Chapter 4 Physical Transformations of Pure Substances 1.

Chapter 4

Physical Transformations of Pure Substances

1

2

Phase Diagrams and Phase Boundaries

Phase Diagram

The phase diagram of a substance showsthe regions of pressure and temperatureat which the various phases are stable.

A typical phase diagram is shown to the right.

Phase Boundaries

These are the lines (combination of p and T) on which two phasesare in equilibrium

3

The critical temperature (Tc) is the highesttemperature at which a substance canbe liquified.

The vapor pressure at Tc is called thecritical pressure (pc).

The Triple Point

The triple point temperature and pressure,T3 and p3, is the one point in a phase diagramat which all three phases (solid, liquid and vapor)are in equilibrium.

The Critical Point

4

The Boiling Point

The boiling point depends upon the applied (external) pressure.

Pext = Vap. Press. Liquid + Gas (boiling point)

Pext > Vap. Press. Liquid

Pext < Vap. Press. Gas

Pext

Liquidor

Gas

The boiling point is the temperature at whichliquid and vapor are in equilibrium.

The curve, representing the combinations of p and T at which solidand liquid are in equilibrium is often called the vaporization curve.

5

The Melting Point

The melting point depends upon the applied (external) pressure, but to a much smaller extent than the boiling point.

The melting (or freezing) point is the temperature at which solid and liquid are in equilibrium.

The Sublimation Curve

This represents the combination of temperatures and pressures atwhich solid and vapor are in equilibrium

A Typical Phase Diagram

Temperature (oC)

Pre

ssur

e (b

ar)

p1

Solid Liquid Vapor

p2

Melting point increases with Pressure

Melting + Vaporization

Sublimation

p3

T3

6

Boiling point increases with Pressure

Pressure dependence of the Boiling Point

Temperature (oC)

Pre

ssur

e (b

ar) Liquid Vapor

100

1

2

120

Liquid Vapor

d(liq) >> d(vap)

V(liq) << V(vap)

Increased pressure shiftsequilibrium in directionof lower volume.

7

Pressure dependence of the Melting Pointin “normal” substances

Temperature (oC)

Pre

ssur

e (b

ar)

1

100 Solid Liquid

d(sol) > d(liq)

V(sol) < V(liq)


Solid Liquid

Tmpo Tmp

8

9

Question:

Water melts and then boils (i.e. it has a liquid phase)Benzene melts and then boilsIron melts and then boilsIodomethane melts and then boilsetc., etc., etc.

Why does CO2 sublime directly from solid to vapor ???

Answer:

Because the triple point pressure of CO2 is 5.1 bar (~5 atm.),which is higher than the pressure of the atmosphere on Earth.

The Phase Diagram of CO2

Temperature (oC)

Pre

ssur

e (b

ar)

p1

Solid Liquid Vapor

p2


Sublimation

5.1bar

-56 oC

10

Melting point increases with Pressure

Boiling point increases with Pressure

Phase Diagram of H2O

P1

Temperature (oC)

Pre

ssur

e (b

ar)

Solid Liquid Vapor

P2

Melting point decreases with Pressure


Sublimation

4.6torr

0.01

11

Temperature (oC)

Pre

ssur

e (b

ar)

0

1

140

-1

Solid (ice) Liquid

d(ice) < d(liq)

V(ice) > V(liq)


Solid Liquid

Pressure dependence of the Melting Pointin Water

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13

Some More Complex Phase Diagrams (FYI)

This is a phase diagram of water over anextended range of pressure and temperature.

It shows the regions of stability of the variousdifferent crystal structures of ice.

Water

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Helium

Even Helium has a fairly sophisticated phasediagram. However, all of the "action" occursat very low temperature; Tc 5 K

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Phase Stability and Phase Transitions

Chemical Potential

As discussed in Chapter 3, the chemical potential of a substance isdefined by:

, , j i

ii p T n n

G

n

However, in a pure (one component) system, the chemical potential

is simply the Molar Gibbs Energy: = Gm

Phase Equilibrium

Consider two phases, and (e.g. solid and liquid). The two phases

will be in equilibrium when: =

When the pressure and temperature are varied, they will stay in

equilibrium when: d = d

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The Temperature Dependence of Phase Stability

From Chapter 3, the temperature and pressure dependence of the Gibbs Energy is given by:

m m mp T

d dG S dT V dp dT dpT p

Therefore, the change in chemical potential (Molar Gibbs Energy) withrising temperature is:

mp

ST

Notes

The entropies of all substances in all phases is positive, Sm > 0.Therefore always decreases with rising temperature.

Relative entropies of different phases are: Sm(gas) >> Sm(liq) > Sm(sol)Therefore, the decrease in with increasing temperature is greatestfor the gas phase, and smallest for the solid.

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Two phases ( and ) are in equilibrium when =

s = l l = g

The phase with the lowest chemical potential at a given temperature isthe most stable phase.

Remember:

mp

ST

Therefore:

gas liq sol

p p pT T T

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The Effect of Pressure on the Melting Point

m m mp T


m

T

Vp

The chemical potential of all phasesincrease with rising pressure.

In most materials, Vm(liq) > Vm(sol)Therefore, (liq) rises more rapidly than (sol) with increasing pressure.

This results in a higher melting pointat higher pressure.

Low pHigh p

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The Effect of Pressure on the Melting Point

m m mp T


m

T

Vp

The chemical potential of all phasesincrease with rising pressure.

In a few materials, most notably water, Vm(sol) > Vm(liq) Therefore, (sol) rises more rapidly than (liq) with increasing pressure.

This results in a lower melting pointat higher pressure.

Low p

High p

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Quantitative Treatment of Phase Equilibria:The Location of Phases Boundaries

Two phases, will be in equilibriumwhen their chemical potentials are equal:

( , ) ( , )p T p T

For the two phases to remain in equilibrium asthe pressure/temperature are changed, theirvariations must be equal:

( , ) ( , )d p T d p T

Remembering that: m md S dT V dp

Thus, the criterion for two phases to remain in equilibrium is:

, , , ,m m m mS dT V dp S dT V dp

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Thus, the criterion for two phases to remain in equilibrium is:

, , , ,m m m mS dT V dp S dT V dp

, , , ,m m m mV V dp S S dT

, ,

, ,

m m

m m

S Sdp

dT V V

trs m

trs m

S

V

The Clapeyron Equation

Because the two phases are in equilibrium, we can use the relation,trsS = trsH/T, to obtain an alternative, equivalent form of the Clapeyron Equation:

trs trs

trs trs

dp S H

dT V T V

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Slopes of the Phase Boundaries

trs trs

trs trs

dp S H

dT V T V

The Clapeyron Equation can be used both qualitatively and quantitatively.

In the next several slides, we will show how it can be used in aqualitative fashion to explain the relative slopes of various equilibrium curves in a phase diagram.

sol

liq

vapE

ntr

op

y (S

)

solliq

vap

Vo

lum

e (V

)

A

D

B

C

Temperature (oC)

Pre

ssur

e (b

ar) Solid Liquid Vapor

AB: Liquid-Vapor Equilibrium

AC: Solid-Liquid Equilibrium

AD: Solid-Vapor Equilibrium

A: Triple Point Solid-Liquid-Vapor

B: Critical Point

Why is Slope(AD) > Slope(AB) ?

AD: Solid-Vapor

V(s-v) = Vvap-Vsol = Vvap

S(s-v) = Svap-Ssol

v ap

solv ap

V

SS

V

S

dT

dP

AB: Liquid-Vapor

V(l-v) = Vvap-Vliq = Vvap

S(l-v) = Svap-Sliq

v ap

liqv ap

V

SS

V

S

dT

dP

Because Svap-Ssol > Svap-Sliq or subS > vapS

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sol

liq

vapE

ntr

op

y (S

)

A

D

B

C

Temperature (oC)

Pre

ssur

e (b






B: Critical Point

Why is Slope(AC) >> Slope(AB) [and Slope(AD)] ?

AC: Solid-Liquid

V(s-l) = Vliq-Vsol

S(s-l) = Sliq-Ssol

solliq

solliq

VV

SS

V

S

dT

dP

AB: Liquid-Vapor

V(l-v) = Vvap-Vliq = Vvap

S(l-v) = Svap-Sliq

v ap

liqv ap

V

SS

V

S

dT

dP

Because Vliq-Vsol << Vvap-Vliq or fusV << vapV

liq

vap

Vo

lum

e (V

)sol

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sol

liq

vapE

ntr

op

y (S

)

Why is Slope(AC) < 0 in H2O ?

AC: Solid-Liquid

V(s-l) = Vliq-Vsol

S(s-l) = Sliq-Ssol

solliq

solliq

VV

SS

V

S

dT

dP

Because Vliq < Vsol

A

D

B

C

Temperature (oC)

Pre

ssur

e (b






B: Critical Point

H2O

H2O

dliq > dsol

Vliq < Vsol

Vliq-Vsol < 0

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Vo

lum

e (V

)liqsol

vap

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Quantitative Application of the Clapeyron Equation

trs trs

trs trs

dp S H

dT V T V

The Solid-Liquid Transition: Melting (aka Fusion)

fus fus

fus liq sol

H Hdp

dT T V T V V

Over small ranges of temperature, T, fusH and fusV are approximatelyconstant. Therefore, one can write:

2 1

2 1

fus

liq sol

Hp p p

T T T T V V

Note: The Clapeyron Equation cannot be used for quantitative calculations on vaporization or sublimation because it cannot be assumed that V is independent of temperature, pressure for these transitions.

or fus

liq sol

Hp T

T V V

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Side Note: A more rigorous (but unnecessary) equation (FYI Only)

2 1

2 1

fus

liq sol

Hp p p

T T T T V V

or

fus

liq sol

Hp T

T V V

It's straightforward to show that if one doesn't make the approximationthat T = Constant, then the result is:

fus fus

fus liq sol

H Hdp

dT T V T V V

2

1

2

1ln

Tfus fus

fus T liq sol

H HdT Tp TV T V V

However, it can also be shown that if T = T2 -T1 << T1 ,then the more rigorous solution reduces to the one we're using.

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2 1

2 1

fus

liq sol

Hp p p

T T T T V V

Example: The melting point of benzene at 1 bar is 6.0 oC. What is the melting point of benzene under an applied pressure of 200 bar?

fusH = 9.8 kJ/molM = 78 g/mold(sol) = 0.95 g/cm3 d(liq) = 0.88 g/cm3 1 kPa-L = 1 J

Tmp (200 bar) = 9.7 oC

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Vaporization and Sublimation

These two phase transitions involve the gas phase and, therefore,one cannot assume that trsV = Vgas -Vliq (or Vsol) = Constant.

trs trs

trs trs

dp S H

dT V T V

However, it is valid to assume that (a) trsV Vgas and (b) Vgas = RT/p (for 1 mole)

Using trsV = RT/p, one has:

2trs trs trs

trs

dp H H p H

dT T V RTRTT p

Separate Variables:2

trsHdpdT

p RT

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Separate Variables:

Now, integrate, assuming that trsH is independent of T:

2 2

1 12

p Ttrs

p T

Hdp dT

p R T

2

2

1

1

1ln

Tp trs trs trsp

Ttrs

H H Hp

T V R TRTT p

The Clausius-Clapeyron Equation

2

1 2 1

1 1ln trs trs trs

trs

H H Hpp T V R T TRTT p

2trsHdp

dTp RT

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2

1 2 1

1 1ln trs trs trs

trs


For vaporization, trsH = vapH

For sublimation, trsH = subH

Note: The text's notation is a little different.

Applications

If one is given the enthalpy of vaporization (or sublimation) +the vapor pressure at one temperature, then one can calculate either

(a) The vapor pressure at a second temperature or(b) The temperature at which the vapor pressure reaches a certain value.

If one knows the vapor pressure at two temperatures, then the enthalpy of vaporization (or sublimation) can be calculated.

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2

1 2 1

1 1ln trs trs trs

trs


Example: The "normal boiling point"** of benzene is 80 oC, and the Enthalpy of Vaporization is 30.8 kJ/mol.

Calculate the temperature at which the vapor pressure of benzene is 150 torr.

1 bar = 750 torrR = 8.31 J/mol-K

**You should know that the "normal" boiling point is the temperature at which the vapor pressure of a substance is 1 bar.

T = 33 oC

Chapter 4 Physical Transformations of Pure Substances 1.

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Transcript of Chapter 4 Physical Transformations of Pure Substances 1.