MATERIALS

THERMODYNAMICS II

Chapter 7

PHASE EQUILIBRIUM IN

A ONE COMPONENT

SYSTEM

Thermodynamic Properties of a

System

1. Extensive Properties, which are additive.

Mass, volume, energy, etc.

2. Intensive Properties, to which a value may be

assigned at each point in the system.

Temperature, pressure, chemical potential, etc.

Intensive Properties

Temperature: Measure of the tendency for heat to leave the system.

Pressure: Measure of the potential of a system to undergo massive movement by expansion or

contraction.

Chemical potential: Measure of the rate of energy change of a system with the change in

the amount of ith component where all other

variables being held constant.

Complete thermodynamic equilibrium is said to prevail when the system under consideration is in mechanical (pressure is same at all points), thermal (no temperature gradient) and chemical equilibrium (Gibbs chemical potential of each component must be equal).

In a closed system of fixed composition, e.g., a one-component system, equilibrium, at the temperature, T, and the pressure, P, occurs when the system exists in that state which has the minimum value of G.

The Gibbs free energy of the system, G, is independent of the

proportions of the ice phase and the water phase present.

at const. T&P

The chemical potential of species in a particular state equals the

molar Gibbs free energy of the species in that particular state.

(P =1 atm, T = 273 K)

If P=1 atm and T > 273 K, 2 () < 2 ()

T < 273 K, 2 () > 2 ()

Hesss Law

273 T (K)

Jou

les

G=H-TS

At all T, H(l) > H(s) S(l) > S(s)

So solid will be stable at

all T.

However,

G=H-TS

Therefore, at T < Tm, H(sl) > TS(sl)

whereas at T > Tm,

H(sl) < TS(sl)

Variation of G with P at Constant T

G

P 1 atm

Const T (273 K)

=

For water: V(s) > V(l)

G as a Function of T & P

For equilibrium to be maintained:

G(l) = G(s) dG(l) = dG(s)

V(l) dP - S(l) dT = V(s) dP - S(s) dT

=()

()=

Clapeyron eqn.

V < 0 for H2O and

H>0 for all materials:

< 0

0 C, 1 atm

0.0075 C ,

0.006 atm

T

P

G

Equilibrium between vapor phase and a

condensed phase

=

Vvapor >> Vcond.phase V Vvapor

Assuming vapor behave as ideal gas: =

=

2

dlnP =

2

Clasius-Clapeyron eqn.

If H is independent from T:

cp(vapor) = cp(cond.phase)

=

+ .

If Cp 0, but independent of T: Cp = const.

= 298 +

298

= 298 + 298

= 298 298 +

2

= 298 298

1

+ + .

=

+ +

For H2O:

,2() =30+10.710-3T+0.33105T-2 J/K (298-2500 K)

,2() =75.44 J/K (273-373 K)

. = 41,090 J

, = ,373 + ()

373

=57,383-45.44T+5.35x10-3T2- 0.33105

dlnP =

2 and R-8.314 J/K mole

= 6,901.64

5.465 + 6.435 104 +

1,984.509

2+ .

Since the normal boiling point of the liquid is defined as the temperature at which the saturated vapor pressure exerted by the liquid is 1 atm, P=1 atm at 373K.

const=50.62

Thus the variation of the saturated vapor pressure of water with T in the range 273-373K:

= 2997

5.465 + 0.279 103 +

862

2+ 21.98

A B

C

1

0.006

0 0.0075 100

m b

O

A

B

C

solid liquid

vapor

Temperature (C)

Pre

ssu

re (

atm

)

Gibbs Phase Rule

Single phase areas have two degrees of freedom since T and P can be altered independently without disturbing the equilibrium.

While the equilibrium through AO , BO , and CO lines, which indicates two phase equilibrium, can be maintained by changing only one variable, either T or P. Therefore, they have only one degree of freedom.

On the other hand, tripoint junction has zero degrees of freedom.

These criteria result in the Gibbs phase rule:

F=C-P+2

Taking F=0 gives the maximum number of phases that can be observed for a given number of species.

a

o

T

P b

P1

s

l

v

T

G

Since G

T P,comp= S :

all of the slopes are negative

and since S(v) > S(l) > S(s) :

slopes are increasing from solid

to vapor.

SKETCH G-P DIAGRAM

FOR GIVEN T WITH THE

BLUE COLOR LINE!!!

s l

v

Example Problem 1

The vapor pressure of solid NaF varies with temperature as

lnP atm =34,450

T 2.01 lnT + 33.74

and the vapor pressure of liquid NaF varies with temperature as

lnP atm =31,090

T 2.52 lnT + 34.66

Calculate the the normal boiling temperature of NaF.

The normal boiling temperature,Tb, is defined as that temperature at which the saturated vapor pressure of the liquid is 1 atm.

ln 1 = 0 =31,090

T 2.52 lnT + 34.66

Tb = 2006K

Calculate the temperature and pressure at the triple point.

The saturated vapor pressures for the solid and liquid phases

intersect at the triple point, Ttp.

34,450

Ttp 2.01 lnTtp + 33.74 =

31,090

Ttp 2.52 lnTtp + 34.66

Ttp = 1239K

The pressure at the triple point can be calculated either from the

vapor pressure of solid or vapor pressure of liquid equation.

Ptp = exp34,450

1239 2.01 ln1239 + 33.74

Ptp = 2.29 1024atm

Calculate the molar heat of evaporation of NaF at its normal boiling temperature.

For vapor in equilibrium with liquid:

lnP atm =31,090

T 2.52 lnT + 34.66

From Clasius-Clapeyron equation:

dlnP

dT=H

RT2=31,090

T22.52

T

Since R=8,314 J/molK and Tb is 2006 K:

H = 31,090 8.314 2.52 8.314 2006

H = 216,500 J

Calculate the molar heat of melting of NaF at the triple point.

For vapor in equilibrium with solid:

lnP atm =34,450

T 2.01 lnT + 33.74

From the previous case,we know that:

H(lv)= 31,090 8.314 2.52 8.314 T

And it is obvious that:

H(sv)= 34,450 8.314 2.01 8.314 T

Since H(sl) + H(lv)= H(sv):

H(sl)= 27,900 + 4.24T

Since Ttp = 1239 K:

H(sl)= 33,150J

Calculate the difference between the constant-pressure molar heat capacities of liquid and solid NaF.

As we know

= ()

Since H(sl)= 27,900 + 4.24T:

()= () = 4.24 /K

Example Problem 2

The vapor pressures of zinc have been written as

lnP atm =15,780

T 0.755 lnT + 19.25

lnP atm =15,250

T 1.255 lnT + 21.79

Which of the two equations is for solid zinc?

Example Problem 3

Below the triple point (-56.2C) the vapor pressure of solid CO2 is given as:

lnP atm =3116

T+ 16.01

The molar latent heat of melting of CO2 is 8330 J and normal melting temperature of CO2 is -57 C.

Calculate the vapor pressure exerted by liquid CO2 at 25 C and explain why solid CO2 is referred to as dry ice.

Homework

Carbon has three allotropes: graphite, diamond, and a metallic form called solid III. Graphite is the stable form of 298 K and 1 atm pressure, and increasing the pressure on graphite at temperatures less than 1440 K causes the transformation of graphite to diamond and then the transformation of diamond to solid III.

Calculate the pressure which, when applied to graphite at 298 K, causes the transformation of graphite to diamond.

Given: H298 K(Graphite) H298 K Diamond = 1900J

S298 K(Graphite) = 5.74 J/K

S298 K(Diamond) = 2.37J/K

298 K(Graphite) = 2.22 g/cm3

298 K(Diamond) = 3.515 g/cm3