Chapter 3 Phase Transitions and Chemical

Reactions

Phase Transitions and Chemical Reactions

1 Gibbs‘ Phase Rule

We now want to return to the important problem of how many state variables are actually necessary to uniquely determine the state of a system.

To this end,we start from an isolated system which contains k different particle species(chemical components)and p different phases (solid,liquid,gaseous,...).Each phase can be understood as a partial system of the total system and one can formulate the first law for each phase,

PidNdVPdSTdU ii

ii

k

i

iiiii ,2,11

ln this formulation of the first law,U(i)of phase i is a function of the extensive state variables ;I.e.,it depends on k+2 variables. i

kiii NNVS ,,, 1

Altogether we therefore have P(K+2)

Pi ,2,1

extensive state variables.If the total system is in thermodynamic equilibrium,we have in addition the following conditions for the intensive state quantities,cf.Equations (2.45-48)

pTTT 21

pppp 21

klplll ,121

Thermal equilibrium

Chemical equilibrium

Mechanical equilibrium

pTTT 21

pppp 21

klplll ,121

Thermal equilibrium

Chemical equilibrium

Each line contains P-1 equations,so that equation is a system of (P-1)(K+2) equations. Thus,we only require

2122 KPKPK

extensive variables to determine the equilibrium state of the total system.As we see, this number is independent of the number of phases.

Mechanical equilibrium

If we now consider that exactly ,so,we can substitute for .Here And

l

i

il NN

ilN

il

l

ili

l N

N

1 il

i

So that U(i)of phase i can be writen a function of the extensive state variables i

kiii VS 11 ,,,

I.e.,it depends on k+1 variables.

Thus,we only require

PKPKPK 2121

PKF 2 independent intensive variables.Equation(3.4)is named after J.W.Gibbs and is called Gibbs'phase rule.

(3.4)

Example 3.1:Clausius-Clapeyron equation

We want to derive a general equation to determine the vapor pressure of a liquid in equilibrium with its vapor. We have the following equilibrium conditions for two partial systems which can exchange energy,volume,and particles:

vli TT

vli pp

vli

Because of the Gibbs-Duhem relation these conditions are not independent from each other: if the equation of state is known and if we assume T and p to be given,we can calculate and · The equation

li v

i

k

iidNVdPSdT

1

0

The equation

(2.74 ）

TPTP vli ,, （ 3.12 ）

yields a dependence between p and T;i.e.,we can calculate the vapor pressure for a given temperature.

If we change the temperature by dT in Equation(3.12), the vapor pressure also has to change by a certain amount dp to account for equilibrium.For the corresponding changes and it must hold that

lid vd

TPdTPd vli ,,

This can be expressed with the Gibbs-Duhem relation in the following way

NdVdPSdT 0

dPN

VdT

N

STPd

li

li

li

lili ,

dPN

VdT

N

STPd

v

v

v

vv ,

(1)

(2)

We assume that

li

lili N

Ss

li

lili N

Vv

and analogously for the vapor:

(1)=(2)

vlivli ssdTvvdP

vli

vli

vv

ss

dT

dP

This is the Clausius-Claperon equation.

T

Q

N

S

N

Sss vli

li

li

v

vliv

'

However,in many cases and for not too large temperature differences, this evaporation heat may be considered to be constant.

With the evaporation heat per particle we thus obtain

vliQ '

liv

vli

vvT

Q

dT

dP

' (3.13)

In many cases and one hasliv vv

vTv

Q

dT

dP '

(3.14)

The intensive quantities and can of course be inserted as measured per mole instead of measured per particle.

'Qvv

Figure 3.1 Phase diagram of water

Exercise 3.2 vapor pressure of a liquid

Determine the vapor pressure of a liquid in equilibrium with its vapor under the assumption that the evaporation heat per particle does not depend on pressure or temperature and that the vapor behaves as an ideal gas.Solution:The best starting point is Equation(3.14):

vTv

Q

dT

dP '

With we obtainPkTNVv vvv //

'2Q

kT

P

dT

dP

We may integrate this after separating variables,for instance from an initial temperature T0 with vapor pressure Po to a final temperature T with vapor pressure p,

00

11'ln

TTk

Q

P

P

Or

000

11'exp

TTk

QTPTP

(3.15)

Thus,the vapor pressure increases strongly with temperature( ).Note that Equation (3.15)holds under the same conditions also for the sublimation pressure of a solid: is correct,but the sublimation heat is larger.Thus the pressure curve for sublimation is steeper than the vapor pressure curve.Both curves cross at the triple point.

0' Q

vsolid vv

2 Phase equilibrium and the Maxwell construction

When we introduced van der Waals‘ equation of state we already mentioned some inconsistencies of this equation.The isotherms of van der Waals’ equation (Figure 3.2) ,

NkTNbVV

NaP

2

Figure 3.2 Isotherms of the van der Waals gas.

show regions of0/ VP

We now want to show that these contradictions can be resolved by considering the phase transition from gas to liquid.

In equilibrium between vapor and liquid, however, a certain vapor pressure is

vP

TPTPTTPP vlivlivli ,, (3.17 ）

The vapor pressure is solely a function of temperature and does not depend on the vapor volume V,so that one obtains a horizontal isotherm in the pV diagram.

TPv

The pressure can be calculated from Equation (3.17), if the temperatures and chemical potentials of the vapor and liquid are known.

vP

dVVPSSTUV

V 2

112

In the first case we simply have ( =T(S2-S1) is the latent heat of the phase transition.

Q

121 VVPQU v

and in the case of the van der Waals isotherm we have

2

2

V

aN

NbV

NkTVP

1

2

1

22

11ln

2 VVaN

NbV

NbVNkTQU (3.21)

From the condition

1

2

1

21221

11ln

2 VVaN

NbV

NbVNkTVVPUU v

(3.22)

Remark:For a given and T the van der Waals isotherm has also a third (unstable)solution at C,Figure3.3.

vP

Figure3.3.Maxwell construction

Figure 3.4 Critical point and critical isotherm for CO2

Because of the importance of the critical point we want to calculate the critical state quantities ,and from van der Waals‘ equation.The critical point is characterized by the fact that both derivatives vanish (saddle point):

crcr PT ,crV

0,0,

2

2

,

crcrcrcr VTVT V

P

V

P

This is the well-known Maxwell construction.

k called critical point

If one brings the negative terms to the other sides of the respective equations and dividesone equation by the other .one obtain ,and thus

0

23

2

2

crcr

cr

V

aN

NbV

NkT

062 4

2

3 crcr

cr

V

aN

NbV

NkT

crcr VNbV3

2

NbVcr 3 (3.25)

If one inserts this into equation (3.24),one gets

kb

aV

kV

aNNbV

kV

aNT cr

cr

cr

cr

cr 27

8

9

422 2

3

2

3

From and it finally follows with van der Waals’ equation that

crVcrT

222

2

279272

8

b

a

Nb

aN

kbbN

aNkPcr

The critical state quantities are therefore uniquely determined by the parameters a and b. Hence ,for all gases one should have

375.08

3

278

273

kbabN

kbbNa

NkT

VP

cr

crcr

Experimentally one finds for Equation(3.28) numbers between 0.25 and 0.35 ,which once again confirms the qualitative usefulness of van der waals‘ equation.

3 Application of the laws of thermodynamics

We want to calculate the internal energy U(V,T)of a real gas.

The exact differential of U reads

{3.49}

dVV

UdT

T

UdU

TV

We have already identified the expression as the heat capacity because of at V=const.

VTCTU VV,/

dTCdUQ V

PdVTdSdU

In most cases,we want to express in terms

of T and p. To this end we denote the exact differential of the entropy S(V,T),

for which on the other hand

dVV

SdT

T

STVdS

TV

,

dVT

PdV

V

U

TdTC

TT

PdVdU

T

QdS

TV

rev

11

(3.51)

PdVTdSdU

VV

V T

U

TdTC

TT

S

11

T

P

V

U

TV

S

TT

1

TVU /

By comparing coefficients one finds

and

Since S has an exact differential it must hold that

VV

V T

U

TdTC

TT

S

11

T

P

V

U

TV

S

TT

1

TVT

U

TVTV

S 12

VT T

P

V

U

TTVT

S

12

the result that

TVT

U

VT

122

111

T

P

T

P

TV

U

TTV

U

T VTT

So that

PT

PT

V

U

VT

(3.56)

Thus we have reached our aim to express by derivatives of the equation of state, since we can readily determine p=p(N,T,V)also for real gases.

TVU /

22

11

T

P

T

P

TV

U

T VT

If we insert Equation (3.56)into Equation (3.49)we have

dVPT

PTdTTVCdU

VV

, (3.57 ）

For example: ideal gases NkTPV

0

PV

NkTP

T

PT

V

0/ T

VU

NkTU2

3

We will see in the next section that such relations are easily derived using the theory of transformations of variables for functions of more than one variable. Since dU is an exact differential one has

dVPT

PTdTTVCdU

VV

, (3.57 ）

VVT

V PT

PT

TV

C

VT

U

TV

U

22

VVT

V PT

PT

TV

C

However,the righthand side can be as well determined from the equation of state,so that we can calculate the volume dependence of the heat capacity.For an ideal gas one has,for instance,

{3.59} V

NkTTVNP ,,

and thus

{3.60} Therefore,the heat capacity of an ideal gas cannot depend on the volume.As we already how,it is even absolutely constant.

00

T

V

V V

CP

T

PT

Exercise 3.8:internal energy of the van der Waals gas

NkTNbVVNaP 2/

Calculate the internal energy of a van der Waals gas as a function of temperature and volume at constant particle number.

SolutionThe equation of state of the van der Waals gas reads

We now evaluate the expression : PT

PT

V

2/,, VNaNbV

NkTTVNP

VT

P

NbV

NkT

PT

PT

V

2/VNa

Hence,as for the ideal gas,the heat capacity of a van der Waals gas cannot depend on the volume because of

Thus we have according to Equation (3.57)

0

,2

VVVT

V aV

N

TP

T

PT

TV

VTC

adVV

NdTTCdU V

2

We can integrate this starting from an initial state T0 and po with the internal energy U0,

aN

VVdTTCTVUTVU

T

T V2

0000

11,,

0

For temperature differences which are not too large Cv(T)is approximately constant and thus

aNVV

TTCTVUTVU V2

00000

11,,

Exercise 3.9:Entropy of the van der Waals gas

Calculate the entropy of a van der Waals gas as a function of temperature and volume at constant particle number.

SolutionAccording to Equations (3.50),(3.51),and (3.56),we have

dVV

SdT

T

SdS

TV

dVV

SdT

T

SdS

TV

dVPV

U

TdTTC

T TV

11

dVT

PdTTC

T VV

1

The quantity for a van der Waals gas was calculated in the preceding Exercise;

VTP /

if we insert Equation (3.61),we obtain

Starting from a state T0 and V0 with entropy So we can integrate this equation:

dVNbV

NkdTTC

TdS V

1

NbV

NbVNkdTTC

TTVSTVS

T

T V

0

000 ln1

,,0

For temperature differences that are not too large ( ≈100 k)we have Cv ≈const.and thus

The entropy of a van der Waals gas is nearly identical to that of an ideal gas;one only has to reduce the volume by the proper volume Nb of the particles. 

NbV

NbVNk

T

TCTVSTVS V

00000 lnln,,

Exercise :

1.There is a kind of material which the equation of state is , Try to prove that its internal energy cannot depend on the volume.

TVfP

2.Calculate the entropy of a ideal gas as a function of temperature and volume at constant particle number.