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不可能把热从低温
物体传到高温物体，
而不引起其它变化
物理化学电子教案 第 3章（中）（添）
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3.7 热力学第二定律的本质和熵的统计意义
气体混合过程的不可逆性
将 N2和 O2放在一盒内隔板的两边，抽去隔板，
N2和 O2自动混合，直至平衡。
这是 混乱度增加 的过程，也是熵增加的过程，
是 自发 的过程，其逆过程决不会自动发生。
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3.8 Concentrating on the system
Entropy is the basic concept for discussing the
direction of natural changes,Nevertheless,in order
to use it we have to investigate both the system and
its surroundings,We have seen that it is always very
simple to calculate the entropy change in the
surroundings,and now we shall show that it is
possible to devise a method of taking the
surroundings into account automatically.
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This focuses attention on the system,
and simplifies thermodynamic discussions.
Consider a system in thermal equilibrium with
its surroundings(so that ),
s u r rs y s TT ?
then the Clausius inequality reads
0?
?
? T
Q
dS
s y s
s y s
The importance of this inequality is that
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it expresses the criterion
for natural,spontaneous change solely in terms
of the properties of the system.
From now on the superscript denoting the system
will be dropped,and everything we do will relate
to the system.
The last equation can be developed in two ways.
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First take the case where the heat
is lost from a constant volume system.
Then dQv can be identified with dU if there is no
non-pV work involved,Putting this into eq.gives
dS - dU/T ≧ 0,or 0 ≧ dU- T dS
dU≤TdS (const,V,no non-pV work)
When the heat is lost from the system under
conditions of constant pressure and when no non-
pV work is involved,
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the only modification is to identify
dQp with dH,the change of enthalpy of the system.
Then dS – dH /T ≧ 0,or 0 ≧ dH - T dS
dH≤ TdS (const,p,no non-pV work)
These expressions can be simplified even more by
introducing two new thermodynamic functions.
They are defined as follows:
Helmholtz function,A = U – T S
Gibbs function,G = H – T S
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How G and A simplify the expressions
Consider what happens to them when the state
of a system changes at constant temperature:
dA = dU –TdS (const.T) dG = dH – TdS (const.T)
Now introduce TdS≥dU (const,V)
TdS≥dH (const,P)
dA≤0 (const,T,V,no non-pV work)
dG≤0 (const,T,P,no non-pV work)
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These inequalities are the most
important conclusions from thermodynamics
in chemistry.
Some remarks on the Helmholtz function.
When the changes in a system are constrained to
occur under conditions of constant temperature
and volume eq,above determines whether they
can occur spontaneously.
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A change corresponding to a decrease
in A can occur spontaneously.
Systems tend to move naturally towards
states of lower A,and the criterion of
equilibrium is (dA)T,V =0,( n,W`=0)
We shall now demonstrate the connection between
△ A and Wmax,
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Helmholtz free energy
亥姆霍兹 （ von Helmholz,H.L.P.,1821~1894,
德国人）定义了一个状态函数
d e f A U T S?
A称为 Helmholz free energy,是状态函数，具有容量性质。
dA =dU-d(TS) = dU-TdS=dU-?QrevT
dU=?Q+??W =?Qrev +?Wrev
dU- ?Qrev= ?Wrev
= ?Wrev
dA=?Wrev
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Gibbs Free Energy d e f G H T S?
Some remarks on the Gibbs function.
The Gibbs function is more common in chemistry
than the Helmholtz function.
This is because we are usually more interested in
the conditions of equilibrium and the direction of
chemical change when systems are at constant
pressure rather than constant volume,
Now dWrev consists of expansion work and
possibly some other kind of work.
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dG=dH – TdS – SdT dH=dU+d(pV)
ef d d Q W W p V V p? ? ? ? ? ? ? ?
f dQ W V p? ? ? ? ?
f,ma xW??
( 可逆）,0d,0d ?? pT
fd d d dG Q W V p T S S T? ? ? ? ? ? ?
所以
,,R f,m ax( d ) TpGW? ? ? ?
或
This means that under conditions of constant pressure
and temperature the change of the Gibbs function
between specified states gives the maximum work
available from that process other than expansion work.
e ( d )W p V? ? ?
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The criterion of Gibbs free energy
At const.T,P,W`=0,closed sys..
0)d( 0,,f ?? ?WpTG
0)d( 0,,f ??WpTG
or
When we want to know whether a reaction will
tend to go in a particular direction (the pressure
and temperature being constant) we have to
determine ?G for the reaction:
?G =G products – G reactants
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Initial state (reactants)? final state(…
If this ?G is negative,then the reaction has a
natural tendency to move spontaneously from
reactants to products.
If ?G is positive,then the reaction as written
will not proceed spontaneously,but the reverse
reaction will be the spontaneous one.
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The criterion of entropy
熵判据 在所有判据中 处于特殊地位,因为所有判
断反应方向和达到平衡的不等式都是由熵的 Clausius不
等式引入的 。但由于熵判据用于隔离体系（保持 U,V
不变），要考虑环境的熵变，使用不太方便。
,( d ) 0UVS ?
""
""
?
?
表示可逆，平衡
表示不可逆，自发
在隔离体系中，如果发生一个不可逆变化，则必
定是自发的,自发变化总是朝熵增加的方向进行 。自
发变化的结果使体系处于平衡状态，这时若有反应发
生，必定是可逆的，熵值不变。
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The criterion of entropy
对于绝热体系
d ( 0S ?绝热）
等号 表示 可逆, 不等号 表示 不可逆,但不能判
断其是否自发。因为绝热不可逆压缩过程是个非自
发过程，但其熵变值也大于零。
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The criteria of free energies
""
""
?
?
表示可逆，平衡
表示不可逆，自发
,,0( d ) 0fT V WA ? ?
(dG)T,P,W`=0 ≤0
“＝” denoting rev.,
equilibrium
“＜” denoting
irrev.,spontaneous
All for a closed system !
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2.10 The significances of G and A
Define the Helmholtz function and the
Gibbs function of a system and use them as
criteria for the direction of spontaneous change.
Relate the Helmholtz function to the maximum
amount of work available from a changing system.
Relate the Gibbs function to the maximum
amount of non-pV work available from a changing
system.
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Calculate the change of G and A
recall H = U + pV △ H=△ U+△ (pV)
then A = U - TS △ A =△ U-△ (TS)
G = H - TS △ G =△ H - △ (TS)
For pg at const,T △ H = △ U
△ A = △ U - T△ S
△ G = △ H - T△ SPay attention:
△ (xy)=x2y2-x1y1 The key is △ S !
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The relations of functions
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Combining the First and Second laws
d d dU T S p V??
(1)
The first law may be written
WQdU ?? ??
If a change of a closed system is brought about
reversible,then in the absence of any kind of work
other than pV-work,?Wrev may be replaced by
- pdV and ?Qrev may be replaced by TdS,
Therefore
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This equation applies when the changes
are reversible,But dU is an exact differential
and is independent of the path,
Therefore,dUrev =dUirrev,
Equation (1) applies irrespective of whether the
change is brought about reversibly or irreversibly,
and so for any change of a closed system,and
where there is no change of composition.
Only in the case of a reversible change may
TdS be identified with dQ and –PdV with dW,
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This we call the fundamental equation.
d d d dH U p V V p? ? ?
VpSTU ddd ??
pVUH ??
pVSTH ddd ??
then
d d dH T S V p??
(2)
definition
The argument to obtain each one runs in the same
way.
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Four fundamental equations
TSSTUA dddd ???
VpSTU ddd ??
TSUA ??as
d d dA S T p V? ? ?
(3)
VpTSA ddd ???
then
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The most important equation
(4)
d d dG S T V p? ? ?
as TSHG ??
TSSTHG dddd ???
pVSTH ddd ??
pVTSG ddd ???then
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Properties of the Gibbs function
The main aim is to obtain the dependence
of the Gibbs function on temperature,
pressure,and the composition of the system.
This chapter also introduces the
important concept of chemical potential
upon which the whole of the subsequent
applications of thermodynamics are based.
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VpSTU ddd ??(1)
pVSTH ddd ??(2)
VpTSA ddd ???(3)
pVTSG ddd ???(4)
( ) ( )VpUHST S??????
( ) ( )STp UAVV??? ? ? ???
( ) ( )STHGpV p??????
( ) ( )VpS AGTT??? ? ? ???
One way of looking at the fundamental equation
This suggests that the
most sensible way of
regarding U is as a
function of S and V.
(1) U = U(S,V)
(4) G = G(T,P)
(3),(2),
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Characteristic function
对于 U,H,S,A,G 等热力学函数，独立变量
选择合适
(,) U S V
这个已知函数就称为 特性函数,所选择的独立变
量就称为该特性函数的 characteristic variables。
常用的特征变量为：
(,) G T p (,) A T V
(,)H S p
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Eular relation
全微分的性质
设函数 z 的独立变量为 x,y,z具有全微分性质
(,)z z x y?
d ( ) d ( ) dyxzzz x yxy?????? ddM x N y??
( ) ( )xyMNyx?????
所以
M 和 N也是 x,y 的函数
22
( ),( )xyM z N z
y x y x x y
? ? ? ???
? ? ? ? ? ?
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( ) ( )xyMNyx?????
Maxwell relations
( ) ( ) VS pTVS ?? ????VpSTU ddd ??(1)
( ) ( ) pSTVpS???pVSTH ddd ??(2)
( ) ( )TVSpVT?????VpTSA ddd ???
(3)
( ) ( ) pTSVpT????
pVTSG ddd ???(4)
Another way of looking at the
fundamental equation.
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Once more deduce the master equation
Having seen the application of some simple
arguments to the fundamental equation for the
thermodynamic energy we can apply the same
techniques to the Gibbs function.
This is important because our discussion of
chemical equilibria will be bases on the
properties of G
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First the fundamental equation
is combined with the definition of G,and then the
new equation is examined in the same way as the
original fundamental equation for U,
The definition of the Gibbs function was given on
p.146,as G = H - TS
When the system changes,G might change because
H,T,and S may change.For infinitesimal changes in
each property,dG = dH – TdS – SdT.
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Since H = U + pV,we have
dH = dU + pdV +Vdp
For a closed system changing without change of
composition dU can be replaced by the fundamental
equation.
Combining all the parts gives
dG = (TdS – pdV) + pdV + Vdp – TdS - SdT
Or dG = Vdp - SdT
This is the new fundamental equation.
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This new equation suggests that
the most sensible way of regarding G is as a function
of p and T,and writing it G(p,T).
This confirms that it is a very important
quantity in chemistry because the pressure and
temperature are variables that are usually under
our control,G,it appears,carries around
the combined consequences of the First and
Second Laws in a way that makes it particularly
suitable for chemical applications,
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Once G is regarded in this way
some conclusions can be drawn very quickly.
The same argument that led to eq,(*) now gives
S
T
G
P
???
?
?
?
?
?
?
?
V
P
G
T
??
?
?
?
?
?
?
?
which show how the Gibbs function varies with
temperature and pressure.
The second conclusion is
TP P
S
T
V ?
?
??
?
?
?
?
????
??
?
?
?
?
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Deduce the Gibbs-Helmholtz eq.
S
T
G
P
???
?
?
?
?
?
?
?
22
T
GTS
T
T
T
G
T
G
T
T
T
G
PP
P
??
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
G = H - TS
2T
H?
?
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How the Gibbs function depends on the T.
2
T
H
T
T
G
P
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?Gibbs-Helmholtz
equation:
initial state (reactants)? final state (products)
△ G = Gf - Gi
?
?
?
?
?
?
????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
22 T
H
T
H
T
G
T
G
T
if
P
if
2T
H
T
T
G
P
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
△ H is the enthalpy of
reaction.
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How the Gibbs function depends on the P.
The foregoing equation indicates that knowing the
volume of the system is sufficient to predict how G
depends on the pressure.
VPG
T
??????? ??
Integration of the equation
leads to
G(p2)=G(p1) +
? 2
1
)(p
p
dppV
There are two cases where the pressure
dependence of the volume is very simple and
allows this expression to be evaluated.
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In the case of a liquid or a solid
the V may be taken outside the integral.
Then for molar amounts
Gm(p2) = Gm(p1) + (p2 – p1) Vm
Except at very high pressure (p2 – p1) Vm is very
small.
For solids,liquids,Gm(p2) ≈ Gm(p1)
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The other simple case is a perfect gas.
In order to find an explicit expression for the
pressure dependence the equation of state V
= nRT/p is put into the integral:
G(p2)=G(p1) + nRT
? 2
1
)/1(
P
P
dPP
Then G(p2)=G(p1) + nRT ln (P2/P1)
If P1= P?,the initial state is the standard state of
a perfect gas.
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The chemical potential of a perfect gas.
So at any other pressure P the Gibbs function is
G(p) = G? + nRT ln ( P/P?)
When the Gibbs function refers to 1 mol of
material the last equation becomes
Gm(p) = Gm? + RT ln ( P/P?)
A special symbol is given to the molar Gibbs
function of a pure substance.
?(p) = ?? + RT ln ( P/P?)
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An interpretation of this name is as follows.
In mechanical systems the direction of
spontaneous change can be predicted simply by
examining the potential in the vicinity of the
particle.
We know that the direction of spontaneous
change when T and P are held constant is towards
a minimum of the Gibbs function for the system.
For this reason the molar Gibbs function
is called the chemical potential.
The end
不可能把热从低温
物体传到高温物体，
而不引起其它变化
物理化学电子教案 第 3章（中）（添）
?上一内容 ?下一内容 ?回主目录 ?返回 2010-5-14
3.7 热力学第二定律的本质和熵的统计意义
气体混合过程的不可逆性
将 N2和 O2放在一盒内隔板的两边，抽去隔板，
N2和 O2自动混合，直至平衡。
这是 混乱度增加 的过程，也是熵增加的过程，
是 自发 的过程，其逆过程决不会自动发生。
?上一内容 ?下一内容 ?回主目录 ?返回 2010-5-14
3.8 Concentrating on the system
Entropy is the basic concept for discussing the
direction of natural changes,Nevertheless,in order
to use it we have to investigate both the system and
its surroundings,We have seen that it is always very
simple to calculate the entropy change in the
surroundings,and now we shall show that it is
possible to devise a method of taking the
surroundings into account automatically.
?上一内容 ?下一内容 ?回主目录 ?返回 2010-5-14
This focuses attention on the system,
and simplifies thermodynamic discussions.
Consider a system in thermal equilibrium with
its surroundings(so that ),
s u r rs y s TT ?
then the Clausius inequality reads
0?
?
? T
Q
dS
s y s
s y s
The importance of this inequality is that
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it expresses the criterion
for natural,spontaneous change solely in terms
of the properties of the system.
From now on the superscript denoting the system
will be dropped,and everything we do will relate
to the system.
The last equation can be developed in two ways.
?上一内容 ?下一内容 ?回主目录 ?返回 2010-5-14
First take the case where the heat
is lost from a constant volume system.
Then dQv can be identified with dU if there is no
non-pV work involved,Putting this into eq.gives
dS - dU/T ≧ 0,or 0 ≧ dU- T dS
dU≤TdS (const,V,no non-pV work)
When the heat is lost from the system under
conditions of constant pressure and when no non-
pV work is involved,
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the only modification is to identify
dQp with dH,the change of enthalpy of the system.
Then dS – dH /T ≧ 0,or 0 ≧ dH - T dS
dH≤ TdS (const,p,no non-pV work)
These expressions can be simplified even more by
introducing two new thermodynamic functions.
They are defined as follows:
Helmholtz function,A = U – T S
Gibbs function,G = H – T S
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How G and A simplify the expressions
Consider what happens to them when the state
of a system changes at constant temperature:
dA = dU –TdS (const.T) dG = dH – TdS (const.T)
Now introduce TdS≥dU (const,V)
TdS≥dH (const,P)
dA≤0 (const,T,V,no non-pV work)
dG≤0 (const,T,P,no non-pV work)
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These inequalities are the most
important conclusions from thermodynamics
in chemistry.
Some remarks on the Helmholtz function.
When the changes in a system are constrained to
occur under conditions of constant temperature
and volume eq,above determines whether they
can occur spontaneously.
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A change corresponding to a decrease
in A can occur spontaneously.
Systems tend to move naturally towards
states of lower A,and the criterion of
equilibrium is (dA)T,V =0,( n,W`=0)
We shall now demonstrate the connection between
△ A and Wmax,
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Helmholtz free energy
亥姆霍兹 （ von Helmholz,H.L.P.,1821~1894,
德国人）定义了一个状态函数
d e f A U T S?
A称为 Helmholz free energy,是状态函数，具有容量性质。
dA =dU-d(TS) = dU-TdS=dU-?QrevT
dU=?Q+??W =?Qrev +?Wrev
dU- ?Qrev= ?Wrev
= ?Wrev
dA=?Wrev
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Gibbs Free Energy d e f G H T S?
Some remarks on the Gibbs function.
The Gibbs function is more common in chemistry
than the Helmholtz function.
This is because we are usually more interested in
the conditions of equilibrium and the direction of
chemical change when systems are at constant
pressure rather than constant volume,
Now dWrev consists of expansion work and
possibly some other kind of work.
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dG=dH – TdS – SdT dH=dU+d(pV)
ef d d Q W W p V V p? ? ? ? ? ? ? ?
f dQ W V p? ? ? ? ?
f,ma xW??
( 可逆）,0d,0d ?? pT
fd d d dG Q W V p T S S T? ? ? ? ? ? ?
所以
,,R f,m ax( d ) TpGW? ? ? ?
或
This means that under conditions of constant pressure
and temperature the change of the Gibbs function
between specified states gives the maximum work
available from that process other than expansion work.
e ( d )W p V? ? ?
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The criterion of Gibbs free energy
At const.T,P,W`=0,closed sys..
0)d( 0,,f ?? ?WpTG
0)d( 0,,f ??WpTG
or
When we want to know whether a reaction will
tend to go in a particular direction (the pressure
and temperature being constant) we have to
determine ?G for the reaction:
?G =G products – G reactants
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Initial state (reactants)? final state(…
If this ?G is negative,then the reaction has a
natural tendency to move spontaneously from
reactants to products.
If ?G is positive,then the reaction as written
will not proceed spontaneously,but the reverse
reaction will be the spontaneous one.
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The criterion of entropy
熵判据 在所有判据中 处于特殊地位,因为所有判
断反应方向和达到平衡的不等式都是由熵的 Clausius不
等式引入的 。但由于熵判据用于隔离体系（保持 U,V
不变），要考虑环境的熵变，使用不太方便。
,( d ) 0UVS ?
""
""
?
?
表示可逆，平衡
表示不可逆，自发
在隔离体系中，如果发生一个不可逆变化，则必
定是自发的,自发变化总是朝熵增加的方向进行 。自
发变化的结果使体系处于平衡状态，这时若有反应发
生，必定是可逆的，熵值不变。
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The criterion of entropy
对于绝热体系
d ( 0S ?绝热）
等号 表示 可逆, 不等号 表示 不可逆,但不能判
断其是否自发。因为绝热不可逆压缩过程是个非自
发过程，但其熵变值也大于零。
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The criteria of free energies
""
""
?
?
表示可逆，平衡
表示不可逆，自发
,,0( d ) 0fT V WA ? ?
(dG)T,P,W`=0 ≤0
“＝” denoting rev.,
equilibrium
“＜” denoting
irrev.,spontaneous
All for a closed system !
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2.10 The significances of G and A
Define the Helmholtz function and the
Gibbs function of a system and use them as
criteria for the direction of spontaneous change.
Relate the Helmholtz function to the maximum
amount of work available from a changing system.
Relate the Gibbs function to the maximum
amount of non-pV work available from a changing
system.
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Calculate the change of G and A
recall H = U + pV △ H=△ U+△ (pV)
then A = U - TS △ A =△ U-△ (TS)
G = H - TS △ G =△ H - △ (TS)
For pg at const,T △ H = △ U
△ A = △ U - T△ S
△ G = △ H - T△ SPay attention:
△ (xy)=x2y2-x1y1 The key is △ S !
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The relations of functions
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Combining the First and Second laws
d d dU T S p V??
(1)
The first law may be written
WQdU ?? ??
If a change of a closed system is brought about
reversible,then in the absence of any kind of work
other than pV-work,?Wrev may be replaced by
- pdV and ?Qrev may be replaced by TdS,
Therefore
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This equation applies when the changes
are reversible,But dU is an exact differential
and is independent of the path,
Therefore,dUrev =dUirrev,
Equation (1) applies irrespective of whether the
change is brought about reversibly or irreversibly,
and so for any change of a closed system,and
where there is no change of composition.
Only in the case of a reversible change may
TdS be identified with dQ and –PdV with dW,
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This we call the fundamental equation.
d d d dH U p V V p? ? ?
VpSTU ddd ??
pVUH ??
pVSTH ddd ??
then
d d dH T S V p??
(2)
definition
The argument to obtain each one runs in the same
way.
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Four fundamental equations
TSSTUA dddd ???
VpSTU ddd ??
TSUA ??as
d d dA S T p V? ? ?
(3)
VpTSA ddd ???
then
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The most important equation
(4)
d d dG S T V p? ? ?
as TSHG ??
TSSTHG dddd ???
pVSTH ddd ??
pVTSG ddd ???then
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Properties of the Gibbs function
The main aim is to obtain the dependence
of the Gibbs function on temperature,
pressure,and the composition of the system.
This chapter also introduces the
important concept of chemical potential
upon which the whole of the subsequent
applications of thermodynamics are based.
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VpSTU ddd ??(1)
pVSTH ddd ??(2)
VpTSA ddd ???(3)
pVTSG ddd ???(4)
( ) ( )VpUHST S??????
( ) ( )STp UAVV??? ? ? ???
( ) ( )STHGpV p??????
( ) ( )VpS AGTT??? ? ? ???
One way of looking at the fundamental equation
This suggests that the
most sensible way of
regarding U is as a
function of S and V.
(1) U = U(S,V)
(4) G = G(T,P)
(3),(2),
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Characteristic function
对于 U,H,S,A,G 等热力学函数，独立变量
选择合适
(,) U S V
这个已知函数就称为 特性函数,所选择的独立变
量就称为该特性函数的 characteristic variables。
常用的特征变量为：
(,) G T p (,) A T V
(,)H S p
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Eular relation
全微分的性质
设函数 z 的独立变量为 x,y,z具有全微分性质
(,)z z x y?
d ( ) d ( ) dyxzzz x yxy?????? ddM x N y??
( ) ( )xyMNyx?????
所以
M 和 N也是 x,y 的函数
22
( ),( )xyM z N z
y x y x x y
? ? ? ???
? ? ? ? ? ?
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( ) ( )xyMNyx?????
Maxwell relations
( ) ( ) VS pTVS ?? ????VpSTU ddd ??(1)
( ) ( ) pSTVpS???pVSTH ddd ??(2)
( ) ( )TVSpVT?????VpTSA ddd ???
(3)
( ) ( ) pTSVpT????
pVTSG ddd ???(4)
Another way of looking at the
fundamental equation.
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Once more deduce the master equation
Having seen the application of some simple
arguments to the fundamental equation for the
thermodynamic energy we can apply the same
techniques to the Gibbs function.
This is important because our discussion of
chemical equilibria will be bases on the
properties of G
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First the fundamental equation
is combined with the definition of G,and then the
new equation is examined in the same way as the
original fundamental equation for U,
The definition of the Gibbs function was given on
p.146,as G = H - TS
When the system changes,G might change because
H,T,and S may change.For infinitesimal changes in
each property,dG = dH – TdS – SdT.
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Since H = U + pV,we have
dH = dU + pdV +Vdp
For a closed system changing without change of
composition dU can be replaced by the fundamental
equation.
Combining all the parts gives
dG = (TdS – pdV) + pdV + Vdp – TdS - SdT
Or dG = Vdp - SdT
This is the new fundamental equation.
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This new equation suggests that
the most sensible way of regarding G is as a function
of p and T,and writing it G(p,T).
This confirms that it is a very important
quantity in chemistry because the pressure and
temperature are variables that are usually under
our control,G,it appears,carries around
the combined consequences of the First and
Second Laws in a way that makes it particularly
suitable for chemical applications,
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Once G is regarded in this way
some conclusions can be drawn very quickly.
The same argument that led to eq,(*) now gives
S
T
G
P
???
?
?
?
?
?
?
?
V
P
G
T
??
?
?
?
?
?
?
?
which show how the Gibbs function varies with
temperature and pressure.
The second conclusion is
TP P
S
T
V ?
?
??
?
?
?
?
????
??
?
?
?
?
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Deduce the Gibbs-Helmholtz eq.
S
T
G
P
???
?
?
?
?
?
?
?
22
T
GTS
T
T
T
G
T
G
T
T
T
G
PP
P
??
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
G = H - TS
2T
H?
?
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How the Gibbs function depends on the T.
2
T
H
T
T
G
P
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?Gibbs-Helmholtz
equation:
initial state (reactants)? final state (products)
△ G = Gf - Gi
?
?
?
?
?
?
????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
22 T
H
T
H
T
G
T
G
T
if
P
if
2T
H
T
T
G
P
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
△ H is the enthalpy of
reaction.
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How the Gibbs function depends on the P.
The foregoing equation indicates that knowing the
volume of the system is sufficient to predict how G
depends on the pressure.
VPG
T
??????? ??
Integration of the equation
leads to
G(p2)=G(p1) +
? 2
1
)(p
p
dppV
There are two cases where the pressure
dependence of the volume is very simple and
allows this expression to be evaluated.
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In the case of a liquid or a solid
the V may be taken outside the integral.
Then for molar amounts
Gm(p2) = Gm(p1) + (p2 – p1) Vm
Except at very high pressure (p2 – p1) Vm is very
small.
For solids,liquids,Gm(p2) ≈ Gm(p1)
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The other simple case is a perfect gas.
In order to find an explicit expression for the
pressure dependence the equation of state V
= nRT/p is put into the integral:
G(p2)=G(p1) + nRT
? 2
1
)/1(
P
P
dPP
Then G(p2)=G(p1) + nRT ln (P2/P1)
If P1= P?,the initial state is the standard state of
a perfect gas.
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The chemical potential of a perfect gas.
So at any other pressure P the Gibbs function is
G(p) = G? + nRT ln ( P/P?)
When the Gibbs function refers to 1 mol of
material the last equation becomes
Gm(p) = Gm? + RT ln ( P/P?)
A special symbol is given to the molar Gibbs
function of a pure substance.
?(p) = ?? + RT ln ( P/P?)
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An interpretation of this name is as follows.
In mechanical systems the direction of
spontaneous change can be predicted simply by
examining the potential in the vicinity of the
particle.
We know that the direction of spontaneous
change when T and P are held constant is towards
a minimum of the Gibbs function for the system.
For this reason the molar Gibbs function
is called the chemical potential.
The end
