FFKKMM Chapter 5 Equilibrium in Thermodynamic …mazlan/?download=Adv Thermo Chapter 5.pdf1 09/19/2001 3-1 CC HH AA PP TT EE RR 33 Advanced Thermodynamics Advanced Thermodynamics --

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Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013

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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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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20015-4

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


dU/sys= [ΤΤΤΤα− ΤΤΤΤβ] dS/α + [V /α − V/β] dV/α + [µµµµα−µµµµβ] dnα



Mechanical: Pα = Pβ


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


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


dH/sys= [ΤΤΤΤα− ΤΤΤΤβ] dS/α + [V /α+V/β] dPα + [µµµµα−µµµµβ] dnα

0



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


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


dF/sys= [S/α+S/β] dTα + [Pα − Pβ] DV/α + [µµµµα−µµµµβ] dnα

0


Thermal: ΤΤΤΤα = ΤΤΤΤβ (assumed)Mechanical: Pα = Pβ


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


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


dG/sys=- [S/α + S/β] dTα + [V /α+V/β] dP/α + [µµµµα−µµµµβ] dnα

0


Thermal: ΤΤΤΤα = ΤΤΤΤβ (assumed)Mechanical: Pα = Pβ (assumed)Chemical: µµµµα = µµµµβ

0

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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..

Documents

FFKKMM Chapter 5 Equilibrium in Thermodynamic …mazlan/?download=Adv Thermo Chapter 5.pdf1 09/19/2001 3-1 CC HH AA PP TT EE RR 33 Advanced Thermodynamics Advanced Thermodynamics --