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Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Chapter 5: Free Energy and ChemicalThermodynamics
Part-2: Phase Transformation
X. Bai
SDSMT, Physics
Fall Semester, 2014
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
1 Phase Transformation of Pure SubstancesShort reviewThe conceptThe Coexistence CurvesThe Clausius-Claypeyron Relation
2 The van der Waals Modelvan der Waals Model and interaction between particlesvan der Waals isotherms and critical point
3 Maxwell constructionGibbs free energy for vdW fluidMaxwell construction
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Free energies
The enthalpy of a system: H = U + PV . The total energy neededto create the system out of nothing and put it in an environmentat pressure P.
The Helmholtz Free Energy: F = U − TS . The total energyneeded to create the system, minus the heat you can get for freefrom an environment at temperature T . It is the energy must beprovided as all work, if you are creating the system out of nothing.All work: including the work done by the the surroundingenvironment.
The Gibbs Free Energy: G = U − TS + PV .The work (energy)you need to do to create a system in an environment withconstant pressure P and temperature P.
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The concept
Up to now we have dealt almost exclusively with systemsconsisting of a single phase, solid, liquid, or gas. In this lecture, wewill learn how more complicated, multi-phase systems can betreated by the methods of thermodynamics. The guiding principleis the minimization of the Gibbs free energy in equilibrium for allsystems, including the multi-phase ones.
G = U − TS + PV and dStotal = − 1T dG
To explain what phase transformation is, in terms ofthermodynamic language, let me show you the generic phasediagram of a substance in the P-T coordinates
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The concept: P-T plot
The P-T plot of phase transformation.
every point of this diagramis an equilibrium state;
different states of the systemin equilibrium are called phases.
lines dividing different phasesare called the coexistence curves.
along these curves, different phasescoexist in equilibrium,
along these curves, the systemis macroscopically inhomogeneous.
all three coexistence curvescan meet at the triple point -all three phases coexist.
critical point: Where no discontinuous change from phase-A to phase-B.
T
P
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The concept: Water
For most normal substances, the slope of the melting curve is positive. Thephase diagram for water shows the characteristic negative slope of thesolid-liquid equilibrium curve.
The ice is less dense than water
As ice melts into water, thechange in entropy (the latentheat) is positive, while thechange in volume is negative,hence the negative slope.
The negative slope of thesolid-liquid coexistence curvemakes ice skating possible:ice melts under the pressureexerted by the skate blade.
Ice I
The Clausius-Clapeyron equation ( dPdT
) provides the connection between ice
skating and the observation that ice floats on water.
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The concept: Helium
The P − T plot of Helium phase transformation: Helium has two isotopes 4He(Boson) and 3He (Fermion). Helium is the only element that remains a liquidat T = 0 K and P = 1 bar because the zero-point oscillations of light atomsare large.
Structure: hexagonal closepacked (hcp), face centeredcubic (fcc), body-centeredcubic (bcc) for solid.
The molar volume of 4He (3He)is more than a factor of 2 (3)larger than classical liquid.
The slope of the phase boundarysolid helium superfluid liquidhelium is essentially 0 at T ∼ 0 K .
Superfluid: zero viscosity, very highconductivity.
4He
3He
(Helium-‐I)
(Helium-‐II)
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
More about phase transformation
Phase transformation caused by other condition changes: Besides the verydifferent details in phase transformations among different substances, phasetransformation caused by other variables:(1) composition: Ice melts when salt is added.(2) external magnetic filed: Superconductor (super conducting and normalmatter), ferromagnet (Magnetized up and down)Classification of phase transformations: according to the abruptness of thechange:
”1st-order” transition: Solid-liquidand liquid-gas transitions. (S and Vare discontinuous at phase boundary,and S and V are the 1st order derivativesof G: dG = −SdT + VdP + µdNEq.(5.23).
”2nd-order” transition: Less abrupt,helium I - helium II transition.
higher orders: derivatives beforeget a discontinuous quantity.
T
S T
G
T
Cp ΔS=Q/T
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The Coexistence Curves
Along the coexistence curves, two different phases 1 and 2 coexist inequilibrium. The system undergoes phase separation each time we cross theequilibrium curve (the system is spatially inhomogeneous along the equilibriumcurves).Any system in contact with the thermal bath is governed by the minimum freeenergy principle. We summarized in our last class:
Table: Quantities that govern the thermal processes
Process What governs the process
constant E and V Entropy Sconstant T and V Helmholtz Free Energy F = U − TSconstant T and P Gibbs Free Energy G = U − TS + PV
The shape of coexistence curves on the P-T phase diagram is determined bythe condition: G1(P,T ) = G2(P,T ), ∵ G = µN, so we have
µ1(P,T ) = µ2(P,T ) (1)
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The Coexistence Curves
Several remarks: dG = −SdT + VdP + µdN
1 The system would be able to decrease its Gibbs free energy bytransforming the phase with a higher µ into the phase with lower µ:crossing the Coexistence Curves.
2 Two phases are in a state of diffusive equilibrium: there are as manymolecules migrating from 1 to 2 as the molecules migrating from 2 to 1=⇒ The equilibrium is dynamic.
3 For equilibrium between the phases: T1 = T2 (for any two sub-systems inequilibrium) and P1 = P2 (the phase boundary does not move.)
4 Though G is continuous changing across the transition, enthalpy H has astep-like behavior:G = Nµ = (U + PV )− TS = H − TS → 0 = δH − TδS(1) different phases have different values of the enthalpy ← heatexchange always EXISTS during phase transformation!(2) µ =
(∂G∂N
)T ,P
.
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Two Phases of Carbon
Carbon has two phases: Diamond and Graphite. See the Gibbs potential in thefigure:
both are solid
graphite is more stable atnormal P: Molar Gibbs FreeEnergy: different by 2.9kJ
pressure dependence of G :(∂G∂P
)T ,N
= V
ρG < ρD , Gg grows fasterthan Gd and dS = − 1
TdG
crossing at about 17 kilo-bars!(1 bar = 105 Pa)
So, natural diamonds must befrom deep underground.
And do not forget the temperature: The right bottom plot.
P (MPa)
G
1 2
diamond
2.9 kJ
G
T (K) 800 1300
diamond 2.9 kJ
300
T = 300K
P = 1 bar
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Clausius-Claypeyron Relation
We have seen how Gibbs Free Energy is correlated with P , T , and V :G = U − TS + PV ; and
(∂G∂P
)T ,N
= V ,(∂G∂T
)P,N
= −S . What determines the
phase boundary? - The Clausius-Clapeyron Relation.
(1799 – 1864) French engineer and physicist, one of the founders of thermodynamics.
(1822 – 1888) German physicist & mathemaBcian, one of the central founders of the science of thermodynamics.
ΔP1 ΔP2
ΔT1
ΔT2 On the red line: ΔP1 = ΔP2 ΔT1 = ΔT2
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Clausius-Claypeyron Relation - cnt.
(∂G∂P
)T ,N
= V ,(∂G∂T
)P,N
= −S . At the phase boundary G1 = G2. For a small
change in P and T , we would have dG1 = dG2 on the phase boundary line.Therefore,
∵ dG = −SdT + VdP + µdN (2)
− S1dT1 + V1dP1 + µ1dN1 = −S2dT2 + V2dP2 + µ2dN2 (3)
− S1dT1 + V1dP1 = −S2dT2 + V2dP2 (4)
Because along the phase boundary line: dT1 = dT2, dP1 = dP2
S2dT − S1dT = V2dP − V1dP (5)
∴dP
dT=δS
δV(6)
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Clausius-Claypeyron Relation - cnt.
If we use latent heat L = Qphase−transition = TδS , Eq. (6) takes the form:
dP
dT=
L
TδV(7)
This is the Clausius-Clapeyron relation, which gives the slope of phaseboundary line on P − T diagram.
Let’s see an examples to help us understand better.
In-class Exercise Ch5-04: Consider the diamond-graphite system. When a mole ofdiamond converted to graphite, its volume increases by 1.9× 10−6 m3. What is theslope of the diamond-graphite phase boundary?Answer: Look at dP
dT= δS
δV, we need δS and δV .
δV is given. We need the change in entropy when a mole of diamond is convertedto graphite.The entropy increases is (p.404): Sg − Sd = 5.74− 2.38 = 3.36 J/KdPdT
= δSδV
= 3.4 J/K
1.9×10−6 m3
dPdT
= 1.8× 106 J/K
m3 = 1.8× 106 N·mK ·m3
dPdT
= 1.8× 106Pa/K = 18 bar/K
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Model for phase transition
To understand phase transformations, can we use Ideal Gas Model?PV = kNT (k is Boltzmann’s constant 1.381× 10−23 J/K)
1 point-like particles
2 zero interaction between particles except elastic collisions with zerocross-section
But in phase transitions, what govern the process are the potentials. So, theanswer is NO.How about the other model we have introduced - the van der Waals Model?(P + aN2
V 2
)(V − Nb) = kNT .
Possible because it takes into account the interactions between particles. Thetwo corrections:
1 point-like particles ← each particle has size by (V − Nb)
2 zero interaction between particles ← short range attraction force whenthey are NOT touching
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
van der Waals Model-1
U(r)
short-distance repulsion
long-distance attraction
-3
-2
-1
0
1
2
3
4
1.5 2.0 2.5 3.0 3.5 4.0distance
Energy
r
Johannes van der Waals, 1873
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
van der Waals Model-2
Why the corrections are V − Nb and P + aN2
V 2 ? Because of the interactions:
1 The strong short-range repulsion: the molecules are ”rigid” as soon asthe molecules touch each other. ”b”: the excluded volume per particle
2 The weak long-range attraction forces between the molecules tend tokeep them closer together: Total potential energy due to attraction:
−αN NV
= −αN2
Vwhere N and V are the total number of particles and
the volume of the solid/liquid/gas.
3 pressure associated with the potential energy: dE = −Pf dV ,
Pf = − dEdV
= −αN2
V 2
The model can help explain phase transformation behavior, but far from preciseenough!
1 inhomogeneous on the microscopic scale when gas becomes denser, withdifferent phases coexist.
2 corrections to the interaction: non-isotropic, may depend on other stateparameters.
Good for QUALITATIVE discussions.
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
van der Waals Model parameters
The constant a and b in the van der Waals equation for typicalsubstances:
Substance
a’ (J. m3/mol2)
b’ (10-5 m3/mol)
Pc (MPa)
Tc (K)
Air
.1358
3.64
3.77
133 K
Carbon Dioxide (CO2)
.3643
4.27
7.39
304.2 K
Nitrogen (N2)
.1361
3.85
3.39
126.2 K
Hydrogen (H2)
.0247
2.65
1.30
33.2 K
Water (H2O)
.5507
3.04
22.09
647.3 K
Ammonia (NH3)
.4233
3.73
11.28
406 K
Helium (He)
.00341
2.34
0.23
5.2 K
Freon (CCl2F2)
1.078
9.98
4.12
385 K
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
van der Waals isotherms
1 The isotherms on P − V plane: Lines with fixed temperatures.
2 The black isotherm is at T = Tc , corresponding to critical behavior.
3 Real isotherms can NOT have negative slope dP/dn or positive slopedP/dV (shaded area!).
4 Critical isotherm: boundary between isotherms along which there are/notphase transformations.
5 Critical point C : The point on isotherm where ∂P/∂V = ∂2P/∂V 2 = 0
0
N·b
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The critical point
Critical point: The point on critical isotherm where ∂P/∂V = ∂2P/∂V 2 = 0.So, what is the critical point for van der Waals gas?
(P +
aN2
V 2
)(V − Nb) = kNT
P =kNT
V − Nb− aN2
V 2
∂P
∂V= − kNT
(V − Nb)2+
2aN2
V 3
∂2P
∂V 2=
2kNT
(V − Nb)3− 6aN2
V 4
At critical point, we have
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
The critical point-cnt.
kNTc
(Vc − Nb)2=
2aN2
V 3c
2kNTc
(Vc − Nb)3=
6aN2
V 4c
Combine to give:(Vc − Nb)
2=
Vc
3⇒ Vc = 3Nb.
and, kTc =8a
27b, Pc =
a
27b2.
Using Vc ,Tc ,Pc to normalize P,V ,T , T = Tc T ,P = Pc P,V = Vc V the vander Waals equation becomes:(
P +3
V 3
)(V − 1
3
)=
8T
3(8)
p =8t
3v − 1− 3
v 2often called the vdW equation in reduced variables (9)
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Gibbs free energy for vdW fluid
When T and P are given, the equilibrium state of a system is determined by itsGibbs free energy G . Let’s see if we can obtain G for vdW fluid.dG = −SdT + VdP + µdN which becomes dG = VdP at fixed T and N.Therefore, we have(
∂G
∂V
)N,T
= V
(∂P
∂V
)N,T
(10)
P and V are correlated by vdW equation:(P +
aN2
V 2
)(V − Nb) = kNT (11)
P =kNT
V − Nb− aN2
V 2(12)(
∂P
∂V
)N,T
= − kNT
(V − Nb)2+
2aN2
V 3(13)(
∂G
∂V
)N,T
= − kNTV
(V − Nb)2+
2aN2
V 2(14)
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Gibbs free energy for vdW fluid - cnt.
G =
∫ (− kNTV
(V − Nb)2+
2aN2
V 2
)dV (15)
G =
∫ (−kNT (V − Nb) + kNTNb
(V − Nb)2+
2aN2
V 2
)dV (16)
G =
∫ (− kNT
V − Nb− kTN2b
(V − Nb)2+
2aN2
V 2
)dV (17)
G = −kNTln(V − Nb) +kTN2b
(V − Nb)− 2aN2
V+ c(T ) (18)
G in reduced variables (off-class exercise):
G
kNTc= −tln(3v − 1) +
t
(3v − 1)− 9
4v+ c(T ) (19)
where c(T ) is the integration constant, independent from V . Here we see:(1) G is a function of T , V , N, and other parameters.
(2) Plotting G as function of V or P: with V and P associated through the
equation of state.
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Gibbs free energy for vdW fluid - cnt.
This is an example for vdW fluid when T = 0.9Tc .
1 States between 2 and 6 (2,3,4,5,6) are unstable: one P has multiple Vvalues.
2 The thermodynamically stable state has minimum G .
3 Point-6 is fluid; point-2 should be gas (by the dP/dV at these points)
4 Point-2 and 6 are equally stable because they have same G .
5 The straight red line connecting point-2 and 6 represents the phasetransformation: V changes while P is fixed.
P/Pc G
V/Vc P/Pc
1 2
3
4 5
6
7
1
2,6
3
4
5 7
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
About plotting:
How to draw the plots of P/Pc vs V /Vc or G vs P/Pc ?Eq. (9): p = 8t
3v−1− 3
v2
Eq. (19): GkNTc
= −tln(3v − 1) + t(3v−1)
− 94v
+ c(T )
cV/V0 0.5 1 1.5 2 2.5
cP
/P
0.6
0.7
0.8
0.9
1
1.1
1.2
cP/P0.7 0.75 0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2
G/k
NT
-2.5
-2.48
-2.46
-2.44
-2.42
-2.4
-2.38
-2.36
c++ code available at the teaching website: plot 5 21b.CC
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Maxwell construction
The question is how to determine the pressure for phase transition.Two ways:
1 From G -plot: P at points 2,4,6.
2 From the PV diagram - invented by James Clerk Maxwell.
Here is why and how it works:The net change in G as the system changes around the loop 2-3-4-5-6 is zero∫
loop
dG =
∫loop
(∂G
∂P
)N,T
dP (20)
dG = −SdT + VdP + µdN (21)∫loop
dG =
∫loop
VdP (22)∫loop
dG = SA + SB = 0 (23)
The straight line that divides the enclosed area into two parts each with half of
the total area - the Maxwell construction.
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Maxwell construction plots
P/Pc
V/Vc
1 2
3
4 5
6
7
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
Outline Phase Transformation of Pure Substances The van der Waals Model Maxwell construction
Vapor pressure and critical point
(1) vapor pressure Pv for each T: at which the liquid-gas transformation takesplace(2) critical pressure Pc , critical temperature Tc , and critical volume Vc : thecritical point.
Pc
Tc Vc
Pv
X. Bai Chapter 5: Free Energy and Chemical Thermodynamics
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