441_06Equilibrium

8/12/2019 441_06Equilibrium

1/24

1

Phase Transformations of

Pure Substances

Phase Diagrams

Phase Stability

-Temperature Dependence

-Pressure Dependence

Classification of Phase Transitions

Surface Tension


2/24

2

Thermodynamically Stable Phases

Usually, only one phase of a given sub-

stance is stable at any given temperatureand pressure.

At some conditions of temperature andpressure, two or more phases may exist in

equilibrium.A slight change in temperature or pressure

will favor one phase over others. Theconversion of one phase to another is aphase transition.

Phase transitions occur with a decrease(spont.) or no change (equil.) in Gibbsenergy.


3/24

3

Example Problem

The standard Gibbs energy of formation of

metallic white tin (a-tin) is 0 at 25 oC andthat of nonmetallic gray tin (b-tin) is +0.13 kJ

mol-1at the same temperature. Which is the

thermodynamically stable phase at 25o

C? Solution:The thermodynamically stable

phase is the one with the lower Gibbs

energy, which would be the a- (white) tin at

25 oC. Note: At lower temperatures, the nonmetallic

gray tin becomes the stable form. The tran-sition is known among metallurgistsas the tin disease.


4/24

4

G-T Phase Diagrams

This is similar to how Fig.5.1 should have looked.

(m/T)p=(Gm/T)p= -Sm The straight lines show that

S is approx. constant withtemp. (Really, they curve.)

Since S is positive for all

phases of all substances,the slopes are all negative.

Note that the gas phasehas the steepest slope;

the solid phase, the leaststee .


5/24

5

p-T Phase Diagrams

At left is a generic

p-T phase diagram. For any of the phase

boundary lines , theslope is given by

dp/dT = DS/DV =DH/TDV For the transitions

s->l, l->g, and s->g,DS > 0

For l->g and s->g,DV > 0, while fors->l, DV is almostalways > 0

This explains thepositive slopes.


6/24

6

Whats In a Phase Diagram

A one-component phasediagram represents the

situation when only thatcomponent is present.

Thus the vapor-pressure curveplots the pressure ONLY of the

vapor (no air) Water filling a container at 50

torr WILL NOT EVAPORATE. If some of the liquid is removed

without decreasing the size of

the container (or letting any airin) vapor will form in the spaceabove the liquid until thepressure of the vapor reachesthe equilibrium vapor pressure

(left).


7/247

Phase Diagrams of ParticularSubstances: H2O, CO2 , He


8/248

Example Problem

What is the minimum pressure at whichliquid is the thermodynamically stablephase of water at 25 oC?

Solution:Start in the liquid region of thephase diagram for water at 25 oC and dropdown in pressure until the vapor pressureline is reached. This occurs at 23.8 torr.Below this pressure liquid water cannotexist at 25 oC .


9/249

Criteria of Equilibrium

At equilibrium, thechemical potential of

a substance is the

same throughout asample, regardless

of how many phases

are present.


10/2410

Temperature Dependence

(m/T)p = -Sm As the temperature is raised, the chemical

potential of a pure substance decreases, always.

When a phase transition occurs, the relative

values of the chemical potentials of the various

phases are modified.


11/2411

Pressure and Melting Point

(m/p)T = Vm. The greater the molarvolume, the more steeply mrises with pressure.


12/2412

Effect of Pressure on Vap. Pres.

(a) and (b) in the figureat left show two ways of

determining the effect of

applied pressure on

vapor pressures.

For a perfect gas,

p=p* exp[Vm

Dp/RT]

shows how the vapor

pressure increases when

the applied pressure

increases.


13/24

13

Slopes of Phase Boundaries

When two phases aand bare in equi-

librium,

ma(

p,T) = mb(

p,T)

dm= -SmdT + Vmdp

-Sa,mdT + Va,mdp =

-Sb,mdT + Vb,mdp

(Vb,m- Va,m)dp =

(Sb,m- Sa,m)dT

dp/dT = DtrsS/DtrsV


14/24

14

The Solid-Liquid Boundary dp/dT = DS/DV =

DH/TDV (the Clapey-ron equation) appliesto any two phases aand b.

For instance, inmelting, dp/dT =DfusH/TDfusV

If DH and DV areapprox. constant,p.p* + (T-T*)

(DfusH/T*DfusV)


15/24

15

The Clausius-Clapeyron Equation The Clausius-Clapeyron

equation applies theClapeyron equation tothe special case wherephase bis the gasphase, and the gas is

assumed perfect. dp/dT = DvapH/TVg=DvapH/T(RT/p)

d(lnp)/dT = DH/RT2

lnp2= lnp1+ DH 1 - 1R T1 T2 Ifpis known at one

temp it can be found atanother temp.


16/24

16

Liquid-Vapor and Solid-Vapor

The Clausius-Clapeyron equation

applied to both

vaporization and

sublimation. For sublimation, sub-

stitute DsubH for DvapH

The assumptions/

approximations are: DV.Vgand

Vg.RT/p


17/24

17

Determiningpat a Given T How do you arrive at the value 23.8 torr for the vapor

pressure of water at 25 oC?

Some possible ways: Read it off ap-T phase diagram (if a diagram that can beread to that precision is available).

Look in a handbook or other reference for vapor pressuretables for water.

Calculate from a known vapor pressure at some other

temperature (C-C eqn, see next slide). Calculate from DG value for the liquid-to-gas transition.

This last approach can be used for water at 25 oC From the Table 2.6, DfG

o= -228.57 kJ/mol for gaseouswater at 25 oC and -237.13 kJ/mol for liquid water at 25 oC.

Therefore DtrGo= + 8.56 kJ/mol for vapn of water at 25 oC.

From DtrGocan be found Ktr , since DtrG

o= - RT ln Ktr= This gives ln Ktr= (+8.56 kJ/mol)/(-2.479 kJ/mol) = -3.453 Therefore Ktr= 0.0317 But for a vapn rxn, Ktris simply equal top/p

o (po= 1 bar)

So we arrive atp= 0.0317 bar = 3.17 kPa= 23.8 torr


18/24

18

Phase Rule For a system at equilibrium,

F = C - P + 2 where C = the number of components (1 so far in

this chapter), P = the number of phases present,and F = the number of degrees of freedom(thenumber of intensive variables such as temp, pres, or molfrac that can be changed without disturbing the number ofphases in equilibrium)

For a one-component system the phase rulebecomes

F = 3 - PWhen only one phase is present bothpand T are

independently variable. (An area on thep-Tdiagram)When two phases are present there is only onep

possible for a given T. (A line on thep-T diagram) Three phases can be present (triple point) but

there is no variation allowed inpor T.


19/24

19

Changes in Various Functions

Most functions(volume, enthalpy,entropy) change dis-continuously at a

phase transition. A few (chemical

potential) changecontinuously.

Cpgoes4and back.Do you see why?


20/24

20

Second-Order Phase Transitions

The previous slide illustrated first-order phasetransitions, in which the chemical potential was acontinuous function with a discontinuous first

derivative. A second-order transition is one in which the first

derivative of mwith respect to temperature is con-tinuous but the second derivation isdiscontinuous.


21/24

21

Lambda Transition

The transitionshown at left is

that of the fluid-

superfluid tran-

sition in liquid

helium.

It is named for

the shape of

the heat

capacity curve.


22/24

22

Surface Tension

Surface tension, orsurface energy, is

the force per unit

length (along h) orenergy per unit

area to increase

the surface area of

a liquid.


23/24

23

Bubbles, Cavities, Droplets

The pressure on theconcave side of an

interface,pinis

always greater than

the pressure on the

convex side,pout.

pin=pout+ 2g/r

This is the Laplace

equation.


24/24

24

Capillary Rise

The pressureexerted by a

column of liquid

of density rand

height h is p= rgh

This matches

the pressure

from 2g/r.

Therefore,

h = 2 rgr

441_06Equilibrium

Documents

Transcript of 441_06Equilibrium