Phase transition - Physics Department UCC · Phase transition Asaf Pe’er1 ... The Clausius-Clapeyron Equation The Clausius-Clapeyron equation is an equation describing the phase

Phase transition

Asaf Pe’er1

November 18, 2013

1. Background

A phase is a region of space, throughout which all physical properties (density, magne-

tization, 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

V1 + V2 = V

N1 +N2 = N

(1)

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.

dS = dS1+dS2 =

(

∂S1

∂E1

dE1 +∂S2

∂E2

dE2

)

+

(

∂S1

∂V1

dV1 +∂S2

∂V2

dV2

)

+

(

∂S1

∂N1

dN1 +∂S2

∂N2

dN2

)

= 0

(2)

1Physics Dep., University College Cork

– 2 –

Fig. 1.— Phase transition.

and since E, V andN are constants, dE1 = −dE2, etc., and thus the condition for equilibrium

is

dS = 0 =

(

∂S1

∂E1

−

∂S2

∂E2

)

dE1 +

(

∂S1

∂V1

−

∂S2

∂V2

)

dV1 +

(

∂S1

∂N1

−

∂S2

∂N2

)

dN1 (3)

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:

1

T≡

(

∂S

∂E

)

V,N

, (4)

Thus the requirement T1 = T2 is written as

T1 = T2 →

∂S1

∂E1

=∂S2

∂E2

(5)

Similarly, equilibrating the pressure using the entropy-based definition of the pressure,

p ≡ T

(

∂S

∂V

)

E,N

, (6)

leads to

p1 = p2 →

∂S1

∂V1

=∂S2

∂V2

(7)

Thus, we are left with the 3rd condition,

∂S1

∂N1

=∂S2

∂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

µi ≡ −Ti

(

∂Si

∂Ni

)

Ei,Vi

(9)

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,

dS =

(

∂S

∂E

)

V,N

dE +

(

∂S

∂V

)

E,N

dV +

(

∂S

∂N

)

E,V

dN (11)

or

dS =1

TdE +

1

TpdV −

1

TµdN (12)

This can be re-arranged to write

dE = TdS − 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 → dF = dE − TdS − SdT

G = E + pV − TS → dG = dE + pdV + V dp− TdS − SdT(14)

We can writedF = −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:

µ =

(

∂F

∂N

)

T,V=const

=

(

∂G

∂N

)

T,p=const

(16)

We now write F and G as functions of a new parameter - N . Out of these two func-

tions, 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) = NG(T, p, 1) ≡ Ng(T, p), (17)

where g is the Gibbs free energy per particle. From Equations 16 and 17 it follows immedi-

ately 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 = N1g1(T, p) +N2g2(T, p) (20)

and

dG = N1dg1 + g1dN1 +N2dg2 + g2dN2 = 0, (21)

and using

dg1,2 =∂g1,2∂T

dT +∂g1,2∂p

dp, (22)

we get

dG =

[

N1

∂g1∂T

+N2

∂g2∂T

]

dT +

[

N1

∂g1∂p

+N2

∂g2∂p

]

dp+ g1(T, p)dN1 + g2(T, p)dN2 (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 temper-

ature 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 = N1g(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 one-

component 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∂T

)

p

dT +

(

∂g1∂p

)

T

dp =

(

∂g2∂T

)

p

dT +

(

∂g2∂p

)

T

dp (28)

or[

(

∂g1∂T

)

p

−

(

∂g2∂T

)

p

]

dT =

[(

∂g2∂p

)

T

−

(

∂g1∂p

)

T

]

dp (29)

and thus the slope of the curve is

dp

dT= −

(

∂g2∂T

)

p−

(

∂g1∂T

)

p(

∂g2∂p

)

T−

(

∂g1∂p

)

T

= −

∆(

∂g

∂T

)

p

∆(

∂g

∂p

)

T

(30)

In order to proceed, we recall that the change in Gibbs free energy is given by Equation

15 for each phase separately:

dGi = −SidT + Vidp+ µidNi. (31)

where µi = gi. Also, G = Ng, from which

dG = Ndg + gdN = −SdT + V dp+ µdN

→ dg = −SNdT + V

Ndp.

(32)

– 8 –

This result enables us to write:(

∂gi∂T

)

p

= −

Si

Ni

;

(

∂gi∂p

)

T

=Vi

Ni

(33)

Using the result of Equation 33 in Equation 30 enables us to write

dp

dT=

S2

N2

−S1

N1

V2

N2

−V1

N1

, (34)

and if we refer to the same amount of substance in each phase, then N1 = N2, and we have

dp

dT=

S2 − S1

V2 − V1

=∆S

∆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

∆S = S2 − S1 =L

T(36)

Using this in Equation 35, we find that in a phase transition

dp

dT=

L

T∆V(37)

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,

dp

dT=

∆S

∆V> 0 (38)

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

dp

dT=

L

T∆V= −

3.35× 105

273.2× 0.0906× 10−3= 1.35× 107 Nm−2K−1 = −134 atmK−1 (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

A= ρgd (41)

with ρice = 917 kgm−3 and depth of ∼ 800 m (e.g., Baring glacier in Alaska), resulting in

p = 7.2× 106 Pa = 71 atm, from which we find that the melting point decreases by

∆T ≈

∆pdp

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

Vgas = 1673× 10−3 m3 kg−1(43)

(note that Vgas ≈ 1000× Vwater!). Thus,

dp

dT=

L

T∆V=

2.257× 106

373.15× 1.672= 3.62× 103 Nm−2K−1 = 27 mmHgK−1 (44)

At the top of the everest mountain (height ≈ 8 km) the pressure is ≈ 3.6 × 104 Nm−2 (as

opposed to p ≈ 1.01 × 105 Nm−2 at sea level). Thus, the temperature difference for water

evaporation at the top of mount everest is

∆T ≈ −

65

3.6= −18◦ C (45)

Thus, water boils at ∼ 80◦C at this height.

– 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

dT=

L

T∆V=

L

TV2

=Lp

nRT 2. (46)

We can write this asdp

p=

L

nR

dT

T 2(47)

orln p = −

LnRT

+ Const,

p = C1e−

L

nRT = C1e−

LM

RT

(48)

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

(

∂p

∂V

)

T=TC

=

(

∂2p

∂V 2

)

T=TC

= 0 (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.