Chap 4 Fundamental Equations of Thermodynamicsmail.scu.edu.tw/~kguo/pchem2/index.files/Lecture09.pdf2005/11/04 1 Chap 4 Fundamental Equations of Thermodynamics Table of Contents 뿽

2005/11/04 1

Chap 4 Fundamental Equations of Thermodynamics

Table of Contents 目錄內容4.1 Fundamental Equation for the Internal Energy for Open Systems 對開放系統內能的基本方程式

4.2 Definition of Additional Thermodynamics Potentials Using LegendreTransformations 以 Legendre 轉換定義新的熱力學位能

4.3 Effect of Temperature on the Gibbs Energy 吉布斯自由能的溫度效應4.4 Effect of Pressure on the Gibbs Energy 吉布斯自由能的壓力效應4.5 Fugacity and Activity 逸壓與活性4.6 The Significance of the Chemical Potential 化學位能的重要性4.7 Additivity of Partial Molar Properties with Applications to Ideal Gases 部分莫耳性質的加成性

4.8 Gibbs-Duhem Equation 吉布斯-都亨方程式4.9 Special Topic: Applications of Maxwell Relations Maxwell 關係式的應用

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• The essential thermodynamic properties: T, U, S.• For isolated system: (q=0)

– Entropy as a good criterion for spontaneity and equilibrium.• For close system (constant composition, but not constant q) :

– Not a good criterion for constant (T, V ) or (T, P ).– Need two more thermodynamic properties as criterion. – Using Legendre transforms to generate two new functions.– Helmholtz free energy (A) as criterion for constant T and V.– Gibbs free energy (G) as criterion for constant T and P.

• For open system:– Chemical potentials for each species of a system at equilibrium is the

same.– Spontaneous mixing of two partially miscible liquids at specific T and P.


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• Combination of 1st and 2nd law for close system– 1st law: dU = ƌq + ƌw.– 2nd law: dS > ƌqirrev/T and dS = ƌqrev/T .– If only PV-work involved, ƌw = - P dV– combine 1st law and 2nd law for close system: dU = T dS - P dV⇨ It is the fundamental equation in a close system involving only PV

work, (an equation involves only state functions U, S, T, P, V ). It applies to both reversible and irreversible processes (state functions).

⇨ Just like -P and V are conjugate variables for work, ƌw = - P dV ,now T and S are also conjugate variables for heat: ƌqrev = T dS .

• Since U (S,V ) is an exact differential:

πΤ : internal pressure


VVUS

SUU

SVddd

∂∂+

∂∂=

VSUT

=

dd

SVUP

=−dd

PVST

VU

TTT −

∂∂=

∂∂=π

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• Derivative of an extensive property with respect to an extensiveproperty gives an intensive property.

• U (T,P ) or U (T,V ) can not be used to calculate all the thermodynamic properties of system, only U (S,V ) will. And S and V are called the natural variables of U

• If U = U (T,V ) =>

• At constant P, (∂U/∂V )T, (∂U/∂T )V are not convenient forms.

PTV

V1α

∂∂= VT

PCαVπ

TU +=

∂∂

⇒

TTU V

VU U

VTddd

∂∂+

∂∂=


VPTP TU

TV

VU

TU

∂∂+

∂∂

∂∂=

∂∂⇒

∂∂

PTx

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Chemical Potential

• In 1876 Gibbs introduced the concept of the chemical potential μ to the fundamental equations in order to discuss phase equilibrium and reaction equilibrium of of a species. For U:

dU = T dS –P dV + μ1dn1 + μ2dn2 + …

• If dni moles of species i are added to system at constant S and V, there is a change in internal energy in terms of μ1dn1.

• If a system contains Ns different species, U is a function of its natural variables S, V and {ni }:

dU = T dS –P dV +

• The natural variables of U are all extensive:

∑=

SN

1iii nµ d

{ } { }∑

=≠

∂∂+

∂∂+

∂∂=

S

ijii

N

1ii

nVSinSnVn

nUV

VUS

SUU dddd

,,,,

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• Fundamental Equation: An expression show how the internal energyU (S,V ) changes with natural variable of S, V

• dU = T dS - P dV + where dU is an exact differential.

• Several important expressions can be derived from this relation,including the Maxwell relations.

• The relations:

•show how measurable properties T, P of a open system can be related to thermodynamic quantities like U. Furthermore,

VSSV VU

SSU

V

∂∂

∂∂=

∂∂

∂∂


∑=

SN

1iii nµ d

ijnVSii n

U

≠

∂∂=

,,

µ{ }inVS

UT,

∂∂=

{ }inSVUP

,

∂∂=−

{ } { }ii nVnS SP

VT

,,

∂∂−=

∂∂

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Chemical Potential

dU = T dS - P dV +

• 若系統性質可依其自然變數的函數決定,則任ㄧ內涵性質(T, P, μi )均可為自身的兩種外延性質的比值.此三式可稱為系統的狀態函數. 對內能的馬克斯威關係式表為:

{ } { }ii nV,

i

nS,V,i Sµ

nT

∂∂=

∂∂

{ } { }ii nVnS SP

VT

,,

∂∂−=

∂∂

{ } { }ii nS,

i

nS,V,i Vµ

nP

∂∂=

∂∂−

jiij S,V,ni

j

S,V,nj

i

nµ

nµ

≠≠

∂∂

=

∂∂

∑=

SN

1iii nµ d

{ } { } { }i

iddd n

nUV

VUS

SU

ijii nVSnSnV≠

∂∂+

∂∂+

∂∂=

,,,,

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Chemical Potential

• If we substitute the second law in the form dS ≥ δq/T and若將熱力學第二定律的形式換成不同的表示: dS ≥ δq/T 與

ƌw=-Pext dV+ , 則內能變化量 dU

• The internal energy remains constant if the infinitesimal change occurs at equilibrium under constant entropy, volume, and {ni } 在恆體積,固定成份與恆熵值下的平衡,其內能維持不變.

(dU )S,V,{nj} ≤ 0 .• The criterion for spontaneous change in the system involving PV-work

and specified amounts of species., U must be approaching the minimum at constant S, V, and {ni }: 在恆體積,固定成份與恆熵值下的自發變化,其內能變化趨向於最低值.

• The integrated form as:均相開放系統的內能以積分形式表示為:

∑=

SN

1iiidnµ

∑=

+−≤SN

1iiiext nVPSTU dd d d µ

∑=

+−=SN

1iiinVPSTU µ

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• For close system, we derive the Maxwell relations • U=U (S,V ), H=H (S,P ), A=A (T,V ), G=G (T,P ); :

Internal energy dU = T dS - P dV.U = H - PV = A + TS = G - PV + TSEnthalpy dH = T dS + V dPH = U + PV = A + PV + TS = G + TSHelmholtz free energy dA = - P dV - S dTA = U - TS = H - PV - TS = G - PVGibbs free energy dG = V dP - S dT.G = U + PV - TS = H - TS = A + PV

pS SV

PT

∂∂=

∂∂

VS SP

VT

∂∂−=

∂∂

TV VS

TP

∂∂=

∂∂

Tp PS

TV

∂∂−=

∂∂


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• To show how measurable properties T, V of a close system (constant composition) can be related to thermodynamic quantities like H :

For enthalpy H = U + PV, dH = dU + P dV + V dP = (T dS – P dV ) + P dV + V dP

As H (S, P ), enthalpy changes with natural variable of S, P,

dH = T dS + V dP

Furthermore, dH as exact differential, the relations:

pSSp PH

SSH

p

∂∂

∂∂=

∂∂

∂∂

PS SV

PT

∂∂=

∂∂

TSH

p=

∂∂ V

PH

S=

∂∂


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For Helmholtz free energy A = U - TS, dA = dU – T dS – S dT = (T dS – P dV) – T dS – S dT

As free energy A (V, T ), changes with natural variable of V, T,

dA = - S dT – P dV

Furthermore, dA as exact differential, we have

• Now the internal pressure πT :

VTTV VA

TTA

V

∂∂

∂∂=

∂∂

∂∂

VT TP

VS

∂∂=

∂∂

PVA

T−=

∂∂S

TA

V−=

∂∂


PTPTP

VST

VUπ

VTTT −

∂∂=−

∂∂=

∂∂=

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For Gibbs free energy G = U + PV – TS = H - TS,dG = dH – T dS – S dT = (T dS + V dP ) – T dS – S dT

As free energy, G (P,T ) changes with natural variable of P, T,

dG = - S dT + V dP

Furthermore, dG as exact differential, we have

pTTp PG

TTG

P

∂∂

∂∂=

∂∂

∂∂

PT TV

PS

∂∂=

∂∂−

VPG

T=

∂∂S

TG

p−=

∂∂


2005/11/04 13

• 在均相系統的外延性質狀態函數U (S,V ), H (S,P ), A (T,V ), G (T,P ) 可表為:

dU = +T dS - P dV dH = +T dS + V dPdA = - S dT - P dV dG = - S dT + V dP


ddUU

ddGGddAA

ddHH--PP ddVV

--SS ddTT

++T T ddSS

++VV ddPP

2005/11/04 14

• Furthermore, from U (S,V ), H (S,P ), A (T,V ), G (T,P ) ) we have the four relations:

VPG

PH

TS=

∂∂=

∂∂

STG

TA

pV−=

∂∂=

∂∂

TSH

SU

pV=

∂∂=

∂∂

PVA

VU

TS−=

∂∂=

∂∂


2005/11/04 15

• 由均相開放系統的狀態方程式得到的性質:

{ } { }ii nTnS PG

PHV

,,

∂∂=

∂∂=

{ } { }ii nPnV TG

TAS

,,

∂∂−=

∂∂−=


{ } { }

ii nPnV SH

SUT

,,

∂∂=

∂∂=

{ } { }ii nTnS VA

VUP

,,

∂∂=

∂∂= --

UU HH

AA GG

TT

VVPP

SS

2005/11/04 16

Legendre transforms:A linear change of variables that involves subtracting the product of conjugate variables from an extensive property of a system.For any state function f (X,Y), since df (X,Y) is exact differential;

df = (∂f /∂X)Y dX + (∂f /∂Y)X dY = df = m dX + n dY

with m = (∂f /∂X)Y and n = (∂f /∂Y)X ; [∂m /∂Y]X = (∂n /∂X)YNow we define a new state function g = f - m X =f - (∂f /∂X)Y Xdg = df - d(mX) = (m dX +n dY ) - (m dX +X dm) = -X dm + n dY

g = g (m,Y) = f (X,Y) - (∂f /∂X)Y X

For the state function g (m,Y):

dg = (∂g /∂m)Y dm + (∂g /∂Y)m dY = -X dm + n dYwith -X = (∂g /∂m)Y and n = (∂g /∂Y)m = (∂f /∂Y)XAlso dg is exact differential => [∂(-X)/∂Y]m = (∂n/∂m)Y

2005/11/04 17

• 均相開放系統的內涵性質的狀態方程式:

dU = T dS - P dV

{ }inPSHT

,

∂∂=

{ }S

SUUSTUA

inV ,

∂∂−=−=

{ }S

SHHSTHG

inP ,

∂∂−=−=

{ }V

VUUVPUH

inS ,

∂∂−=+=


∑=

+SN

1iii ndµ

{ }

inVSUT

,

∂∂=

{ }inSVUP

,

∂∂=-

∑=

++=SN

1iiinPVSTH µd d d

{ }inSPHV

,

∂∂=

{ }P

PHHPVHU

inS ,

∂∂−=−=

UU HH

AA GG

TT

VVPP

SS

2005/11/04 18

• 均相開放系統的內涵性質的狀態方程式:dA = -S dT - P dV

{ }inTPGV

,

∂∂=

{ }P

PGGPVGA

inT

,

∂∂−=−=

{ }T

TGGTSGH

inP ,

∂∂−=+=


∑=

+SN

1iii ndµ

{ } -

inPTGS

,

∂∂=

∑=

++=SN

1iiinPVTSG µd d -d

{ }inTVAP

,

∂∂−=

{ }V

VAAVPAG

inT

,

∂∂−=+=

{ }inVTAS

,

∂∂−=

{ }T

TAATSAU

inV

,

∂∂=+= -

UU HH

AA GG

TT

VVPP

SS

2005/11/04 19

• From U (S,V ), H (S,P ), A (T,V ), G (T,P ) we have


∑=

=SN

1iiinµG

TSPVGU +−=

∑=

+−=SN

1iiinPVTS µ

TSGPVUH +=+=

∑=

+=SN

1iiinTS µ

∑=

+−=SN

1iiinPV µ

PVGU - TSA −==

PVAH - TSG +==

∑=

=+=SN

1iiinPVU-TS µ

2005/11/04 20

• 均相開放系統在平衡下有固定的內涵性質(T, P, μi ) :

{ } { }ii

S

nPnT

N

1iii T

GTPGPGTSPVGnPVTSU

,,

∂∂−

∂∂−=+−=+−= ∑

=

µ

{ }i

S

nP

N

1iii T

GTGTSGnTSH,

∂∂−=+=+= ∑

=

µ

{ }i

S

nT

N

1iii P

GPGPVGnPVA,

∂∂−=−=+−= ∑

=

µ

∑=

=SN

1iiinG µ


{ }ijnPSii n

H

≠

∂∂=

,,

µ{ }ijnVTi

i nA

≠

∂∂=

,,

µ{ }ijnVSi

i nU

≠

∂∂=

,,

µ{ }ijnPTi

i nG

≠

∂∂=

,,

µ

2005/11/04 21

Example 4.2 Calculation of molar thermodynamic properties for an ideal gas. Since the molar Gibbs energy of an ideal gas is given by

Derive the corresponding expressions for Ans: Using equation,

where

Note that the internal energy U and enthalpy H of an ideal gas are independent of pressure and volume.

°+°=

PPRT lnG G

A, and S, H, U, V .

P

RTPGV

T

=

∂∂=

°=−°=°=−°+°=−+= URTH-PVHVPSTGVPSTGU

°=°+°=+= HSTGSTGH

°°=

°−

∂

°∂−=

∂∂−=

PP-R lnS

PPlnR

TG

TGS

PP

°+°=−=

PPRT lnAVPGA

PTG-S

∂

°∂=° -RTSTGUand °+°=°


2005/11/11 22

• 均相開放系統的平衡或自發的要件與熱力學基本關係:

{ }∑

=≠

∂∂+−=

S

ij

N

1ii

nVSin

nUVPSTU dd d d

,,

{ }∑

=≠

∂∂++=

S

ij

N

1ii

nPSin

nHPV ST H dddd

,,

{ }∑

=≠

∂∂+−−=

S

ij

N

1ii

nVTin

nAVPTSA dd d d

,,

∑=

≠

∂∂++−=

S

ij

N

1ii

nPTin

nGPVTSG dd d d

,,

( ) { } 0UinVS ≤,,d

( ) { } )(,, extnPS PP0Hi

=≤d

( ) { } 0AinVT ≤,,d

( ) { } ),(,, surrextnPT TTPP0Gi

==≤d


2005/11/11 23

Table 4.1 Criteria for Irreversibility and Reversibility for Processes Involving No Work or Only Pressure-Volume Work.

( dU )V,S,{nj} = 0( dU )V,S,{nj} < 0

( dG )T,SP,{nj} = 0( dG ) T,P,{nj} < 0( dA )T,V,{nj} = 0( dA )T,V,{nj} < 0

( dH )P,S,{nj} = 0( dH )P,S,{nj} < 0

( dS )V,U,,{nj} = 0( dS )V,U,{nj} > 0

For Reversible Processes

For Irreversible Processes


2005/11/11 24


Fig 4.1 When a system undergoes spontaneous change at constant T and P, the Gibbs energy decreases until equilibrium is reached.

2005/11/11 25

Determine partial molar volume

• 對其他的非體積功也參與的系統的Gibbs自由能

• 非體積功對內能的貢獻,會有多出一個項. 當延伸到Gibbs自由能時也是一樣.

• 例如: extension功與 surface功,參與Gibbs自由能時,表為:

dG=-S dT + V dP• 其中f 為伸張力, L 為長度, γ為表面張力.而 As 為表面積大小.

• 因此,當變數改變時伴隨的G變化, 可由Legendre Transform 衍伸出的更多的Maxwell方程式:

Si

ii ALnµ ddd γf +++∑

2005/11/11 26

• When work other than PV work occurs in the system:

{ } Si AnPTLG

,,,

f

∂∂−=

S

N

1iii ALdnµdPVdTSdG

S

d d γ++++−= ∑=

f

{ } L,P, inTSAG

,

∂∂=γ

{ }ii

,LAnPTiG

nG

Sij

==

∂∂

≠

µ,,,


f: the force of extension.

γ: the surface tension.

μi : the chemical potential of species i.

2005/11/11 27

以功來推演自由能函數的定義

• 可將熱力學第ㄧ定律與第二定律的形式換成以下的表示: dS ≥ ƌq/TsurrdU =ƌ q + ƌ w ≤ T dS + ƌ w ƌ w ≥ dU – T dS = d (U-TS ) at constant T, which means ƌ w ≥ (ΔA )T

where A is the Helmholtz energy. The symbol A actually comes from arbeit, the German word for work.

• Thus, in a reversible process at constant temperature, the work done on the system is equal to the increase in the Helmholtzenergy.

• -d (U-TS ) ≥ - ƌ w : the decrease in A is an upper bound on the total work done by the system to the surroundings at constant temperature .


2005/11/11 28

Chemical Potential

• 若有非體積功的参與, 可將熱力學第ㄧ定律與第二定律的形式換成以下的表示:

dS ≥ ƌ q/Tsurr

dU = ƌ q + ƌ w + ƌ wnonPV ≤ T dS – Pext dV + ƌ wnonPV

or-dU – Pext dV + T dS ≥ - ƌ wnonPV

at constant T and P = Pext = constant, which means

-d (U + PV - TS) ≥ - ƌ wnonPV

(ΔG )T,P ≤ ƌ wnonPV

2005/11/11 29

Chemical Potential

• For a reversible process at constant T and P, the change in Gibbs energy is equal to the non-PV work done on the system by the surroundings.

• Thus, when work is done on the system, the Gibbs energy increases, and when the system done work on the surroundings, the Gibbs energy decreases. In general, the decrease in G is an upper bound on the non-PV work done on the surroundings at constant temperature.

• When the system does work on the surroundings, the work done is less than the decrease in Gibbs energy.

2005/11/11 30

Chap 4.3 Effect of T on the Gibbs Energy

The variation of the Gibbs energy of a system with (a) temperature at constant pressure and (b) pressure at constant temperature. The slope of the former is equal to the negative of the entropy of the system and that of the latter is equal to the volume.

2005/11/11 31

The temperature variation of the Gibbs energyis determined by the entropy. Becausethe entropy is largest for the gaseous phaseof a substance, the Gibbs energy changesmost steeply in the gas phase, followed bythe liquid phase, and by the solid phase.


2005/11/11 32

• Gibbs-Helmholtz equation

{ }

−

∂∂=

TG

TG

T1

inP ,

{ };

THG

TGS

inP

−=

∂∂=

, -

{ } TH

TG

TG

inP

−=−

∂∂

,

{ } { } { }2

nPnPnP TG

TG

T1

T1

dTdG

TG

T1

TG

Tiii

−

∂∂=

+

∂∂=

∂∂

,,,

{ }

( )( ) { }

( ) ( )( ) { }

−

∂∂=

∂∂=

∂∂

2nPnPnP T

1T1TG

dTT1d

T1TG

TG

Tiii ,,,

( )( ) { }

HT1TG

inP

=

∂∂

⇒,

{ }

−=

∂∂

2nP T

HTG

Ti,

The Variation of the G with Temperature

PTGTGH

∂∂−=

{ } -

inPTGS

,

∂∂=

{ } - T SH

TGTHG

inP=

∂∂+=

,

2005/11/11 33

The Variation of the G with Temperature

• The Gibbs-Helmholtz equation for a change between state 1 and state 2, the equation can be written as:

{ } { }jj np2

np TG

T1

TG

TG

T ,,

∂∂+−=

∂∂

{ }jnPTGTHTSHG

,

∂∂+=−=

{ }jnpTG

TTH

,

2

∂∂−=

{ }jnpTG

TTH

,

2

∆

∂∂−=∆

{ }jnV

2TA

TTU

,

∂∂−=

{ }jnV

2TA

TTU

,

∂∂−= ∆∆or

2005/11/11 34

Gibbs-Helmholtz Equation

This equation is very useful because:1. If we can determine ΔG for a process or a reaction as a

function of temperature, the ΔH for the process or reaction can be calculated without using calorimeter.

2. If ΔH and ΔG are known at one temperature, the equation can be integrated to calculate ΔG at another temperature assuming that ΔH is independent of temperature.

3. A plot of [ΔG (T )/T ] versus (1/T ) can give a linear line with a slope of ΔH :

{ }jnpT

TG

H

,

1

∂

∆∂

=∆

2005/11/11 35

• 均相開放系統的理想氣體化學勢能與部分莫耳性質:

∫∫ =22

1

p

p

G

GpVG

1

d d{ }inTp

GV,

∂∂= ∫+=

2p

p12 pVGG

1

d { }

;TGS

inp ,

∂∂=−

( ) ( )

°+=+= ∫

° PP nRT G p nRTGG o

p

p

o lnln d

The Variation of the G with Pressure

( )

°+=

PP RTGG o lnmm

°+=

PP RTµµ io

ii ln

∑=

++−=SN

1iii nPVTSG d d d d µ

iii Gµn =

2005/11/11 36


Example 4.2 Calculation of changes in thermodynamic properties in the reversible isothermal expansion of an ideal gas.

Ans: Since the internal energy of an ideal gas is not affected by a change in volume,

1-mol J 5746 d -101RT lnPVG∆ 1

10=

== ∫

( ) 000mol J 5746 57460 w -q 0; -1

=+=+=

=+===

VP∆U∆H∆U∆U∆

1-1--1

rev molK J 19.14K 300.15

mol J 5746T

qS ===∆

1-1--1

molK J 19.14K 300.15mol J 57460 =+==

TG-∆H∆S∆or

2005/11/11 37


Example 4.3 Calculation of changes in thermodynamic properties in the irreversible isothermal expansion of an ideal gas.

Ans: An ideal gas expands isothermally at 27 ℃ into an evacuated vessel so that the pressure drops from 10 to 1 bar; that is, it expands from a vessel of 2.463 L into a connecting vessel such that total volume is 24.63 L. Calculate the change in thermodynamic quantities.

This process is isothermal, but is not reversible.

All state functions are the same as in Example 4.2 because the initial and final states are the same

-1mol J 0 0-0 w -q 0; 0;w ===== U∆U∆

S, and ∆A, ∆G, ∆H, ∆U∆

2005/11/11 38


The variation of the Gibbs energy with the pressure is determined by the molar volume of the Gibbs energy changes most steeply for the gas phase, followed by the liquid phase, and then the solid phase of the substance.. Because the volumes of the solid and liquid phase are similar, they vary by similar amount as the pressure is changed.

2005/11/11 39


Example 4.4 Calculation of the Gibbs energy of formation of gaseous and liquid methanol as a function of pressure. The pressure effect on the Gibbs energy of a gas is much larger than the liquid due to the molar volume difference. The standard Gibbs energy of formation for liquid methanol ∆fGº(CH3OH, l) at 298.15 K is -166.27 kJ mol-1, and that for gaseous CH3OH ∆fGº(CH3OH, g) is -161.96 kJ mol-1. The density of liquid methanol at 298.15 K is 0.7914 g cm-3. (a) Calculate ∆fG(CH3OH, g) at 10 bar at 298.15 K assuming methanol vapor is an ideal gas. (b) Calculate ∆fG(CH3OH, l) at 10 bar at 298.15 K.

Ans: (a)= -161.96 kJ mol-1 + (0.0083145 kJ K-1 mol-1) (298.15 K) ln 10= -156.25 kJ mol-1

(b)V ̿ = (32.04 g mol-1) /(0.7914 g cm-3) (10-2 m cm-1)3= 40.49x10-6 m3 mol-1

∆fG(CH3OH, l)=-166.27 kJ mol-1+(40.49x10-6 m3 mol-1)(9x105 Pa/ 103 J kJ-1) = -166.23 kJ mol-1

( ) ( )°+= PPRTGPG o∆∆ ff ln

( ) ( )oo∆∆ P-PVGPG ff +=

2005/11/11 40


The difference in Gibbs energy for a ideal gas at two pressures is equal to the area shown below the ideal-gas isotherm.

2005/11/11 41

• dG = V dP - S dT

dT = 0 => dG = V dP , for gases, a logarithmic dependence of the molar Gibbs energy on the pressure:

• The deviation from idealized behavior still can preserve the form of expression by replacing the true pressure (P ) with an effective pressure (f ) called the fugacity:

• An integration between any two pressures to obtain:


( )

°+=

PPnRTGPG o ln

°+=

PfRTGG o

mm ln

==

1

212 -

PPnRTGGG ln∆

2005/11/11 42


Fig 4.2 Dependence of the Gibbs energy of formation of an ideal gas on the pressure of the gas relative to the Gibbs energy of the gas at the standard pressure of 1 bar.

2005/11/11 43


The difference in Gibbs energy for a solid or liquid at two pressures is equal to the rectangular area shown. We have assumed that the variation of volume with pressure is negligible.

( ) ( ) ( ) PVPGdpVPGPG mim

p

pmimfm

f

i

∆+=+= ∫

2005/11/11 44

Fugacity

• Fugacity is a function of pressure and temperature. Also a characteristics of a fugitive.

• The name comes from the Latin for “fleetness” in the sense of “escaping tendency”.

• Fugacity has the same dimension as pressure.• Define as:

–

• If we write the fugacity as f = Φp, Φ is the dimensionless fugacity coefficient, related to the compression factor, Z of the gas between P =0 and the actual pressure of interest by

( ) ( ) ( )

°+=

fT,Pf RT T,PGT,PG o

mm ln

2005/11/11 45

Fugacity Coefficient Φ

• For any gas at different pressures

°

=−pfRT ln GG o

mm

=−=∫ f'fRT ln 'GGdpV mm

p

p'm

=−=∫ p'pRT ln 'GGdpV mm

p

p'

idm

( )

−

=−∫ p'

pln f'fln RT dpVV

p

p'

idmm

( )∫ −=

p

p'

idmm p VV

RT1

f'p'x

pfln d

1p'f0, p'

'→

→

( ) ln φp VVRT1

pfln

p

0

idmm =−=

⇒ ∫ d

2005/11/11 46

Fugacity

• When attractive forces are dominant, the molar Gibbs energy and the chemical potential are lower than those of a ideal gas. At high pressure, when repulsions are dominant, the molar Gibbs energy and the chemical potential are higher than those for the ideal gas.

2005/11/11 47


( )

∫∫

∫

−=

−=

−=

=

p

0

p

0

p

0

idmm

dpP

Z dpP

RTP

RTZRT1

dpVVRT1

pfln ln φ

1

1pf →

→0P

lim0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

10-2

10-1

100

101

102

103

104

P/bar

Vm

/m3

Vm

Vmid

Nitrogen gas at 300 K

2005/11/11 48


Fig 4.4 Plot of fugacity versus pressure for a real gas. The dashed line is for an ideal gas. The standard state is the pure substance at a pressure of 1 bar in a hypothetical state in which it exhibits ideal gas behavior.

2005/11/15 49

– True for all gases• For Ideal Gas:• Take difference:

• Rearrange:


( )

°== ∫ f

T,Pf RT pVG mT lnd∆

°=∫ P

Pd RT pVid ln

( ) ( )

°

°=∫ P

PfT,Pf RT pVV idm lnln -d-

( )∫= pVV idm d-RT1 γln

2005/11/15 50

The fugacity of Nitrogen

2005/11/15 51


Example 4.5 Expression of the fugacity in terms of virial coefficients.Given the expression for the compressibility factor Z as a powerseries in P, what is the expression for the fugacity in terms of the virial coefficients?

Ans:

( ) ...2

C'PB'P dP...C'PB'Pf ln

2P

0++=++= ∫

dPP

Z-1Pfln P

0∫

= ...C'PB'P1 Z 2 +++=

2005/11/15 52


Example 4.6 The fugacity of a van der Waals gas.Using the expression for the compressibility factor Z of a van der Waals gas given in equation, what is the expression for fugacityof a van der Waals gas?

Ans: As an approximation, terms in P 2 and higher in the series expansion are omitted.

RTP

RTab-

dPRT1

RTab- dP

PZ-1

Pf ln P

0

P

0

=

=

= ∫∫

RTP

RTab-1...C'PB'P1 Z 2

+=+++=

RTP

RTab-P f

= exp

2005/11/15 53


Example 4.7 Estimating the fugacity of nitrogen gas at 50 bar and 298 K.Given that the van der Waals constants of nitrogen are a = 1.408 L2 bar mol-2 and b = 0.03913 L mol-1, estimate the fugacity of nitrogen gas at 50 bar and 298 K.

Ans:

( ) ( )( ) ( )( )bar 48.2

2980.0831550

2980.083151.408-0.03913 bar 50

-

=

=

=

RTP

RTPf abexp

0.964=

==

RTP

RTab-

Pf expΦ

2005/11/15 54

The molar Gibbs energy of a real gas coincides with the perfect gas value as P → 0, When the attractive force dominant, f<p, and the molecules have a lower ‘escaping tendency’.


2005/11/15 55


Fig 4.6 Chemical potential μiof an ideal gas as a function of pressure relative to the chemical potential μºi of the gas at the standard pressure of 1 bar.

°

+=pPRT io

ii lnµµ

2005/11/15 56

The molar Gibbs energy of a perfect gas is proportional to ln p, and the standard state is reached at po . Note that as p-> 0, the molar Gibbs energy becomes negatively infinite.


2005/11/15 57


Example 4.8 Calculating the activity of liquid water at 10 and 100 bar.What is the activity of liquid water at 1, 10 and 100 bar at 25 ℃, assuming that V ̿ is constant?

Ans: for liquid

At P = 1 bar, a = 1

At P = 10 bar,

At P = 100 bar, a = 1.075

( )

( )( )( )( )

1.007K 298mol K bar 0.083145

bar 9mol kg 0.018

1-1-

1-

=

=

°=

exp

expaRTP-PV

( ) ( ) ( ) ( ) ( )°+°=+°= P-PV TRT TT,P µµµ aln( )

°=RTP-PVexpa

2005/11/15 58

4.6 Significance of Chemical Potential

• Gibbs以化學勢能(chemical potential)證明在等溫等壓下，如果一個反應能被用來作功，則該反應是自發的，反之為非自發.對內能 U 來說:

dU = T dS –P dV + μ1dn1 + μ2dn2 + …• μi：對應於第 i 類物種的化學勢能，每單位dn i 莫耳粒子進入系統所引起的內能增加為μidni.

• 在均相混合物中物種 i 的化學勢能μi的函數是系統內能U對物種 i 的變化量偏微分. 內能的自然變數 (natural variables) S, V 與 {ni } 全為系統的外延性質:

• 物種 i 的化學勢能μi的函數可用其他三種方式得到:

{ } { } { } { }ijijijij nPTinVTinPSinS,Vii n

GnA

nH

nU

≠≠≠≠

∂∂=

∂∂=

∂∂=

∂∂=

,,,,,,,

µ

{ } { }∑

=≠

∂∂+

∂∂+

∂∂=

S

ijii

N

1ii

nVSinSnVn

nUV

VUS

SUU dddd

,,,,

2005/11/15 59


Fig 4.5 Two phases (α,β) at the same temperature and pressure. Many species may be present, but we will focus on species i.

( ) ( ) ( ) ( ) ( )[ ]αµβµβµαµ iiiiiiiT, P nnnG -ddd-d =+=

( ) ( ) ( )αµβµ iiT, PG == or 0,dAt equilibrium,

2005/11/15 60


• 部分莫耳自由能(partial molar Gibbs energies, μJ)混合物中J 成分的化學勢能(μJ )是 J 成分改變時的部分莫耳Gibbs自由能.μJ可視為當J 成分改變時與伴隨的G變化量的比值

jinTpjj n

G

≠

∂∂=

,,µ

∑∑ =

∂∂=

≠j

jjjj nTPj

nnnGG

ji

d dd µ,,

BBAA

BBAA

BBAA

nµnµwGnµnµGnµnµTSPVG

d d d dd d d ,dd dd d

maxPV,-non +==+=++−=

在定溫定壓下

2005/11/15 61

• 均相開放系統的馬克斯威關係式與部分莫耳性質:

{ } { }ii nPnT TV

PS

,,

∂∂=

∂∂−

{ } { }i

nP

i

nPTiS

TnS

ijij

−=

∂∂=

∂∂−

≠≠,,,

µ

∑=

++−=SN

1iii nPVTSG dd d d µ

{ } { }i

nT

i

nPTiV

PnV

ijij

=

∂∂=

∂∂

≠≠,,,

µ

{ }ii

nPTiG

nG

ij

==

∂∂

≠

µ,,


: the partial molar entropy of species i.

: the partial molar volume of species i.

: the partial molar Gibbs energy of species i.

iS

iV

iG

2005/11/15 62

The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest. In general, the chemical potential varies with composition, as shown for the two values at a and b. In this case, both chemical potentials are positive.

jinTpjj n

G

≠

∂∂=

,,µ


2005/11/15 63

• 對任一混合在理想氣體的物種的部份莫耳體積,等於混合物的莫耳體積.

{ } { } { } i

i ijijij nT,i

i

nT,

ii

nT,

ii P

µxPP

Pµ

PµV

≠≠≠

∂∂=

∂∂

∂∂=

∂∂=

{ }ijnP

ii T

S≠

∂∂=−

,

µ∑

=

++−=SN

1iii nPVTSG d d d d µ

{ }ijnT

ii P

V≠

∂∂=

,

µ


{ } ii

i

nT,i

i

PRT

xV

Pµ

ij

==

∂∂

≠

∫∫ °°= i

i

i

i

P

Pi

iµ

µ i PPRTµ dd

PRTVV i ==

2005/11/15 64

• 上式Pº表示標準壓力1 bar=105 Pa. 對理想氣體混合物的部分莫耳熵值:

°+°=

PPRT µµ i

ii ln


∫∫ °°= i

i

i

i

P

Pi

iµ

µ i PPRTµ dd

{ }ijnP

ii T

S≠

∂∂=

,

µ- since

°=

PP-R SS iii lno

{ }ijnP

ii T

S≠

∂°∂=°

,

µ- with

2005/11/15 65

• Since i

N

1ii

N

1iii GnnµG

SS

∑∑==

==

{ }∑∑

==

=

∂∂=

≠

SS

ij

N

1iii

N

1ii

nP,

i nS-nTµ-S

( ){ }

∑∑==

=

∂∂=

≠

SS

ij

N

1iii

N

1ii

nP,

i nHnT

Tµ-TH /2

{ }ijnPTii n

ΗΗ≠

∂∂=

,,

{ }∑∑

==

=

∂∂=

≠

SS

ij

N

1iii

N

1ii

nT,

i nVnPµV


The partial molar enthalpy of species i

2005/11/15 66


Example 4.9 Derivations of relations between partial molar propertiesTake the derivatives of G = H – T S and –S = (∂G/∂T )P,{ni} with respect to ni to obtain the corresponding equations for the partial molar properties.

Ans:

{ } { } { }ijijij nP,TinP,TinP,Ti nST

nH

nG

≠≠≠

∂∂−

∂∂=

∂∂

,,,iiii STHG −==µ

{ } { } { }{ }ij

ijij

nP,ijnT,P,inT,P,nP,inT,P,i n

GTT

Gnn

S

≠

≠≠

∂∂

∂∂=

∂∂

∂∂=

∂∂−

≠

{ } { }ijijnP

i

nP

ii TT

GS≠≠

∂∂=

∂∂=−

,,

µ

This last equation is actually a Maxwell relation from dG.

Note that whereas V and S for a system are always positive Vi ̿ and Si̿ may be negative.

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