Phase transition
Asaf Pe’er1
November 18, 2013
1.
Background
A phase is a region of space, throughout which all physical properties (density, magnetization, etc.) of a material (or thermodynamic system) are essentially uniform. Well known
examples are gaseous phase, liquid phase and solid phase.
During a phase transition of a given medium certain properties of the medium change,
often discontinuously, as a result of some external condition, such as temperature, pressure,
and others. For example, a liquid may become gas upon heating to the boiling point.
We are interested in understanding the conditions under which a material undergoes a
phase transition. In other words, we are interested in finding the equilibrium conditions
between two phases of a medium.
2.
Equilibrium conditions
Let us look at an isolated system, which contains a medium which is in an equilibrium
state between two phases. Being isolated, the energy, volume and number of particles are
conserved,
E1 + E2 = E
(1)
V1 + V2 = V
N1 + N2 = N
where the subscripts 1 and 2 represent the two phases (see Figure 1).
Since the system is isolated, the equilibrium is achieved when the entropy S = S1 (E1 , V1 , N1 )+
S2 (E2 , V2 , N2 ) is maximized.
∂S1
∂S1
∂S2
∂S2
∂S2
∂S1
dE1 +
dE2 +
dV1 +
dV2 +
dN1 +
dN2 = 0
dS = dS1 +dS2 =
∂E1
∂E2
∂V1
∂V2
∂N1
∂N2
(2)
1
Physics Dep., University College Cork
–2–
Fig. 1.— Phase transition.
and since E, V and N are constants, dE1 = −dE2 , etc., and thus the condition for equilibrium
is
∂S1
∂S1 ∂S2
∂S1
∂S2
∂S2
dS = 0 =
dE1 +
dV1 +
dN1
(3)
−
−
−
∂E1 ∂E2
∂V1 ∂V2
∂N1 ∂N2
Since E1 , V1 and N1 are independent variables, each of the terms in parenthesis must vanish
for the system to be in equilibrium.
The first two are familiar: they follow immediately from equilibrating the temperature
and pressure using the entropy-based definitions:
∂S
1
≡
,
(4)
T
∂E V,N
Thus the requirement T1 = T2 is written as
T1 = T2 →
∂S2
∂S1
=
∂E1
∂E2
(5)
Similarly, equilibrating the pressure using the entropy-based definition of the pressure,
∂S
p≡T
,
(6)
∂V E,N
leads to
p1 = p2 →
∂S2
∂S1
=
∂V1
∂V2
(7)
Thus, we are left with the 3rd condition,
∂S2
∂S1
=
∂N1
∂N2
(8)
–3–
Its physical meaning is that there is no passing of molecules between one phase and the
other. We can use this to define the chemical potential µi of each phase by
∂Si
(9)
µi ≡ −Ti
∂Ni Ei ,Vi
Thus, the condition in Equation 8 is simply
µ1 = µ2
(10)
In equilibrium, the chemical potential of the two phases must be the same. Note
that the chemical potential is really a measure of the energy that is associated with the
change in number of molecules in the system. Such a change can result from chemical
reactions, hence the name.
2.1.
Extension of the thermodynamic relations
Let us leave for the moment the two-phase system, and look at a single phase only. We
can write S = S(E, V, N ). Thus,
∂S
∂S
∂S
dS =
dE +
dV +
dN
(11)
∂E V,N
∂V E,N
∂N E,V
or
dS =
1
1
1
dE + pdV − µdN
T
T
T
(12)
This can be re-arranged to write
dE = T dS − pdV + µdN
(13)
Equation 13 is a generalization of the fundamental thermodynamic relation for a system in
which the particle number can change.
Using this result in the definitions of Helmholtz and Gibbs free energies,
F = E − TS
G = E + pV − T S
→
→
dF = dE − T dS − SdT
dG = dE + pdV + V dp − T dS − SdT
(14)
We can write
dF = −SdT − pdV + µdN
dG = −SdT + V dp + µdN.
(15)
–4–
We can thus express the chemical potential in various ways depending on which of the
variables (T , p or V ) are held constant:
∂F
∂G
µ=
=
(16)
∂N T,V =const
∂N T,p=const
We now write F and G as functions of a new parameter - N . Out of these two functions, the Gibbs free energy, G = G(T, p, N ) is an extensive variable: namely, it is linearly
proportional to the particle number, N .
G(T, p, N ) = N G(T, p, 1) ≡ N g(T, p),
(17)
where g is the Gibbs free energy per particle. From Equations 16 and 17 it follows immediately that
µ = g(T, p)
(18)
Thus, the chemical potential is the Gibbs free energy per particle, and the condition for
phase equilibrium is
µ1 = µ2 ↔ g1 (T, p) = g2 (T, p)
(19)
3.
Equilibrium conditions (II)
Experimentally, it is difficult to deal with isolated systems. It is easier to handle systems
at constant pressure and temperature. Let us determine the condition for phase equilibrium
under these conditions.
We know that for a system at fixed temperature T and pressure p, the condition for
equilibrium is that the Gibbs free energy is minimized. Thus,
G = G1 + G2 = N1 g1 (T, p) + N2 g2 (T, p)
(20)
dG = N1 dg1 + g1 dN1 + N2 dg2 + g2 dN2 = 0,
(21)
and
and using
dg1,2 =
∂g1,2
∂g1,2
dT +
dp,
∂T
∂p
(22)
we get
∂g1
∂g1
∂g2
∂g2
dG = N1
dT + N1
dp + g1 (T, p)dN1 + g2 (T, p)dN2
+ N2
+ N2
∂T
∂T
∂p
∂p
(23)
–5–
In equilibrium G is a minimum, and dG = 0, under the conditions that T and p are constants,
thus dp = dT = 0. This leaves us with
dG = g1 (T, p)dN1 + g2 (T, p)dN2 = 0.
(24)
Furthermore, the total number of particles is conserved, N1 +N2 = N , and thus dN1 = −dN2 .
This implies that
dG = [g1 (T, p) − g2 (T, p)] dN1 = 0.
(25)
We thus retrieve again the Equilibrium condition in Equation 19, g1 = g2 .
We thus find that the phase equilibrium condition for a system held at constant temperature and pressure is the same as that of an isolated system. This should not be surprising:
for a system to be in equilibrium between two phases, they must be at the same pressure,
temperature and chemical potential, irrespective of the applied (external) constraints.
4.
Implications of the equilibrium conditions
The equilibrium condition in Equation 19, g1 = g2 defines a curve in p − T plane.
Recall that the condition for equilibrium is a minimization of Gibbs free energy. Thus, if the
system is in a point that does not lie on this curve, it means that the minimum of Gibbs free
energy is achieved if all the substance molecules are in phase 1 (namely, N1 = N , N2 = 0,
G = N1 g(T, p)), or phase 2. The curve g1 = g2 thus divides the (p, T ) plane into regions
where one or the other phase represents a stable equilibrium state. In is only on the curve
that the two phases can coexist in equilibrium (see Figure 2). This curve is called a phase
equilibrium curve.
Let us now consider equilibrium of 3 different phases (solid, liquid and vapor) of a onecomponent system. Repeating the same calculation as we have done before, we obtain the
equilibrium condition
g1 (T, p) = g2 (T, p) = g3 (T, p)
(26)
Equations 26 represent the intersection of two curves: g1 = g2 and g2 = g3 in the (T, p)
diagram. This is known as the triple point. It is shown in Figure 3, which is known as
the phase diagram of the system
Clearly, at the triple point, all three phases are in equilibrium with each other.
Pure substances may be capable of existing in more than one allotropic form (e.g.,
diamond and coal), in which case they will have several triple points. This is illustrated in
Figure 3 (right).
–6–
Fig. 2.— Temperature - Pressure diagram of a phase equilibrium curve (g1 = g2 ).
The three phase equilibrium curves divide the (T, p) plane into three regions in which
the solid, liquid and gaseous phases respectively are the stable state. There are in addition
meta-stable states (e.g., supercooled liquids), but these are not stable.
5.
The Clausius-Clapeyron Equation
The Clausius-Clapeyron equation is an equation describing the phase equilibrium
curve, namely the slope dp/dT at any point along the curve.
Think of two nearby points along the phase equilibrium curve. We know that
g1 (T, p) = g2 (T, p)
g1 (T + dT, p + dp) = g2 (T + dT, p + dp).
(27)
–7–
Fig. 3.— Left: A phase diagram of a one-component system possessing one triple point.
Right: Schematic phase diagram for a sulphur. A sulphur can exist in two different crystalline
forms, rhombic and monoclinic, and has three triple points.
By Taylor expanding the second Equation and subtracting the first one,
∂g1
∂g2
∂g2
∂g1
dT +
dp =
dT +
dp
∂T p
∂p T
∂T p
∂p T
or
"
∂g1
∂T
−
p
∂g2
∂T
#
dT =
p
∂g2
∂p
−
T
∂g1
∂p
dp
(28)
(29)
T
and thus the slope of the curve is
∂g2
∂T p
∂g1
∂T p
∆
∂g
∂T p
dp
=− = − ∂g2
dT
1
∆ ∂g
− ∂g
∂p
∂p
∂p
T
−
T
(30)
T
In order to proceed, we recall that the change in Gibbs free energy is given by Equation
15 for each phase separately:
dGi = −Si dT + Vi dp + µi dNi .
(31)
where µi = gi . Also, G = N g, from which
dG = N dg + gdN = −SdT + V dp + µdN
V
dp.
→ dg = − NS dT + N
(32)
–8–
This result enables us to write:
Si
∂gi
;
=−
∂T p
Ni
∂gi
∂p
=
T
Vi
Ni
(33)
Using the result of Equation 33 in Equation 30 enables us to write
dp
=
dT
S2
N2
V2
N2
−
−
S1
N1
V1
N1
,
(34)
and if we refer to the same amount of substance in each phase, then N1 = N2 , and we have
dp
S2 − S1
∆S
=
=
dT
V2 − V1
∆V
(35)
Equation 35 is known as the Clausius-Clapeyron equation.
For every phase change which is accompanied by a change in entropy ∆S, there is
emission or absorption of heat - known as the latent heat, L. The entropy change in phase
transition at temperature T is
L
(36)
∆S = S2 − S1 =
T
Using this in Equation 35, we find that in a phase transition
dp
L
=
(37)
dT
T ∆V
where recall again that L and ∆V refer to the same amount of substance.
The difference in entropy between the two phases imply that phase change involves
a latent heat. This type of phase change is called first order transition. Many known
examples are like that - e.g., solid-liquid-vapor phase change, or allotropic transitions (e.g.,
grey to white tin). There are phase changes in which the entropy continuously changes, so
there is no latent heat involved. These do involve higher order derivatives of gi and other
thermodynamic quantities, such as the heat capacity. We will not consider those here.
For the processes of melting, evaporation and sublimation, ∆S > 0. This is easily
understood as the change is from an ordered phase to a less ordered one.
In vaporization or sublimation, the density decreases, and so ∆V > 0. Thus, from
the Clausius-Clapeyron equation, for vaporization and sublimation,
∆S
dp
=
>0
(38)
dT
∆V
Most substances expand in melting; however, there are exceptions - the most notable
one is, of course, water, which contract when melting. For these, dp/dT < 0. The phase
diagrams thus looks like presented in Figure 4.
–9–
Fig. 4.— General appearance of a phase diagram. Left: Solid expands upon melting,
∆V > 0. Right: Solid contracts upon melting, ∆V < 0.
6.
Applications of the Clausius-Clapeyron Equation
6.1.
Pressure dependence of the melting point
Consider the transition between ice and water. We know that at 0◦ C, the latent heat
is
L = 3.35 × 105 J/kg
Furthermore, the volume (per gram) of ice and water are
Vice = 1.0907 × 10−3 m3 kg−1
Vwater = 1.0013 × 10−3 m3 kg−1
(39)
and therefore in melting ∆V = Vwater − Vice = −0.0906 × 10−3 m3 kg−1 . Using these in the
Clausius-Clapeyron Equation, we find
L
3.35 × 105
dp
=
=−
= 1.35 × 107 N m−2 K−1 = −134 atm K−1
dT
T ∆V
273.2 × 0.0906 × 10−3
(40)
This means, that as the pressure increases, the melting point decreases. For example,
an increase in 1000 Atmospheres, lowers the melting point by ∼ 7.5 ◦ C.
It is this effect that is responsible for the motion of glaciers. Consider a glacier of
– 10 –
depth (thickness) d. The pressure at the bottom is
p=
F
= ρgd
A
(41)
with ρice = 917 kg m−3 and depth of ∼ 800 m (e.g., Baring glacier in Alaska), resulting in
p = 7.2 × 106 P a = 71 atm, from which we find that the melting point decreases by
∆T ≈
∆p
dp
dT
≈ 0.5◦ K
(42)
This implies that the deeper parts of the glacier melt due to the pressure, enabling the glacier
to flow. They freeze again when the pressure decreases.
6.2.
Pressure dependence of the boiling point
Since the volume of the gas is always larger than that of liquid, in evaporation ∆V is
always positive. Thus, increasing the pressure always increases the boiling point.
Consider again water as an example, the latent heat of vaporization is
L = 2.257 × 106 J/kg.
At T = 373.15◦ K, and p = 1 atm, the volume (per gram) of liquid water and gas (water
vapor) are
Vwater = 1.043 × 10−3 m3 kg−1
(43)
Vgas = 1673 × 10−3 m3 kg−1
(note that Vgas ≈ 1000 × Vwater !). Thus,
L
2.257 × 106
dp
=
=
= 3.62 × 103 N m−2 K−1 = 27 mmHg K−1
dT
T ∆V
373.15 × 1.672
(44)
At the top of the everest mountain (height ≈ 8 km) the pressure is ≈ 3.6 × 104 N m−2 (as
opposed to p ≈ 1.01 × 105 N m−2 at sea level). Thus, the temperature difference for water
evaporation at the top of mount everest is
∆T ≈ −
Thus, water boils at ∼ 80◦ C at this height.
65
= −18◦ C
3.6
(45)
– 11 –
6.3.
Evaporation and sublimation
We can use the following approximations when calculating evaporation and sublimation
using the Clausius-Clapeyron equation:
(i) Since the volume of a gas is so much larger than that of a solid or liquid, we may
approximate the change in volume as ∆V = V2 − V1 ≈ V2 .
(ii) We can assume that the vapor behaves like a perfect gas, for which the equation of
state is pV = nRT .
Combined into the Clausius-Clapeyron equation, one finds
dp
L
L
Lp
=
=
=
.
dT
T ∆V
T V2
nRT 2
(46)
dp
L dT
=
p
nR T 2
(47)
L
+ Const,
ln p = − nRT
LM
L
p = C1 e− nRT = C1 e− RT
(48)
We can write this as
or
where LM = L × M is the latent heat per mole. Thus, for small temperature change,
equation 48 gives the corresponding vapor pressure. For large change in T , the latent heat
may change, and the approximation no longer holds.
7.
The critical point
As the pressure and temperature are increased along the transition curve between liquid
and vapor (the vapor pressure curve), one reaches a critical point.
The increase of the pressure and temperature results in a decrease of the latent heat
and the volume change ∆V between the two phases, and thus as one reaches the critical
point they become zero (see Figure 5).
At temperatures below the critical temperature Tc (corresponding to the critical point),
the fluid can co-exist in two states with different specific volumes (liquid phase and gas
phase). Above Tc the substance exists in one fluid phase only.
In order to understand that, let us consider the isotherms in a pV diagram (see Figure 6).
Assume that our system is composed of a material initially at temperature T1 , corresponding
to point A in Figure 6.
– 12 –
Fig. 5.— General appearance of a phase diagram. Shown are the vapor pressure curve and
the critical point.
1. If the system is compressed isothermally, then the the pressure and the density increase
where the volume decreases until the vapors reaches saturation; this occurs at point
A2 .
2. If the volume is continued to be reduced, condensation occurs at constant pressure. At
this state, there are two coexisting phases, gas and liquid.
3. Once the material reaches point A1 , all the vapor is condensed (became liquid).
4. Further compression (volume reduction) requires enormous pressure, due to the low
compressibility of liquids.
Now, if the initial temperature is higher than T1 , the condensation interval becomes
shorter, and eventually, for an isotherm at the critical temperature, Tc , it disappears.
Mathematically, the critical point is defined by
2 ∂p
∂ p
=
=0
∂V T =TC
∂V 2 T =TC
(49)
– 13 –
Fig. 6.— Schematic m of the isotherms of a fluid. V is the specific volume (=volume per
gram). C is the critical point.
Above TC , the isotherms monotonically decrease in the (p, V ) plane, implying that there
is no distinct phase change. The properties of the system change continuously.