10/17/2008
Lecture 18
Phase Diagrams
Definition of a phase diagram
A phase diagram is a representation of the states of matter,
solid, liquid, or gas as a function of temperature and pressure.
In the Figure shown below the regions of space indicate
the three phases of carbon dioxide. The curved lines
indicate the coexistence curves. Note there is a unique
triple point.
Pressure (atm)
Chemistry 433
NC State University
Degrees of freedom
Within any one of the single-phase regions both temperature
and pressure must be specified. Because two thermodynamic
variables can be changed independently we say that the
system has two degrees of freedom. Along any of the
coexistence curves the pressure and temperature are coupled,
i.e. any change in the temperature implies a change in
pressure to remain on the line
line. Thus
Thus, along the curves there
is only one degree of freedom. The triple point is a unique
point in phase space and there is only one set of values of
pressure and temperature consistent with the triple point.
Thus, we say that at the triple point the system has zero
degrees of freedom. If we follow the liquid-vapor coexistence
curve towards higher temperature we find that it ends at the
critical point. Above the critical point there is no distinction
between liquid and vapor and there is a single fluid phase.
Degrees of freedom
How many degrees of freedom are there in the liquid phase?
A. 0
B. 1
C 2
C.
D. 3
Degrees of freedom
How many degrees of freedom are there in the liquid phase?
A. 0
B. 1
C 2
C.
D. 3
Free energy dependence
along the coexistence curve
In a system where two phases (e.g. liquid and gas) are in
equilibrium the Gibbs energy is G = Gl + Gg, where
Gl and Gg are the Gibbs energies of the liquid phase and the
gas phase, respectively. If dn modes (a differential amount
of n the number of moles) are transferred from one phase to
another at constant temperature and pressure, the differential
Gibbs energy for the process is:
g
dG = ∂Gg
∂n
P,T
l
dn g + ∂Gl
∂n
dn l
P,T
The rate of change of free energy with number of moles is
called the chemical potential.
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The significance of chemical
potential of coexisting phases
We can write the Gibbs free energy change using the following
notation:
dG = μ gdn g + μ ldn l
Note that if the system is entirely composed of gas molecules
the chemical potential μg will be large and μl will be zero.
Under these conditions the number of moles of gas will
decrease dng < 0 and the number of moles of liquid will
increase dnl > 0. Since every mole of gas molecules converted
results in a mole of liquid molecules we have that:
dng = -dnl
Solid-liquid coexistence curve
Coexistence criterion
In terms of chemical potential, the Gibbs energy for the phase
equilibrium is:
dG = μ g – μ l dn g
Since the two phases are in equilibrium dG = 0 and since
plain language,
g g , if two p
phases of a
dng ≠ 0 we have μg = μl. In p
single substance are in equilibrium their chemical potentials
are equal.
If the two phases are not in equilibrium a spontaneous
transfer of matter from one phase to the other will occur in
the direction that minimizes dG. Matter is transferred from a
phase with higher chemical potential to a phase with lower
chemical potential consistent with the negative sign of Gibb's
free energy for a spontaneous process.
Degrees of freedom
What are the degrees of freedom in the liquid phase?
To derive expressions for the coexistence curves on the
phase diagram we use the fact that the chemical potential
is equivalent in the two phases. We consider two phases
α and β and write
μα(T,P) = μβ(T,P)
Now we take the total derivative of both sides
∂μ α
∂μ α
∂μ β
∂μ β
dP +
dT =
dP +
dT
∂P T
∂T P
∂P T
∂T P
A. Temperature and Volume
B. Temperature and Pressure
C Pressure and Volume
C.
D. All of the above
The appearance of this equation is quite different from
previous equations and yet you have seen this equation
before. The reason for the apparent difference is the
symbol μ. Remember that μ for a single substance is
just the molar free energy.
Degrees of freedom
What are the degrees of freedom in the liquid phase?
A. Temperature and Volume
The Clapeyron equation
Substituting these factors into the total derivative above
we have
VmαdP – SmαdT = VmβdP – Smβ dT
Solving for dP/dT gives
B. Temperature and Pressure
C Pressure and Volume
C.
D. All of the above
β
α
dP = Sm – Sm = Δ trsSm = Δ trsHm
dT Vmβ – Vmα Δ trsVm TΔ trsVm
This equation
Thi
ti iis kknown as th
the Cl
Clapeyron equation.
ti
It gives
i
the two-phase boundary curve in a phase diagram with
ΔtrsH and ΔtrsV between them. The Clapeyron equation can
be used to determine the solid-liquid curve by integration.
P2
dP =
P1
Δ trsHm
Δ trsVm
T2
T1
dT
T
Starting with a known point along the curve (e.g. the triple
point or the melting temperature at one bar) we can calculate
the rest of the curve referenced to this point.
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The Clapeyron equation
The integrated form of the Clapeyron equation is:
Δ trsHm T2
ln
Δ trsVm T1
P
Δ H
T
B. ln 2 = trs m ln 2
P1
Δ trsVm T1
Δ H
C. P2 – P1 = trs m 1 – 1
Δ trsVm T1 T2
A. P2 – P1 =
P2
Δ H
= trs m 1 – 1
P1
Δ trsVm T1 T2
D. ln
The liquid-vapor and solidvapor coexistence curves
The Clapeyron equation cannot be applied to a phase
transition to the gas phase since the molar volume of a gas
is a function of the pressure. Making the assumption that
Vmg >> Vml we can use the ideal gas law to obtain a new
expression for dP/dT.
dP = Δ trsHm = PΔ trsHm
dT
TVmg
RT 2
The integrated form of this equation
P2
P1
dP =
P
T2
T1
Δ trsHm
dT
RT 2
yields the Clausius-Clapeyron equation.
ln
P2
Δ H
Δ H T – T1
= trs m 1 – 1 = trs m 2
P1
R
R
T1T2
T1 T2
The Clapeyron equation
The integrated form of the Clapeyron equation is:
Δ trsHm T2
ln
Δ trsVm T1
P
Δ H
T
B. ln 2 = trs m ln 2
P1
Δ trsVm T1
Δ H
C. P2 – P1 = trs m 1 – 1
Δ trsVm T1 T2
A. P2 – P1 =
D. ln
P2
Δ H
= trs m 1 – 1
P1
Δ trsVm T1 T2
Applying the ClausiusClapeyron equation
If we use ΔH of evaporation the C-C equation can be used
to describe the liquid-vapor coexistence curve and if we
use ΔH of sublimation this equation can be used to describe
the solid-vapor curve.
The pressure derived from the C-C equation is the vapor
pressure at the given temperature. Applications also include
determining the pressure in a high temperature vessel
containing a liquid (e.g. a pressure cooker). If you are given
an initial set of parameters such as the normal boiling point,
for example you may use these as T1 and P1. Then if you
are given a new temperature T2 you can use the C-C to
calculate P2.
Conceptual Question
Conceptual Question
The key assumptions for the Clausius-Clapeyron equation
that defines the liquid-vapor and solid-vapor coexistence
curves are:
The key assumptions for the Clausius-Clapeyron equation
that defines the liquid-vapor and solid-vapor coexistence
curves are:
A. Vmg >> Vml ,Vms >> Vmg
A. Vmg >> Vml ,Vms >> Vmg
B. Vml >> Vmg ,Vmg >> Vms
B. Vml >> Vmg ,Vmg >> Vms
C. Vms >> Vmg ,Vml >> Vmg
C. Vms >> Vmg ,Vml >> Vmg
D. Vmg >> Vml ,Vmg >> Vms
D. Vmg >> Vml ,Vmg >> Vms
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Conceptual Question
Which expression can be used to determine the coexistence
boundary of two phases:
Δ S
A. dP = trs m
dT Δ trsVm
Δ H
B. dP = trs m
dT Δ trsVm
Δ S
C. dG = trs m
dT Δ trsVm
Δ H
D. dG = trs m
dT Δ trsVm
Which expression can be used to determine the coexistence
boundary of two phases:
Δ S
A. dP = trs m
dT Δ trsVm
Δ H
B. dP = trs m
dT Δ trsVm
Δ S
C. dG = trs m
dT Δ trsVm
Δ H
D. dG = trs m
dT Δ trsVm
Constructing the phase diagram for CO2
We can use the Clapeyron and Clausius-Clapeyron equations
to calculate a phase diagram.
For example, we can begin with the CO2 diagram shown above.
The triple point for CO2 is 5.11 atm and 216.15 K.
The critical point for for CO2 is 72.85 atm and 304.2 K.
We also have the following data
Transition
Fusion
Sublimation
Conceptual Question
ΔtrsHo (kJ/mol)
8 33
8.33
25.23
Ttrs (K)
217 0
217.0
194.6
Note that we can calculate the enthalpy of sublimation from
ΔvapHo = ΔsubHo - ΔfusHo = 16.9 kJ/mol.
ρsolid = 1.53 g/cm3 and ρliquid = 0.78 g/cm3, respectively.
The density ρ = m/V = nM/V so the molar volume is
Vm = V/n = M/ρ where M is the molar mass.
In units of L/mole we have
Vsm = 44 g/mole/[1530 g/L] = 0.0287
Vlm = 44 g/mole/[780 g/L] = 0.0564
ΔfusV = Vlm - Vsm = 0.0564 - 0.0287 = 0.0277 L/mole
Constructing the liquid-vapor curve
Constructing the phase
diagram for CO2
Starting with the triple point we use the Clausius-Clapeyron
equation to calculate the liquid-vapor coexistence curve.
P = 5.11exp{ΔvapH/R[T – 216.15]/216.15T}
P = 5.11exp{2,032[T – 216.15]/216.15T}
Notice that if we were to calculate the critical pressure
using this formula we would obtain 77.3 atm which is about
5 atm larger than the experimental number. There are several
sources of inaccuracy including mainly our neglect of the
temperature dependence of the enthalpy.
We can also begin a the critical point
P = 72.8 exp{ΔvapH/R[T – 304.2]/304.2T}
P = 72.8 exp{2,032[T – 304.2]/304.2T}
Constructing the liquid-vapor curve
T (K)
216.15
220
230
Pressure ((atm)
P = 5.11exp{2032[T – 216.15]/216.15T}
Liquid-vapor
P (atm)
5.11
6.03
90
9.0
Pressure ((atm)
P = 5.11exp{2032[T – 216.15]/216.15T}
Liquid-vapor
P (atm)
T (K)
5.11
216.15
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P = 5.11exp{2032[T – 216.15]/216.15T}
P = 5.11exp{2032[T – 216.15]/216.15T}
Liquid-vapor
P (atm)
5.11
6.03
90
9.0
13.0
Liquid-vapor
P (atm)
5.11
6.03
90
9.0
13.0
24.9
T (K)
216.15
220
230
240
Constructing the liquid-vapor curve
T (K)
216.15
220
230
240
260
Pressure ((atm)
Constructing the liquid-vapor curve
Pressure ((atm)
Constructing the liquid-vapor curve
Constructing the liquid-vapor curve
T (K)
216.15
220
230
240
260
280
Constructing the solid-vapor curve
T (K)
216.15
220
230
240
260
280
300
304
Pressure ((atm)
P = 5.11exp{2032[T – 216.15]/216.15T}
Liquid-vapor
P (atm)
5.11
6.03
90
9.0
13.0
24.9
41.0
63.7
72.8
Pressure ((atm)
P = 5.11exp{2032[T – 216.15]/216.15T}
Liquid-vapor
P (atm)
5.11
6.03
90
9.0
13.0
24.9
41.0
Constructing the solid-vapor curve
Solid-vapor
P (atm)
T (K)
5.11
216.15
Solid-vapor
P (atm)
5.11
3.38
1.64
T (K)
216.15
210
200
Pressure (atm
m)
Starting again at the triple point
P = 5.11exp{ΔvapH/R[T – 216.15]/216.15T}
P = 5.11exp{3034[T – 216.15]/216.15T}
Pressure (atm
m)
Starting again at the triple point
P = 5.11exp{ΔsubH/R[T – 216.15]/216.15T}
P = 5.11exp{3034[T – 216.15]/216.15T}
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Constructing the solid-vapor curve
Constructing the solid-vapor curve
Solid-vapor
P (atm)
5.11
3.38
1.64
0.725
0.298
Solid-vapor
P (atm)
5.11
3.38
1.64
0.725
0.298
0.111
T (K)
216.15
210
200
190
180
Constructing the solid-liquid curve
T (K)
216.15
210
200
190
180
170
Pressure (atm
m)
Starting again at the triple point
P = 5.11exp{ΔvapH/R[T – 216.15]/216.15T}
P = 5.11exp{3034[T – 216.15]/216.15T}
Pressure (atm
m)
Starting again at the triple point
P = 5.11exp{ΔvapH/R[T – 216.15]/216.15T}
P = 5.11exp{3034[T – 216.15]/216.15T}
Constructing the solid-liquid curve
Solid-liquid
P (atm)
5 11
5.11
Solid-liquid
P (atm)
5 11
5.11
58.0
T (K)
216 15
216.15
Constructing the solid-liquid curve
T (K)
216.15
216
15
220
Pressure (atm
m)
Using the Clapeyron equation we calculate:
P = 5.11 + [ΔfusH/ΔfusV] ln{T/216.15}
P = 5.11 + 2,967 ln{T/216.15}
Pressure (atm
m)
Using the Clapeyron equation we calculate:
P = 5.11 + [ΔfusH/ΔfusV] ln{T/216.15}
P = 5.11 + 2,967 ln{T/216.15}
Constructing the solid-liquid curve
Using the Clapeyron equation we calculate:
P = 5.11 + [ΔfusH/ΔfusV] ln{T/216.15}
P = 5.11 + 2,967 ln{T/216.15}
Solid-liquid
P (atm)
5 11
5.11
58.0
124.2
Solid-liquid
P (atm)
5 11
5.11
58.0
124.2
319
559
780
987
1180
T (K)
216 15
216.15
220
230
Pressure (atm
m)
Using the Clapeyron equation we calculate:
P = 5.11 + [ΔfusH/ΔfusV] ln{T/216.15}
P = 5.11 + 2,967 ln{T/216.15}
T (K)
216.15
216
15
220
230
240
260
280
300
320
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