1版权所有：华东理工大学物理化学教研室
Physical Chemistry
Peter Atkins
(Sixth edition)
Bilingual Program
2版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
5,The Second Law:the machinery
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In this chapter:
First,to find relations between properties that might not
be thought to be related; to derive expressions for
the variation of the G with T and p,
Second,to introduce the chemical potential,a property
that will be at the center of discussions in the remaining
chapters of this part of the text; to derive expression of
fugacity,
The 'chemical potential',the quantity on which almost
all the most important applications of thermodynamics
to chemistry are based.
5,The Second Law,the machinery
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Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
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The fundamental equations
Five state functions in the First and Second Laws,
The internal energy,U
The entropy,S
The enthalpy,H
The Helmholtz energy,A
The Gibbs energy,G
H
UpV
pV A TS
TSGH=U+pV A=U-TS
G= H-TS = U+pV-TS =A+pV
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The First Law,dU =dq+dw H
UpV
pV A TS
TSG
For a reversible change in a
closed system of constant
composition,and in the
absence of any non-expansion
work:
d w = -pdV and dq =TdS
dU=TdS-pdV
dU is an exact differential,its value is independent of
path.
The fundamental equations
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H
UpV
pV A TS
TSG
dH = dU+ pdV + Vdp
=(TdS-pdV)+ pdV + V dp
H=U+pV,A=U-TS,G=U+pV-TS
dA= dU-TdS-SdT
= (TdS -pdV) -TdS- SdT
dG = dU+pdV+ Vdp –TdS- SdT
= (TdS-pdV) +pdV+ Vdp –TdS- SdT
= TdS +V dp
= Vdp -SdT
= -p dV -SdT
The fundamental equations
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dU=TdS -pdV
dH=TdS+Vdp
dG=Vdp-SdT
dA=-pdV-SdT
H
UpV
pV A TS
TSG
The fundamental equations
The fundamental equations
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5.1 Properties of the internal energy
1),The Maxwell relations
dd yhxg
)( y,xfz? y
y
zx
x
zz
xy
d dd
xy y
zh
x
zg
,w h e r e
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1),The Maxwell relations
yx
z
y
g
x
2
xy
z
x
h
y
2
yx x
h
y
g
ddd yhxgz
The first derivative of g with respect to y,and h to x
5.1 Properties of the internal energy
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1),The Maxwell relations
ddd yhxgz
V
V
US
S
UU
SV
d dd?
VpSTU d dd
p
V
UhT
S
Ug
SV
,
VS S
p
V
T?
yx x
h
y
g
5.1 Properties of the internal energy
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1),The Maxwell relations
VpSTU d dd
VS S
p
V
T?
pS S
V
p
T?
TV V
S
T
p?
dH=TdS + V dp
dA= -p dV -SdT
dG= Vd p -SdT
Tp p
S
T
V
Maxwell
relations
yx x
h
y
g?
x
y
yzh
xzg
ddd yhxgz
5.1 Properties of the internal energy
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VpSTU d dd
VS S
p
V
T?
pS S
V
p
T?
TV V
S
T
p?
dH=TdS + V dp
dA= -p dV -SdT
dG= Vd p -SdT
Tp p
S
T
V
pVUTSU
SV
,
VpHTSH
Sp
,
STApVA
VT
,
STAVpG
pT
,
5.1 Properties of the internal energy
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2),The variation of internal energy with volume
T
T V
U?
πThe internal pressure is defined as
Since V
V
US
S
UU
SV
d dd?
If divide both sides of equation by dV with the constraint
of constant T
STVT V
U
V
S
S
U
V
U?
5.1 Properties of the internal energy
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2),The variation of internal energy with volume
STVT V
U
V
S
S
U
V
U?
p
V
UT
S
U
SV
,
pVST
T
TV V
S
T
p?
pTpT
V
Therefore,p
T
pT
V
T
π
5.1 Properties of the internal energy
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Example - Deriving a thermodynamic relation
Show thermodynamically that πT = 0 for a perfect
gas,and compute its value for a van der Waals gas
Method,Proving a result 'thermodynamically' means
basing it entirely on general thermodynamic relations
and equations of state,
Answer,1),For a perfect gas,pV = nRT,and
V/nRTp V
From the equation of p
T
pT
V
T
π
0 pVnR TTπ
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2),The equation of state of a van der Waals gas is
That is
2
2
2
2
V
n
a
V
n
a
nbV
n R T
nbV
n R T
p
T
p
T
V
T
π
2
2
V
na
nbV
nRp?
nbV
nR
T
p
V?
Example 5.1 Deriving a thermodynamic relation
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Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
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TSG d d pVd
For a closed system doing non-expansion work and at
constant composition
),( TpGG?
Tp p
S
T
V
STGVpG
pT
,
1),The outline of the Gibbs energy with T and p
5.2 Properties of the Gibbs energy
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The entropy of the gaseous phase
of a substance is greater than
that of the liquid phase,and the
entropy of the solid phase is
smallest,the Gibbs energy
changes most steeply for the gas
phase,followed by the liquid
p hase,and th en t he soli d
phase of the substance.The variation of the Gibbs energy with the temperature is
determined by the entropy
TSG d d pVd),( TpGG?
1),The outline of the Gibbs energy with T and p
STG
p
5.2 Properties of the Gibbs energy
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The volume of the gaseous phase
of a substance is greater than that
of the same amount of liquid phase,
and the volume of the solid phase is
smallest,the Gibbs energy changes
most steeply for the gas phase,
followed by the liquid phase,and
t h en th e sol id ph ase of th e
substance,The variation of the Gibbs energy with the pressure determined by
the volume of the sample
),( TpGG? d V d pG TSd
1),The outline of the Gibbs energy with T and p
VpG
T
5.2 Properties of the Gibbs energy
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Example - Calculating the effect of pressure on the Gibbs energy
Calculate the change in the molar Gibbs
energy of,
(a) liquid water treated as an incompressible
fluid,and
(b) water vapour treated as a perfect gas,when
the pressure is increased isothermally from
1.0 bar to 2.0 bar at 298 K.
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Method,In each case,the change in molar Gibbs
energy can be obtained by integration of the
equation dG= Vd p –SdT with the temperature held
constant:
pVpGpG pp df
i
mimfm
Example - Calculating the effect of pressure on the Gibbs energy
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For an incompressible fluid,the molar volume is
independent of the pressure so Vm can be treated
as a constant,For the incompressible liquid,Vm
is constant at 18.0 cm3mol-1,then
Answer,
(a)
ifmmmm df
iif
ppVpVpGpG pp
Pa1001m o lm10018 5136,.
113 J m o l8.1m o l m Pa8.1
Example- Calculating the effect of pressure on the Gibbs energy
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For a perfect gas,the molar volume varies with
pressure as Vm = RT/p,so this expression must
be used in the integrand,and the integration
performed treating RT as a constant:
Answer,
(b)
i
ff
i
if
lndmm
p
p
RTp
p
RTpGpG p
p
1
1
k Jm ol7.1
0.2lnk Jm ol48.2
Example - Calculating the effect of pressure on the Gibbs energy
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(a) For an incompressible
1mm J m o l 1,8if pGpG
(b) For a perfect gas
1mm k J m o l 1,7if pGpG
Example - Calculating the effect of pressure on the Gibbs energy
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2),The temperature dependence of the Gibbs energy
TV
G?
SV
Up?
VS
UT?
pS
H?
pT
G?
VT
G?
S
Sp
H
V
Tp
G
STG
p
TGHS /)(
G=H-TS
T
HG
T
G
p
T
H
T
G
T
G
p
T
G
T
G
T
T
G
T
G
T
TT
G
T
G
TT
G
T
p
p
pp
1
d
d
2
1
11
5.2 Properties of the Gibbs energy
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2),The temperature dependence of the Gibbs energy
TV
A?
SV
Up?
VS
UT?
pS
H?
pT
G?
VT
A?
S
Sp
H
V
Tp
G
2T
H
T
G
T p
the Gibbs-Helmholtz (G-H) equation
2T
H
T
G
T p
5.2 Properties of the Gibbs energy
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2),The temperature dependence of the Gibbs energy
TV
A?
SV
Up?
VS
UT?
pS
H?
pT
G?
VT
A?
S
Sp
H
V
Tp
G
2
2
1
1
1
1
1
)(
T
H
TT
TG
T
T
T
TG
T
TG
p
pp
d
d
pT
TGH
1
5.2 Properties of the Gibbs energy
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3),The pressure dependence of the Gibbs energy
At constant temperature,)TSpVG d d(d
fiif dpp pVpGpG )()(
For a liquid or solid,the volume changes only slightly as
the pressure changes,so V may be treated as a constant,
For molar quantities:
pVpG
ppVpGpG
mim
ifmimfm
At normal lab,conditions VmΔp is small,so it suggests
that the G of solid and liquid are independent of p.
5.2 Properties of the Gibbs energy
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3),The pressure dependence of the Gibbs energy
For a perfect gas,the Gibbs energy may depend strongly
on the pressure; and the volume also varies markedly
with the pressure,Then V = nRT/p:
f
i
d
i
p
p p
pn R TpG
fiif dpp pVpGpG )()(
i
f
i p
pn R TpG ln
5.2 Properties of the Gibbs energy
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3),The pressure dependence of the Gibbs energy
i
f
if p
pn R TpGpG ln
p
p
n R TGpG ln
If set pi=,then the Gibbs energy of a perfect gas at a
pressure p is related to its standard value by
p
5.2 Properties of the Gibbs energy
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Calculate the change in the molar Gibbs
energy of hydrogen when its pressure is
increased isothermally from 1.0 atm to
100.0 atm at 298K,
Example
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Answer:
1.00mol hydrogen (id.g)
298K,1.0atm,V1
1.00mol hydrogen (id.g)
298K,100.0atm,V2
△ G
i
f
if p
pn R TpGpG ln
)pp(RTG
i
f
m ln
1
11
kJ m o l 1,4 11
)
1,0 0 a t m
1 0 0,0 0 a t m
l n (K)298()m o lJK3148(
,
Example
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Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
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5.3 The chemical potential of a pure substance
V,U,H,S,A and G(general X) are extensive properties,X
depends on not only T and p,but also n.
1) Partial molar quantities
For a single phase system of a pure substance,Xm is certain
when T and p are specific
X = nXm
For a mixture of pure substances at constant T and p
X = n1 Xm,1+ n2 Xm,2 + ···?
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For perfect gases,ΔV=ΔU =ΔH =0; ΔS? 0,norΔA orΔG.
For real gases,the change in extensive properties in a mixture
is generally ΔX? 0.
Substance 1 Substance 2
n1 n2
T,p T,p
V1= n1 Vm,1 V2= n2 Vm,2
U1= n1 Um,1 U2= n2 Um,2
H1= n1 Hm,1 H2= n2 Hm,2
S1= n1 Sm,1 S2= n2 Sm,2
A1= n1 Am,1 A2= n2 Am,2
G1= n1 Gm,1 G2= n2Gm,2
Mixture
n2+ n1
T,p
V
U
H
S
A
G
5.3 The chemical potential of a pure substance
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Generally,X = f (T,p,n1,n2 ··· ),and
1) Partial molar quantities
1
1 322121
nnXppXTTXX
n,n,p,Tn,n,Tn,n,p
dddd
the partial molar quantity
jn,p,Ti
n
X
Note that,1) X,extensive properties
2) at constant T and p
5.3 The chemical potential of a pure substance
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1) Partial molar quantities
jn,p,T
i
i n
VV
jn,p,T
i
i n
UU
jn,p,T
i
i n
SS
jn,p,T
i
i n
GG
The chemical potential of
component i in the mixture.
5.3 The chemical potential of a pure substance
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The chemical potential,?,of a pure substance is defined as
For a pure substance,the Gibbs energy is G = nGm
pTn
Gμ
,
m
m G
n
nGμ
pT
,
The chemical potential,?,is the same as the molar
Gibbs energy Gm.
2) The definition of chemical potential of a pure substance
5.3 The chemical potential of a pure substance
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m
m V
p
G
T
mGμ?pVG dd mm?
pV dd m
3) The chemical potential
5.3 The chemical potential of a pure substance
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For a perfect gas:
p
RTV?
m
p
pRT dd
Let denotes the standard chemical potential,the
molar Gibbs energy of the pure gas at (1 bar)
p
p
p
RT ln
pV dd m
5.3 The chemical potential of a pure substance
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The variation of chemical
potential of a perfect gas
with p
p
p
RT ln
p
The chemical potential,?,of a
perfect gas is proportional to
1n p,and the standard state is
reached at,Note that,as
p? 0,? becomes negatively
infinite.
5.3 The chemical potential of a pure substance
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Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
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Real gases,the fugacity
At various stages in the development of physical
chemistry it is necessary to switch from considera-
tion of idealized systems to real systems,In many
cases it is desirable to preserve the form of the
expressions that have been derived for the idealized
system,Then deviations from the idealized behavior
can be expressed most simply,We shall illustrate
such procedure in this section by considering how
the expressions that have been derived for perfect
gases,particularly the equation for the chemical
potential of a perfect gas,are adapted to describe
real gases.
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5.4 The definition of fugacity
For a real gas,replace the true pressure,p,by an effective
pressure,f,then the fugacity of a real gas is defined as,
The name 'fugacity' comes from the Latin for 'fleetness'
in the sense of 'escaping tendency ',Fugacity has the
same dimension as pressure,
p
f
RT ln
p
p
RT ln
The chemical potential of a perfect gas:
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at higher p,
the repulsive forces are dominant and
the `escaping tendency' is increased.
at lower p,
the attractive forces are dominant and
the molecules have a lower `escaping
tendency',
as p?0,? coincides with the value for
a perfect gas,
p e r f e c tr e a l
p e r f e c tr e a l
The? of a real gas
For the chemical potential of a real gas:
5.4 The definition of fugacity
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A perfect gas is in its standard state when its pressure
is (1 bar),the pressure arises solely from the kinetic
energy of the molecules and there are no intermo-
lecular forces to take into account,
p
The standard state of a real gas is a hypothetical state in
which the gas is at a pressure and behaving perfectly.?p
p
f
RT lnpV dd m
5.5 Standard states of real gases
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p
fφ?
The fugacity coefficient,?,is defined as:
pφf?o r
The fugacity coefficient,,
1),Dimensionless;
2),depends on the identity of the gas,the pressure,
and the temperature.
φ
p
pφ
RT ln?
p
f
RT ln
5.6 The relation between fugacity and pressure
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φRT
p
pRTμμ lnln?
In terms of fugacity coefficient:
The terms of the and are the same as those in
the above equation,Therefore,the term must
express the entire effect of all intermolecular forces.
φRT ln
μ
p
pln
For all gases,f = p as p? 0,1 lim
0 p
f
p
For a prefect gas,f = p and? =1; for a real gas,f? p
and 1,? is a measure of a real gas derived from
the perfect gas,
5.6 The relation between fugacity and pressure
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The equation is true for all gases,Let f denotes the
fugacity when the pressure is p and f' the fugacity when the
pressure is p',from the definition of fugacity we have:
pVμ dd m?
p
fRTμpVμ f
p m
lnd
p
fRTμpVμ f
p m
'ln' ' d
5.6 The relation between fugacity and pressure
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If the gas were perfect:
'ln'' f
fRTμμpVp
p m
d
'
ln'
' p
pRTμμpVp
p p e r f e c tp e r f e c tm p e r f e c t,
d
The difference of the two equations is:
''' p
p
f
fRTpVVp
p
nllndmp e r f e c t,m
5.6 The relation between fugacity and pressure
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p
p
pVV
RTf
p
p
f
'
1
'
'ln d
mp e r f e c t,m
When p'?0,the gas behaves perfectly and f 'becomes
equal to the pressure p',Therefore,p'/f '?1 as p'?0,
If we take this limit,this equation becomes:
p pVV
RTp
f
0
1ln d
mp e r f e c t,m
With? = f / p
5.6 The relation between fugacity and pressure
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At a general pressure p,the fugacity coefficient of a
gas is given by:
p
p
Zφ p d?
0
1ln
where Z is the compression factor of the gas,This
equation is an explicit expression for the fugacity
coefficient at any pressure p,The fugacity of the
gas at that pressure can be obtained by,φ pf?
φRT
p
p
RTμμ lnln?
5.6 The relation between fugacity and pressure
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Real gases,the fugacity
ppZφ p d 0 1ln
The repulsive interactions are dominant and tend to
drive the particles apart,and the chemical potential of
the gas is greater than that of a perfect gas
The molecules tend to stick together and the chemical
potential of the gas is less than that of a perfect gas;
If Z> 1 throughout the rage of integration(higher pressure):
If Z < 1 throughout the rage of integration：
The integrand < 0,? < 1,f < p,
The integrand > 0,? > 1,f > p,
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ppZφ p d 0 1ln
The chemical potential of a real gas and the fugacity,f:
p
fRT ln
The fugacity coefficient,? (f,p),? (Z,p):
pfφ? φpf?o r
The chemical potential of a perfect gas,
p
pRT ln
pVG ddd mm
The expression for chemical potential:
lnln RT
p
pRT?
Real gases,the fugacity
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Example - Calculating a fugacity
Suppose that the attractive interactions
between gas particles can be neglected,
and find an expression for the fugacity
of a van der Waals gas in terms of the
pressure,Estimate its value for
ammonia at 10.00 atm and 298.15 K.
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Answer,the van der Waals equation is as followings:
2
mm V
a
bV
RTp?
bV
RTp
m
and
RT
bpZ 1
bpRTpVm
hence
RT
pVZ m?
Method,The starting point for the calculation is equation
ppZφ p d
0
1ln
Example - Calculating a fugacity
版权所有：华东理工大学物理化学教研室 59
b p / R Te
ppZp d1ln
0
= f / p
pf
a t m210a t m0010 015150,e.f,
RTbpp e?
RT
bpp
RT
bp
d
0
For ammonia at 10.00 atm and 298.15 K
Example - Calculating a fugacity
版权所有：华东理工大学物理化学教研室 60
The fugacity coefficient of a certain gas at 290
K and 2.1MPa is 0.68,Calculate the difference
of its chemical potential from that of a perfect
gas in the same state,
pfφ?
Method:
p
pRTμμ lnid
p
fRTμμ ln
φRTpfRT lnlnid
Example
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Answer:
φRT lnid
1m o lJ 9 3 0
0,6 8ln2 9 0 Km o l8,3 1 4 J K 11
Example
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Consider a system consisting of 1.5 mol CO2 (g),
initially at 15℃ and 9.0 atm and confined to a
cylinder of cross-section 100.0 cm2,The sample is
allowed to expand adiabatically against an external
pressure of 1.5 atm until the piston has moved
outwards through 15 cm,Assume that carbon
dioxide may be considered a perfect gas with
CV,m=28.8 J K-1 mol-1,and calculate (a) q,(b) w,(c)
△ U,(d) △ T,and (e) △ S,
Example
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(a),q:
(b),w,Vpw
ex
36 325 cm10 1m1 5 c m1 0 0 c ma t m Pa101,0 11,5 a t mw
Method and answer:
0?q(adiabatic)
J 2 2 7,2
(c),△ U,wqUΔ J2227,
Example
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(d),△ T,TnCU
m,V
m
ΔΔ
,VnC
UT?
V
V
CTU
11 m o l2 8,8 J K1,5 m o l
2 2 7,2 J
K 5,2 8
Example
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i
i
i p
nR TV?
L 5,4 41 0 0 0 c m1L1 5 c m1 0 0 c m3,9 4 2 L 32
-1KJ 23,S
TTT Δif 2 8 2,9 K5,2 8 K2 8 8,1 5 K
3,9 4 2 L
a t m 9,0
2 8 8,2 Km o lL a t m K108,2 0 61,5 a t m 112
VVV Δif
Example
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
Physical Chemistry
Peter Atkins
(Sixth edition)
Bilingual Program
2版权所有：华东理工大学物理化学教研室
Part 1,Equilibrium
Bilingual Program
5,The Second Law:the machinery
版权所有：华东理工大学物理化学教研室 3
In this chapter:
First,to find relations between properties that might not
be thought to be related; to derive expressions for
the variation of the G with T and p,
Second,to introduce the chemical potential,a property
that will be at the center of discussions in the remaining
chapters of this part of the text; to derive expression of
fugacity,
The 'chemical potential',the quantity on which almost
all the most important applications of thermodynamics
to chemistry are based.
5,The Second Law,the machinery
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Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
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The fundamental equations
Five state functions in the First and Second Laws,
The internal energy,U
The entropy,S
The enthalpy,H
The Helmholtz energy,A
The Gibbs energy,G
H
UpV
pV A TS
TSGH=U+pV A=U-TS
G= H-TS = U+pV-TS =A+pV
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The First Law,dU =dq+dw H
UpV
pV A TS
TSG
For a reversible change in a
closed system of constant
composition,and in the
absence of any non-expansion
work:
d w = -pdV and dq =TdS
dU=TdS-pdV
dU is an exact differential,its value is independent of
path.
The fundamental equations
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H
UpV
pV A TS
TSG
dH = dU+ pdV + Vdp
=(TdS-pdV)+ pdV + V dp
H=U+pV,A=U-TS,G=U+pV-TS
dA= dU-TdS-SdT
= (TdS -pdV) -TdS- SdT
dG = dU+pdV+ Vdp –TdS- SdT
= (TdS-pdV) +pdV+ Vdp –TdS- SdT
= TdS +V dp
= Vdp -SdT
= -p dV -SdT
The fundamental equations
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dU=TdS -pdV
dH=TdS+Vdp
dG=Vdp-SdT
dA=-pdV-SdT
H
UpV
pV A TS
TSG
The fundamental equations
The fundamental equations
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5.1 Properties of the internal energy
1),The Maxwell relations
dd yhxg
)( y,xfz? y
y
zx
x
zz
xy
d dd
xy y
zh
x
zg
,w h e r e
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1),The Maxwell relations
yx
z
y
g
x
2
xy
z
x
h
y
2
yx x
h
y
g
ddd yhxgz
The first derivative of g with respect to y,and h to x
5.1 Properties of the internal energy
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1),The Maxwell relations
ddd yhxgz
V
V
US
S
UU
SV
d dd?
VpSTU d dd
p
V
UhT
S
Ug
SV
,
VS S
p
V
T?
yx x
h
y
g
5.1 Properties of the internal energy
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1),The Maxwell relations
VpSTU d dd
VS S
p
V
T?
pS S
V
p
T?
TV V
S
T
p?
dH=TdS + V dp
dA= -p dV -SdT
dG= Vd p -SdT
Tp p
S
T
V
Maxwell
relations
yx x
h
y
g?
x
y
yzh
xzg
ddd yhxgz
5.1 Properties of the internal energy
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VpSTU d dd
VS S
p
V
T?
pS S
V
p
T?
TV V
S
T
p?
dH=TdS + V dp
dA= -p dV -SdT
dG= Vd p -SdT
Tp p
S
T
V
pVUTSU
SV
,
VpHTSH
Sp
,
STApVA
VT
,
STAVpG
pT
,
5.1 Properties of the internal energy
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2),The variation of internal energy with volume
T
T V
U?
πThe internal pressure is defined as
Since V
V
US
S
UU
SV
d dd?
If divide both sides of equation by dV with the constraint
of constant T
STVT V
U
V
S
S
U
V
U?
5.1 Properties of the internal energy
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2),The variation of internal energy with volume
STVT V
U
V
S
S
U
V
U?
p
V
UT
S
U
SV
,
pVST
T
TV V
S
T
p?
pTpT
V
Therefore,p
T
pT
V
T
π
5.1 Properties of the internal energy
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Example - Deriving a thermodynamic relation
Show thermodynamically that πT = 0 for a perfect
gas,and compute its value for a van der Waals gas
Method,Proving a result 'thermodynamically' means
basing it entirely on general thermodynamic relations
and equations of state,
Answer,1),For a perfect gas,pV = nRT,and
V/nRTp V
From the equation of p
T
pT
V
T
π
0 pVnR TTπ
版权所有：华东理工大学物理化学教研室 17
2),The equation of state of a van der Waals gas is
That is
2
2
2
2
V
n
a
V
n
a
nbV
n R T
nbV
n R T
p
T
p
T
V
T
π
2
2
V
na
nbV
nRp?
nbV
nR
T
p
V?
Example 5.1 Deriving a thermodynamic relation
版权所有：华东理工大学物理化学教研室 18
Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
版权所有：华东理工大学物理化学教研室 19
TSG d d pVd
For a closed system doing non-expansion work and at
constant composition
),( TpGG?
Tp p
S
T
V
STGVpG
pT
,
1),The outline of the Gibbs energy with T and p
5.2 Properties of the Gibbs energy
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The entropy of the gaseous phase
of a substance is greater than
that of the liquid phase,and the
entropy of the solid phase is
smallest,the Gibbs energy
changes most steeply for the gas
phase,followed by the liquid
p hase,and th en t he soli d
phase of the substance.The variation of the Gibbs energy with the temperature is
determined by the entropy
TSG d d pVd),( TpGG?
1),The outline of the Gibbs energy with T and p
STG
p
5.2 Properties of the Gibbs energy
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The volume of the gaseous phase
of a substance is greater than that
of the same amount of liquid phase,
and the volume of the solid phase is
smallest,the Gibbs energy changes
most steeply for the gas phase,
followed by the liquid phase,and
t h en th e sol id ph ase of th e
substance,The variation of the Gibbs energy with the pressure determined by
the volume of the sample
),( TpGG? d V d pG TSd
1),The outline of the Gibbs energy with T and p
VpG
T
5.2 Properties of the Gibbs energy
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Example - Calculating the effect of pressure on the Gibbs energy
Calculate the change in the molar Gibbs
energy of,
(a) liquid water treated as an incompressible
fluid,and
(b) water vapour treated as a perfect gas,when
the pressure is increased isothermally from
1.0 bar to 2.0 bar at 298 K.
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Method,In each case,the change in molar Gibbs
energy can be obtained by integration of the
equation dG= Vd p –SdT with the temperature held
constant:
pVpGpG pp df
i
mimfm
Example - Calculating the effect of pressure on the Gibbs energy
版权所有：华东理工大学物理化学教研室 24
For an incompressible fluid,the molar volume is
independent of the pressure so Vm can be treated
as a constant,For the incompressible liquid,Vm
is constant at 18.0 cm3mol-1,then
Answer,
(a)
ifmmmm df
iif
ppVpVpGpG pp
Pa1001m o lm10018 5136,.
113 J m o l8.1m o l m Pa8.1
Example- Calculating the effect of pressure on the Gibbs energy
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For a perfect gas,the molar volume varies with
pressure as Vm = RT/p,so this expression must
be used in the integrand,and the integration
performed treating RT as a constant:
Answer,
(b)
i
ff
i
if
lndmm
p
p
RTp
p
RTpGpG p
p
1
1
k Jm ol7.1
0.2lnk Jm ol48.2
Example - Calculating the effect of pressure on the Gibbs energy
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(a) For an incompressible
1mm J m o l 1,8if pGpG
(b) For a perfect gas
1mm k J m o l 1,7if pGpG
Example - Calculating the effect of pressure on the Gibbs energy
版权所有：华东理工大学物理化学教研室 27
2),The temperature dependence of the Gibbs energy
TV
G?
SV
Up?
VS
UT?
pS
H?
pT
G?
VT
G?
S
Sp
H
V
Tp
G
STG
p
TGHS /)(
G=H-TS
T
HG
T
G
p
T
H
T
G
T
G
p
T
G
T
G
T
T
G
T
G
T
TT
G
T
G
TT
G
T
p
p
pp
1
d
d
2
1
11
5.2 Properties of the Gibbs energy
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2),The temperature dependence of the Gibbs energy
TV
A?
SV
Up?
VS
UT?
pS
H?
pT
G?
VT
A?
S
Sp
H
V
Tp
G
2T
H
T
G
T p
the Gibbs-Helmholtz (G-H) equation
2T
H
T
G
T p
5.2 Properties of the Gibbs energy
版权所有：华东理工大学物理化学教研室 29
2),The temperature dependence of the Gibbs energy
TV
A?
SV
Up?
VS
UT?
pS
H?
pT
G?
VT
A?
S
Sp
H
V
Tp
G
2
2
1
1
1
1
1
)(
T
H
TT
TG
T
T
T
TG
T
TG
p
pp
d
d
pT
TGH
1
5.2 Properties of the Gibbs energy
版权所有：华东理工大学物理化学教研室 30
3),The pressure dependence of the Gibbs energy
At constant temperature,)TSpVG d d(d
fiif dpp pVpGpG )()(
For a liquid or solid,the volume changes only slightly as
the pressure changes,so V may be treated as a constant,
For molar quantities:
pVpG
ppVpGpG
mim
ifmimfm
At normal lab,conditions VmΔp is small,so it suggests
that the G of solid and liquid are independent of p.
5.2 Properties of the Gibbs energy
版权所有：华东理工大学物理化学教研室 31
3),The pressure dependence of the Gibbs energy
For a perfect gas,the Gibbs energy may depend strongly
on the pressure; and the volume also varies markedly
with the pressure,Then V = nRT/p:
f
i
d
i
p
p p
pn R TpG
fiif dpp pVpGpG )()(
i
f
i p
pn R TpG ln
5.2 Properties of the Gibbs energy
版权所有：华东理工大学物理化学教研室 32
3),The pressure dependence of the Gibbs energy
i
f
if p
pn R TpGpG ln
p
p
n R TGpG ln
If set pi=,then the Gibbs energy of a perfect gas at a
pressure p is related to its standard value by
p
5.2 Properties of the Gibbs energy
版权所有：华东理工大学物理化学教研室 33
Calculate the change in the molar Gibbs
energy of hydrogen when its pressure is
increased isothermally from 1.0 atm to
100.0 atm at 298K,
Example
版权所有：华东理工大学物理化学教研室 34
Answer:
1.00mol hydrogen (id.g)
298K,1.0atm,V1
1.00mol hydrogen (id.g)
298K,100.0atm,V2
△ G
i
f
if p
pn R TpGpG ln
)pp(RTG
i
f
m ln
1
11
kJ m o l 1,4 11
)
1,0 0 a t m
1 0 0,0 0 a t m
l n (K)298()m o lJK3148(
,
Example
版权所有：华东理工大学物理化学教研室 35
Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
版权所有：华东理工大学物理化学教研室 36
5.3 The chemical potential of a pure substance
V,U,H,S,A and G(general X) are extensive properties,X
depends on not only T and p,but also n.
1) Partial molar quantities
For a single phase system of a pure substance,Xm is certain
when T and p are specific
X = nXm
For a mixture of pure substances at constant T and p
X = n1 Xm,1+ n2 Xm,2 + ···?
版权所有：华东理工大学物理化学教研室 37
For perfect gases,ΔV=ΔU =ΔH =0; ΔS? 0,norΔA orΔG.
For real gases,the change in extensive properties in a mixture
is generally ΔX? 0.
Substance 1 Substance 2
n1 n2
T,p T,p
V1= n1 Vm,1 V2= n2 Vm,2
U1= n1 Um,1 U2= n2 Um,2
H1= n1 Hm,1 H2= n2 Hm,2
S1= n1 Sm,1 S2= n2 Sm,2
A1= n1 Am,1 A2= n2 Am,2
G1= n1 Gm,1 G2= n2Gm,2
Mixture
n2+ n1
T,p
V
U
H
S
A
G
5.3 The chemical potential of a pure substance
版权所有：华东理工大学物理化学教研室 38
Generally,X = f (T,p,n1,n2 ··· ),and
1) Partial molar quantities
1
1 322121
nnXppXTTXX
n,n,p,Tn,n,Tn,n,p
dddd
the partial molar quantity
jn,p,Ti
n
X
Note that,1) X,extensive properties
2) at constant T and p
5.3 The chemical potential of a pure substance
版权所有：华东理工大学物理化学教研室 39
1) Partial molar quantities
jn,p,T
i
i n
VV
jn,p,T
i
i n
UU
jn,p,T
i
i n
SS
jn,p,T
i
i n
GG
The chemical potential of
component i in the mixture.
5.3 The chemical potential of a pure substance
版权所有：华东理工大学物理化学教研室 40
The chemical potential,?,of a pure substance is defined as
For a pure substance,the Gibbs energy is G = nGm
pTn
Gμ
,
m
m G
n
nGμ
pT
,
The chemical potential,?,is the same as the molar
Gibbs energy Gm.
2) The definition of chemical potential of a pure substance
5.3 The chemical potential of a pure substance
版权所有：华东理工大学物理化学教研室 41
m
m V
p
G
T
mGμ?pVG dd mm?
pV dd m
3) The chemical potential
5.3 The chemical potential of a pure substance
版权所有：华东理工大学物理化学教研室 42
For a perfect gas:
p
RTV?
m
p
pRT dd
Let denotes the standard chemical potential,the
molar Gibbs energy of the pure gas at (1 bar)
p
p
p
RT ln
pV dd m
5.3 The chemical potential of a pure substance
版权所有：华东理工大学物理化学教研室 43
The variation of chemical
potential of a perfect gas
with p
p
p
RT ln
p
The chemical potential,?,of a
perfect gas is proportional to
1n p,and the standard state is
reached at,Note that,as
p? 0,? becomes negatively
infinite.
5.3 The chemical potential of a pure substance
版权所有：华东理工大学物理化学教研室 44
Combing the First and Second Laws
5.1 Properties of the internal energy
5.2 Properties of the Gibbs energy
5.3 The chemical potential of a pure substance
Real gases:the fugacity
5.4 The definition of fugacity
5.5 Standard states of real gases
5.6 The relation between fugacity and pressure
5,The Second Law,the machinery
版权所有：华东理工大学物理化学教研室 45
Real gases,the fugacity
At various stages in the development of physical
chemistry it is necessary to switch from considera-
tion of idealized systems to real systems,In many
cases it is desirable to preserve the form of the
expressions that have been derived for the idealized
system,Then deviations from the idealized behavior
can be expressed most simply,We shall illustrate
such procedure in this section by considering how
the expressions that have been derived for perfect
gases,particularly the equation for the chemical
potential of a perfect gas,are adapted to describe
real gases.
版权所有：华东理工大学物理化学教研室 46
5.4 The definition of fugacity
For a real gas,replace the true pressure,p,by an effective
pressure,f,then the fugacity of a real gas is defined as,
The name 'fugacity' comes from the Latin for 'fleetness'
in the sense of 'escaping tendency ',Fugacity has the
same dimension as pressure,
p
f
RT ln
p
p
RT ln
The chemical potential of a perfect gas:
版权所有：华东理工大学物理化学教研室 47
at higher p,
the repulsive forces are dominant and
the `escaping tendency' is increased.
at lower p,
the attractive forces are dominant and
the molecules have a lower `escaping
tendency',
as p?0,? coincides with the value for
a perfect gas,
p e r f e c tr e a l
p e r f e c tr e a l
The? of a real gas
For the chemical potential of a real gas:
5.4 The definition of fugacity
版权所有：华东理工大学物理化学教研室 48
A perfect gas is in its standard state when its pressure
is (1 bar),the pressure arises solely from the kinetic
energy of the molecules and there are no intermo-
lecular forces to take into account,
p
The standard state of a real gas is a hypothetical state in
which the gas is at a pressure and behaving perfectly.?p
p
f
RT lnpV dd m
5.5 Standard states of real gases
版权所有：华东理工大学物理化学教研室 49
p
fφ?
The fugacity coefficient,?,is defined as:
pφf?o r
The fugacity coefficient,,
1),Dimensionless;
2),depends on the identity of the gas,the pressure,
and the temperature.
φ
p
pφ
RT ln?
p
f
RT ln
5.6 The relation between fugacity and pressure
版权所有：华东理工大学物理化学教研室 50
φRT
p
pRTμμ lnln?
In terms of fugacity coefficient:
The terms of the and are the same as those in
the above equation,Therefore,the term must
express the entire effect of all intermolecular forces.
φRT ln
μ
p
pln
For all gases,f = p as p? 0,1 lim
0 p
f
p
For a prefect gas,f = p and? =1; for a real gas,f? p
and 1,? is a measure of a real gas derived from
the perfect gas,
5.6 The relation between fugacity and pressure
版权所有：华东理工大学物理化学教研室 51
The equation is true for all gases,Let f denotes the
fugacity when the pressure is p and f' the fugacity when the
pressure is p',from the definition of fugacity we have:
pVμ dd m?
p
fRTμpVμ f
p m
lnd
p
fRTμpVμ f
p m
'ln' ' d
5.6 The relation between fugacity and pressure
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If the gas were perfect:
'ln'' f
fRTμμpVp
p m
d
'
ln'
' p
pRTμμpVp
p p e r f e c tp e r f e c tm p e r f e c t,
d
The difference of the two equations is:
''' p
p
f
fRTpVVp
p
nllndmp e r f e c t,m
5.6 The relation between fugacity and pressure
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p
p
pVV
RTf
p
p
f
'
1
'
'ln d
mp e r f e c t,m
When p'?0,the gas behaves perfectly and f 'becomes
equal to the pressure p',Therefore,p'/f '?1 as p'?0,
If we take this limit,this equation becomes:
p pVV
RTp
f
0
1ln d
mp e r f e c t,m
With? = f / p
5.6 The relation between fugacity and pressure
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At a general pressure p,the fugacity coefficient of a
gas is given by:
p
p
Zφ p d?
0
1ln
where Z is the compression factor of the gas,This
equation is an explicit expression for the fugacity
coefficient at any pressure p,The fugacity of the
gas at that pressure can be obtained by,φ pf?
φRT
p
p
RTμμ lnln?
5.6 The relation between fugacity and pressure
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Real gases,the fugacity
ppZφ p d 0 1ln
The repulsive interactions are dominant and tend to
drive the particles apart,and the chemical potential of
the gas is greater than that of a perfect gas
The molecules tend to stick together and the chemical
potential of the gas is less than that of a perfect gas;
If Z> 1 throughout the rage of integration(higher pressure):
If Z < 1 throughout the rage of integration：
The integrand < 0,? < 1,f < p,
The integrand > 0,? > 1,f > p,
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ppZφ p d 0 1ln
The chemical potential of a real gas and the fugacity,f:
p
fRT ln
The fugacity coefficient,? (f,p),? (Z,p):
pfφ? φpf?o r
The chemical potential of a perfect gas,
p
pRT ln
pVG ddd mm
The expression for chemical potential:
lnln RT
p
pRT?
Real gases,the fugacity
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Example - Calculating a fugacity
Suppose that the attractive interactions
between gas particles can be neglected,
and find an expression for the fugacity
of a van der Waals gas in terms of the
pressure,Estimate its value for
ammonia at 10.00 atm and 298.15 K.
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Answer,the van der Waals equation is as followings:
2
mm V
a
bV
RTp?
bV
RTp
m
and
RT
bpZ 1
bpRTpVm
hence
RT
pVZ m?
Method,The starting point for the calculation is equation
ppZφ p d
0
1ln
Example - Calculating a fugacity
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b p / R Te
ppZp d1ln
0
= f / p
pf
a t m210a t m0010 015150,e.f,
RTbpp e?
RT
bpp
RT
bp
d
0
For ammonia at 10.00 atm and 298.15 K
Example - Calculating a fugacity
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The fugacity coefficient of a certain gas at 290
K and 2.1MPa is 0.68,Calculate the difference
of its chemical potential from that of a perfect
gas in the same state,
pfφ?
Method:
p
pRTμμ lnid
p
fRTμμ ln
φRTpfRT lnlnid
Example
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Answer:
φRT lnid
1m o lJ 9 3 0
0,6 8ln2 9 0 Km o l8,3 1 4 J K 11
Example
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Consider a system consisting of 1.5 mol CO2 (g),
initially at 15℃ and 9.0 atm and confined to a
cylinder of cross-section 100.0 cm2,The sample is
allowed to expand adiabatically against an external
pressure of 1.5 atm until the piston has moved
outwards through 15 cm,Assume that carbon
dioxide may be considered a perfect gas with
CV,m=28.8 J K-1 mol-1,and calculate (a) q,(b) w,(c)
△ U,(d) △ T,and (e) △ S,
Example
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(a),q:
(b),w,Vpw
ex
36 325 cm10 1m1 5 c m1 0 0 c ma t m Pa101,0 11,5 a t mw
Method and answer:
0?q(adiabatic)
J 2 2 7,2
(c),△ U,wqUΔ J2227,
Example
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(d),△ T,TnCU
m,V
m
ΔΔ
,VnC
UT?
V
V
CTU
11 m o l2 8,8 J K1,5 m o l
2 2 7,2 J
K 5,2 8
Example
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i
i
i p
nR TV?
L 5,4 41 0 0 0 c m1L1 5 c m1 0 0 c m3,9 4 2 L 32
-1KJ 23,S
TTT Δif 2 8 2,9 K5,2 8 K2 8 8,1 5 K
3,9 4 2 L
a t m 9,0
2 8 8,2 Km o lL a t m K108,2 0 61,5 a t m 112
VVV Δif
Example
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1,The properties of gases
2,The First Law,the concepts
3,The First Law,the machinery
4,The Second Law,the concepts
5,The Second Law,the machinery
6,Physical transformations of pure substances
7,Simple mixtures
8,Phase diagrams
9,Chemical equilibrium
10,Electrochemistry
Part 1,Equilibrium
