Upload
milo-mynatt
View
238
Download
8
Tags:
Embed Size (px)
Citation preview
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)
Increased pressure shiftsequilibrium in directionof lower volume.
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
Melting + Vaporization
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
Melting + Vaporization
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)
Increased pressure shiftsequilibrium in directionof lower volume.
Solid Liquid
Pressure dependence of the Melting Pointin Water
12
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
14
Helium
Even Helium has a fairly sophisticated phasediagram. However, all of the "action" occursat very low temperature; Tc 5 K
15
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
16
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.
17
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
18
The Effect of Pressure on the Melting Point
m m mp T
d dG S dT V dp dT dpT p
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
19
The Effect of Pressure on the Melting Point
m m mp T
d dG S dT V dp dT dpT p
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
20
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
21
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
22
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
23
sol
liq
vapE
ntr
op
y (S
)
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(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
24
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
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
H2O
H2O
dliq > dsol
Vliq < Vsol
Vliq-Vsol < 0
25
Vo
lum
e (V
)liqsol
vap
26
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
27
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.
28
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
29
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
30
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
31
The Clausius-Clapeyron Equation
2
1 2 1
1 1ln trs trs trs
trs
H H Hpp T V R T TRTT p
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.
32
The Clausius-Clapeyron Equation
2
1 2 1
1 1ln trs trs trs
trs
H H Hpp T V R T TRTT p
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