Chapter 7 Fundamentals of
Thermodynamics
热力学基础
§ 7-1 Internal Energy Heat & Work 内能 功 热量
§ 7-2 The First Law of Thermodynamics 热力学 第一
定律
§ 7-3 Application of the First Law of thermodynamics
热力学 第一定律对理想气体等容、等压和等温过
程的应用
§ 7-4 The Heat Capacities of an Ideal Gas 理想 气体
的热容
§ 7-5 Application of the First Law to Adiabatic
Processes 热力学 第一定律对理想气体绝热过程
的应用
§ 7-6 Cyclical Processes Thermal Efficiency
Carnot Cycle Reverse Cycle 循环过程 热机
的效率 卡诺循环 逆循环
§ 7-7 The Second Law of Thermodynamics热力学
第二定律
§ 7-8 Reversible & Irreversible Process
可逆过程与不可逆过程、卡诺定理
§ 7-9 Statistical Meaning of The Second Law
热力学第二定律的统计意义
§ 7-10 Entropy 熵
1.理解内能，功和热量的概念。掌握热力学第一定
律和它在理想气体各种等值过程中的应用；
2.理解热容量的概念，掌握能量按自由度均分定理；
3.明确循环的意义，能对简单循环进行计算。了解
熵增加原理和热力学第二定律的统计意义。
基本要求
its surroundings
1,Two basic concepts(系统与外界）,
is a system that can interact
with its surroundings in at least two ways,one of
which must be heat transfer.
Thermodynamic system:
,are everything outside the system,The surroundings
Thermodynamic system
§ 7-1 Internal Energy Heat &Work 内能
功 热量
§ 7-1 Internal Energy Heat &Work 内能 功 热量
Thermodynamic systems in engineering:
a gas,such as air,a vapor,such as steam; a
liquid,such as freon(氟利昂, 一种致冷剂 ） ; a solid,
such as semiconductor(半导体 ） ; ……,,They are also
called as working substance（ 工作物质 ）,
Thermodynamic system
working substance
A system of gas will be studied in this chapter.
Man is the most complicated system which needs to eat,
sleep,work,do many and many things………………,.
The concept of internal energy of a system
have been introduced in the last chapter.
RTiTE 2?M)( ? (6-18)
It means that internal energy is a single value
function of the gas,对于理想气体, 由于分子间无相互
作用力, 所以, 理想气体的内能等于所有气体分子的动
能之和,系统的内能取决于系统的状态, 是温度的单值函
数
2,Internal energy of a system(系统的内能）
内能理想气体
In general,the internal energy of thermodynamic
system which concludes（ 包括 ） the kinetic energy
and potential energy of all the molecules depends on
not only the temperature but also the volume of the
system,That is
)V,T(EE ?
E has a definite value,decided only by the state of
system.
系统
its surroundings
3,Heat and Work 热与功的等效性
A thermodynamic system can interact with its
surroundings in at least two ways:
(1) transfer of heat;
(2) Doing the mechanical work on the system by its
surroundings or on its surroundings by the
system.
Thermodynamic system
交换热量
做功物质
There are two ways of energy exchange（ 能量交换 ）
between the system and its surroundings,Their effects
are that the state of system may change.
its surroundings
Thermodynamic system
交换热量
做功物质
状态改变
Heat is energy that flows（ 流动 ） from one
body to another because of a temperature
difference between them.
Hot surrounding
Cold system
Question,Is the heat something having mass?
Heat（ Q）
Work（ W） is related to
forces and mechanical
energy.
W
W
Mechanical energy heat energy
Heat energy Mechanical energy
加热 搅拌作功
热与功的等效性
传热 （ Q） 和作功 （ W） 是系统与外界交换能
量的两种形式, 从能量传递的角度来讲, 它们是
等价的 。
但是, 系统与外界进行热交换和作功的, 方
式, 截然不同, 这将带来许多, 美妙, 奇特,
的现象 。
Jc a l 18.41 ?
热的单位,cal卡 (传统的原因 calorie）, 现统一为
焦尔 J(Joule)。
当 系统同时通过传热和 作功与外界交换能量, 改变
状态 （ 内能变化 ） 时, 遵守能量守恒与转化定律 — 热
力学第一定律 。
Q
W
?E
1,The First Law of Thermodynamics
The system,changes from an initial state E1 to a
final state E2,In this process,heat Q is transferred into
the system and work W is done on the surrounding by
the system.
1E 2E
吸热 Q
作功
§ 7-2 The First Law of Thermodynamics
热力学第一定律
吃吃吃！！
玩玩玩学学
工作！！！
Q
W
?E
The experiments have shown
that
WEEQ ??? )( 12
It is the first law of thermodynamics(热力学第一定理）,
The first law of thermodynamics is the law of
conservation of energy,generalized to include energy
transfer through heat as well as mechanical energy.
（ 能量守恒推广到热学领域 ）
The following conventions（ 约定 ） have been adopted:
（ 1） Q>0,which is added to system(吸热） ;
Q<0,which is removed from system（放热） ;
（ 2） W>0,done on the surrounding by the system;
W<0,done on the system by its surroundings;
(3) If E2-E1>0,the internal
energy of system increases;
If E2-E1<0,the internal
energy of system decreases;
Q
W
?E
（ 1 ） A process involving only infinitesimal
changes in the thermodynamic parameters of a
system is known as an infinitesimal（ 无穷小的 ）
process,For such a process,the fist law becomes
WddEQd ??
It represents the mathematical relation of
transformation among the internal energy,heat and
work.
Note:
and represents only a small amount of heat
and work,having no differential meanings（ 微
分 ）,
Qd Wd
(2)There is not any system(machine) that can do
work forever without the supply of heat(第一类永
动机是不可能的 ）,
WEEQ ??? )( 12
劳逸结合
热力学第一定律指出：
第一类永动机是不可能实现的。
第一类 永动机：
E2 - E1= 0 （循环）
Q = 0 （外界不供给能量）
W > 0 (对外界作功 )
（ 3 ）, heat” is energy that a given system has
absorbed from or released（ 释放 ） to its surroundings;
“Work” is work that a given system has done on its
surroundings or its surrounding have done on the
system.
Heat Worksystem
2.The work and heat of a process
We do not intend to discuss the calculation of heat
and work in a complicated process,and only deal
with a equilibrium process of a gas such as an ideal
gas as in shown in the following figure in which the
expansion of gas in a piston（ 以活塞理想气体的膨胀
为例 ） 。
f pS?
dx
or Fex
The work done by
system is equal to
P d VP S d xf d xdW ???
for an infinitesimally equilibrium process.
f pS?
dx
or Fex
? ??? 21 dd VV VPWW
which is equal to the total area under the process curve
connecting the initial state and final state,Therefore,
work depends on the path of process.
P
V
),( 11 PV
),( 22 PV
The total work during the
process is obtained by integration
(积分 ） of
结论：
体积功与路径有关，是一 过程量，其值 等于
P-V曲线 与 V轴包围的面积。
P
V
),( 11 PV
),( 22 PV
内能 E 是状态量, 只与
始, 末态有关, Q,W是过
程量, 与过程进行的方式
有关 。
We can rewrite the first law as
???? 21 d12 VV VPEEQ )(
(7-3)
Hence,heat depends on the path of process.
功和热只有联系到某一具体
的过程才能计算出它们来,
系统在某一状态时, 我们不
能说它有多少热和多少功 。
P
V
),( 11 PV
),( 22 PV
§ 7-3 Application of the First Law to isochoric,
isobaric & isothermal Processes理想气体
的等容、等压和等温过程
1,Isochoric process（等容过程）
When gas undergoes(经历 ）
a process in which its volume
remains unchanged,the
process is called isochoric
process,热源
fixed
(1) Equation of
path(过程方程 ）,
tt a nc o n sV ? or
0?dV
P
V
(V,P1)
(V,P2)
热源
fixed
(2) Equation of state:
RTMPV ??
(3) Work done in this process:
0?W
(4) The increment（增量） of internal energy:
)VPVP(i)TT(RiMEEE 121212 22 ??????? ?
(5) Heat,
)VPVP(i)TT(RiMEEQ V 121212 22 ?????? ?
P
V
(V,P1,T1)
(V,P2,T2)
2,Isobaric process（等压过程）
A process taking place at constant pressure is called
an isobaric process.
P
V
(V2,P,T2)
(V1,P,T1)
热源
No
change
(1) Equation of path:
tt a nc o n sP ?
or
0?dP
P
V
(V2,P,T2)
(V1,P,T1)
热源
No
change
(2) Equation of state:
RTMPV ??
(3) Work done in this process:
)TT(RM)VV(PW 1212 ???? ?
(4) The increment of internal energy:
)PVPV(i)TT(RiMEEE 121212 22 ??????? ?
(5) Heat,
)PVPV(i)TT(RiMWEEQ P 121212 2 22 2 ????????? ?
P
V
(V2,P,T2)
(V1,P,T1)
3.Isothermal Processes (等温过程 )
A process taking place at constant temperature is
said to be isothermal
P
V
(V2,P2,T)
(V1,P1,T)
恒温热源
temperature
tt a nc o n sT ?
(1) Equation of path:
or
tt a nc o n sPV ?
P
V
(V2,P2,T)
(V1,P1,T)
恒温热源
temperature
(2) Equation of state:
RTMPV ??
(3) Work done in this process:
2
1
1
22
1
2
1 P
PlnRTM
V
VlnRTM
V
dVRTMP d VW V
V ???
???? ??
(4) The increment of internal energy:
???? ?? EEE?
(5) Heat,
2
1
1
2
P
PRTM
V
VRTMWQ
T lnln ?? ???
P
V
(V2,P2,T)
(V1,P1,T)
在解决实际问题时，注意应用下列四个公式：
（ 1）热力学第一定理；
（ 2）内能公式（对理想气体）；
（ 3）状态方程（对理想气体）；
（ 4）过程方程。
一个实际过程可能是几个分过程的组合, 求解时
将整个过程分解为几个分过程 。
Example 7-1,一定量的理想气体从初态 a(P1,V1)经等
温过程到达体积为 4V1的 b态, 再经过等压过程到达 c，
最后经等容过程回到 a点 。 求整个过程系统对外所作
的功和吸收的热量 。
解：（ 1）画出 P-V图
a(P1,V1)
b(Pb,4V1)
c(Pc,V1)
V1 4V1 V
P(2)整个过程由三个过程组成：
等温、等压和等容，因此
VPT QQQQ ???
VPT WWWW ???
显然：
QW ???? 0E
(3)计算每个过程的热量和功：
等温：
???? ?? lnVPVVlnRTMWQ
i
f
TT ?
等压：
???
?
??
?? ?
????
?????? VPVV
VP)VV(PW
bP
等容：
0?VW
（ 4）整个吸收的热量和功：
)( l nVPVPlnVPQW 434434 111111 ?????
例题 7-1 设质量一定的单原子理想气体开始时
压强为 3.039× 105Pa，体积为 1L，先作等压膨胀至
体积为 2L，再作等温膨胀至体积为 3L，最后被等
体冷却到压强为 1.013× 105Pa。求气体在全过程中
内能的变化、所作的功和吸收的热量。
1 2 3
3
1
0 V/L
P/（ 1.013× 105Pa）
a b
c
d
解 (1)如
图,ab,bc、
及 cd分别表示
等压膨胀、等
温膨胀及等体
冷却过程。
PaPa ?
?
????????
?
???????????,.
V
Vpp
c
bb
c
在状态 d,压强为 pd=1.013× 105Pa，体积为 Vd= 3L
1 2 3
3
1
0 V/L
P/（ 1.013× 105Pa）
a b
c
d
由图可求出
（ 2） 在全过程中
内能的变化 △ E 为末
状态内能减去初状态
内能, 有理想气体内
能公式及理想气体状
态方程得：
)VpVp(
i
)TT(R
i
M
m
EEE
aadd
ad
ad
?
?
?
?
?
?
???
1 2 3
3
1
0 V/L
P/（ 1.013× 105Pa）
a b
c
d
??
（ 3）在全过程中所作
的功等于在各分过程中所
作的功之和，即：
W = Wp + WT + WV
得由 )( 122
1
VVpp d VW VVp ??? ?
Wp = pa(Vb-Va)=3× 1.013× 105× 10-3J
=304J
1 2 3
3
1
0 V/L
P/（ 1.013× 105Pa）
a b
c
d
得由
1
2ln
V
VRT
M
mW
T ?
J246
2
3
ln10013.123
lnln
2
?
????
??
b
c
bb
b
c
bT
V
V
Vp
V
V
RT
M
m
W
在等体过程中气体不作功，即 WV = 0
所以 W = Wp + WT + WV =304J+246J+0J=550J
（ 4） 吸收的热量？
1,Heat capacity(热容） and Specific Heat capacity(比热容）
As shown in the figure,when
heat is added to an object,its
temperature will increase.
Experiments have indicated the
heat that needs to be added
to make the object have an
increment of its temperature
is satisfied to
Q?
Q?
T?
)TT(McTMcQ if ?????
M Ti
Tf T?
Q?
§ 7-4 Heat Capacities of an Ideal Gas
理想气体的摩尔热容
where:
is called the heat capacity of the object.Mc
is called the specific heat capacity(比热容 ） of this
kind of material,That is the heat capacity per unit
mass.
c
For one mole of an object,its heat capacity is
labeled by C and equal to
cC ?? （摩尔热容）
m ol eon eT
QC
?
??
One mole
QC ??
)KorC(T ??? 1
,and obviously depend on the conditions
imposed during the heat transfer.
Mc c C
Air Conditioning（空调设备）
Hence,a substance has many different molar heat
capacities among which two heat capacities are of
greatest practical usefulness,at constant pressure or
constant volume（ 定压和定容 ）,
2,The molar heat capacity of an ideal
gas at constant volume（ 定容摩尔热容 ）
定容摩尔热容,is the amount of
heat per mole needed to cause a
unit rise in temperature at
constant volume
即：理想气体的定容摩尔热容是一个只与分子自由度有
关的量。
One mole
V=constant
QC ??
)KorC(T ??? 1
∵ TRiTRiMQ V ????? 22?
∴ Ri
T
QC V
V 2??
??
Using CV,the change of the internal energy can
be expressed as
)()( 12122 TTCMTTRiME V ????? ??
In a word:
?
?
?
?
?
?
?
?
?
?
多原子理想气体
双原子理想气体
单原子理想气体
R
R
R
C
V
3
2
5
2
3
3,The molar heat capacity of an ideal
gas at constant pressure（ 定压摩尔热
容 ）,
定压摩尔热容, is the amount
of heat per mole needed to cause
a unit rise in temperature at
constant pressure
即：理想气体的定压摩尔热容是一个只与分子自由度有
关的量。
One mole
P=constant
QC P ??
)KorC(T ??? 1
∵ TRiTRiMQ P ??????? 2 22 2?
∴ Ri
T
QC P
P 2
2??
?
??
In summary:
?
?
?
?
?
?
?
?
?
?
多原子理想气体
双原子理想气体
单原子理想气体
R
R
R
C
P
4
2
7
2
5
)TT(CMQ PP 12 ??? ?
Therefore,the heat in an isobaric processes(等压过程）
is given by
For the same material,why is greater than?
PC VC
This is because that in a constant-pressure the
temperature change,and however,the volume must
increase---otherwise the pressure could not remain
constant and as the material expands it does work.
Free enoughNo free
4,The relation between and
VC
PC
RCC VP ??
Here,we introduce the values of the ratio,
denoted by the Greek letter ?(gamma):
VP CC
i
i
C
C
V
P 2????
which is called 热容比, The quantity ? plays an
important role in adiabatic processes （ 绝热过程 ） for
an ideal gas,which we will study in the next section.
RiC P 2 2?? RiC
V 2?
（单、双和多？）
常温下理想气体
定体摩尔热容 定压摩尔热容 热容比实验值表
气
体
CP,m
(J/mol.K)
CV,m
(J/mol.
K)
CP,m-CV,m ?
单原子气体 氦氩 20.920.9 12.612.5 8.38.4 1.661.67
双原子气体
氢
氮
一氧化碳
氧
28.8
28.6
29.3
28.9
20.4
20.4
21.2
21.0
8.4
8.2
8.1
7.9
1.41
1.41
1.40
1.40
多原子气体
水蒸汽
甲烷
氯仿
乙醚
36.2
35.6
72.0
87.5
27.8
27.2
63.7
79.2
8.4
8.4
8.3
8.2
1.31
1.30
1.13
1.11
1,Adiabatic process(绝热过程）
An adiabatic process is a process in
which no heat transfer takes place
between a system and its surrounding.
A completely adiabatic process is an
idealization,but a process may be
approximately adiabatic if the system
is well insulated(被绝热 ） or if the
process takes place so rapidly that
there is no time for appreciable heat
flow to occur.
0??Q
§ 7-5Application of the first Law of thermodynamics to
Adiabatic Process热力学第一定律对理想绝热过程的应用
Examples:
保暖内衣
过程太快！！
2,Work and adiabatic curve
Because the heat in an
adiabatic process is zero,the
work is equal to the negative
value of increment of the
internal energy:
)TT(CM)TT(RiM)EE(W VQ 121212 2 ????????? ??
)TT(CM)TT(RiME V 12122 ????? ??
0??Q
which implies that if a system expands under
adiabatic conditions,is positive,and the internal
energy decreases; when a system is compressed
adiabatically,is negative and increases.
QW
QW E
EW Q ???
在绝热过程中，消耗内能系统对外作功；
或外界对系统作功，系统内能增加！
不吃不行！！
气功
V
P
O
1
2
P1
V1
P2
V2
绝热曲线
It can be proved that the
equation of path of the
adiabatic process is
tt a nc o n sPV ??
tt anc on sTV ?? 1? or tt a nc o n sTP ??? ?? 1
Using the equation of state,we have
important!!!
4.The comparison of an adiabatic curve with an isotherm
(等温线）
The adiabatic curves,at any point,have a
somewhat steeper（ 陡的 ） slope（ 斜率 ） than
the isothermal curve passing through the same
point.
V
P
0
绝热线 Adiabatic curve
等温线 Isotherm
a
dV
dPTdP
Q
Why?
物理原因：在等温膨胀中, 压力降低的因素只有一个,
即体积增大使 n 减小 （ P=nkT） ；在绝热过程中, 压
力降低的因素有两个, 即体积增大使 n 减小, 而温
度 T也降低, 故对应于相同的体积变化 dV, 绝热过程
的压强变化 dPQ大于等温过程的压强变化 dPT。
绝热线
Va
P
dV
dP a??? 等温线
Va
P
dV
dP a ??
The adiabatic curve is steeper than the isotherm,绝热线
斜率的绝对值比等温线的大, 故 绝热线比等温线陡 。
Prove:
由绝热方程 和等温线 PV = C得斜率CPV ??
Example 7-2,有 1 mole刚性多原子分子理想气体, 原来的
压强为 1.0atm,温度为 27?C。 若经过一绝热过程, 使其压
强增加到 16atm.试求 （ 1） 气体内能的增量； （ 2） 该过程
中气体所作的功； （ 3） 终态时气体的分子数密度 。
解：（ 1）从绝热方程可求得终态的温度：
??
?
?
?
?
???
?? CTP
i
??
?
1
3
46
所以：
J)TT(RiE 7 4 792 12 ????
K
P
PTT 600
1
1
2
12 ???
?
?
???
??
?
?
?
V
P
O
2
1
绝热曲线
2P
1P
2V 1V
（ 2）功等于：
JEW Q 7 4 7 9?????
（ 3）由,可得
22 n k TP ?
3
26
2
2 10961
m
.
kT
Pn 个???
V
P
O
2
1
绝热曲线
2P
1P
2V 1V
Example 7-3:一定量的某单原子分子理想气体装在封闭
的汽缸里, 此汽缸有可活动的活塞 （ 不计摩擦且无漏
气 ） 。 已知气体的初压强 P1=1 atm,体积 V=1L。 现将
该气体在等压下加热直到体积为原来的两倍, 然后在等
容下加热到压强为原来的两倍, 最后作绝热膨胀直到温
度下降到初温为止 。 试求,（ 1） 画出 P-V图； （ 2） 在
整个过程中气体内能的改变； （ 3） 在整个过程中气体
所吸收的热量； （ 4） 在整个过程中气体所作的功； （ 1
atm=1.013*105Pa)
解,（ 1） 作出 P-V图
如图所示
1 2
V（ L）
P (atm)
1
2
（ 2） 因初末两态温度相等,则
0??E
（ 3）整个过程吸收的热量
等于
VP QQQ ??
1112112 2
5
2
5
2
5 VP)VV(P)TT(RQ
P ?????
1112223 32
3
2
3 VP)PP(V)TT(RQ
V ?????
J.VPQQQ VP 211 1065211 ?????
（ 4）由热力学第一定理，有
J.QW 21065 ???
1 2
V（ L）
P (atm)
1
2
1,Cyclical Processes 循环过程
In order to withdraw(提取 ）
heat from a high temperature
source and convert( 转化 ） as
large a fraction as possible to
mechanical energy,we must have
a heat engine(热机 ）,
§ 7-6 Cyclical Processes Thermal efficiency
循环过程 热机的效率
The heat engines work usually as the following way:
system Q吸withdraw system E system W
system
E
Q放 W’
Recover（恢复）
Cyclical process
Cyclical process
The system(here called as working substance) is
carried through a cyclical process,that is a
sequence（ 一连串 ） of processes in which it
eventually returns to its original state as shown in
below figure。
a
P
V
A cyclical process can be
represented by a closed curve
( 闭合曲线 ） on the p-V
diagram.
顺时针的, 叫正循环,
系统对外作功 。 按这种方
式工作的机器叫热机 。
逆时针的, 叫逆循环, 外
界对系统作功 。 按这种方
式工作的机器叫制冷机 。
P
Vo
A
B
CD
P
o
A
B
CD
V
热水
蒸汽
冷水
锅炉 进气阀
排气阀
泵
冷凝器
热机工作原理图
冰箱（制冷机）
工作原理图
冷凝器
蒸
发
器高压蒸气
高温液体
低压蒸气
压缩机
节流阀
传
感
器
When a system is carried through a cyclical
process,its initial and final internal energy are equal.
That is:
0?? E
WQ ?净
W =net work= the area
enclosed by the curve
representing the process in
the P-V diagram.
P
V
W
吸Q
放Q
Conventions(约定 ):
=the heat absorbed by the system per cycle>0
吸Q
=the heat rejected by the system per cycle>0
放Q
P
V
W
吸Q
放Q
Therefore,we have
WQQQ ??? 放吸净
P
V
W
吸Q
放Q
2,The efficiency of heat engines
（热机的效率）
The efficiency of engine is
defined as
吸
放
吸
放吸
吸 Q
Q
Q
Q
W
??
?
?? 1?
Thus ? represents the fraction of that is converted
to work.
吸Q
P
V
W
吸Q
放Q
One hope that the work output of heat engine is as
large as possible and the heat thrown away is as
small as possible.
In 1794~1840,? was about 3%~4%,It was very low!!!
In general,a heat engine works between a hot
reservoirs(仓库, 储藏所 ) at T1 and cold reservoir at T2
as shown in below figure:
吸Q
放Q
W
Cold reservoir
at T2
Heat engine
Hot reservoir
at T1
Example 7-4:一定量的某理想气体进行了如图所示的循
环过程, 已知气体在状态 A的温度为 TA=300K。 求：
（ 1） 气体在状态 B和 C的温度； （ 2） 各过程中气体对
外所作的功； （ 3） 经过整个过程, 气体对外吸收的总
热量 （ 各过程吸收热量的代数和 ） 。
V(m3)
P(Pa)
A
BC
1 3
100
300
解：（ 1）用理想气
体的状态方程，可
求得：
KT B 3 0 0?
KT c 1 0 0?
（ 2）各过程所作的功：
JW AB 4001330010021 ????? )()(
JW BC 20031100 ????? )(
JW CA 0?
（ 3）整个过程气体对外吸收的热量：
JWWWQ CABCAB 2 0 0????
（因为 ）0??E
V(m3)
P(Pa)
A
BC
1 3
100
300
Example 7-5,One mole of a monatomic ideal gas is caused
to go through the cycle as shown in the figure,Process bc is
adiabatic expansion,Pb=10 atm,Vb=1.0m3 and Vc=8.0m3.
Calculate (a) the heat added to the gas; (b) the heat leaving
the gas; (c) the net work done by the gas; (d) the efficiency
of the cycle.
V
P b
ca
Solution,(a) In process ab,
heat is added to the gas,
which is given by
? ?
? ? bcb
babbab
VPP
VPVP)TT(RQ
??
????
2
3
2
3
2
3
吸
atm.VVPP
c
b
bc 320????
?
???
?? ? J...).(Q 65 10510110013132010
2
3 ???????
吸
(2) The heat leaving the gas in the process ca is,
? ?
? ? J.PVV
VPVP)TT(RQ
cac
acccca
51075
2
5
2
5
2
5
????
????放
(3) The net work done by the gas is equal to
J.QQW 51039 ???? 放吸
(4) The efficiency of the
cycle is
%QQ 621 ???
放
吸?
V
P b
ca
3,Refrigerators(致冷机 ） and performance coefficient
（ 致冷系数 ）
We can think of a
refrigerator as a heat engine
operating in reverse as shown
in Fig.**,It takes heat from a
cold place and give it off to a
warmer place.
Performance coefficient is
defined as
吸放
吸吸
Q
W
Qw
???
放Q
吸Q
W
Hot reservoir
at T1
Cold reservoir
at T2
Refrigerator
Although their efficiencies differ from one another,
none of the heat engines has an efficiency of 100%.
放Q
吸Q
W
Hot reservoir
at T1
Cold reservoir
at T2
Refrigerator
吸Q
放Q
W
Cold reservoir
at T2
Hot reservoir
at T1
Heat engine
Question,What is the maximum attainable efficiency
of the heat engines that work between two same
reservoirs?
m ax ??
It was answered by卡
诺,Carnot,1796~1832，
法国物理学家，工程师，
热力学奠基人，提出卡诺
循环和卡诺定理。
吸Q
放Q
W
Cold reservoir
at T2
Hot reservoir
at T1
Heat engine
3.Carnot cycle
The Carnot cycle is shown
in a P-V diagram in which:
BC, adiabatic expansion（ 绝热 ） and done works on
external world;
V
P A
B
C
D
T1
T2
吸Q
放Q
AB,isothermal expansion
（ 等温 膨胀 ） and done
works on external world;
吸Q
放QCD,isothermal compression（ 压缩 ） and done
works on the system;
DA, adiabatic compression（ 压缩 ） and done works
on external world;
When a heat engine finish
a Carnot cycle,work is done
on the external world by
system,That is
放吸 QQW ??
in which:
a
b
V
VlnRTMQ
?? ?吸
d
c
c
d
V
VlnRTM
V
VlnRTMQ
?? ?? ??放
V
P A
B
C
D
T1
T2
吸Q
放Q
Using the adiabatic equations of path
??
?
??
?
??
?
??
?
?
?
??
??
da
cb
VTVT
VTVT
d
c
a
b
V
V
V
V ? V
P A
B
C
D
T1
T2
吸Q
放Q
Therefore,the efficiency
of heat engine whose cycle
is Carnot cycle is (卡诺热
机 ）,
1
211
T
T
Q
Q
????
吸
放?
吸Q
放Q
W
Cold reservoir
at T2
Hot reservoir
at T1
Heat engine
Similarly,the performance
coefficient of refrigerators
operating Carnot cycle in
reverse is
21
2
TT
T
w
?
?
V
P A
B
C
D
T1
T2
放Q
吸Q
放Q
吸Q
W
Hot reservoir
at T1
Cold reservoir
at T2
Refrigerator
Notes:
(1) There are two heat sources（热源） in Carnot cycle;
(2) ? or w are decided only by the temperatures of
two heat source,and independent of the working
substance（ 工作物质 ） ;
(3) The higher T1 is,the greater ?; the lower T2,
the greater ?,
(4) ?<1(100%);
讨论：
卡诺循环必须有高温和低温两个热源。
卡诺循环的效率只与两个热源的温度有关，与工
作物质无关。
卡诺循环的效率总是小于 1的。
T2愈低或 T1愈高，卡诺循环的效率愈大。
工程上一般采取提高高温热源温度的方
法。
Example 7-6:一卡诺循环热机, 高温热源的温度是 400K，
每一循环从此热源吸进 100J热量并向一低温热源放出 80J
热量 。 求 （ 1） 这循环的热机的效率； （ 2） 低温热源的
温度 。
解：（ 1） 这循环的热机的效率为：
%QQ ???? ? ?????????
吸
放?
（ 2）设低温热源的温度 T2,有
%TQQ ???? ? ??????? ?
吸
放?
KT 3 2 02 ?
吸Q
放Q
W
Cold reservoir
at T2
Hot reservoir
at T1
Heat engine
Example 7-7:卡诺致冷机的低温热源温度为 T2=300K，
高温热源温度为 T1=450K,每一循环从低温热源吸进
400J热量 。 求 （ 1） 致冷机的制冷系数； （ 2） 每一循
环中外界必须作的功 。
解,(1)致冷机的致冷系数等于：
????????? ??????
??
?
TT
Tw
（ 2）每一循环中外界必须作的功
W
Q
TT
Tw 吸?
?? ??
?
JW 2 0 0?
放Q
吸Q
W
Hot reservoir
at T1
Cold reservoir
at T2
Refrigerator
The efficiency of heat engines
（热机的效率）,
吸
放
吸
放吸
吸 Q
Q
Q
Q
W ????? 1?
Is it possible that so that ?
reaches 100%?
0?放Q
良好的愿望！！！
吸Q
放Q
W
Cold reservoir
at T2
Hot reservoir
at T1
Heat engine
0?放Q
§ 7-7 The Second Law of Thermodynamics
热力学第二定律
Performance coefficient
吸放
吸吸
Q
W
Qw
???
??
or is no work by the external
word needed to operate
refrigerator?
放Q
吸Q
W
Hot reservoir
at T1
Cold reservoir
at T2
Refrigerator
There is another principle,independent of the first law
and not derivable from it,that is the second law of
thermodynamics.
在 热力学第一定律之外还有其它定律制约着热功
转换的效率。那就是热力学第二定理。
There are,in general,two statements of the second law
of thermodynamics(热力学第二定律 ）,
Kelvin statement（开尔文说法）,
不可能从单一热源吸取热量,
使它完全变为有用的功而不引
起其它变化 。 In other words,it
is impossible in principle for
any heat engine to have a
thermal efficiency of 100%.
If the second law were not true,it would be possible
to power an automobile or to run a power plant(发电
厂 ） by extracting（ 提取, 采掘 ） heat from the
surrounding air or water in the Pacific Ocean.
Clausius statement（克劳修斯说法）：
不可能把热量从低温物体
传到高温物体而不引起其它
变化 (other effects)。 That is
heat flows spontaneously
（ 自发地 ） from hotter to
colder bodies,never the
reverse（ 倒转 ）,
不需要电或其它能源的
冰箱还未问世，而且永远
不可能出现；如可能，那
就是南极洲了！
放Q
吸Q
W
Hot reservoir
at T1
Cold reservoir
at T2
Refrigerator
节能
Note:
1.What’s meaning of,no other effects”?
3.The two statements are completely equivalent（ 等
价的 ）,That is 热力学第二定律的两种表述形式是等
效的, 若其中一种说法成立, 则另一种说法也成立；
反之亦然 （ 见书上证明 ） 。
2.The second law is not a deduction（ 推论 ） from
the first but stands by itself as a separate（ 独立 ）
law of nature.
W
热源
4.The directions of nature processes,that is the
processes take place naturally in only one direction.
The second law of thermodynamics is an expression
of the inherent(内在的, 固有的 ） one-way aspect
of processes 除非 ……,.。
(不能相信李洪志的歪理邪说 ; 人出生，即意味着走向
死亡的开始）
成长
5,Other statements of the second law of thermodynamics.
热力学第二定律不是推出来的，而是从大量
客观实践中总结出来的规律，因此，不能直接验
证其正确性。
非常重要, 记住两种说法 !!!!!!
如：说出的话，泼出去的水是收不回来的除非 ………,。
A
定义：
1,Reversible Process & Irreversible Process
Irreversible Process,Process P(A?B) if there is
no way to make system back to the initial state
without other effects.
P(A?B)
There is no way(other processes)
B
§ 7-8 Reversible Process & Irreversible Process
可逆过程与不可逆过程 Carnot Theorem卡诺定理
A
P(A?B)
B
Reversible Process,Process P(A?B) if there is a
way to make system back to the initial state
without other effects.
There is a way(other process)
则：过程 P(A?B)叫可逆过程。
Irreversible Processes,examples
（ 1） The process in which work transfers into
heat is irreversible(功转换为热,摩擦生热 );
搅拌作功
2,The process in which heat flows from a hotter
body to a colder one is irreversible(热传导 );
hot cold
3.The free expansion（自由膨胀） of gas is
irreversible;
真空
A B
4,The rapid expansion（快速） of gas is
irreversible;
A B
△ V
………………….
Reversible processes,(1) The process must be
equilibruim(平衡 );(2) there is no friction(摩擦 )
in each step of process.
● 过程无限缓慢，为准静态过程；
● 没有耗散力作功，即没有摩擦发生。
It is concluded that all the real
processes are irreversible(与热现象有关
的实际过程都是不可逆的 ).
过去的就让它过去吧 ! 但你可以调节下一步的方向 !!!
Angry smile
future
(1) 在温度为 T1 和 T2 两个热源之间工作的
任意可逆卡诺热机具有相同的效率
(2) 不可逆卡诺机的效率不可能大于可逆卡诺
机的效率。
卡诺定理指出了提高热机效率的途径,(1)过程
尽可能接近准静态 ;(2)提高两热源的温度差,
2,Carnot Theorem 卡诺定理
1
21
T
T???
1
21
T
T???
1.Thermodynamic Probability ? 热力学几率 ?
Take the free expansion of
gas as an example,As
shown in Fig.**,there are
four molecules in the vessel.
In beginning,all the four
molecules lie in the par t A
of vessel,When the middle
plank is removed,the four
molecules may be moving
in the whole vessel.
A
B
麦克斯韦妖
§ 7-9 Statistical Meaning of the Second Law of
Thermodynamics 热力学第二定律的统计意义
波耳兹曼关系式
A
B
We cannot see the
molecules that are moving
random(无序 ).We imagine
there would be a
balance(天平 ） that can
weigh the weight of the
molecule,The results of
two MEN are listed in the
table.
微观 abcd dabc abdc abdc acdb abcd acbd adbc ad ac ab bcd acd abd abc
abcd
dcbadcbdbc
A1 A2 A3 A4 A5
A
B
1 4 6 4 1
宏观
A
B
微观 abcd dabc abdc abdc acdb abcd acbd adbc ad ac ab bcd acd abd abc
abcd
dcbadcbdbc
A1 A2 A3 A4 A5
A
B
1 4 6 4 1
宏观
任一宏观状态所对应的微观状态数称为该宏观状态的
热力学几率，记为 ?。 对于孤立系统，各个微观状态出现
的可能性（或概率）是相同的。
显然, 一个宏观态的 ?越大, 这个 一个宏观态
出现的概率 。
11 ?? 42 ?? 63 ?? 44 ?? 15 ??
16
1
16
4
16
6
16
4
16
1
6
If the number of molecules is very great,it will be
found that the system lie in the macrostate A3 in
which there are same number of molecules in the
part A and part B,The number of microstate for A3 is
very greater than that for other macrostate.
A
B
所以, 系统总是由几率小的宏观状态向几率大
的宏观状态进行, 在没有外界的影响下, 反过程
是不可能的 。 这就是热力学第二定理的统计意义 。
Important!!!
2,Boltzmann Relation & The Principle of the
entropy increasing 波耳兹曼关系式和熵增加原理
这就是著名的波耳兹曼关系式。
?? lnkS
Boltzmann Relation 波耳兹曼关系式
对一个大数粒子组成的系统，如 N =1023, 热力学几
率 ?将是一个巨大的数字，不便使用,引入了一个新的
热力学函数 —熵，记为 S
S,熵是一个态函数。
A
B
The principle of the entropy increasing 熵增加原理
?
?
?? ??? Ω
ΩlnkSSS?
脚标 2表示末态，脚标 1表示初态，既然一切自然过程总是
沿着热力学几率增大的方向进行，即
???
?
?
?? Ω
ΩΩΩ 或
0 12 ?? SS
A
B
称为 熵增加原理，它表明 一切自然过程（不可逆过程）
总是沿着 熵增加 的方向进行。
0 ?? S
在绝热过程中系统的 熵永不减少, 对于可逆绝热
过程, 系统的熵不变, 对于不可逆绝热过程, 系统的
熵增加, 这一结论称为熵增加原理 。
Important!!!!
我们可以根据熵的变化判断实际过程进行的方
向和限度，基于此，我们把熵增加原理看作是热力
学第二定律的另一叙述形式。
孤立系中不可逆过程总是朝着熵增加的方向进
行，直至达到熵最大。
一个不受外界影响的封闭系统, 其内部发生的
过程, 总是由概率小的状态向概率大的状态进行,
由包含微观状态数目少的状态向包含微观状态数目
多的状态进行 。 这是熵增加原理的实质, 也是热力
学第二定律的统计意义之所在 。
无序性问题 ………… 。
Entropy 熵,1854年 R.克劳修斯引入的判断过程进
行方向的物理量, 1865年他给这个物理量定名为 熵 。
1948年 C.E.香农将 熵 的概念同信息论联系起来,
这样, 便可以对信息的计量, 传输和储存进行理论研
究 。 此外, 熵 在控制论, 概率论, 数论, 天体物理乃
乃至生命科学等领域, 有广泛的应用 。
