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Transcript
ME 083
Thermodynamic Aside:
Gibbs Free Energy
Professor David M. Stepp
Mechanical Engineering and Materials Science
189 Hudson Annex
[email protected]
549-4329 or 660-5325
24 February 2003
From Last Time….
Gibbs Free Energy: G = H – T*S
Recall our Arrhenius relationship for the equilibrium number
of vacancies (defects):
Nv  Ntot  e
Q 
 v 
 kB T 
Frenkel defects (vacancy plus interstitial defect)
THERMODYNAMIC ASIDE:
GIBBS FREE ENERGY
Energy of a System: A State Function
(Depends only on the current condition of the system)
Three Categories of Energy:
Kinetic (motion), Potential (position),
and INTERNAL (molecular motions)
State Function Internal Energy:
– How can we change a material’s (or system’s)
internal energy?
∆U = Q + W + W’
State Function Internal Energy (U):
∆U = Q + W + W’
(The First Law of Thermodynamics)
Q: Heat flow (into system)
W: P-V work (on system)
W’: Other work (on system)
Energy Conservation
Processes in nature have a natural direction of change
SPONTANEOUS
DOES NOT
OCCUR
State Function Entropy (S) ∆SP ≥ 0
The Entropy of a system may increase or decrease
during a process.
The Entropy of the universe, taken as a system plus
surroundings, can only increase.
(The Second Law of Thermodynamics)
“Entropy is Time’s Arrow”
Note: The laws of thermodynamics are empirical,
based on considerable experimental evidence.
One Can Show That:
For Reversible Processes QREV =
 TdS
– QREV denotes the maximum heat absorbed
– Note the cyclic path integral (reversibility)
With this, the combined Notation for the First and
Second Law can be expressed:
dU =TdS + dWREV + dW’
BUT dWREV = F*dx = F*(A/A)*dx = F/A*(-dV) = -PdV
dU = TdS – PdV + dW’
Combined statement of
First and Second Laws
The Third Law of Thermodynamics:
The Entropy of all substances is the same at 0 K
Both Entropy and Temperature Have Absolute Values
(Both have an empirically observed zero point)
State Function Enthalpy (H) H = U + P*V
So: dH = dU + PdV + VdP
Consider the special case where dP = 0 and dW’ = 0:
dHP = TdSP = dQREV,P
State Function Gibbs Free Energy
G = H – TS = U + PV - TS
dG = dU + PdV + VdP – TdS – SdT
= (TdS – PdV + dW’) + PdV + VdP – TdS – SdT
= VdP – SdT + dW’
Special Case of dT = 0 and dP = 0: dGT,P = dW’T,P
Or, if W’ = 0: dGP = -SdT
At constant Temperature and Pressure, the (change
in) Gibbs Free Energy reflects all non-mechanical
work done on the system.
Back to Crystals…
Consider Frenkel Defects (vacancy + interstitial)
The Free Energy of the Crystal can be written as
– the Free Energy of the perfect crystal (G0)
– plus the free energy change necessary to create n
interstitials and vacancies (n*∆g)
– minus the entropy increase that derives from the different
possible ways in which defects can be arranged (∆SC)
∆G = G-G0 = n*∆g - T∆SC
n = the number of defects
∆g = ∆h - T∆S (the energy to create defects)
The Configurational Entropy, ∆SC, is proportional to
the number of ways in which the defects can be
arranged (W).
∆SC = kB* ln(W)
(The Boltzmann Equation)
Note: this constitutes a connection between atomistic
and phenomenological descriptions (thru statistics).
Perfect Crystal:
N atoms which are indistinguishable can only be
placed in one way on the N lattice sites:
∆SC = k*ln(N!/N!) = k*ln(1) = 0
Crystal with Vacancies and Interstitials
With N normal lattice sites (and equal interstitial sites),
ni (number of interstitial atoms) can be arranged in:
N!
(N  ni )!ni! distinct ways
Similarly, the vacant sites can be arranged in:
N!
distinct
ways
(N  nv )! nv !
Recalling that the equilibrium number of vacancies
obeys an Arrhenius relationship:
n
e
N
 g 


 2kT 
e
 s 
 
 2k 
e
 h 


 2kT 
Remember: ∆g = ∆h - T∆s
n
e
Therefore:
N
 h 


 2kT 
Assuming configurational entropy
in creating defects is negligible
Now compare this with our
equation for Frenkel Defects:
NV NI

e
NT NT
Q

 sV / I 
 2kT 
The Configurational Entropy of these noninteracting defects is:

 

N!
N!

SC  k  ln  








N

n
!
n
!
N

n
!
n
!
i
i  
v
v 

Probability theory for statistically independent events:
PAB = PA * PB
Stirling’s Approximation for Large Numbers:
ln(x!) ≈ x * ln(x) - x
Recalling that ni = nv = n (for Frenkel defects):

 N 
 N  n 
SC  2  k N  ln 
  n  ln 

Nn
 n 

At equilibrium, the free energy is a minimum with
respect to the number of defects
Note: Entropy (per defect) is a maximum at this point
Recall: ∆G = G – G0 = n * ∆g – T∆SC
G = G0 + n * (∆h – T∆s) - T∆SC
Next Time…
The Statistical Interpretation of Entropy
∆SC = k * ln(W)
Configurational Entropy is proportional to the number
of ways in which defects can be arranged