5Phase transitions

II Statistical Physics

5.1 Liquid-gas transition

We are now going to understand the liquid-gas transition. This is sometimes

known as boiling and condensation. We will employ the van der Waal’s equation

of state,

p =

kT

v −b

−

a

v

2

,

where

v =

V

N

is the volume per particle. This is only valid for low density gases, but let’s now

just ignore that, and assume we can use this for any density. This provides a

toy model of the liquid-gas phase transition.

Let’s look at isotherms of this system in the

pv

diagram. Note that in the

equation state, there is a minimum value of

v

, namely at

v

=

b

. Depending on

the values of T , the isotherms can look different:

v

p

T > T

c

T < T

c

b

There is a critical temperature

T

C

that separates two possible behaviours. If

T > T

C

, then the isotherms are monotonic, whereas when

T < T

C

, then there

are two turning points. At

T

=

T

C

, the two extrema merge into a single point,

which is a point of inflection. To find this critical point, we have to solve

∂p

∂v

T

=

∂

2

p

∂v

2

T

= 0.

Using the equation of state, we can determine the critical temperature to be

kT

c

=

8a

27b

.

We are mainly interested in what happens when

T < T

C

. We see that there

exists a range of pressures where there are three states with the same

p, T

but

different V .

v

p

L U G

At U, we have

∂p

∂v

T

> 0.

Thus, when we increase the volume a bit, the pressure goes up, which makes

it expand further. Similarly, if we compress it a bit, then

p

decreases, and it

collapses further. So this state is unstable.

How about

L

? Here we have

v ∼ b

. So the atoms are closely packed. Also,

the quantity

∂p

∂v

T

is large. So this is very hard to compress. We call this a liquid .

There are no surprise to guessing what

G

is. Here we have

v b

, and

∂p

∂v

is small. So we get a gas.

Note that once we’ve made this identification with liquids and gases, we see

that above T

C

, there is no distinction between liquids and gases!

Phase equilibriums

What actually happens when we try to go to this region? Suppose we fix a

pressure

p

and temperature

T

. Ruling out the possibility of living at

U

, we

deduce that we must either be at

L

or at

G

. But actually, there is a third

possibility. It could be that part of the system is at

L

and the other part is at

G

.

We suppose we have two separate systems, namely

L

and

G

. They have the

same

T

and

p

. To determine which of the above possibilities happen, we have to

consider the chemical potentials µ

L

and µ

G

. If we have

µ

L

= µ

G

,

then as we have previously seen, this implies the two systems can live in equilib-

rium. Thus, it follows that we can have any combination of

L

and

G

we like,

and we will see that the actual combination will be dictated by the volume v.

What if they are not equal? Say, suppose

µ

L

> µ

G

. Then essentially by

definition of

µ

, this says we would want to have as little liquid as possible. Of

course, since the number of liquid molecules has to be non-negative, this implies

we have no liquid molecules, i.e. we have a pure gas. Similarly, if

µ

G

> µ

L

, then

we have a pure liquid state.

Now let’s try to figure out when these happen. Since we work at fixed

T

and

p, the right thermodynamic potential to use is the Gibbs free energy

G = E + pV − T S.

We have previously discussed this for a fixed amount of gas, but we would now

let it vary. Then

dE = T dS − p dV + µ dN.

So we have

dG = −S dT + V dp + µ dN.

So we have

G

=

G

(

T, p, N

), and we also know

G

is extensive. As before, we can

argue using intensivity and extensivity that we must have

G = f(T, p)N

for some function f. Then we have

f =

∂G

∂N

T,p

= µ.

Therefore

µ(T, p) = g(T, p) ≡

G

N

.

This is the Gibbs free energy per particle. Of course, this notation is misleading,

since we know

T

and

p

don’t uniquely specify the state. We should have a copy

of this equation for L, and another for G (and also U, but nobody cares).

Using the first law, we get

∂µ

∂p

T

=

1

N

∂G

∂p

T,N

=

V

N

= v(T, p). (†)

This allows us to compute

µ

(

T, p

) (up to a constant). To do so, we fix an

arbitrary starting point

O

, and then integrate (

†

) along the isotherm starting at

O

. At some other point

Q

, we can write an equation for the chemical potential

µ

Q

= µ

O

+

Z

Q

0

dp v(T, p).

Geometrically, this integral is just the area between the isotherm and the

p

axis

between the two endpoints.

We will pick

O

to be a gas state of high volume (and hence low pressure).

Referring to the upcoming diagram, we see that as we go from

O

to

N

, the

pressure is increasing. So the integral increases. But once we reach

N

, the

pressure starts decreasing until we get

J

. It then keeps increasing again on

JM

.

v

p

J

N

O

M

K

We can now sketch what µ looks like.

p

µ

unstable

N

J

X

p(T )

p

1

p

2

K

M

gas

liquid

We see that at a unique point

X

, we have

µ

L

=

µ

S

. We define the vapour

pressure p(T ) to be the pressure at X.

Geometrically, the equilibrium condition is equivalent to saying

Z

L

G

dp v = 0.

Thus,

p

(

T

) is determined by the condition that the shaded regions have equal

area. This is known as the Maxwell construction.

v

p

We write N

L

for the number of particles in liquid phase. Then we have

N

G

= N − N

L

= number of particles in G phase.

Then we have

G = N

L

g

L

+ (N − N

L

)g

G

= N

L

µ

L

+ (N − N

L

)µ

G

.

The second law tells us we want to try to minimize

G

. Consider the part of

the plot where

p < p

(

T

). Since

µ

G

< µ

L

, we find that

G

is minimized when

N

L

= 0. What does this mean? If we live in the bit of the liquid curve where

p < p

(

T

), namely

JX

, then we are not on an unstable part of the curve, as the

liquid obeys the stability condition

∂p

∂v

T

< 0.

However, it has higher Gibbs free energy than the gas phase. So it seems like we

wouldn’t want to live there. Thus, the liquid is locally stable, but not globally

stable. It is meta-stable. We can indeed prepare such states experimentally.

They are long-lived but delicate, and they evaporate whenever perturbed. This

is called a superheated liquid.

Dually, when

p > p

(

T

), global stability is achieved when

N

G

= 0. If we have

a gas living on XN, then this is supercooled vapour.

The interesting part happens when we look at the case

p

=

p

(

T

). Here if we

look at the equation for

G

, we see that

N

L

is undetermined. We can have an

arbitrary portion of L and G. To have a better picture of what is going on, we

go back to the phase diagram we had before, and add two curves. We define

the coexistence curve to be the line in the

pV

plane where liquid and gas are in

equilibrium, and the spinodal curve to be the line where

∂p

∂v

T

= 0.

v

p

The unstable states are the states inside the spinodal curve, and the metastable

ones are those in between the spinodal curve and coexistence curve. Now suppose

we only work with stable states. Then we want to remove the metastable states.

v

p

It seems like there is some sort of missing states in the curve. But really, there

aren’t. What the phase diagram shows is the plots of the pure states, i.e. those

that are purely liquid or gas. The missing portion is where

p

=

p

(

T

), and we

just saw that at this particular pressure, we can have any mixture of liquid and

gas. Suppose the volume per unit particle of the liquid and gas phases are

v

L

= lim

p→p(T )

−

v(T, p), v

G

= lim

p→p(T )

+

v(T, p).

Then if we have

N

L

many liquid particles and

N

G

many gas particles, then the

total volume is

V = V

L

+ V

G

= N

L

v

L

+ N

G

+ v

G

.

or equivalently,

v =

N

L

N

v

L

+

N − N

L

N

v

G

.

since

N

L

is not fixed,

v

can take any value between

v

L

and

v

G

. Thus, we

should fill the gap by horizontal lines corresponding to liquid-gas equilibrium at

p = p(T ).

v

p

Thus, inside the coexistence curve, the value of

v

(size of container) determines

N

L

inside the coexistence curve.

Practically, we can take our gas, and try to study its behaviour at different

temperatures and pressures, and see how the phase transition behaves. By doing

so, we might be able to figure out

a

and

b

. Once we know their values, we can

use them to predict the value of

T

C

, as well as the temperature needed to liquefy

a gas.

Clausius–Clapeyron equation

We now look at the liquid-gas phase transition in the (p, T) plane.

T

p

critical point

p = p(T )

T

c

liquid

gas

Crossing the line

p

=

p

(

T

) results in a phase transition. The volume changes

discontinuously from

v

L

(

T

) to

v

G

(

T

), or vice versa. On the other hand,

g

=

G

N

changes continuously, since we know g

L

= g

G

.

In general, we can write

dg =

∂g

∂T

p

dT +

∂g

∂p

T

dp.

Using

dG = −S dT + V dp + µ dN,

we find

dg = −s dT + v dp,

where

s =

S

N

is the entropy per particle.

Along p = p(T ), we know that g

L

= g

G

, and hence

dg

L

= dg

G

.

Plugging in our expressions for dg, we find

−s

L

dT + v

L

dp = −s

G

dT + v

G

dp,

where

s

G

and

s

L

are the entropy (per particle) right below and above the line

p = p(T ).

We can rearrange this to get

dp

dT

=

s

G

− s

L

v

G

− v

L

.

This is the Clausius–Clapeyron equation. Alternatively, we can write this as

dp

dT

=

S

G

− S

L

V

G

− V

L

.

There is another way of writing it, in terms of the latent heat of vaporization

L = T (S

G

− S

L

),

the heat we need to convert things of entropy

S

L

to things of entropy

S

G

at

temperature T . Then we can write it as

dp

dT

=

L

T (V

G

− V

L

)

.

This is also called the Clapeyron–Clapeyron equation.

Example.

Suppose

L

does not depend on temperature, and also

V

G

V

L

. We

further assume that the gas is ideal. Then we have

dp

dT

≈

L

T V

G

=

Lp

NkT

2

.

So we have

p = p

0

exp

−

L

NkT

for some p

0

.