zero temperature.
4.9.1 MOTIVATION
Application of first law and second law of thermodynamics to reactive systems
become difficult due to the non availability of a standard reference entropy value
of various substances. There is a need to have a reference entropy value for all
substances for evaluating the efficiency of a reactive system. Third law of
thermodynamics provides a base value for the entropy.
80
The third law of thermodynamics was formulated during the early part of
twentieth century. The initial work was done primarily by W. H. Nernst [18641941] and Max Planck [1858-1947].
4.9.2 ATTAINING LOW TEMPERATURE
1. Below 5 K is possible by Joule Kelvin expansion, by producing liquid
helium.
2. Still lower temperature can be attained by adiabatic demagnetization of a
paramagnetic salt.
3. Temperature as low as 0.001 K has been achieved by magnetic cooling.
4.9.3 MAGNETIC PROPERTY
1. Diamagnetic: substance is repelled by magnet
2. Paramagnetic: substance attracted by magnet, such as Iron, Gadolinium
sulphate
Adiabatic Demagnetization of a paramagnetic salt:
Gadolinium sulphate is used here.
Salt is hung by a fine nylon thread inside the salt tube such that it does not
touch the sides.
To pump
Radiation shield
Exhaust gas
Liquid nitrogen
Liquid helium
Magnet
Measuring coils
Salt
Supporting threads
Dewar flask
Fig: 4.7 Adiabatic demagnetization of a paramagnetic salt
81
The salt is first cooled slightly below 1 K by surrounding it with liquid
helium boiling under reduced pressure.
Next, the salt is exposed to a strong magnetic field of about 25000
Gauss. Heat produced due to magnetization of the salt is transferred to
the liquid helium without causing an increase in salt temperature.
With the magnetic field still present, the inner chamber containing the
salt is evacuated of gaseous helium.
Finally, the magnetic field is removed. The molecules disalign
themselves, which require energy. This energy is obtained by the salt
getting cooler in the process.
Repetition of the process lowers down the temperature of the salt.
Temperature as low as 0.001 K have been achieved so far.
4.9.4 MEASUREMENT OF TEMPERATURE
In the neighborhood of absolute zero, all ordinary methods of temperature
measurement fail. Curie’s law gives the most convenient method for
measurement of at low temperature (approximately).
(4.101)
c
χ=
T
where χ is the magnetic susceptibility of the salt
T is the absolute temperature
c is the Curie’s constant.
The fundamental features of all cooling process are that the lower the
temperature achieved, the harder it is to go still lower.
Interpretations of adiabatic demagnetization:
Final temperature T f ∞ Ti (initial temperature)
1 ⎞
⎛
First demagnetization produces a temperature half that at start ⎜ T f = Ti ⎟
2 ⎠
⎝
1⎛1 ⎞ 1
Second demagnetization produces a temperature T f = ⎜ Ti ⎟ = Ti
2⎝2 ⎠ 4
1⎛1 ⎞ 1
Third demagnetization produces a temperature T f = ⎜ Ti ⎟ = Ti
2⎝4 ⎠ 8
……………………………………………………………………………………
1
rth demagnetization produces a temperature T f = r Ti
2
82
Infinite number of adiabatic demagnetizations will be required to attain absolute
zero temperature.
FOWLER-GUGGENHEIM STATEMENT OF THIRD LAW:
It is impossible by any procedure, no matter how idealized, to reduce any
condensed system to the absolute zero of temperature in a finite number of
operations.
Any isothermal magnetization from 0 to H (magnetic intensity) such as k − i1 ,
f1 − i2 etc. is associated with the release of heat, i.e., a decrease in entropy (Fig.
4.8).
H=0 H=Hi
T
T
i1
l
H=Hi
l1
S(0,Hi) -S(0,0)=0
k
1
k
T
f1
f1
2
f
1
l3
i
3
f2
f3
(0,0)
k
l2
l2
i
H=0
f3
H
S(0,0)
{SS((00,,H0))
f2
S(0,Hi)
(0, Hi )
i
f2
S
T=
0
Fig. 4.8 T-H and T-S diagrams of a paramagnetic substance to show the
equivalence of three statements of the third law
The processes i1 − f1 , i2 − f 2 , i3 − f 3 etc. represent reversible adiabatic
demagnetizations with temperatures getting lower and lower. Repeated cycles of
isothermal magnetization and adiabatic demagnetization would bring about a
very low temperature. It is seen that ⎡⎣ S (T , H ) − S (T , 0 ) ⎤⎦ decreases as the
temperature decreases, i.e., ∆S III < ∆S II < ∆S I . Thus entropy change associated
83
s
with an isothermal reversible process of a condensed system approaches zero. It
is called Nernst-Simon statement of third law of thermodynamics.
NERNST-SIMON STATEMENT:
The entropy change associated with any isothermal reversible process of a
condensed system approaches zero.
lim ∆ST = 0
T →0
Another statement of third law can be given like this:
It is impossible by any procedure, no matter how idealized, to reduce the entropy
of a system to zero point value in a finite number of operations.
Physical and chemical facts which substantiate the third law:
1. For any phase change that takes place at low temperature, Clausius-Clayperon
equation hold good,
(4.102)
dp s f − si
=
dT v f − vi
From third law of thermodynamics, lim ( s f − si ) = 0 , since v f − vi is not zero. It
T →0
shows that
lim
δ x →0
(4.103)
dp
=0
dT
This is substantiated by all known sublimation curves.
H
H
G
G
T
Fig.4.9 The temperature dependence of the change in the Gibbs function and in
the enthalpy for an isobaric process
84
2.
∆G = ∆H − T ∆s
At low temperature, T ∆s is very small.
∴ ∆G = ∆H
Which confirms that lim ∆s = 0 .
(4.104)
(4.105)
T →0
3. From the Gibbs-Helmholtz equation
⎡ ∂∆G ⎤
∆G = ∆H + T ⎢
⎣ ∂T ⎥⎦ p
(4.106)
∵ lim ( ∆G − ∆H ) → 0 (Fig. 4.9)
T →0
⎡ ∂∆G ⎤
lim ⎢
=0
T →0
⎣ ∂T ⎥⎦
∆G − ∆H
∂∆G
⇒ lim
= lim
T →0
T
→
0
∂T
T
∂G
⎡ ∂G ∂H ⎤
⇒ lim ⎢
−
= lim
− lim C p
⎥
T →0
⎣ ∂T ∂T ⎦ T →0 ∂T T →0
∴ lim C p = 0
(4.107)
T →0
Similarly, using Gibbs Helmholtz equation containing U and F
lim Cv = 0 as T → 0
T →0
∴ C p = Cv = 0
as T → 0
85
(4.108)
(4.109)