ffkkmm chapter 5 equilibrium in thermodynamic …mazlan/?download=adv thermo chapter 5.pdf1...
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA5-1
Chapter 5
Equilibrium in Thermodynamic Systems
Assoc. Prof. Dr. Mazlan Abdul WahidFaculty of Mechanical EngineeringUniversiti Teknologi Malaysiawww.fkm.utm.my/~mazlan
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State of EquilibriumState of Rest --- Time independent. No
changes occur in a system at equilibriumexcept by outside action.
State of Balance --- If the system is perturbed,it returns to its equilibrium condition.
Steady State versus Equilibrium State --- Ifisolating a system at rest from itssurroundings (rigid, impervious, insulatingboundary), does not cause changes in thesystem, the system is at equilibrium.
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States of Equilibrium
StableEquilibrium
UnstableEquilibrium
MetastableEquilibrium
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Steady State versus
Equilibrium
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Initial State
Fix @ T2
Fix @ T1
Tem
pera
ture
(T)
T2
T1
Conductor
Insulator
Distance (X)
LXTXT O <<= 0;)0,(
TO
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Transient State
Fix @ T2
Fix @ T1
Tem
pera
ture
(T)
T2
T1
Conductor
Insulator
Distance (X)
),();,( tXfXT
tXfT ====∂∂∂∂∂∂∂∂====
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Steady StateEquilibrium with Surroundings
Fix @ T2
Fix @ T1
Tem
pera
ture
(T)
T2
T1
Conductor
Insulator
q, constant heat flux
dT/dX, constant gradient
Distance (X)
)();( tfTXfT ≠≠≠≠====
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Initial StateIsolate System from Surroundings
Tem
pera
ture
(T
)
T2
T1
Conductor
Insulator
q, heat flux
dT/dX, constant gradient
Distance (X)
(((( ))))1212
12 TT
XXXX
TT −−−−−−−−−−−−−−−−====
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Equilibrium StateIsolated System, Internal Equilibrium
Tem
pera
ture
(T)
T2
T1
Conductor
Insulator
Distance (X)
0;0;2
12 ====∂∂∂∂∂∂∂∂====
∂∂∂∂∂∂∂∂
++++====tT
XTTT
Teq
Teq
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Extremum in Isolated System
Criterion for Equilibrium
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J. W. Gibbs StatementsIf a system is in equilibrium both internally
and with its surroundings, then isolating it from its surroundings produces no change in the internal state of the system. (J. W. Gibbs)
Thus, the internal state of any system that achieves equilibrium with its surroundings is identical with the condition of some other system that comes to the same final state but is isolated from its surroundings during its approach to equilibrium.
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Extremum Principle for Entropy
(Not necessarily the maximum entropy possible under any condition.)
0
In an isolated system the equilibrium state is the statethat has the maximum value of entropy that thesystem can exhibit.The conditions for equilibrium that are derived for anisolated system are the same as those that hold for anysystem that achieves an equilibrium state, no matterthe history of the system.
For any real process that occurs within an isolatedsystem, the total entropy of the system can onlyincrease.
0 S S S transferproductionsys >∆+∆=∆
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Entropy a Maximum at Equilibrium.Entropy production is always positive.In an isolated system with no entropy transfer,
the change in entropy is limited to entropy production.
In an isolated system, the entropy can only increase.
As the system evolves toward its final state of rest, its entropy continually increases.
The final resting state, the equilibrium state, must have the highest entropy that the system, once isolated from its surroundings can exhibit.
When an isolated system reaches its equilibrium state, its entropy is a maximum.
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Definition of Symbols &
Terms
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Extensive Properties, Z/
Molar Properties, ZUnary Two Phase System
ββββαααα /// ZZZ sys ++++====
ννννββββαααα //// ... ZZZZ sys ++++++++++++====
In general, for a multicomponent, multiphase (νννν-phase) system:
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Isolated Systems
Boundaries are
Rigid --- ∆Vsys= 0
Impervious --- ∆nsys= 0
Insulating --- ∆Usys= 0
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Chemical Potential
µια = (dU/α/dnι
α)S/α,V/α
µιβ = (dU/β/dnι
β)S/β,V/β
µιγ = (dU/γ/dnι
γ)S/γ,V/γ
µjα = (dU/α/dnj
α)S/α,V/α
µjβ = (dU/β/dnj
β)S/β,V/β
µjγ = (dU/γ/dnj
γ)S/γ,V/γ
µkα = (dU/α/dnk
α)S/α,V/α
µkβ = (dU/β/dnk
β)S/β,V/β
µkγ = (dU/γ/dnk
γ)S/γ,V/γ
Components: i, j, k. Phases: α, β, γ.
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Apply Extremum Condition with Constraints to
Entropy for Unary Two Phase System
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Unary Two Phase System
Phase boundary is open --- matter may flow between phases.
If the system is isolated ---
dV/α + dV/β = 0, dU/α + dU/β = 0, dnα + dnβ = 0
dV/α = −dV/β, dU/α = −dU/β, dnα = −dnβ
Chemical Potential (single component system)
µα = (dU/α////dn)S/α,V/α, µβ = (dU/β////dn)S/β,V/β
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Find Conditions for EquilibriumConstrained Extremum
• Write a differential expression for the change inentropy that the system may experience. Includeall possible independent variables.
• Write differential equations that describe theconstraints for an isolated system.
• Use the isolation constraints to eliminate dependentvariables. Resulting differential equation appliesexplicitly to changes in independent variables forisolated system.
• Set coefficients of differentials to zero. Results areconditions for equilibrium.
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Application of the Entropy Criterion for Equilibrium
Expression for entropy change of the system:dS/
sys= dS/α + dS/β
Using combined 1st & 2nd laws:dU/α = TαdS/α - Pα dV/α + µµµµαdnα
dU/β = TβdS/β - Pβ dV/β + µµµµβdnβ
Rearrange:dS/α = dU/α/ΤΤΤΤα + Pα dV/α/ΤΤΤΤα - µµµµαdnα/ΤΤΤΤα
dS/β = dU/β/ΤΤΤΤβ + Pβ dV/β/ΤΤΤΤβ - µµµµββββdnββββ/ΤΤΤΤβ
Sum:dS/
sys= dS/α + dS/β = dU/α/ΤΤΤΤα + Pα dV/α/ΤΤΤΤα - µµµµαdnα/ΤΤΤΤα
+ dU/β/ΤΤΤΤβ + Pβ dV/β/ΤΤΤΤβ - µµµµββββdnββββ/ΤΤΤΤβ
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Application of the Entropy Criterion for Equilibrium
Constraints for an isolated system:
dU/α = −dU/β dV/α = −dV/β dnα = −dnβ
Eliminate variables:
dS/sys= [1/ΤΤΤΤα−1/ ΤΤΤΤβ] dU/α + [Pα/ΤΤΤΤα−Pβ/ΤΤΤΤβ] dV/α
− [µµµµα/ΤΤΤΤα−µµµµβ/ΤΤΤΤβ] dnα
Solve for conditions for equilibrium:
Thermal: ΤΤΤΤα = ΤΤΤΤβ
Mechanical: Pα = Pβ
Chemical: µµµµα = µµµµβ
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Generalize
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Criteria for Spontaneous ChangeIn an isolated system the entropyfunction increasesduring every spontaneous change.In a system constrained to constant entropy and volume the internal energyfunction decreasesduring every spontaneous change.In a system constrained to constant entropy and pressure the direction of spontaneous change is monitored by a decreasein the enthalpy function.In a system constrained to constant temperature and volume the Helmholtz free energyfunction decreasesduring every spontaneous change.In a system constrained to constant temperature and pressure the Gibbs free energyfunction decreasesduring every spontaneous change.
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Criteria & Constraints for Equilibrium
Equilibrium Criterion Constant
S is a maximumat equilibrium. U, V, n
U is a minimumat equilibrium. S, V, n
H is a minimumat equilibrium. S, P, n
F is a minimumat equilibrium. T, V, n
G is a minimumat equilibrium. T, P, n
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Application of the Internal Energy Criterion for EquilibriumExpression for internal energy change of the
system: dH/sys= dH/α + dH/β
Using combined 1st & 2nd laws:
dU/α = TαdS/α - PαdV/α + µµµµαdnα
dU/β = TβdS/β - PβdV/β + µµµµβdnβ
Sum:
dU/sys= dU/α + dU/β = TαdS/α - PαdV/α
+ µµµµαdnα + TβdS/β - PβdV/β + µµµµβdnβ
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Constraints: S, V, ndS/α = −dS/β dV/α = −dV/β dnα = −dnβ
Eliminate variables:
dU/sys= [ΤΤΤΤα− ΤΤΤΤβ] dS/α + [V /α − V/β] dV/α + [µµµµα−µµµµβ] dnα
Solve for conditions for equilibrium:
Thermal: ΤΤΤΤα = ΤΤΤΤβ
Mechanical: Pα = Pβ
Chemical: µµµµα = µµµµβ
Application of the Internal Energy Criterion for Equilibrium
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Application of the Enthalpy Criterion for Equilibrium
Expression for enthalpy change of the system:
dH/sys= dH/α + dH/β
Using combined 1st & 2nd laws:
dH/α = TαdS/α + V/α dPα + µµµµαdnα
dH/β = TβdS/β + V/β dPβ + µµµµβdnβ
Sum:
dH/sys= dH/α + dH/β = TαdS/α + V/αdPα
+ µµµµαdnα + TβdS/β + V/βdPβ + µµµµβdnβ
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Application of the Enthalpy Criterion for Equilibrium
Constraints: S, P, n
dS/α = −dS/β dPα = dPβ = 0 dnα = −dnβ
Eliminate variables:
dH/sys= [ΤΤΤΤα− ΤΤΤΤβ] dS/α + [V /α+V/β] dPα + [µµµµα−µµµµβ] dnα
0
Solve for conditions for equilibrium:
Thermal: ΤΤΤΤα = ΤΤΤΤβ
Mechanical: Pα = Pβ (assumed)Chemical: µµµµα = µµµµβ
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Application of the Helmholtz Free Energy Criterion for EquilibriumExpression for enthalpy change of the system:
dF/sys= dF/α + dF/β
Using combined 1st & 2nd laws:
dF/α = S/αdTα + Pα dV/α + µµµµαdnα
dF/β = S/βdTβ + Pβ dV/β + µµµµβdnβ
Sum:
dF/sys= dF/α + dF/β = S/αdTα + PαdV/α
+ µµµµαdnα + S/βdTβ + PβdV/β + µµµµβdnβ
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Constraints: T, V, n
dTα = dTβ = 0 dV/α = −dV/β dnα = −dnβ
Eliminate variables:
dF/sys= [S/α+S/β] dTα + [Pα − Pβ] DV/α + [µµµµα−µµµµβ] dnα
0
Solve for conditions for equilibrium:
Thermal: ΤΤΤΤα = ΤΤΤΤβ (assumed)Mechanical: Pα = Pβ
Chemical: µµµµα = µµµµβ
Application of the Helmholtz Free Energy Criterion for Equilibrium
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Application of the Free Energy Criterion for Equilibrium
Expression for free energy change of the system:
dG/sys= dG/α + dG/β
Using combined 1st & 2nd laws:
dG/α = -S/αdTα + V/α dPα + µµµµαdnα
dG/β = -S/βdTβ + V/β dPβ + µµµµβdnβ
Sum:
dG/sys= dG/α + dG/β = -S/αdTα + V/α dPα
+ µµµµαdnα - S/βdTβ + V/β dP/β + µµµµβdnβ
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Application of the Free Energy Criterion for Equilibrium
Constraints: T, P, n
dT/α = dT/β = 0 dP/α = dP/β = 0 dnα = −dnβ
Eliminate variables:
dG/sys=- [S/α + S/β] dTα + [V /α+V/β] dP/α + [µµµµα−µµµµβ] dnα
0
Solve for conditions for equilibrium:
Thermal: ΤΤΤΤα = ΤΤΤΤβ (assumed)Mechanical: Pα = Pβ (assumed)Chemical: µµµµα = µµµµβ
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(a) Give three illustrative examples ofan equilibrium state.
• A mixture of reacting gas molecules constrained to a fixed temperature.
• Ice floating in water isolated from its surroundings.
• A foil birthday balloon brought into a hot room.
EXAMPLE
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(b) Give three illustrative examples ofa steady state.
• Molten steel poured into a continuous caster.
• A hose supplies water to a bucket with a hole.
• Current passing through a tungsten wire heat it to incandescence..