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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 —第三章（上）
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Chapter 3 The second law of thermodynamics
3.1 Common feature of spontaneous change
3.2 The second law of thermodynamics
3.3 Carnot cycle and Carnot principle
3.4 Definition of entropy
3.5 Clausius inequality ; principle of
increase of entropy
3.6 Calculate the change of entropy
3.7 Statistical view of entropy
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The second law of thermodynamics
3.9 Direction of change and conditions
in equilibrium
3.10 Calculate of ?G
3.11 Relations of thermodynamic functions
3.12 Clapeyron equation
3.13 The third law and conventional entropy
3.8 Helmholtz function and Gibbs function
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3.1 Common feature
Spontaneous change,Some things happen
spontaneously,some things don`t.
Common feature —irreversible A gas expands to
fill the available volume; it does not spontaneously
contract into something smaller.
A hot body cools to the temperature of its
surroundings;it does not spontaneously get hotter at
their expense.
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What determines the direction of …?
Can the direction of the spontaneous change be
related to some aspect of the distribution of energy?
We shall see that this is so,Spontaneous changes
are always accompanied by a reduction in the
?quality‘ of energy,in the sense that it is degraded
into a more dispersed,chaotic form.
Spontaneous,natural changes are simply manifestation
of the natural tendency of the universe towards greater
chaos.
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3.2 The Second Law of Thermodynamics
Clausius statement：, it is impossible for any device
to operate in a cycle in such a manner that the sole
effect is the transfer of heat from one body to another
body at a higher temperature.‖
Kelvin-Planck statement：, no process is possible
whose sole result is the absorption of heat from a
reservoir and the conversion of this heat into work.‖
The second perpetual motion machine is no way!
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Measuring dispersal, the entropy
The First Law led to the introduction of the internal
energy,a property of state,The Second Law also
leads to a property,the entropy,
The First Law uses the internal energy to identify
the permissible changes; the Second Law uses the
entropy to identify the natural changes among
these permissible changes.
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The statistical view of the entropy
One way of introducing the entropy develops the
view that the extent of the dispersal of energy can
be calculated,this leads to the statistical definition
of entropy.
Another develops the view that the dispersal can
be related to the amount of heat involved in a
process,this leads to the thermodynamic
definition.
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Carnot Cycle
Of particular interest is an
idealized engine,known as
the Carnot engine,which
employs two heat
reservoirs at different
temperatures and operates
on the following cycle.
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Carnot cycle
Process 4,A reversible
adiabatic compression
Process 3,A reversible
isothermal
compression at T2
Process 2,A reversible
adiabatic expansion.
Process 1,A reversible isothermal expansion at Th
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The efficiency of engine
hc
hh
QQW
QQ
? ???? )0( c ?Q
1? ?hc
h
c
h
1 TT TT T ????
The thermal efficiency of the cycle,defined as the
ratio of the net work output to the heat input,
is given by
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Carnot principle:
No engine can be more efficient than a
reversible engine operating between the
same temperature limits,
and all reversible engines operating
between the same temperature limits have
the same efficiency.
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Arrive at a conclusion from Carnot C
h c h c
h h h
Q Q T TW
Q Q T
? ???? ? ?
h
c
h
c 11
T
T
Q
Q ???
h
h
c
c
T
Q
T
Q ??
c h
c h
0
QQ
TT
??
or：
For a reversible engine,
the sum of the ratio of
the heat to the temperature is equal to zero.
?Irrev.
less than
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任意可逆循环的热温商
用相同的方法把任意可逆
循环分成许多 首尾连接的小卡
诺循环,前一个循环的等温可
逆膨胀线就是下一个循环的绝
热可逆压缩线，如图所示的虚
线部分，这样两个过程的功恰
好抵消。
从而使众多小卡诺循环的 总效应 与任意可逆循
环的 封闭曲线 相当，所以任意可逆循环的热温商的
加和等于零，或它的 环程积分等于零 。
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任意可逆循环的热温商
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Consider a system undergoing
A reversible cycle as represented by the continuous
curve of Fig,It is possible to subdivide this cycle into
a number of small Carnot cycles as indicated.
As the number of Carnot cycles is increased,the
zigzag curve can be made to approach the original
cycle to any desired degree.
0
2
2
1
1 ????
T
Q
T
Q 0
`
2
`
2
`
1
`
1 ????
T
Q
T
Q etc.
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Adding the preceding equations gives
?????????,,,`
2
`
2
`
1
`
1
2
2
1
1
T
Q
T
Q
T
Q
T
Q
0??? TQ
In the limit,upon replacement of the summation
of finite terms by a cyclic integral,we obtain
0?
??
T
Q r e v
where the subscript ―rev‖
has been used to recall that
the original cycle as well as the small Carnot cycles
must all be reversible.
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The foregoing equation states that
the integral of dQ/T when carried out over a
reversible cycle is equal to zero.
It follows that the differential dQrev/T is a
perfect differential of a state function.
This property,called entropy,is denoted by
the symbol S.
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This is the defining expression for
the entropy.
T
Q
dS r e v
?
?
Integrating along a reversible path between two
equlibrium states 1 and 2 gives
? ?????
2
1
1212 T
Q
SSS r e v
Since entropy is a point function,it does not matter
what particular reversible path is followed in the
integration as long as it is reversible.
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Deduce entropy
用一闭合曲线代表任意可逆循环。
R( ) 0
Q
T
? ??
12
BA
RRAB( ) ( ) 0
QQ
TT
?? ????
可分成两项的加和
在曲线上任意取 A,B两点，把循环分成 A?B和
B?A两个可逆过程。
根据任意可逆循环热温商的公式：
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熵的引出
说明任意可逆过程的热温
商的值决定于始终状态，而
与可逆途径无关,这个热温
商具有状态函数的性质。
移项得：
12
BB
RRAA( ) ( )
QQ
TT
?? ???
任意可逆过程
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Clausius`s inequality
设温度相同的两个高、低温热源间有一个可逆
机和一个不可逆机。
h
c
h
ch
R 1 T
T
T
TT ?????
IR R???
根据卡诺定理：
0
h
h
c
c ??
T
Q
T
Q
i
IR
i i
( ) 0QT? ??
推广为与多个热源接触的任 意不可逆过程得：
h
c
h
ch
IR 1 Q
Q
Q
QQ ?????则：
h
l
h
l
T
T
Q
Q ??? 11
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Clausius inequality
A
I R,A B RB
i
( ) ( ) 0QQTT??? ??? ?
then
For a cycle,if A ?B is a irreversible,B ?A is a rev...
0???
??
?
? ?
???
??
?
? ? ??
r e v
A
Bir
B
A T
Q
T
Q
0??????? ???????? ? ??
r e v
B
Air
B
A T
Q
T
Q
ir
B
Ar e v
B
A T
Q
T
Q ?? ?
?
??
?
? ???
?
??
?
? ?△ S = SB – SA=
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Clausius inequality
These are Clausius inequality, mathematic
expression of thermodynamic second law.
A B A B
i
( ) 0QS T?? ?? ? ??
d QS T??d0
QS
T
???
Where the sign of equality applies to a reversible
process,and the sign of inequality applies to an
irreversible process.
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The principle of increase of entropy
0Q??
d0S ?
The entropy of an isolated system
increases for irreversible processes and
remains constant in the course of
reversible processes,The entropy of an
isolated system never decreases.
For an isolated system
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The implication of Clausius inequality
using Clsusius inequality to determine the
direction of spontaneous change,
d QS
T
??
,>” applies to irreversible processes
,=” applies to reversible processes
0d is o ?S
,>” a spontaneous change takes place
,=” is in an equilibrium state
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For a big isolated system
Sometimes take into account the system
and surroundings which has relation with
the system as a big isolated system.
is o ( ( 0S S S? ? ? ? ? ?体系） 环境）
,>” spontaneous
,=” equilibrium
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Calculate the change of entropy for Pg
P1V1T1 → P2V2T2
↘
P2V2 `T1
↗
Ⅰ
Ⅱ Ⅲ T
QdS r e v??
r e vr e v WQWQdU ????????
P d VdUQ r e v ???
T
P d VdUdS ?? ??? ????? 2
1
2
1
2
1 T
P dV
T
dU
T
P dVdU
S
`
2
2
1
2
,
1
`
2 lnlnln
V
VnR
T
TnC
V
VnRS
mv ????
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Important calculation equation
1
2
,
1
2
,
1
2
1
2
,
1
2
1
2
,
lnln
lnln
lnln
V
V
nC
P
P
nCS
P
P
nR
T
T
nCS
V
V
nR
T
T
nCS
mPmV
mP
mV
???
???
???
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Entropy change in isothermal processes
(1) Isothermal change of perfect gases
)ln (
1
2
V
VnRS ?? )ln (
2
1
p
pnR?
(2) Reversible phase transition
相变）
相变）相变）
(
((
T
HS ???
dTT
C
T
dH
T
Q
dS mpr e v,??
?
?
(For solid and liquid)
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The entropy change arising from
a phase transition.
Under the conditions of constant pressure the heat
of a phase transition is its enthalpy of transition,
△ H t.At the temperature of the transition,Tt the
system is in equilibrium (for instance,at the boiling
point liquid and vapour are in equibrium and the
system‘s temperature is the same as its
surroundings),and so heat is transferred reversibly
from the surroundings.
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等温过程的熵变
例 1,1mol理想气体在等温下,(1)可逆膨胀,
(2)真空膨胀，体积增加到 10倍，分别求其熵变。
解,（ 1）可逆膨胀
m a x
R( ( )
WQS
TT
?? ? ?体系）
1
2ln
V
VnR?
1l n 1 0 1 9, 1 4 J KnR ?? ? ?
0?????? （环境）（体系）（隔离） SSS
（ 1）为可逆过程。
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熵是状态函数，始终态相同，体系熵变也相同，
所以：
等温过程的熵变
（ 2）真空膨胀
11 9, 1 4 J KS ?? ? ?（体系）
但环境没有熵变，则：
11 9, 1 4 J K 0SS ?? ? ? ? ? ?（隔离） （体系）
（ 2）为不可逆过程
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等温过程的熵变
例 2：求下述过程熵变。已知 H2O（ l） 的汽化热
为
-14 0, 6 2 0 k J m o l?
22H O ( 1 m o l,l,,3 7 3, 1 5 K ) H O ( 1 m o l,g,,3 7 3, 1 5 K)pp$$
R）（（体系） T
QS ??
b
mv a p
T
H??
14 0, 6 2 0 k J m o l
3 7 3, 1 5 K
??
? 111 0 8, 9 J K m o l??? ? ?
解,
如果是不可逆相变，可以设计可逆相变求 值。S?
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等温过程的熵变
例 3：在 273 K时，将一个 的盒子用隔板一分为
二,一边放,另一边放 。
322.4 dm
20,5 m o l O (g ) 20,5 m o l N (g )
解法 1：
1
2
2 ln)O( V
VnRS ??
2.12
4.22ln5.0 R?
2
2 2, 4( N 0, 5 l n
1 2, 2
SR?? )
)N()O( 22m i x SSS ????? 2ln
2.12
4.22ln RR ??
求抽去隔板后，两种气体混合过程的熵变？
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等温过程的熵变
解法 2：
????
B
BBm i x ln xnRS
22
11( O ) l n ( N ) l n
22
R n n??? ? ???
??
1ln
2
R??
ln 2R?
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The change of entropy when a system is heated
(1)物质的量一定的 等容 变温过程
??? 2
1
dm,T
T
V
T
TnCS
??? 2
1
dm,T
T
p
T
TnC
S
(2)物质的量一定的 等压 变温过程
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变温过程的熵变
(4)没有相变的两个恒温热源之间的热传导
)()( 21 TSTSS ????? )11(
12 TT
Q ??
*(5)没有相变的两个变温物体之间的热传导，首先要
求出终态温度 T
21
2211 )(
CC
TCTCT
?
??
21 SSS ?????
2
2
1
1 lnln T
TC
T
TC ??
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Entropy change in chemical reactions
(1)在标准压力下,298.15 K时，各物质的 标准摩尔
熵值有表可查 。根据化学反应计量方程，可以计算
反应进度为 1 mol时的熵变值。
r m B m
B
( B )SS ??? ?$$
B,m
B
r m r m 2 9 8, 1 5 K
( B ) d
( ) ( 2 9 8,1 5 K )
pT CT
S T S
T
?
? ? ? ?
?
?$$
(2)在标准压力下，求反应温度 T时的熵变值。
298.15K时的熵变值从查表得到：
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Entropy changes in the surroundings
(1)任何可逆变化时环境的熵变
Rd ( ) ( ) / ( )S Q T??环 环 环
(2)体系的热效应可能是不可逆的，但由于环境很
大，对环境可看作是可逆热效应
d ( ) ( ) / ( )S Q T? ??环 体系 环
The temperature of the surroundings does
not change however much heat is injected.
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How to determine the entropy change
in an irreversible process.
We have already emphasized that as entropy is a
state function,any change in the entropy of a
system is independent of the path between the
specified initial and final states.
In contrast,the overall entropy change (of the
system and its surroundings) will depend on the
path,for if the change is reversible the overall
entropy change is zero,but if it is irreversible…
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3.7 热力学第二定律的本质和熵的统计意义
热与功转换的不可逆性
热 是分子 混乱运动 的一种表现，而 功 是分子
有序运动 的结果。
功转变成热 是从规则运动转化为不规则运动，
混乱度增加，是 自发 的过程；
而要将无序运动的 热转化为 有序运动的 功 就
不可能自动 发生。
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 —第三章（上）
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Chapter 3 The second law of thermodynamics
3.1 Common feature of spontaneous change
3.2 The second law of thermodynamics
3.3 Carnot cycle and Carnot principle
3.4 Definition of entropy
3.5 Clausius inequality ; principle of
increase of entropy
3.6 Calculate the change of entropy
3.7 Statistical view of entropy
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The second law of thermodynamics
3.9 Direction of change and conditions
in equilibrium
3.10 Calculate of ?G
3.11 Relations of thermodynamic functions
3.12 Clapeyron equation
3.13 The third law and conventional entropy
3.8 Helmholtz function and Gibbs function
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3.1 Common feature
Spontaneous change,Some things happen
spontaneously,some things don`t.
Common feature —irreversible A gas expands to
fill the available volume; it does not spontaneously
contract into something smaller.
A hot body cools to the temperature of its
surroundings;it does not spontaneously get hotter at
their expense.
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What determines the direction of …?
Can the direction of the spontaneous change be
related to some aspect of the distribution of energy?
We shall see that this is so,Spontaneous changes
are always accompanied by a reduction in the
?quality‘ of energy,in the sense that it is degraded
into a more dispersed,chaotic form.
Spontaneous,natural changes are simply manifestation
of the natural tendency of the universe towards greater
chaos.
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3.2 The Second Law of Thermodynamics
Clausius statement：, it is impossible for any device
to operate in a cycle in such a manner that the sole
effect is the transfer of heat from one body to another
body at a higher temperature.‖
Kelvin-Planck statement：, no process is possible
whose sole result is the absorption of heat from a
reservoir and the conversion of this heat into work.‖
The second perpetual motion machine is no way!
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Measuring dispersal, the entropy
The First Law led to the introduction of the internal
energy,a property of state,The Second Law also
leads to a property,the entropy,
The First Law uses the internal energy to identify
the permissible changes; the Second Law uses the
entropy to identify the natural changes among
these permissible changes.
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The statistical view of the entropy
One way of introducing the entropy develops the
view that the extent of the dispersal of energy can
be calculated,this leads to the statistical definition
of entropy.
Another develops the view that the dispersal can
be related to the amount of heat involved in a
process,this leads to the thermodynamic
definition.
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Carnot Cycle
Of particular interest is an
idealized engine,known as
the Carnot engine,which
employs two heat
reservoirs at different
temperatures and operates
on the following cycle.
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Carnot cycle
Process 4,A reversible
adiabatic compression
Process 3,A reversible
isothermal
compression at T2
Process 2,A reversible
adiabatic expansion.
Process 1,A reversible isothermal expansion at Th
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The efficiency of engine
hc
hh
QQW
? ???? )0( c ?Q
1? ?hc
h
c
h
1 TT TT T ????
The thermal efficiency of the cycle,defined as the
ratio of the net work output to the heat input,
is given by
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Carnot principle:
No engine can be more efficient than a
reversible engine operating between the
same temperature limits,
and all reversible engines operating
between the same temperature limits have
the same efficiency.
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Arrive at a conclusion from Carnot C
h c h c
h h h
Q Q T TW
Q Q T
? ???? ? ?
h
c
h
c 11
T
T
Q
Q ???
h
h
c
c
T
Q
T
Q ??
c h
c h
0
TT
??
or：
For a reversible engine,
the sum of the ratio of
the heat to the temperature is equal to zero.
?Irrev.
less than
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任意可逆循环的热温商
用相同的方法把任意可逆
循环分成许多 首尾连接的小卡
诺循环,前一个循环的等温可
逆膨胀线就是下一个循环的绝
热可逆压缩线，如图所示的虚
线部分，这样两个过程的功恰
好抵消。
从而使众多小卡诺循环的 总效应 与任意可逆循
环的 封闭曲线 相当，所以任意可逆循环的热温商的
加和等于零，或它的 环程积分等于零 。
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任意可逆循环的热温商
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Consider a system undergoing
A reversible cycle as represented by the continuous
curve of Fig,It is possible to subdivide this cycle into
a number of small Carnot cycles as indicated.
As the number of Carnot cycles is increased,the
zigzag curve can be made to approach the original
cycle to any desired degree.
0
2
2
1
1 ????
T
Q
T
Q 0
`
2
`
2
`
1
`
1 ????
T
Q
T
Q etc.
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Adding the preceding equations gives
?????????,,,`
2
`
2
`
1
`
1
2
2
1
1
T
Q
T
Q
T
Q
T
Q
0??? TQ
In the limit,upon replacement of the summation
of finite terms by a cyclic integral,we obtain
0?
??
T
Q r e v
where the subscript ―rev‖
has been used to recall that
the original cycle as well as the small Carnot cycles
must all be reversible.
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The foregoing equation states that
the integral of dQ/T when carried out over a
reversible cycle is equal to zero.
It follows that the differential dQrev/T is a
perfect differential of a state function.
This property,called entropy,is denoted by
the symbol S.
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This is the defining expression for
the entropy.
T
Q
dS r e v
?
?
Integrating along a reversible path between two
equlibrium states 1 and 2 gives
? ?????
2
1
1212 T
Q
SSS r e v
Since entropy is a point function,it does not matter
what particular reversible path is followed in the
integration as long as it is reversible.
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Deduce entropy
用一闭合曲线代表任意可逆循环。
R( ) 0
Q
T
? ??
12
BA
RRAB( ) ( ) 0
TT
?? ????
可分成两项的加和
在曲线上任意取 A,B两点，把循环分成 A?B和
B?A两个可逆过程。
根据任意可逆循环热温商的公式：
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熵的引出
说明任意可逆过程的热温
商的值决定于始终状态，而
与可逆途径无关,这个热温
商具有状态函数的性质。
移项得：
12
BB
RRAA( ) ( )
TT
?? ???
任意可逆过程
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Clausius`s inequality
设温度相同的两个高、低温热源间有一个可逆
机和一个不可逆机。
h
c
h
ch
R 1 T
T
T
TT ?????
IR R???
根据卡诺定理：
0
h
h
c
c ??
T
Q
T
Q
i
IR
i i
( ) 0QT? ??
推广为与多个热源接触的任 意不可逆过程得：
h
c
h
ch
IR 1 Q
Q
Q
QQ ?????则：
h
l
h
l
T
T
Q
Q ??? 11
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Clausius inequality
A
I R,A B RB
i
( ) ( ) 0QQTT??? ??? ?
then
For a cycle,if A ?B is a irreversible,B ?A is a rev...
0???
??
?
? ?
???
??
?
? ? ??
r e v
A
Bir
B
A T
Q
T
Q
0??????? ???????? ? ??
r e v
B
Air
B
A T
Q
T
Q
ir
B
Ar e v
B
A T
Q
T
Q ?? ?
?
??
?
? ???
?
??
?
? ?△ S = SB – SA=
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Clausius inequality
These are Clausius inequality, mathematic
expression of thermodynamic second law.
A B A B
i
( ) 0QS T?? ?? ? ??
d QS T??d0
QS
T
???
Where the sign of equality applies to a reversible
process,and the sign of inequality applies to an
irreversible process.
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The principle of increase of entropy
0Q??
d0S ?
The entropy of an isolated system
increases for irreversible processes and
remains constant in the course of
reversible processes,The entropy of an
isolated system never decreases.
For an isolated system
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The implication of Clausius inequality
using Clsusius inequality to determine the
direction of spontaneous change,
d QS
T
??
,>” applies to irreversible processes
,=” applies to reversible processes
0d is o ?S
,>” a spontaneous change takes place
,=” is in an equilibrium state
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For a big isolated system
Sometimes take into account the system
and surroundings which has relation with
the system as a big isolated system.
is o ( ( 0S S S? ? ? ? ? ?体系） 环境）
,>” spontaneous
,=” equilibrium
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Calculate the change of entropy for Pg
P1V1T1 → P2V2T2
↘
P2V2 `T1
↗
Ⅰ
Ⅱ Ⅲ T
QdS r e v??
r e vr e v WQWQdU ????????
P d VdUQ r e v ???
T
P d VdUdS ?? ??? ????? 2
1
2
1
2
1 T
P dV
T
dU
T
P dVdU
S
`
2
2
1
2
,
1
`
2 lnlnln
V
VnR
T
TnC
V
VnRS
mv ????
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Important calculation equation
1
2
,
1
2
,
1
2
1
2
,
1
2
1
2
,
lnln
lnln
lnln
V
V
nC
P
P
nCS
P
P
nR
T
T
nCS
V
V
nR
T
T
nCS
mPmV
mP
mV
???
???
???
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Entropy change in isothermal processes
(1) Isothermal change of perfect gases
)ln (
1
2
V
VnRS ?? )ln (
2
1
p
pnR?
(2) Reversible phase transition
相变）
相变）相变）
(
((
T
HS ???
dTT
C
T
dH
T
Q
dS mpr e v,??
?
?
(For solid and liquid)
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The entropy change arising from
a phase transition.
Under the conditions of constant pressure the heat
of a phase transition is its enthalpy of transition,
△ H t.At the temperature of the transition,Tt the
system is in equilibrium (for instance,at the boiling
point liquid and vapour are in equibrium and the
system‘s temperature is the same as its
surroundings),and so heat is transferred reversibly
from the surroundings.
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等温过程的熵变
例 1,1mol理想气体在等温下,(1)可逆膨胀,
(2)真空膨胀，体积增加到 10倍，分别求其熵变。
解,（ 1）可逆膨胀
m a x
R( ( )
WQS
TT
?? ? ?体系）
1
2ln
V
VnR?
1l n 1 0 1 9, 1 4 J KnR ?? ? ?
0?????? （环境）（体系）（隔离） SSS
（ 1）为可逆过程。
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熵是状态函数，始终态相同，体系熵变也相同，
所以：
等温过程的熵变
（ 2）真空膨胀
11 9, 1 4 J KS ?? ? ?（体系）
但环境没有熵变，则：
11 9, 1 4 J K 0SS ?? ? ? ? ? ?（隔离） （体系）
（ 2）为不可逆过程
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等温过程的熵变
例 2：求下述过程熵变。已知 H2O（ l） 的汽化热
为
-14 0, 6 2 0 k J m o l?
22H O ( 1 m o l,l,,3 7 3, 1 5 K ) H O ( 1 m o l,g,,3 7 3, 1 5 K)pp$$
R）（（体系） T
QS ??
b
mv a p
T
H??
14 0, 6 2 0 k J m o l
3 7 3, 1 5 K
??
? 111 0 8, 9 J K m o l??? ? ?
解,
如果是不可逆相变，可以设计可逆相变求 值。S?
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等温过程的熵变
例 3：在 273 K时，将一个 的盒子用隔板一分为
二,一边放,另一边放 。
322.4 dm
20,5 m o l O (g ) 20,5 m o l N (g )
解法 1：
1
2
2 ln)O( V
VnRS ??
2.12
4.22ln5.0 R?
2
2 2, 4( N 0, 5 l n
1 2, 2
SR?? )
)N()O( 22m i x SSS ????? 2ln
2.12
4.22ln RR ??
求抽去隔板后，两种气体混合过程的熵变？
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等温过程的熵变
解法 2：
????
B
BBm i x ln xnRS
22
11( O ) l n ( N ) l n
22
R n n??? ? ???
??
1ln
2
R??
ln 2R?
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The change of entropy when a system is heated
(1)物质的量一定的 等容 变温过程
??? 2
1
dm,T
T
V
T
TnCS
??? 2
1
dm,T
T
p
T
TnC
S
(2)物质的量一定的 等压 变温过程
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变温过程的熵变
(4)没有相变的两个恒温热源之间的热传导
)()( 21 TSTSS ????? )11(
12 TT
Q ??
*(5)没有相变的两个变温物体之间的热传导，首先要
求出终态温度 T
21
2211 )(
CC
TCTCT
?
??
21 SSS ?????
2
2
1
1 lnln T
TC
T
TC ??
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Entropy change in chemical reactions
(1)在标准压力下,298.15 K时，各物质的 标准摩尔
熵值有表可查 。根据化学反应计量方程，可以计算
反应进度为 1 mol时的熵变值。
r m B m
B
( B )SS ??? ?$$
B,m
B
r m r m 2 9 8, 1 5 K
( B ) d
( ) ( 2 9 8,1 5 K )
pT CT
S T S
T
?
? ? ? ?
?
?$$
(2)在标准压力下，求反应温度 T时的熵变值。
298.15K时的熵变值从查表得到：
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Entropy changes in the surroundings
(1)任何可逆变化时环境的熵变
Rd ( ) ( ) / ( )S Q T??环 环 环
(2)体系的热效应可能是不可逆的，但由于环境很
大，对环境可看作是可逆热效应
d ( ) ( ) / ( )S Q T? ??环 体系 环
The temperature of the surroundings does
not change however much heat is injected.
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How to determine the entropy change
in an irreversible process.
We have already emphasized that as entropy is a
state function,any change in the entropy of a
system is independent of the path between the
specified initial and final states.
In contrast,the overall entropy change (of the
system and its surroundings) will depend on the
path,for if the change is reversible the overall
entropy change is zero,but if it is irreversible…
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3.7 热力学第二定律的本质和熵的统计意义
热与功转换的不可逆性
热 是分子 混乱运动 的一种表现，而 功 是分子
有序运动 的结果。
功转变成热 是从规则运动转化为不规则运动，
混乱度增加，是 自发 的过程；
而要将无序运动的 热转化为 有序运动的 功 就
不可能自动 发生。
The end
