1b_ Physical Transformations of Pure Substances.ppt

Physical Transformations of Pure Substances

1. Chapter 6: Physical Transformations of Pure
Substances
Homework:
Exercises(a only):4, 5, 8, 12, 15
Problems:4, 6,10, 15, 21

2. Phase Diagrams
Phase - a form of matter that is uniform throughout in chemical
composition and physical state (J. Willard Gibbs)
»Homogeneous - one phase present, e.g., glass of cracked ice (not including
the glass)
»Heterogeneous - more than one phase, e.g., glass of cracked ice with water
Phase transition - spontaneous conversion of one phase to a another
»Occurs at characteristic temperature for a given pressure, transition temperature, T
trs
» At T
trs
, phases are in equilibrium and Gibbs energy minimized
Example ice and water at 0°C
»This says nothing about the rate the transition occurs
Graphite is the stable phase of carbon at R.T., but diamonds exist because rate of
thermodynamically “spontaneous” to low
Such kinetically persistent unstable phases are called metastable phases
Phase Diagram is a P,T plot showing regions of
thermodynamically stable phases
»Lines separating phases are called phase boundaries
Vapor-liquid boundary shows variation of vapor pressure with temperature
Solid-liquid boundary shows variation of sublimation pressure with temperature

3. Phase Boundaries (cont.)
Boiling Points
»Temperature at which vapor pressure equals external pressure
 Normal boiling point (T
b
) external pressure = 1 atm (H
2
O: 100.0°C)
Standard boiling point external pressure = 1 bar (0.987 atm) (H
2O: 99.6°C)
Critical Point
»In closed vessel boiling does not occur. Pressure increases and fluid level drops.

liquid
decreases and 
vapor
increase as T increases
At some point 
liquid
= 
vapor
and boundary disappears: critical temperature
(T
c
),vapor pressure is critical pressure, p
c
Above a single uniform phase exists, supercritical fluid
Highest temperature liquid can exist
Melting Points
»Melting temperature both solid and liquid phases exist.
Equivalent to freezing temperature
Normal freezing(melting) point (T
f
) external pressure = 1 atm (H
2
O: 0°C)
Standard freezing point external pressure = 1 bar (0.987 atm)
Triple Point (T
3
) - place at which 3 phase boundaries meet
»All 3 phases exist simultaneously
»Typically solid, liquid and gas
Lowest pressure and temperature liquid can exist
»Invariant, property of substance (H
2O: 273.15K, 6.11 mbar)
p
c
T
b
1 atm = p
ex
supercritical fluid

4. Phase Diagrams - Water
Slope of solid-liquid boundary means
large p necessary to significantly change
melting temp.
»Negative slope means T
melt decreases as p
increase
Due to structure of water
Liquid has lower volume than solid
(
liquid
>
solid
)
High pressure phases
»Different crystal structure and density
Ice I (hexagonal); Ice III (Tetragonal)
At -175°C and 1 atm: Ice I (= 0.94 g/cc);
Ice VI (= 1.31 g/cc);
»Can melt at higher temperatures than Ice I
Ice VII melts at 100°C but only exists at
p>25 kbar
8 triple points (6 plotted)
»Only one between solid (Ice I), liquid and
gas
Ice VI,VII, VIII
21 kbar, ~5 °C
Ice VI,VII, liquid
22 kbar, 81.6 °C

5. Phase Diagrams - Carbon Dioxide
Melting temperature increases a p
increases
»Unlike water& more typical
T
3 > 1 atm
»Liquid CO
2 doesn’t exist at 1 atm
High pressure CO
2 tanks contain liquid
At 25°C (298.15 K) and 67 atm, gas and
liquid co-exist

6. Phase Diagrams - Helium
Note temperature scale (< 6K)
Solid and gas never in equilibrium
Two liquid phases
»Phase boundary -line (type of phase transition)
»Higher temperature liquid (He-I) is regular
liquid
»Low temperature liquid (He-II) superfluid (zero
viscosity)
It rather than solid exists close to 0K
3
He
4
He
Phase diagram depends on nuclear spin
»
3
H - non-zero spin,
4
H -zero spin
»3
H diagram different than 4
3
H, esp. low
temperature
3
H S
liquid < Ssolid, melting exothermic

7. Gibbs Energies and Phase Diagrams
Chemical potential (µ) - for one component system µ =Gm
»Measure of the potential of a substance to bring about change
»More detail in Chapters 7 & 9
At equilibrium, µ is the same throughout the system
»Regardless of number of phases
»If µ
1 is chem. potential of phase 1 and µ
2 is chem. potential of phase 2, at
equilibrium µ
1 = µ
2
Consequence of 2
nd
Law of Thermodynamics
If dn is transferred from one location (phase) to another, -µ
1dn, is the
change in Gibbs energy in that phase and µ
2dn is change in free energy in
the second phase.
dG = -µ
1dn + µ
2dn = (µ
2 - µ
1)dn
If µ
1 > µ
2 dG < 0 and change spontaneous
If µ
1 = µ
2 dG = 0 and system at equilibrium
Transition temperature is that temperature where µ
1 = µ
2

Chemical potentials of phases change with temperature
»At low T and reasonable p, solid has lowest µ and, hence, most stable
»As T raised, µ of other phase may become less than solid(at that temp.) so it
becomes stable phase

8. T and p Dependence on Melting
Recall, (∂G/∂T)
p
= -S so (∂µ/∂T)
p
= -S
m
»Since S
m
> 0 for all pure substances, (∂µ/∂T)
p
< 0 or a plot of µ
vs. T will have a negative slope
»Because S
m
(gas)> S
m
(liquid)> S
m
(solid)> slopes going from gas
 solid increasingly negative
Transitions (melting, vaporization) occur when µ of one phase
becomes greater than another so substances melt when
µ(liquid)>µ(solid)
Phase change means modifying relative values of µ for each phase
Given differences in slopes, change in T easiest way to do it
Similarly, (∂G/∂p)
T = V so (∂µ/∂T)
p = V
m
»Because V
m
increases with p, graph of µ vs. T translates
upward as p increases
For most substances V
m(l)> V
m(s), µ(l) increases more than µ(s),
T
f increases as p increases (a)
Consistent with observations and physical sense higher p retards
movement to lower density
Water exception since V
m(s)> V
m(l), as p increases µ(s) increases
more than µ(l), so T
f decreases as p increases (b)
V
m
(l)> V
m
(s)
V
m
(s)> V
m
(l)

9. Assessing Effect of Pressure on Melting
Example 6.1
Calculate µ for each state over pressure range, µ (water)  p = 1 bar; µ (ice)  p = 1 bar

µ (water) = 1.80 J mol
-1
and µ (ice) = 1.97 J mol
-1

J mol
-1
are units of µ just like G

µ (water) < µ (ice), so tendency for ice to melt
Look at Self Test 6.2 (opposite of water)

p






T
V
m,whereV
misthemolarvolume
But,V
mM/,wheredensityandMmolarmass

p






T
M/
Overfinitechangeinpresure
 M/dp
p1
p2

M/p

10. Effect of Pressure on Vapor Pressure
When pressure applied to a condensed phase
(solid or liquid), vapor pressure increases, i.e.,
molecules move to gas phase
»Increase in p can be mechanical or with inert gas
Ignore dissolution of pressurizing gas in liquid
Ignore gas solvation, attachment of liquid molecules
to gas-phase species
»Vapor pressure in equilibrium with condensed
phase is the partial vapor pressure of the
substance, p*
p = p*e
VmP/RT
[1]
Math Moment: e
x
= 1+ x + 1/2x
2
+….
If x<<1, .e ≈ 1 + x
Since(g)(l)atequilibrium,
d(g)d(l)whenadditionalexternalpressuredPappplied
d(l)V
m
(l)dPandd(g)V
m
(g)dP
Foridealgas,V
m
(g)
RT
p
so,
V
m(l)dp
RT
p
dp
IntegratingfromP
1
toP
2
PP
2
P
1 ,
V
m
(l)dp
P1
P2


RT
p
dp
P1
P2

ifnoadditonalpressure,pP
1p*,
withexternalpressure,P,ppP
pPp(effectofpressuresmall),sointegralsbecome
V
m(l)dp
p*
p*P


RT
p
dp
p*
p

V
mPRTln
p
p*






exponetiate
p
p*
e
V
mP
RT
orpp*e
V
mP
RT
Proof 
[1] becomes
p = p*(1 + V
m
P/RT)
If V
mP/RT<<1
orpp*
p*

V
mP
RT

11. Location of Phase Boundaries
Phase boundaries occur when chemical potentials are equal and phases are in
equilibrium, or for phases  and :
µ
(p,T) = µ
(p,T)
»Need to solve this equation for p, p= f(T)
»On plot of p vs. T, f(T) is a gives the phase boundary
Slopes of phase boundaries
»Slope is dp/dT
If p and T are changed such that two phases,  and  , are in equilibrium, dµ
= dµ

But, dµ = -S
mdT + Vdp
So, -S
m,
dT + V
m
dp = -S
m,
dT + V
m
dp
Or (S
m,-S
m,dT + = (V
m
- V
m )dp
But (S
m,- S
m, = 
trsS and (V
m - V
m
) = 
trsV
This rearranges to:
dp/dT = 
trs
S / 
trs
V
where and are the entropy and volume of transition
»This is called the Clapeyron Equation
Exact expression for the phase boundary at equilibrium
Can be used to predict appearance of phase diagrams and form of boundaries
µ 
=
µ 

12. Solid-Liquid Boundary
For solid-liquid boundary, Clapeyron Equation becomes
dp/dT = 
fusS / 
fusV
where 
fusV is the change in molar volume on melting

fus
S is always positive (except for
3
He) and 
fus
V is usually small so dp/dT is large (steep
slope) and positive
Formula for phase boundary comes from integrating Clapeyron equation
dp

fusS

fus
V
dT
recall
trsS

trsH
T
,so
dp

fusH
T
fus
V
dT
Integrating
dp
p*
p



fusH

fusV
1
TT*
T

dT,whereT*ismeltingtemperature@p*
pp*

fusH

fusV
ln
T
T*






IfTclosetoT*,ln
T
T*





ln1
TT*
T






pp*

fusH

fus
V
ln1
TT*
T






Math Moment: ln(1+x) = x - 1/2x
2
+….
If x<<1, ln(1 + x) = x
pp*

fus
H

fusV
ln1
TT*
T*






pp*

fus
H

fusV
TT*
T*





or
pp*

fus
H

fusV
TT*
T*






pp*

fus
H
T*
fusV
TT* 
This is straight line of slope
[
fus
H / (T*
fus
V)]

13. Liquid Vapor Boundary
Again, Clapeyron equation can be used
» 
vap
V is large and positive so dp/dT is positive, but smaller than for solid-liquid transition
» 
vap
H/T is Trouton’s constant
»Because V
m(gas) >> V
m(liquid), 
vapV ≈ V
m(gas)
For ideal gas, V
m
(gas) = RT/p so 
vap
V ≈ RT/p
Clapeyron equation becomes
dp
dT


vap
H
T
vap
V
dp
dT


vapH
T
vap
V


vapH
T
RT
p







p
vapH
RT
2
1
p






dp
dT


vapH
RT
2
Recall
dx
x
dlnx
d(lnp)
dT


vapH
RT
2
ClausiusClapeyronEquation
Integrating Clausius-Clapeyron equation gives variation of vapor pressure with temperature
As s umes 
v a p
H is in depend ent of T and
p* is v apor pres s ure at T*
and p the vap or pres s ure a t T
»This is a curve, not a line
»Does not extend beyond T
c
pp*e

, 

vap
H
R
1
T

1
T*







14. Solid-Vapor Boundary
Solid-vapor boundary same as
liquid vapor boundary except
use 
sub
H instead of 
vap
H
 Since 
sub
H > 
vap
H, slope of
curve is steeper
Curves coincide at triple point
along with solid-liquid
boundary

15. Classification of Phase Transitions
Ehrenfest Clasification
We’ve been talking a lot about the slopes of phase transitions, (∂µ/∂T)
p
or (∂µ/∂V)
T

»Transitions are accompanied by changes in entropy and volume
At transition from a phase, , to another phase, 
(∂µ
 /∂T)
p - (∂µ
 /∂T)
p = -S
,m + S
,m = - S
trs = -
trsH/T
(∂µ
 /∂p)
T - (∂µ
 /∂p)
T = V
,m - V
,m = -
trsV
1st Order Transitions (e.g., melting, vaporization)
»Since 
fus
H, 
vap
H and 
fus
V, 
vap
V are non-zero, the changes in µ {(∂µ/∂T)
p
or (∂µ/∂p)
T
} as you
approach the transition are different.
There is a discontinuity at the transition
A transition in which the slope of µ, (∂µ/∂T)
p , is discontinuous is called a 1st order
transition
C
p
is slope of plot of H vs. T (∂H/∂T)
p
at 1st order transition is infinite
Infinitesimal change in T produces finite change in H
T
trs
V,
H.
S
T
trs
µ
T
trs
C
p

16. Classification of Phase Transitions
Ehrenfest Classification (continued)
2nd Order Transition (glass transition, superconducting to conducting transition)
»(∂µ/∂T)
p
, is continuous
Volume and entropy don’t change at transition
»(∂
2
µ/∂T
2
)
p
is discontinuous
»Heat capacity is discontinuous, but not infinite
T
trs
C
p
T
trs
µ
T
trs
V,
S,
H
-Transition (
4
He super-fluid to liquid transition,
order-disorder transition in -brass)
»Not 1st order, (∂µ/∂T)
p
, is continuous
»Heat capacity is discontinuous, and infinite at
transition temperature

17. Liquid Surfaces
Surface effects can be expressed in terms of thermodynamic functions since
work is required to change the surface area of a liquid
»If  is the surface area of a liquid, d is its infinitesimal change when an amount of
work, dw is done
»dw is proportional to the surface area of the liquid, i.e.,
dw = d
the proportionality constant is defined as the surface tension
Dimensions of  : energy/area (J/m
2
or N/m)
To calculate the work needed to create or change a surface by a particular area increment
you only need to calculate the area since is a constant
Look at Self Test 6.4
At constant volume and temperature the work is the Helmholtz energy (A)
dA = d
»As d decreases the Helmholtz energy minimizes
This is the direction of spontaneous change so surfaces of liquids tend to contract
Often minimizing of dA results in curved surfaces

18. Curved Liquid Surfaces
Bubble - a region in which vapor (+ air) is trapped by thin
film
»Bubbles have 2 surfaces one on each side of the film
»Cavity is a vapor-filled hole in liquid (commonly called
bubbles) but 1/2 the surface area
»Droplet - small volume of liquid surrounded by vapor (+air)
Pressure inside a concave surface (p
in
) is always greater
than pressure outside (p
out
)
»Difference depends on surface area and surface tension
p
in = p
out + 2/r Laplace
eqn
forspheretheoutwardforceis
p
inarea4r
2
p
in
theinwardforcecomesfrom
Fexternalpressure Fsurfacetension 
sufacetension
dwd
d4rdr
2
4r
2
d4r
2
2rdrdr
2
 
2
4r
2
d8rdr4dr
2
Ignore4dr
2
(small)sod8rdr
dw8rdr
SincewFdorFw
d
F
in
surfacetension 
dw
dr
8r
externalpressure
F
inpressure 4r
2
p
out
Atequilibrium,F
outF
inor
4r
2
p
in4r
2
p
out8r
rp
in
rp
out
2
p
inp
out
2
r
Laplaceequation
As r  ∞ (flat surface), p
in
= p
out

As r  small(small bubbles), 2/r

important
»Earlier we saw, the vapor pressure in the presence of external pressure p is
p = p*e
VmP/RT
So, for bubble since p = 2/r
p = p*e
2Vm/rRT
Kelvin Equation (bubble)
In a cavity, p
out < p
in, so sign of exponential term is reversed
p = p*e
-2Vm/rRT
Kelvin Equation (cavity)

19. Nucleation, Superheating &Supercooling
Nucleation
»For water droplets
r =1µm; @25°C, p/p* = 1.001
Small effect but mat have important consequences
r =1nm; @25°C, p/p* = 3
»Clouds form when water droplets condense. Warm moist air rises, condenses at colder altitude
Initial droplets small so Kelvin equation tells us vapor pressure of droplet increases
Small droplets tend to evaporate
Unless large numbers of molecules congregate (spontaneous nucleation)
Air becomes supersaturated and thermodynamically unstable
Nucleation centers (dust particles, sea salt) allow clouds to form by allowing condensation
to occur on larger surfaces
Superheating - liquids persist above boiling point
»Vapor pressure inside a small cavity in liquid is low so cavities tend to collapse
Spontaneous nucleation causes larger more stable cavities (and bumping!)
Nucleation centers allow for stabilization of cavity
Basis of bubble chamber
»Supercooling, persistence of liquids past freezing point, is analogous

20. Capillaries
Capillary action - tendency of liquids to rise up/fall in narrow
bore (capillary) tubes
Capillary rise/fall
»If liquid has tendency to adhere to the tube walls (e.g. water),
energy lowest when most surface is covered
Liquid creeps up wall (concave meniscus)
Pressure beneath curve of meniscus is lower than atmosphere
by 2/r, where r is radius (Kelvin equation, cylindrical tube)
Pressure at flat surface = atmospheric pressure (r  ∞)
»Liquid rises in capillary until hydrostatic equilibrium is reached
As liquid rises p increases by gh ( = density; h= height)
At equilibrium, p
capillary
= p
external
or

2/r = gh
Height of capillary rise h = 2/gr
As tube gets smaller, h gets higher
Can be used to measure surface tension of liquids
is temperature dependent
»If liquid has a tendency not to adhere (e.g. Hg) liquid will fall in
capillary because pressure less under meniscus
Treatment the same, except sign reversed

21. Contact Angle
Contact angle is the angle between edge of meniscus and wall
»If 
c ≠ 0, then the equation for capillary rise becomes
h = 2Cosc/gr
Arises from balance of forces at the point of contact between liquid and solid
»The surface tension is essentially the energy needed to create a unit area of each of the interfaces

sg
= Energy to create unit area at gas-solid interface

lg = Energy to create unit area at gas-liquid interface

sl = Energy to create unit area at solid-liquid interface
»At equilibrium, the vertical forces in capillary are in balance so

sg
= 
sl
+ Cos(
c
) 
lg
Or Cos(
c
) =( 
sg
-
sl
)/ 
lg
See diagram
»
Work of adhesion (w
ad
) of liquid to solid is 
sg
+ 
lg - 
sl

»
So Cos(
c
) =( 
sg
-
sl
)/ 
lg
=(w
ad
-
lg
)/ 
lg =(w
ad
/ 
lg
) - 1
If liquid wets surface 0° < 
c < 90° , 1> Cos(
c)  0 so 1< (wad / lg ) 2
If liquid doesn’t wet surface 90° < 
c
< 180° , 0> Cos(
c
)  -1 so 1> (wad / lg ) 0
Takes more work to overcome cohesive forces in liquid
“wets”
“doesn’t wet”