Download Outline Introduction Introduction Gibbs Free Energy

Introduction
Combined statement of 1st and 2nd Laws of Thermodynamics:
dU = TdS – PdV. This equation provides:
 Relationship between U versus V and S:
U = U(S,V) or dU = (∂U/∂S)VdS + (∂U/∂V)SdV
 The criteria for equilibrium: in a system of constant V and S; the
internal energy has its ……….. value, or, in a system of constant U
and V, the entropy has its ………… value.
Outline
• Introduction
• Enthalpy and entropy dependence on P and T
• Gibbs free energy dependence on P and T
• Clapeyron equation
• Gibbs-Duhem equation
The problem is that the pair of independent variables (S,V) is rather
inconvenient since entropy is hard to measure or control. Therefore
it is desired to have fundamental equations with independent
variables that are easier to control. The two convenient choices are:
• Unary phase diagrams
• Gibbs phase rule
• Driving force for a phase transition
• P and T The practical usefulness of thermo of materials is determined by the
• First order and second-order phase transitions
• V and T
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/1
MECH6661 lecture 5/2
Gibbs Free Energy
 From practical point of view, P and T pair is the best choice because
they are easy to control and measure.
- For systems with constant pressure the most suitable state function
is the Gibbs free energy G = H - TS
 On the other hand, V and T pair is easy to examine in statistical
mechanics.
- For systems with constant volume (and variable pressure), the most
suitable state function is the ………….. free energy A = U – TS
In principle, any state function can be used to describe any
system at equilibrium, but for a given system some are more
convenient than others. The most convenient one for materials
(usually under constant pressure) is the ……… free energy.
Mech. Eng. Dept. - Concordia University
Mech. Eng. Dept. - Concordia University
Dr. M. Medraj
Introduction
Dr. M. Medraj
practicality of the equations of state for the system (or the relationships which can be established among the thermo properties).
MECH6661 lecture 5/3
G = H – TS = U + PV – TS
⇒ dG = dU+PdV+VdP–TdS–SdT
From 1st law:
dU = TdS – PdV
Combine both:
⇒ dG = VdP – SdT
G = G(P,T) ⇒ dG = (∂G/∂P)TdP + (∂G/∂T)PdT
Comparing the two equations we see that:
V = (∂G/∂P)T and S = – (∂G/∂T)P
For isothermal-isobaric system the equilibrium state corresponds
to the minimum of the Gibbs Free Energy (dG=0).
- From G = H – TS, we see that low values of G are obtained with
low values of H and high values of S.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/4
The Helmholtz Free Energy
The Helmholtz Free Energy
A = U – TS
⇒ dA = dU–TdS–SdT
From 1st law:
dU = TdS – PdV
• This figure illustrates the criterion for
equilibrium in a closed solid-vapor system at
constant V and T.
• The transfer of one atom to vapor increases
U by a sublimation energy, whereas increase
of entropy is slowing down with increasing
number of atoms in the vapor phase, nv.
Combine both:
⇒ dA = -PdV – SdT
A = A(V,T) ⇒ dA = (∂A/∂V)TdV + (∂A/∂T)VdT
Comparing the two equations we see that:
P = – (∂A/∂V)T and S = – (∂A/∂T)V
T1 < T2
Figure 5.3 - Gaskell
At constant T and V the equilibrium state corresponds to the
minimum of the Helmholtz Free Energy (dA=0).
- From A = U – TS, we see that low values of A are obtained
with low values of U and high values of S.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/5
Often, entropy and energies are opposed to each other:
High entropies mean low energies and vice versa. And the
entropy part becomes more important at high temperatures.
Dr. M. Medraj
Gibbs Free Energy
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/6
Chemical Potential
G = H – TS Equilibrium is a tradeoff between enthalpy and entropy
• A change to a lower enthalpy state (H < 0, exothermic) usually
decreases the randomness (S < 0). (e.g. solidification and oxidation)
• A change to a higher entropy state (S > 0) usually increases the
enthalpy (H > 0, endothermic). (e.g. melting and evaporation processes)
• All of these processes are characterized by a lowering of the Gibbs free
energy: ∆G = ∆(H - TS) < 0
• A crystal at equilibrium has its …………. Gibbs energy.
• If the composition of the system changes during a process, then two
independent variables are …. sufficient to describe the state of the system.
• Chemical reactions or exchange with surroundings can lead to the
change in composition (number of moles of different species, ni, nj, nk, ….).
• The Gibbs free energy (as well as other thermodynamic potentials) is an
extensive property, i.e. depends on …… of the system and on …………..
……….. of different species, G = G(T,P,ni,nj,nk,…..).
Fundamental equations for a closed system:
1st law: U=q – w
 Enthalpy: H=U+PV
 Gibbs energy: G = H – TS
dU = TdS – PdV
dH = TdS + VdP
dG = VdP – SdT
 Helmholtz energy: A = U – TS
dA = -PdV – SdT
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/7
If composition is not changing,
dG = VdP – SdT
⇒ S = – (∂G/∂T)P,ni,nj
and V = (∂G/∂P)T,ni,nj,…
Dr. M. Medraj
If ni varies:
⇒
nj
where nj is the number of moles of
every species other than the ith species.
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/8
Chemical Potential
Chemical Potential
This is called the chemical potential of the species i
⇒
• The chemical potential of the species i is the rate of increase of G with
ni when species i are added to the system at constant T, P and number of
moles of all the other species.
• The above equation can be applied to open systems that exchange both
matter and heat with their surroundings, as well as to closed systems
where changes in composition may occur due to chemical reactions.
Also, we can rewrite the 1st law for a closed system undergoing a
reversible change in composition due to a chemical reaction:
Also, we can obtain the following relations:
Similar equations can be written for U, H, and A:
These fundamental
relations can be
used to derive
Maxwell equations.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/9
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/10
Maxwell’s Equations: Example 1
Maxwell’s Equations
Consider the dependence of the entropy of an ideal gas on
the independent variables T and V: S = S(T,V)
Since U is a state function, the order of differentiation for its second
derivative does not matter:
For a reversible constant volume process TdS = δqv = dU = CvdT and
Therefore, for a system with constant composition:
Using the following Maxwell equation
And from other fundamental equations we have:
and the ideal gas law, PV = RT, which gives
Maxwell’s equations allow one to derive dependences
among experimentally measurable quantities.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/11
Integration between states 1 and 2 gives
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/12
Maxwell Equations: Example 2
Maxwell Equations: Example 2
Considering V = V(P,T):
Recall that
While P and
can be directly measured experimentally,
……..
• It would be useful to express this term through other variables.
• Differentiating the combined 1st and 2nd laws, dU = TdS - PdV by V
at constant T we have
and differentiating by T at
constant V we have:
⇒
since
(∂V/∂T)P and (∂V/∂P)T are extensive properties
It would be more convenient to express cp - cv through intensive properties.
For isotropic materials we can define
Using Maxwell’s equation:
this is valid for any system, we have
not used any approximation.
coefficient of ………………….
α is small for solids
(e.g. 1.7×10-5 K-1 for
isothermal ……………….
Cu, 1.0×10-6 K-1 for
diamond at T = 300 K)
⇒
Dr. M. Medraj
MECH6661 lecture 5/13
Mech. Eng. Dept. - Concordia University
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/14
Gibbs-Duhem Equation
For a homogenous phase of two components, A and B, the fundamental
equation becomes: dG   SdT  VdP   A dn A   B dnB
If we now specify equilibrium at constant T and P: dG   Adn A   B dnB  0
Now, we have shown that G   An A   B nB
Differentiating this, we obtain dG   Adn A   B dnB  n Ad A  nB d B
At equilibrium dG   Adn A   B dnB  n Ad A  nB d B  0
Substituting the previous expression dG   Adn A   B dnB  0
we obtain n Ad A  nB d B  0
n
In the general case we get the Gibbs-Duhem equation  ni di  0
10 minutes Break
i 1
  
  
 nB  B 
0
If we divide through both sides by dXA we get: n A  A 
 X A T ,P ,n
 X A T ,P ,n
B
And now dividing by nA + nB we get:
Dr. M. Medraj
B
  
  
X A  A 
0
 1  X A  B 
 X A T ,P ,nB
 X A T ,P ,nB
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/15
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/16
G as Function of T at Constant P
H and S as Function of T at Constant P
• In a closed one-component system equilibrium, at temperature T and
pressure P, corresponds to the state with ………… Gibbs free energy G.
• Therefore, in order to predict what phases are stable under different
conditions, we have to examine the dependence of G on T and P.
• Let us use thermodynamic relations to predict the temperature
dependence of H, S, and G at constant P.
G = H - TS
dG = -SdT +VdP
Therefore for constant P:
………….
……………
For H(T) we have
For S(T) we have
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/17
G(T) for two phases
at P = const
Dr. M. Medraj
G(T) for a single phase at
P = const
• At all temperatures, the liquid has a higher
internal energy U and enthalpy H as compared to
the solid. Therefore Gl > Gs at low T.
• The liquid phase, however, has a higher entropy
S than the solid phase at all T. Therefore Gl
decreases more rapidly with T as compared to Gs.
• At Tm Gl(T) crosses Gs(T) and both liquid and
solid phases can co-exist in equilibrium (Gl = Gs)
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/18
P-T Phase Diagram (PD) for a Pure Material
G as Function of T at Constant P
The solid lines (phase boundary lines)
on the phase diagram show the
conditions where different phases
coexist in equilibrium: Gphase1 = Gphase2
G(T) for two phases
at P = const
At Tm the heat supplied to the system will not rise its
temperature but will be used to supply the …………... of
melting ∆Hm that is required to convert solid into liquid.
At Tm the heat capacity Cp = (∂H/∂T)P is infinite,
i.e. addition of heat does not increase T.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/19
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/20
The Relationship between P and V for Pure Substance
Equilibrium Between Liquid and Solid Phases
 As pressure is increased at T1, the volume of the gas decreases.
 When the pressure reaches a certain value, the volume suddenly
reduces as the gas liquefies
 A further increase in pressure causes little further reduction in volume as
liquid is generally not compressible to a large extent.
 When the P-V relationship is examined at different T, one
may obtain the results shown in the diagram below:
A
T5
dG = -SdT +VdP
As we can see from this fundamental equation, the free energy of
a phase increases with pressure:
This means that if the two phases have …………… molar
volumes, their free energies will increase by different amounts
when pressure changes at a fixed T.
Vl < Vs for water
At the point A the liquid and vapor are
indistinguishable and their densities are identical.
Vl > Vs for most materials
T4
Tc
T2
T1
Dr. M. Medraj
What is the curvature of the G(P) at const. T?
This is called isothermal
………………..
All gases exhibit this type of behavior, but values of
critical pressure and critical temperature vary from one
substance to another.
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MECH6661 lecture 5/21
This is the …………….
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Equilibrium Between Two Phases
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/22
G as Function of T and P for Two Phases
• We have seen that if two phases in equilibrium have different molar
volumes, their free energies will increase by different amounts when P
changes at certain T.
• The equilibrium, therefore will be disturbed by the change in pressure.
• The only way to maintain equilibrium at different pressures is to
change temperature as well.
Schematic representation of the
equilibrium surfaces of the solid and
liquid phases of water in G-T-P space.
For two phases (say liquid and solid) in equilibrium Gl = Gs
and dGl = dGs for infinitesimal change in T and P (so that the
system remains in equilibrium):
The planes show the free energies of
liquid and solid phases, the intersections
of the planes correspond to the (P, T)
conditions needed for maintaining
equilibrium between the phases, Gl = Gs
We know that at equilibrium ∆G = 0 ⇒ ∆H – T ∆S = 0 ∴ ∆S = ∆H/T
Clapeyron equation: it gives the relationship
between the variations of P and T required to
⇒
maintain the equilibrium between two phases.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/23
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/24
Unary System - Density
Clapeyron’s Equation – Example 1
For liquid to gas transition:
∆V = Vg - Vl >> 0
∆H = Hg - Hl > 0
i.e. we have to add heat to convert liquid to gas.
Pressure
Liquid
Solid
Vapor
⇒
⇒
A typical phase diagram
for a pure material
Liquid
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Dr. M. Medraj
Vapor
MECH6661 lecture 5/25
Dr. M. Medraj
Temperature
Example 2: Gibbs Phase Rule
MECH6661 lecture 5/26
Mech. Eng. Dept. - Concordia University
Example 3: Gibbs Phase Rule
The following is the phase diagram for a pure substance.
Calculate the number of degrees of freedom at the triple point c
Calculate the number of degrees of freedom
at the triple point A
Gibbs phase rule:
Temperature
Vapor
This is the case for H2O (ice floats in liquid water), Ga, Ge, diamond
Liquid
Solid1
Solid
For some materials, however, ∆V = Vs - Vl > 0
⇒
Solid2
Pressure
The slope of the equilibrium
lines in a P-T phase diagram
of a pure material reflects the
relative densities of the two
phases.
Note what happens along the
isotherm when you increase the
pressure.
Temperature
Pressure
For liquid to solid transition:
∆V = Vs - Vl < 0  for most materials
∆H = Hs - Hl < 0
i.e. heat is released upon crystallization.
Which is denser liquid or solid?
Gibbs phase rule:
f=c–p+2-r
f=c–p+2-r
1
(liquid=vapor)
c
A
This special condition is established only at
definite T and P which are characteristic of the
substance and outside our control.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/27
The critical point is ………....
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
The vapor–liquid
critical point
denotes the
conditions above
which distinct
liquid and gas
phases do not
exist.
MECH6661 lecture 5/28
Vapor–Liquid Critical Point for Pure Substances
Clapeyron’s Equation – Example 4
Some materials may exist in more than one crystal structure, this is called
………...……. If the material is an elemental solid, it is called ……….
Close-packed FCC γ-Fe has a smaller
molar volume than BCC α-Fe:
∆V = Vγ - Vα < 0
And at the same time ∆H = Hγ - Hα > 0
⇒
Diagram shows the stable phases
for pure iron at varying T and P.
The effect of increasing P is to
increase the area of the PD over
which the phases of the smaller
molar V (higher density) are stable.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 4/29
Phase Diagrams of One-Component Materials
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/30
The Driving Force for the Phase Transformation
• If solid and liquid are in equilibrium, Gs = Gl and a slow addition of
heat leads to the melting of some part of the solid, but does not change
the total G of the system: G = nl Gl + ns Gs = const, where nl and ns are the
numbers of moles of liquid and solid phases, and Gl and Gs are the molar Gibbs
free energies.
graphite
From the phase diagram, what phase has higher
density near two-phase equilibrium conditions:
- Diamond or graphite?
- Diamond or liquid carbon?
- Graphite or liquid carbon?
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
diamond
MECH6661 lecture 5/31
• However, if energy is added or removed quickly, the system can be
brought out of equilibrium (overheated or undercooled), the
melting/freezing process is spontaneous/irreversible and G is decreasing.
For small undercooling ∆T, we can neglect the
difference in Cp of liquid and solid phases and
assume that ∆H and ∆S are independent of T.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
The driving force
for ……………..
MECH6661 lecture 5/32
First-Order and Second-Order Phase Transitions
First-Order and Second-Order Phase Transitions
The classification of phase transitions proposed by Ehrenfest is based on
the behavior of G near the phase transformation.
First-order phase transition: first derivatives of G are
discontinuous.
Second-order phase transition: first derivatives of G are
continuous, but second derivatives of G are discontinuous.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/33
Dr. M. Medraj
Summary
• Second order phase transitions include the ferromagnetic
phase transition in materials such as iron, where the
magnetization (which is the first derivative of the free
energy with the applied magnetic field strength) increases
continuously from zero as the temperature is lowered below
the Curie temperature. The magnetic susceptibility (the
second derivative of the free energy with the field) changes
discontinuously.
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/34
Summary
• The various solid/liquid/gas transitions are classified as
first-order transitions because they involve a discontinuous
change in density, which is the first derivative of the free
energy with respect to chemical potential.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 4/35
• Just knowing the internal energy U of a system with a constant volume
and temperature is not enough to tell us what the equilibrium
configuration will be.
• There could be many macrostates with the same U. That is why just
minimizing U (or H) is not good enough, we have to minimize A = U TS or G = H - TS to find the equilibrium configuration of the system.
• Of all the macrostates possible for a given U (or H) the one with the
largest entropy at the given temperature will be the one that the system
will adopt.
• Equilibrium occurs when the balance between a small energy and a
large entropy is achieved. High entropies often mean small energies and
vice versa - both quantities are opposed to each other.
• The entropy of a certain macrostate can be calculated by the statistical
definition of the entropy S, using the “Boltzmann entropy equation”:
S = k ln Ω
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/36
Next time:
Binary Solutions
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH6661 lecture 5/37