Thermodynamics

Chapter 13

Phase Diagrams

NC State University

Definition of a phase diagram A phase diagram is a representation of the states of matter,

solid, liquid, or gas as a function of temperature and pressure.

In the Figure shown below the regions of space indicate

the three phases of carbon dioxide. The curved lines

indicate the coexistence curves. Note there is a unique

triple point.

Pre

ssu

re (

atm

)

Degrees of freedom Within any one of the single-phase regions both temperature

and pressure must be specified. Because two thermodynamic

variables can be changed independently we say that the

system has two degrees of freedom. Along any of the

coexistence curves the pressure and temperature are coupled,

i.e. any change in the temperature implies a change in

pressure to remain on the line. Thus, along the curves there

is only one degree of freedom. The triple point is a unique

point in phase space and there is only one set of values of

pressure and temperature consistent with the triple point.

Thus, we say that at the triple point the system has zero

degrees of freedom. If we follow the liquid-vapor coexistence

curve towards higher temperature we find that it ends at the

critical point. Above the critical point there is no distinction

between liquid and vapor and there is a single fluid phase.

Free energy dependence

along the coexistence curve In a system where two phases (e.g. liquid and gas) are in

equilibrium the Gibbs energy is G = Gl + Gg, where

Gl and Gg are the Gibbs energies of the liquid phase and the

gas phase, respectively. If dn modes (a differential amount

of n the number of moles) are transferred from one phase to

another at constant temperature and pressure, the differential

Gibbs energy for the process is:

The rate of change of free energy with number of moles is

called the chemical potential.

dG = Gg

ngP,T

dng + Gl

n lP,T

dn l

The significance of chemical

potential of coexisting phases We can write the Gibbs free energy change using the following

notation:

Note that if the system is entirely composed of gas molecules

the chemical potential mg will be large and ml will be zero.

Under these conditions the number of moles of gas will

decrease dng < 0 and the number of moles of liquid will

increase dnl > 0. Since every mole of gas molecules converted

results in a mole of liquid molecules we have that:

dng = -dnl

dG = mgdng + mldn l

Coexistence criterion

In terms of chemical potential, the Gibbs energy for the phase

equilibrium is:

Since the two phases are in equilibrium dG = 0 and since

dng 0 we have mg = ml. In plain language, if two phases of a

single substance are in equilibrium their chemical potentials

are equal.

If the two phases are not in equilibrium a spontaneous

transfer of matter from one phase to the other will occur in

the direction that minimizes dG. Matter is transferred from a

phase with higher chemical potential to a phase with lower

chemical potential consistent with the negative sign of Gibb's

free energy for a spontaneous process.

dG = m g – m l dng

Solid-liquid coexistence curve

To derive expressions for the coexistence curves on the

phase diagram we use the fact that the chemical potential

is equivalent in the two phases. We consider two phases

a and b and write

ma(T,P) = mb(T,P)

Now we take the total derivative of both sides

The appearance of this equation is quite different from

previous equations and yet you have seen this equation

before. The reason for the apparent difference is the

symbol m. Remember that m for a single substance is

just the molar free energy.

ma

P T

dP +ma

T P

dT =mb

P T

dP +mb

T P

dT

The Clapeyron equation Substituting these factors into the total derivative above

we have

Solving for dP/dT gives

This equation is known as the Clapeyron equation. It gives

the two-phase boundary curve in a phase diagram with

DtrsH and DtrsV between them. The Clapeyron equation can

be used to determine the solid-liquid curve by integration.

Starting with a known point along the curve (e.g. the triple

point or the melting temperature at one bar) we can calculate

the rest of the curve referenced to this point.

Vm

adP – Sm

adT = Vm

bdP – Sm

bdT

dPdT

=Sm

b– Sm

a

Vm

b– Vm

a=D trsSm

D trsVm

=D trsHm

TD trsVm

dPP1

P2

=D trsHm

D trsVm

dTTT1

T2

The liquid-vapor and solid-

vapor coexistence curves The Clapeyron equation cannot be applied to a phase

transition to the gas phase since the molar volume of a gas

is a function of the pressure. Making the assumption that

Vmg >> Vm

l we can use the ideal gas law to obtain a new

expression for dP/dT.

The integrated form of this equation

yields the Clausius-Clapeyron equation.

dPdT

=D trsHm

TVm

g =PD trsHm

RT2

dPPP1

P2

=D trsHm

RT2

dTT1

T2

lnP2

P1

=D trsHm

R1T1

– 1T2

=D trsHm

R

T2 – T1

T1T2

Applying the Clausius-

Clapeyron equation If we use DH of evaporation the C-C equation can be used

to describe the liquid-vapor coexistence curve and if we

use DH of sublimation this equation can be used to describe

the solid-vapor curve.

The pressure derived from the C-C equation is the vapor

pressure at the given temperature. Applications also include

determining the pressure in a high temperature vessel

containing a liquid (e.g. a pressure cooker). If you are given

an initial set of parameters such as the normal boiling point,

for example you may use these as T1 and P1. Then if you

are given a new temperature T2 you can use the C-C to

calculate P2.

Constructing the phase diagram for CO2 We can use the Clapeyron and Clausius-Clapeyron equations

to calculate a phase diagram.

For example, we can begin with the CO2 diagram shown above.

The triple point for CO2 is 5.11 atm and 216.15 K.

The critical point for for CO2 is 72.85 atm and 304.2 K.

We also have the following data

Note that we can calculate the enthalpy of sublimation from

DvapHo = DsubH

o - DfusHo = 16.9 kJ/mol.

rsolid = 1.53 g/cm3 and rliquid = 0.78 g/cm3, respectively.

The density r = m/V = nM/V so the molar volume is

Vm = V/n = M/r where M is the molar mass.

In units of L/mole we have

Vsm = 44 g/mole/[1530 g/L] = 0.0287

Vlm = 44 g/mole/[780 g/L] = 0.0564

DfusV = Vlm - V

sm = 0.0564 - 0.0287 = 0.0277 L/mole

Transition DtrsHo (kJ/mol) Ttrs (K)

Fusion 8.33 217.0

Sublimation 25.23 194.6

Constructing the phase

diagram for CO2 Starting with the triple point we use the Clausius-Clapeyron

equation to calculate the liquid-vapor coexistence curve.

P = 5.11exp{DvapH/R[T – 216.15]/216.15T}

P = 5.11exp{2,032[T – 216.15]/216.15T}

Notice that if we were to calculate the critical pressure

using this formula we would obtain 77.3 atm which is about

5 atm larger than the experimental number. There are several

sources of inaccuracy including mainly our neglect of the

temperature dependence of the enthalpy.

We can also begin a the critical point

P = 72.8 exp{DvapH/R[T – 304.2]/304.2T}

P = 72.8 exp{2,032[T – 304.2]/304.2T}

Constructing the liquid-vapor curve

Liquid-vapor

P (atm) T (K)

5.11 216.15

P = 5.11exp{2032[T – 216.15]/216.15T}

Pre

ssure

(atm

)

Constructing the liquid-vapor curve

Liquid-vapor

P (atm) T (K)

5.11 216.15

6.03 220

9.0 230

P

ressure

(atm

)

P = 5.11exp{2032[T – 216.15]/216.15T}

Constructing the liquid-vapor curve

Liquid-vapor

P (atm) T (K)

5.11 216.15

6.03 220

9.0 230

13.0 240

Pre

ssure

(atm

)

P = 5.11exp{2032[T – 216.15]/216.15T}

Constructing the liquid-vapor curve

Liquid-vapor

P (atm) T (K)

5.11 216.15

6.03 220

9.0 230

13.0 240

24.9 260

Pre

ssure

(atm

)

P = 5.11exp{2032[T – 216.15]/216.15T}

Constructing the liquid-vapor curve

Liquid-vapor

P (atm) T (K)

5.11 216.15

6.03 220

9.0 230

13.0 240

24.9 260

41.0 280

Pre

ssure

(atm

)

P = 5.11exp{2032[T – 216.15]/216.15T}

Constructing the liquid-vapor curve

Liquid-vapor

P (atm) T (K)

5.11 216.15

6.03 220

9.0 230

13.0 240

24.9 260

41.0 280

63.7 300

72.8 304

Pre

ssure

(atm

)

P = 5.11exp{2032[T – 216.15]/216.15T}

Constructing the solid-vapor curve

Starting again at the triple point

P = 5.11exp{DsubH/R[T – 216.15]/216.15T}

P = 5.11exp{3034[T – 216.15]/216.15T}

Solid-vapor

P (atm) T (K)

5.11 216.15

Pre

ssure

(atm

)

Constructing the solid-vapor curve

Starting again at the triple point

P = 5.11exp{DsubH/R[T – 216.15]/216.15T}

P = 5.11exp{3034[T – 216.15]/216.15T}

Solid-vapor

P (atm) T (K)

5.11 216.15

3.38 210

1.64 200

Pre

ssure

(atm

)

Constructing the solid-vapor curve

Starting again at the triple point

P = 5.11exp{DsubH/R[T – 216.15]/216.15T}

P = 5.11exp{3034[T – 216.15]/216.15T}

Solid-vapor

P (atm) T (K)

5.11 216.15

3.38 210

1.64 200

0.725 190

0.298 180

Pre

ssure

(atm

)

Constructing the solid-vapor curve

Starting again at the triple point

P = 5.11exp{DsubH/R[T – 216.15]/216.15T}

P = 5.11exp{3034[T – 216.15]/216.15T}

Solid-vapor

P (atm) T (K)

5.11 216.15

3.38 210

1.64 200

0.725 190

0.298 180

0.111 170

Pre

ssure

(atm

)

Constructing the solid-liquid curve Using the Clapeyron equation we calculate:

P = 5.11 + [DfusH/DfusV] ln{T/216.15}

P = 5.11 + 2,967 ln{T/216.15}

Solid-liquid

P (atm) T (K)

5.11 216.15

Pre

ssure

(atm

)

Constructing the solid-liquid curve Using the Clapeyron equation we calculate:

P = 5.11 + [DfusH/DfusV] ln{T/216.15}

P = 5.11 + 2,967 ln{T/216.15}

Solid-liquid

P (atm) T (K)

5.11 216.15

58.0 220

Pre

ssure

(atm

)

Constructing the solid-liquid curve Using the Clapeyron equation we calculate:

P = 5.11 + [DfusH/DfusV] ln{T/216.15}

P = 5.11 + 2,967 ln{T/216.15}

Solid-liquid

P (atm) T (K)

5.11 216.15

58.0 220

124.2 230

Pre

ssure

(atm

)

Constructing the solid-liquid curve Using the Clapeyron equation we calculate:

P = 5.11 + [DfusH/DfusV] ln{T/216.15}

P = 5.11 + 2,967 ln{T/216.15}

Solid-liquid

P (atm) T (K)

5.11 216.15

58.0 220

124.2 230

319 240

559 260

780 280

987 300

1180 320

Solving Problems

What is the vapor pressure of water above a lake on a day

When the temperature of both water and air is 32 oC?

Solving Problems

What is the melting point of ice underneath an ice skater

Who has a mass of 100 kg assuming that the area of the

Skate blades is 10-7 m2.