Chap 4 Fundamental Equations of Thermodynamicsmail.scu.edu.tw/~kguo/pchem2/index.files/Lecture10.pdf · 2005/11/04 1 Chap 4 Fundamental Equations of Thermodynamics Table of Contents

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

2005/11/04 4

• 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 .

iG


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.

2005/11/18 67


•理想混合物 (ideal mixture)–熱力學不保證理想氣體的混合物表現得像理想氣體, 必須增加更多的假設.–低壓的真實氣體混合物有可能表現得像理想氣體 . 如同分子間無作用力一般,稱之為理想混合物.

–在裡想氣體混合物中的任ㄧ成分i的分壓為Pi=yiP,其中yi為氣體莫耳分率,而P為總壓力.

–而成分 i 的化學勢能μi可表示為: °+°=

PPyRT µµ i

ii ln

∑∑∑∑ °++°==

= PPnRTynRTµnµnG iiiii

N

1iii

S

lnln

GnPPRTyyRTµyn totiiiitot =

°++°= ∑∑ lnln

2005/11/18 68


• 理想混合物 (ideal mixture)– 在裡想氣體混合物中的熵值S可表示為:

{ }∑∑∑ °

−−°=

∂∂−=

PPnRynRSn

TGS iiiii

nP i

lnln,

SnPPRyyRSyn totiiiitot =

°−−°= ∑∑ lnln

–在裡想氣體混合物中的熱焓值 H與壓力無關, H 可表示為:

( ) [ ] °=°=°=°+°= ∑∑∑ HnHynHnSTnH totiitotiiiii µ

°+°=° iii STH µ where

2005/11/18 69


• 理想混合物 (ideal mixture)– 在理想氣體混合物中的體積 V 可表示為各成分的莫耳量與莫耳體積貢獻的總和:

– 在理想混合物中各成分的莫耳體積都是相等:– 在理想混合物中的每種氣體對其他氣体而言都可視如真空ㄧ樣.

{ }∑∑ ==

∂∂= iii

nTVn

PRTn

PGV

i,

PRTV i =

2005/11/18 70


Example 4.10 Calculation of changes in thermodynamic properties on mixing ideal gases. If consider the mixing of two ideal gases that are initially at the same temperature and pressure but are separated from each other by a partition. Calculate the changes in Gibbs energy, enthalpy,entropy and volume when the partition is withdrawn. Note that final pressure is the same as the initial pressure of each gas.

Ans: The initial values of these quantities are:

( ) °°° ++°+°=

+°+

+°=

PPRTnnnn

PPRTn

PPRTnGinit lnlnln 2122112211 µµµµ

( ) °°° +−°+°=

−°+

−°=

PPRnnSnSn

PPRSn

PPRSnSinit lnlnln 2122112211

°+°= 2211 HnHnHinit

( )P

RTnnP

RTnP

RTnVinit 2121 +=+=

2005/11/18 71


Example 4.10 Calculation of changes in thermodynamic properties on mixing ideal gases. (continue)

Ans: The changes upon pulling out the partition are the difference between initial and final values after mixing. The values of these quantities are:

( ) ( ) 0lnlnlnln 22112211 <+=+=−=∆ yyyyRTnynynRTGGG totinitmix

( ) ( ) 0lnlnlnln 22112211 >+−=+−=−=∆ yyyyRnynynRSSS totinitmix

0=−=∆ initmix HHH

0=−=∆ initmix VVV

2005/11/18 72

• The Gibbs energy of mixing Let two ideal gases in two containers be nA and nB; both at the same T and p.

where µ° is the standard chemical potential for the pure gas at 1 bar.The initial Gibbs energy of the binary system is Ginit = nAμA + nBμB

++

+=

°° PPRTn

PPRTnG BBAAinit lnln oo µµ

°° +=

PPRT lnµµ


2005/11/18 73

After mixing, the partial pressure of the gases are pAand pB, with p1 + p2 = p. The total Gibbs energy change to

The difference of the Gibbs energy Gfinal - Ginit, is the Gibbs energy of mixing △mixG

= ntot RT ( y1 ln y1 + y2 ln y2 ) < 0

++

+=

°° PPRTn

PPRTnG 2

221

11final lnln oo µµ


PPRTn

PPRTnG 2

21

1mix lnln +=∆

2005/11/18 74

Fig 4.7 The Gibbs free energy of mixing ΔmixG of two ideal gases to form an ideal mixture. The total amount of gas is represented by nt .

ΔmixG = RT(n1 ln y1 + n2 ln y2 )

The Gibbs energy of mixing is negative for all compositions and temperatures, so perfect gases mix spontaneously in all proportions.


2005/11/18 75

• Other thermodynamic mixing functionsBecause (∂G/∂T)p,n = -S, for a mixture of ideal gases, the entropy of mixing, ΔmixS is:ΔmixS = - (∂ΔmixG/∂T)p,nA,nB

= -nR (yAln yA+ yBln yB) > 0for a isothermal, isobaric mixing of ideal gases, since no interactions between the molecules, the enthalpy of mixing is zero,ΔmixH = ΔmixG + TΔmixS = 0no entropy of the surrounding been created, the driving force for mixing comes from the increase in entropy of the system.


2005/11/18 76

Fig 4.7 The entropy of mixing ΔmixS for two ideal gases to form an ideal mixture. The total amount of gas is represented by nt.

ΔmixS=-R(n1 ln y1 + n2 ln y2 )

The entropy increases for all compositions and temperatures, so ideal gases mix spontaneously in all proportions.


2005/11/18 77

Example: Calculate Gibbs energy of mixingA container is divided into two equal compartments. One contains 3.0 mol H2 at 25 ℃; the other contains 1.0 mol N2 at 25 ℃. Calculate the Gibbs energy of mixing when the partition is removed. Assume perfect behavior.

Method:We proceed by calculating the initial Gibbs energy from the chemical potentials. To do so, we need the pressure of each gas. Write the pressure of nitrogen as p; then the pressure of hydrogen as a multiple of p can be found from the gas laws. Next, calculate the Gibbs energy for the system when the partition is removed. The volume of each gas doubles, so its partial pressure falls by a factor of 2.


2005/11/18 78

The initial and final states considered in the calculation of the Gibbs energy of mixing of gases at different initial pressures.


2005/11/18 79

Answer: Given that the pressure of nitrogen is p, the pressure of hydrogen is 3p, therefore, the initial Gibbs energy Gi is

After mixing, the pressure of nitrogen is p/2, the pressure of hydrogen is 3p/2, therefore, the Gibbs energy changes to

( )( ) ( )( )pRTµnp3RTµnG2222 NNHHi lnln +++= oo

( ) ( )

++

+= PRTµPRTµG NHf 2

1mol 1.023mol 3.0

22lnln oo


2005/11/18 80

• Answer:Gibbs energy of mixing is the difference between Gf - Gi

( ) ( ) ln2mol 1.0ln2 mol 3.0 RTRT −−=

( ) ( )

+

=p2pmol 1.0

3p23pmol 3.0 lnln RTRT∆Gmix

( )kJ 6.9

ln2mol 4.0−=−= RT


2005/11/18 81

• Comment:In this example, the value of △mixGi is the sum of two contributions: the mixing itself, and the changes in pressure of the two gases to their final pressure, 2p. When 3.0 mol H2 mixes with 1.0 mol N2 at the same pressure, the change of Gibbs energy is

= -5.6 kJ and is independent of the initial common pressure.

( )

( ) ( )

+

=

+=

41mol 1.0

43mol 3.0

BBAA

lnln

lnln

RTRT

ynynRT∆Gmix


2005/11/18 82


• Gibbs-Duhem equation在開放系統中, G = U + PV - TS = ∑μi ni, 或表示為:

U + PV - TS - ∑μi ni = 0當成分改變時雙成分的系統的微量變化可表示為;dU+P dV+V dP-T dS-S dT -nAdμA -nBdμB -μAdnA - μBdnB=0,其中 dU = T dS –P dV –μA dnA –μB dnB

抵銷後 V dP – S dT- nA dμA – nB dμB = 0 在常溫常壓下nA dμA + nB dμB = 0 對ㄧ莫耳的物質,xA dμA + xB dμB = xA dμA + (1-xA) dμB = 0

• ㄧ般而言, Gibbs-Duhem equation 可寫成:

( ) ( ) AA

AA

AB

AB n

nn

nnn µµµµ dddd

-1-1

AA=−=−=或

∑ =j

jjn 0d µ

2005/11/18 83

• Gibbs-Duhem equation

As G = μA nA+μB nB, 當J 成分改變時雙成分的系統 G 的微量變化可為;

dG = μAdnA + μAdnA + nAdμA + nBdμB , but

dG = μAdnA + μAdnA , we havenAdμA + nBdμB = 0 under constant T and p.

In general, Gibbs-Duhem equation as:

AB

AB

jjj

nn

n

µµ

µ

dd

0d

−=

=∑


2005/11/18 84

Example : Using Gibbs-Duhem equation

The experimental partial molar volume of K2SO4(aq) at 298 K are given by the expression:

VK2SO4/(cm3 mol-1) = 32.280 + 18.216 m½

where m is the numerical value of the molality of K2SO4. Using Gibbs-Duhem equation to derive an expression for the partial molar volume of water in the solution. The molar volume of pure water at 298 K is 18.079 cm3 mol-1


2005/11/18 85

Method: Let A denote K2SO4 and B denote H2O, the solvent. Using Gibbs-Duhem equation for partial molar volumes of A & B.nAdVA + nBdVB = 0 under constant T and p. or

Therefore VB can be found by integration:

where VB* is the molar volume of pure B. The first step is to

change the variable dVA to the molality m, and then to integrate the right-hand side between m=0 (pure B) and the molality of interest.

AB

AB V

nnV dd −=

AB

ABB V

nnVV d

m

0∫−= ∗


2005/11/18 86

Answer: With A denote K2SO4 and B denote H2O, VA/(cm3 mol-1) = 32.280 + 18.216 m½

dVA = (½) 18.216 m-½ dmTherefore VB can be found by integration:

however, the molality m is related to the amounts of A and B by the expression: m = nA /(nB MB)where MB is the molar mass of water (in kg/mol). Therefore

VB/(cm3 mol-1) = 18.079 – 0.1094 b3/2

dm m 9.108 1/2m

0−∗∗ ∫∫ −=−=

B

ABA

B

ABB n

nVdVnnVV

( )3/2BB

m1/2

BBB m9.10832dmm9.108 MVMVV

0−=−= ∫ **


2005/11/18 87

• For Binary mixture:

• 莫耳體積為個別組成的部分莫耳體積乘莫耳分率的總和. • 是否與液體的原體積大小有關?• 在任一組成下的斜率為

Example: If , xi for molar fraction, find V1=?Answer:

22112n2

1n1

dnVnVnnVn

nVdV

12

+=

∂∂+

∂∂= ddd

2211 dxVdxVVd +=2211 xVxVV +=

iV( )12 +++= ii xcbxaV

1xTP1

1 cx2bxVV

2

+=

∂∂=

,,

12

12 dd V

nnV −=( ) 2112 xVxVV -=

2211 nVnVV +=


2005/11/18 88

The partial molar volume of ethanol (V E) as expressed by the polynomial :

V E /mL mol-1 = (∂ V/∂ m )p,T,nw= 54.6664 –2(0.36394)m +3(0.028256)m2


2005/11/18 89

Example At 20 ℃, the density of a 20 percent by mass ethanol-water solution is 968.7 kg m-3. Given that the partial molar volume of ethanol in the solution is 52.2 cm3 mol-1. Calculate the partial molar volume of the water.

Answer: Let A denote water and B ethanol. The total volume of the solution is V = nAVA + nBVB

We know VB; we need to determine nA and nB in order to solve for VA.Assume we have 100 cm3 of solution; then the mass ism = ρV = (0.9687 g cm3) × (100 cm3) = 96.87 g of which (0.20) × (96.87 g) = 19.4 g is ethanol and (0.80) × (96.87 g) = 77.5 g is water.nA =77.496 g/18.02 g mol-1 = 4.30 mol H2OnB =19.374 g/46.07 g mol-1 = 0.4205 mol ethanolVA = (V – nBVB)/nA

= [100 cm3 - (0.4205 mol) × (52.2 cm3 mol-1)]/4.3.0 mol = 18.15 cm3 = 18 cm3


2005/11/18 90

The partial molar volumes of the components of an aqueous solution of potassium sulfate.


2005/11/18 91

Special Topic:

• Additional applications of Maxwell relations:• The fundamental equations for U, H, A and G are:

dU = +T dS - P dV dH = +T dS + V dPdA = - S dT - P dV dG = - S dT + V dP• For a mole of a substance, the Maxwell relations are:

VT SP

VT

∂∂=

∂∂ -

VT TP

VS

∂∂=

∂∂ --

PT TV

PS

∂∂=

∂∂-

PS SV

PT

∂∂=

∂∂

2005/11/18 92

Expansion Coefficient (α) and compressibility (κ)• The expansion coefficient αis a measure of the variation of

the volume on change in temperature at constant pressure.α = (1/V ) (∂V/ ∂T )P

By substituting V=nRT/p into equation, we may obtain α = 1/T ,

• The isothermal compressibility κ is a measure of the variation of the volume on change in pressure at constant temperature.

κ = -(1/V ) (∂V/ ∂P )T• By substituting V=nRT/p into equation, we may obtain κ = 1/P , which shows that it is easier to compress a perfect gas at a low pressure than at a high pressure.


2005/11/18 93

PκαTP

TPTP

VST

VU

VTTT −

=−

∂∂=−

∂∂=

∂∂=π

TTSTVT

PVST

VU

VS

SU

VU π=−

∂∂=

∂∂+

∂∂

∂∂=

∂∂

VVUS

SUU

SVddd

∂∂+

∂∂=

TSVPTTAV

VAA

VTd d ddd −−=

∂∂+

∂∂=

T

P

TVPV

V1

TV

V1

κα

VS

TP

∂∂−

∂∂

==

∂∂=

∂∂

⇒

The variation of internal energy with volume

2005/11/18 94

Example For ideal gas, πT=0, for van der Waals gas πT = a/ V̿ 2

( ) 2 (b)V

--bV

RT P a=

0=−

∂∂=

∂∂=⇒ P

TPT

VU

VTTπ

VnR

TP

V=

∂∂ (a)


( )-bVR

TP

V=

∂∂

PTPT

VU

VTT −

∂∂=

∂∂=⇒π

( ) ( ) ( )

== 2V

--V

RT--V

RT-P-V

RT abbb

2Va=

2005/11/18 95


Example 4.11 Calculation of the change in molar internal energy in the expansion of propane gas, assuming it is a van der Waals gasPropane gas is allowed to expand isothermally from 10 to 30 L. What is the change in molar internal energy?

Ans: The change in internal energy for a given change in volume at constant temperature is given by:

( ) ( ) ( ) 1-621-33-1-52-2 mol m Pa 0.8779 L m 10 bar Pa 10 mol bar L 8.779 ==a

2

1

2

1

2

1

-1 dd 2

V

V

V

V

U

U VV

VU

== ∫∫ aa

=

21

1-1∆VV

U a

According to Table 1.3 a = 8.779 L2 bar mol -2, but we need to convert to SI unit,

( )1-

1-33-1-33-1-6

21

mol J 58.5 mol m 30x10

1-mol m 10x10

1mol m Pa 0.87791-1∆

=

=

=

VVU a

2005/11/18 96


Example 4.12 Calculation of the molar entropy of the isothermal expansion of a van der Waals gas

Ans: Derive the molar entropy of isothermal expansion of a van der Waals gas:

dA = -P dV -S dT results in a Maxwell relation:

bVR

TP

VS

VT −=

∂∂=

∂∂

2Va

bVRTP −−

=

∫∫ −=

2

1

2

1

V

V

S

SV

bV1RS d d

bVbVRS

−−=∆

1

2ln

2005/11/18 97


To deal with derivatives of the volume of fluid:

PP TV

V1

TV

V1

∂∂=

∂∂=αα tcoefficien expansion cubic for

TT PV

VPV

V

∂∂−=

∂∂−= 11κκ ility,compressib isothermal for

κα=

∂∂

∂∂−=

∂∂

TPV PV

TV

TP

κκαπ PTP

TPTP

VST

VU

VTTT

−=−

∂∂=−

∂∂=

∂∂=

VPdSTdU d - as =

2005/11/18 98

κ

κ

2005/11/18 99


TCV πT TU V

VU Uas d VT

VT

ddd d

+=

∂∂+

∂∂=

( ) VTV

PT

P

CVCTV

TU +=+

∂∂=

∂∂ αππ

V

PPVP

VP CTVP

TU

TU

THCC −

∂∂+

∂∂=

∂∂−

∂∂=−

( )[ ] ( ) ( )( )( )

καα

κα

απααπ2VTVT

VPCVPCV TVVT

=

=

+=−++=

2005/11/18 100


( )TVVTVTV

PST

PH

PTT

α−=+

∂∂−=+

∂∂=

∂∂ 1

PVSTH dd das +=

pT TV

PS

∂∂−=

∂∂

PVTSG dd - das +=

καTPP

TPT

VVP

VU

VH

VTTT

=+−

∂∂=

∂∂+

∂∂=

∂∂

TV VS

TP

∂∂=

∂∂

TSVPA d ddas −−=

For ideal gas, V/n = RT/P , (∂V /∂T )P = R/P, so (∂H /∂P )T = -V +V = 0

(∂H /∂V )T = P

2005/11/18 101


( ) ( )( ) ( )

P

PPPT

PP

PVTV

PPPT

C

CCCPVTVC

CTP

PH

TH

TCPCTTHp

PHH

d d ddd

JT

JTJT

JT

−=

+

−=+

∂∂∂∂

−−=

+

∂∂

∂∂=

∂∂

+−=

∂∂+

∂∂=

καµ

καµµ

µ

1

2005/11/18 102


[ ]( )

( )TCV

PH

C1

TCPT1VTCPVTV

TCPTVTVH

PTP

P

P

PP

αµ

αα

−=

∂∂−=

+−=+−=

+

∂∂−=

1-

d d d d

d d d

JT

Also μJT is a state function:

2005/11/18 103

Example

VTTVSV TS

VS

SV

TS

VT

SP

∂∂

∂∂=

∂∂

∂∂=

∂∂−=

∂∂ 1

VVpTVVV TU

US

TV

Vp

TU

US

TP

∂∂

∂∂

∂∂

∂∂−=

∂∂

∂∂

∂∂=

VVTVp CT

TU

pV

SU

TV

κα=

∂∂

∂∂

∂∂

∂∂−=


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Chap 4 Fundamental Equations of Thermodynamicsmail.scu.edu.tw/~kguo/pchem2/index.files/Lecture10.pdf · 2005/11/04 1 Chap 4 Fundamental Equations of Thermodynamics Table of Contents