Chapter 18 Second Law of Thermodynamics
The second law of thermodynamics
Reversible and irreversible process (可逆过程与不可逆过程 ):
Carnot cycle(卡诺循环 )
Chapter 18,Second Law of
Thermodynamics（ 热力学第二定律 ）
Entropy(熵 )
Chapter 18 Second Law of Thermodynamics
1 开尔文说法：不可能制造出这样一种 循环 工作的热机，它只使 单一 热源冷却来做功，而不 放出热量给其他物体，或者说 不 使 外 界发生任何变化,
第二定律的提出
1 功热转换的条件第一定律无法说明,
2 热传导的方向性、气体自由膨胀的不可逆性问题第一定律无法说明,
一 热力学第二定律的两种表述
Chapter 18 Second Law of Thermodynamics
New words
Engine Efficiency 热机效率
Reversible and Irreversible Process
可逆和不可逆过程 p433
Entropy 熵 p440
Chapter 18 Second Law of Thermodynamics
§ 18-1 The second Law of Thermodynamics P429-430
1 功热转换 p429
通过摩擦而使功变热的过程是不可逆的 。
例如：飞轮转动,焦耳实验 （ 重物下落 ） 。
The initial potential energy of the rock changes to
kinetic energy as the rock falls,but you will never
see the reverse happen-a rock at rest on the
ground suddenly rise up in the air because the
internal energy is transformed into kinetic energy.
Some one-way processes P429-430
自然过程是不可逆的，是按一定的方向进行的。
Chapter 18 Second Law of Thermodynamics
Heat flows naturally from a hot object to a cold
object; heat will not flow spontaneously a cold
object to a hot object.
2 热传导现象 P429
热量由高温物体 自动地 传向低温物体的过程是不可逆的。
Second Law of Thermodynamics,p430
Heat flows naturally from a hot object to a cold
object; heat will not flow spontaneously form a
cold object to a hot object.
Chapter 18 Second Law of Thermodynamics
§ 18-3 Reversible and irreversible process (可逆过程与不可逆过程 ),p433
Reversible process(可逆过程 )
----无摩擦的准静态过程,
Reversible process is carried out infinitely
slowly,so that the process can be considered a
series of equilibrium states,and the whole
process could be done in reverse with no change
in magnitude of the work done or heat
exchanged.
Chapter 18 Second Law of Thermodynamics
准静态无摩擦过程为可逆过程可逆过程,在系统状态变化过程中,如果逆过程能重复正过程的每一状态,而不引起其他变化,这样的过程叫做可逆过程,
Chapter 18 Second Law of Thermodynamics
Irreversible process is one that can not be
reversed by means of small changes in the
environment.
在 不引起其他变化 的条件下,不能使逆过程重复正过程的每一个状态的过程称为不可逆过程,
(自然过程都是不可逆的 ) All real process are
irreversible.
非 准静态过程为不可逆过程,
Chapter 18 Second Law of Thermodynamics
§ 18-2 Heat Engines P430
The working substances absorb heat from high
temperature resource;
one part of energy
becomes work,and
the other part release
to the low
temperature resource.
After a cycle,the
working substances
resume their original
state.
Chapter 18 Second Law of Thermodynamics
历史上热力学理论最初是在研究热机工作过程的基础上发展起来的。热机是利用热来做功的机器。 1698
年萨维利和 1705年纽可门先后发明了 蒸汽机,当时蒸汽机的效率极低,1765年瓦特进行了重大改进,大大提高了效率,人们一直在为提高热机的效率而努力，从理论上研究热机效率问题,一方面指明了提高效率的方向,另一方面也推动了热学理论的发展,
各种热机的效率液体燃料火箭 柴油机汽油机 蒸汽机
%48
%8
%37
%25
1 热机发展简介
Chapter 18 Second Law of Thermodynamics
热机,持续地将热量转变为功的机器,
工作物质 （工质）：热机中被利用来吸收热量并对外做功的物质,
Chapter 18 Second Law of Thermodynamics
高温热源 T1
低温热源 T2
热机
W
放Q
吸Q
热机从高温热源吸取热量,一部分转变成功,另一部分放到低温热源 。 The working
substances absorb heat
from high temperature
resource; one part of
energy becomes work,
and the other part
release to the low
temperature resource.
2 Schematic working process of steam engine
Chapter 18 Second Law of Thermodynamics
In general,if a system
starts from a certain initial
state,through a closed
series of thermodynamic
processes,the system
returns to its initial state,
then these processes are
called a cycle.
3 循环过程 Cycle
系统经过一系列变化状态过程后，又回到原来的状态的过程叫 热力学循环过程,
p
Vo
A
B
AV
BV
c
d
Chapter 18 Second Law of Thermodynamics
4 循环过程的特点
important features of cyclic process
经过一个循环，内能不变。
循环曲线所包围的面积为系统做的净功。
循环曲线为闭合曲线。
DU = 0
p
Vo
W
A
B
AV
BV
c
d
Chapter 18 Second Law of Thermodynamics
p
Vo
W
A
B
AV
BV
c
d
从状态 A经状态 c达到状态 B，系统对外界做功 W1，
系统对外做 净 功的数值为 W=W1-W2
BA A c B VVSW?1
从状态 A经状态 d达到状态
B，外界对系统做功 W2，
BA A d B VVSW?2
A d B cSW?
Chapter 18 Second Law of Thermodynamics
热力学第一定律 WQ?
2Q
总放热 （ 取绝对值）
QQQW 21
净功
1Q
总吸热即：工质以传热的方式从 高温热库 得到能量，
有一部分仍以传热的方式放给 低温热库,二者的差额等于工质对外做的 净功 。
Chapter 18 Second Law of Thermodynamics
p
Vo
A
B
AV
BV
c
d
正循环 热机循环过程沿 顺时针 方向进行,系统对外做功,返回时,系统放热,对外做负功；
循环面积为正值,这种循环叫正循环,或热循环 。
热机高温热源低温热源
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
逆循环 制冷机循环过程沿 逆时针 方向进行，外界对系统做功，
返回时，系统放热，对外做负功；循环面积为负值。
这种循环叫逆循环，或 致冷循环 。
W
致冷 机高温热源低温热源
p
Vo
A
B
AV
BV
c
d
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
5 Efficiency of a Engine P432
对于正循环,其效率是在一次循环过程中工质对外做的净功占它从高温热库吸收的热量的比率 。 其是热机效能的一个重要标志 。
热机高温热源低温热源
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
热机高温热源低温热源
1Q
2Q
W
热机（ 正 循环） 0?W
吸Q
W
e?
The efficiency e of any engine is
defined by
.
||
||
|h e at a b s o r b e d|
| w o r kdt r an s f o r m e|
HQ
We
Chapter 18 Second Law of Thermodynamics
1
21
Q
Qe
热机高温热源低温热源
1Q
2Q
W
,|| 放吸 QQW由能量守恒吸放吸
Q
QQe ||
吸放
Q
Q ||1
1?
Using Ther.-I-law to the whole cyclic process,
LHne t QQQUU,021?
c y c l en e tn e t AWQ
LH QWQ
ne tLH WQQ
Chapter 18 Second Law of Thermodynamics
P432,Example 18-1
Chapter 18 Second Law of Thermodynamics
致冷机致冷系数
21
22
QQ
Q
W
Q
e
致冷机（ 逆 循环） 0?W
coefficient of performance
Chapter 18 Second Law of Thermodynamics
1 Kelvin-Planck statement,不可能制造出这样一种 循环 工作的热机,它只使单一 热源冷却来做功,而 不 放出热量给其他物体,或者说 不 使 外 界发生任何变化,即热全部转变为功的过程是不可能的 。
1851年开尔文总结出热力学过程进行的限度。
§ 18-3 Second Law of Thermodynamics –
further discussion P432
No device is possible whose sole effect is totransform a given amount of heat completely into
work.
Chapter 18 Second Law of Thermodynamics
高温热源 T1
热机 W
吸Q吸 放吸 Q
QQe || 1?
由于 e=1是不可能的，
第二类永动机（单热机）不能制成。
热 功?
%10 0
Chapter 18 Second Law of Thermodynamics
No device is possible whose sole effects is to
transfer heat from one system at one
temperature into a second system at a higher
temperature.
2 Clausius statement of the second law of
Thermodynamics:
不可能把热量从低温物体 自动 传到高温物体而 不 引起 外界的变化,
p438
Chapter 18 Second Law of Thermodynamics
高温热源 T1
低温热源 T2
吸Q
放Q
Chapter 18 Second Law of Thermodynamics
3 两种表述的统一性
1.从开尔文表述入手假定单热机是可以造成的，则高温源 低温源Q
高温热源 T1
低温热源 T2
1Q WQ?2
单热机 2Q
致冷机高温热源 T1
低温热源 T2
2Q
W )( 1Q
Chapter 18 Second Law of Thermodynamics
从克劳修斯表述入手反之,假定热量能自动地从低温源传到高温源,
则单热机也能造成 。
高温热源 T1
低温热源 T2
2Q 热机 2Q
W1Q
高温热源 T1
21 QQ?
W
单热机
Chapter 18 Second Law of Thermodynamics
永动机
1、第一类永动机不需要能量输入而能继续做功的机器叫第一类用动机。
第一类永动机违反了热力学第一定律，因此是不可能的。
1、第二类永动机有能量输入的单热源热机叫第二类用动机。
第二类永动机违反了热力学第二定律，因此是不可能的。
Chapter 18 Second Law of Thermodynamics
永动机的设想图
Chapter 18 Second Law of Thermodynamics
非 自发传热自发传热高温物体 低温物体? 热传导
热功转换完全功不 完全热自然界一切与热现象有关的实际宏观过程都是不可逆的,
热力学第二定律的实质无序有序 自发非均匀、非平衡 均匀、平衡自发
Chapter 18 Second Law of Thermodynamics
§ 18-5 Refrigerators,Air Conditioners,and
Heat Pumps P438-440
1 致冷循环,
如果工质做逆循环,即沿着与热机循环相反的方向进行循环，则在一次循环中，工质从低温库中吸热，向 高温热库放热,而外界必须对工质做功 W,由于工质 从低温库中吸热使它的温度降低，这种循环叫 致冷循环 。
Chapter 18 Second Law of Thermodynamics
高温热源 1T
低温热源 2T
致冷机
1Q
2Q
W 这就是说 工质 把从低温库中吸收的热和外界对它做的功一并以热量的形式传给 高温热库 。
由热力学第一定律
21 QQW WQQ 21
Chapter 18 Second Law of Thermodynamics
工质用较易液化的物质,如氨 。 氨 气 在 压 缩 机
(compressor motor)内被急速压缩,它的压强增大 (high
pressure,而且温度升高,进入冷凝器 (heat exchanger
or condenser)（ 高温热库 ） 后,由于冷却水 （ 或周围空气 ） 放热而凝结为液态氨 (cool to become liquid)。 液态氨经节流阀 (low-pressure tube)的小口通道后,降压降温,再进入蒸发器 。 此处由于压气机的抽吸作用因而压强很低 。 液态氨将从冷库 （ 低温热度 ） 中吸热,
使冷库温度降低而自身全部蒸发为蒸气 。 此氨蒸气最后被吸入压气机进行下一循环 。
第七节 致冷循环 / 四、冰箱工作原理2 冰箱工作原理 (p438) figure 18-9
Chapter 18 Second Law of Thermodynamics
冰箱循环示意图
Chapter 18 Second Law of Thermodynamics
3 Coefficient of performance (CP)p438
QL为工质从低温热库吸收的热量,W是外界对工质做的功，吸热越多，做功越少，则致冷机性能越好。
致冷机致冷系数
LH
LL
QQ
Q
W
Q
e
致冷 机高温热源低温热源
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
||||
||
LH
L
QQ
Q
W
QCP L?
O
p
V
The efficiency of any refrigerator (致冷系数
coefficient of performance) can be defined as
Based on the Ther-I-law,
|W | = |QH| - |QL|
Chapter 18 Second Law of Thermodynamics
New words and expressions
Engine Efficiency 热机效率
Reversible 可逆过程 p433
Irreversible Process 不可逆过程 p433
Entropy 熵 p440
Review
Chapter 18 Second Law of Thermodynamics
Two kinds of statement of the
second law of thermodynamics
1 R.J.E.Clausius Statement (P430)
Heat flows naturally from a hot object to a
cold object; heat will not flow spontaneously
form a cold object to a hot object.
热量由高温物体 自动地 传向低温物体的过程是不可逆的。
Chapter 18 Second Law of Thermodynamics
不可能制造出这样一种 循环 工作的热机,它只使单一 热源冷却来做功,而 不 放出热量给其他物体,或者说 不 使 外 界发生任何变化,即热全部转变为功的过程是不可能的 。
2 Kelvin-Planck statement,(P438)
No device is possible whose sole effect is to
transform a given amount of heat completely
into work.
Chapter 18 Second Law of Thermodynamics
Efficiency of a Engine
热机高温热源低温热源
1Q
2Q
W
The efficiency of any engine is
defined by
.
||
||
|h eat ab s o r b ed|
| w o r kdt r an s f o r m e|
1Q
We
1
21
Q
Q
e
Chapter 18 Second Law of Thermodynamics
21
22
QQ
Q
W
Q
e
致冷 机高温热源低温热源
1Q
2Q
W
The coefficient of performance (CP)p438
Chapter 18 Second Law of Thermodynamics
The work done in isothermal process
1
2
V
Vn R T ln
2
1ln
p
pn R T?dV
V
RTnWQ V
V
T
2
1
绝热方程
TV 1?
pV
Tp 1
常量常量常量
Equations of adiabatic process
Chapter 18 Second Law of Thermodynamics
§ 18-3 Carnot Engine and Its Efficiency P433-436
French scientist and engineer
N,L,Sadi Carnot first
proposed the ideal engine’s
concept in 1824.
1824 年法国的年青工程师卡诺提出一个工作在 两 热源之间的 理想 循环 —卡诺 循环,给出了热机效率的理论极限值 ; 他还提出了著名的卡诺定理,
Chapter 18 Second Law of Thermodynamics
萨迪,卡诺是法国青年工程师,热力学的创始人之一,是第一个把热和动力联系起来的人 。 他出色地,创造性地用,理想实验,的思维方法,提出了最简单,但有重要理论意义的热机循环 —— 卡诺循环,并假定该循环在准静态条件下是可逆的,与工质无关,创造了一部理想的热机 （ 卡诺热机 ） 。
卡诺的目标是揭示热产生动力的真正的,
独立的过程和普遍的规律 。 1824年卡诺提出了对热机设计具有普遍指导意义的卡诺定理,指出了提高热机效率的有效途径,揭示了热力学的不可逆性,被后人认为是热力学第二定律的先驱 。
Chapter 18 Second Law of Thermodynamics
1 Carnot Engine (p433)
Carnot engine is ideal engine,all processes
are reversible and no wasteful energy transfers
occur due to friction.
Carnot cycle consists
of two isothermal
processes and two
adiabatic processes.
卡诺 循环是由两个准静态 等温 过程和两个准静态 绝热 过程组成,理想气体的卡诺循环
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
abQ
cdQ
Chapter 18 Second Law of Thermodynamics
Operation of Carnot engine
低温热源 2T
高温热源 1T
卡诺热机
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
abQ
cdQ
Isothermal
Adiabatic
2 Efficiency of Carnot Engine (p433-435)
A--B expand isothermally
B--C expand adiabatically
C--D compress at constant
temperature
D--A compress adiabatically
卡诺循环
Chapter 18 Second Law of Thermodynamics
1,a?b,isothermal process T1,feature is DU = 0.
2
1
2
1
d
d
1
V
1
V
V
V
ab
V
V
n R T
VpWQQ
1
2
1 ln V
Vn R T?
1
2
11 ln V
Vn R TWQ
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
abQ Isothermal
A B,使气缸和温度为 T1 的高温库接触,气体 等温膨胀,体积由 V1增到 V2，它从高温库中 吸收热量 Q1
(1)
Chapter 18 Second Law of Thermodynamics
2,b?c the adiabatic process
(2)
V
p
2T
1T
A
B
CD
1p
2p
4p
3p
21 TT?
Adiabatic
B C,将气缸从高温库移开，使气体做绝热膨胀，
体积变为 V3，温度降到 T2.
213112 TVTV
Chapter 18 Second Law of Thermodynamics
4
3
22 ln V
Vn R TQQ
cd
3,c?d,Isothermal process T2,DU = 0
(3)
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
cdQ
Isothermal
C D：使气缸和温度为 T2 的低温库接触,使气体等温压缩,体积由 V3减小到 V4,状态 D和 A位于同一条绝热线上,气体向低温库中 放出热量 Q2.
Chapter 18 Second Law of Thermodynamics
4,d?a the adiabatic process
214111 TVTV
(4)
4
3
1
2
V
V
V
V?
214111 TVTV
由 213112 TVTV
(3)
V
o
2T
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
Adiabatic
D A：将气缸从低温库移开，沿 绝热线压缩 气体，
直到它回到起始状态 A，体积变为 V1,完成一个循环。
p
Chapter 18 Second Law of Thermodynamics
1
2
4
3
1
2
1
2
ln
ln
11
V
V
V
V
T
T
Q
Q
e
卡诺 循环的效率
1
21
T
T
the efficiency of the Carnot engine is given by
1
21
T
T
e
Chapter 18 Second Law of Thermodynamics
3 Carnot Theorem (P435)
卡诺热机 效率与工作物质无关,只与两个热源的温度有关,两热源的温差越大，则卡诺循环的效率越高,
All reversible engines operating between the
same two constant temperatures TH and TL
have the same efficiency,Any irreversible
engine operating between the same two fixed
temperatures will have an efficiency less than
this.
Chapter 18 Second Law of Thermodynamics
1） 在 相同 高温热源和低温热源之间工作的任意工作物质的 可逆机 都具有 相同 的效率,
2） 工作在 相同 的高温热源和低温热源之间的一切 不 可逆机的效率都 不可能 大于可逆机的效率,
1
21
1
21
T
TT
Q
QQe
( 不 可逆机 )
（ 可逆 机 ）
Chapter 18 Second Law of Thermodynamics
1) The efficiency of the cycle of Carnot engine
depends only on T1,and T2.
卡诺机必须有两个热源 。 热机效率与工作物质无关,
只与两热源温度有关 。
Summary:
2) The efficiency of Carnot engine is always less
than 1,
热机效率不能大于 1 或等于 1，只能小于 1。
例如：波音飞机不用价格较贵的高标号汽油作燃料，
而采用航空煤油作燃料。
Chapter 18 Second Law of Thermodynamics
A phony claim,A engine manufacturer maker the
following claims,The heat input per second of the
engine is 9.0kJ at 475K,The heat output per second
is 4.0kJ at 325K,Do you believe these claims?
SOLUTION,The efficiency of the engine is:
56.0
0.9
0.411
1
2
Q
Qe
However,the maximum possible efficiency is given
by the Carnot efficiency,
Example 18-3 on page 435
32.0
4 7 5
3 2 511
1
2
T
Te
So the manufacturer’s claims violate the second law
of thermodynamics and cannot be believed.
Chapter 18 Second Law of Thermodynamics
Example,One mole of a monatomic ideal gas is
taken through the reversible cycle shown in Fig.
Process bc is an adiabatic expansion,with
pb=10.0atm and Vb=1.00?10-3 m3,Find (a ) the
energy added to the gas as heat,(b) the energy
leaving the gas as heat,(c) net work done by the
gas,and (d) the efficiency of the cycle.
Solution:
(a) a?b,V=const,W=0
)J(1047.1)(23 3D babVin VPPTnCUQ
Vb 8Vb V
p
idiabatic
o
c
b
a
Chapter 18 Second Law of Thermodynamics
( J)109,1 8||Q||QQW( c) 2o u tin
%4.62)(
inQ
Wed
( J )105,54)V(Vp25ΔTnCQ( b) 2cacpo u t
Chapter 18 Second Law of Thermodynamics
Carnot Refrigerator
The basic elements of an ideal refrigerator that
operates in the reverse of the Carnot engine.
4 Carnot Refrigerators (致冷机 ) and Its Efficiency (439)
V
o
p
2T
W
1T
A
B
CD
21 TT?
2Q
1Q 高温热源 1T
低温热源 2T
卡诺致冷机
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
Adiabatic
Isotherm
O
p
VV1
ap1
p2
p3
p4
V2 V3V4
b
cdQ
L
QH
Since Carnot refrigerator
is a Carnot engine
operating in reverse,
so the coefficient of
performance of Carnot’s
cycle is:
LH
L
TT
TCP
||||
||
LH
L
QQ
QCP
or
Chapter 18 Second Law of Thermodynamics
§ 18-5 &18-6 Entropy & The Second Law of
Thermodynamics (P440-445)
We have not arrived at a general statement of
the second law of thermodynamic,This general
statement can be made in terms of a quantity,
called entropy (熵 ).
1 Entropy and Change in Entropy (440-441):
In the study of the Carnot cycle,we found that
H
L
H
L
T
T
Q
Q?
||
||
H
H
L
L
T
Q
T
Q ||||?
Chapter 18 Second Law of Thermodynamics
Recalling our original convention as used before,
0
L
L
H
H
T
Q
T
Q
Now consider any reversible
cycle,as represented by the
smooth curve,Any
reversible cycle can be
approximated as a series of
Carnot cycles.
0 TQ
So we can write
Chapter 18 Second Law of Thermodynamics
0 TdQ
The symbol means take the integral around a
closed path.?
系统经历任意可逆循环一周后，
其热温比之和为零。
In the case of infinitely Carnot cycles,the eqution
becomes
Chapter 18 Second Law of Thermodynamics
Let’s divide the cycle of Fig,into two parts,then
0III abba TdQTdQ
0
III
b
a
b
a T
dQ
T
dQ
baba TdQTdQ III
the integral of dQ/T between any two
equilibrium states does not depend on the path
of the process.
Chapter 18 Second Law of Thermodynamics
We define a new quantity — entropy S
T
dQdS?
From above Eq.,for reversible cycle,we have
0dS
The change of entropy DS should be
D babaab TdQdSSSS
It is independent of the path between the two
points a and b,This is an important result.
— entropy is a state variable!
Chapter 18 Second Law of Thermodynamics
2 Entropy and the 2nd law of Thermodynamics:
热力学第二定律的统计表述 —— 熵增加原理,
The entropy of an isolated system never
decreases,It either stays constant (reversible
process) or increases (irreversible process).
孤立系统中的熵永不减少,
Since all real processes are irreversible,we can
state the second law as:
The total entropy of any system plus that of its
environment increases as a result of any natural
process:
0?D?D?D e n vs y s t SSS
Chapter 18 Second Law of Thermodynamics
在 封闭 系统中发生的任何 不可逆 过程,都导致了整个系统熵的 增加,系统的总熵只有在 可逆 过程中 才是不变的,
孤立系统 不 可逆过程 0?DS
孤立系统 可逆 过程 0?DS0?D S
孤立系统中的 可逆 过程，其熵不变；孤立系统中的不 可逆过程，其熵要增加,
平衡态 A 平衡态 B （ 熵不变）可逆 过程非平衡态 平衡态（熵增加）不可逆 过程自发过程
Chapter 18 Second Law of Thermodynamics
熵增加原理的应用,
给出自发过程进行方向的判据,
Entropy tells us the direction of process go,
Hence it has been called ―time’s arrow‖.
热力学第二定律亦可表述为,一切自发过程总是向着熵增加的方向进行,
熵增加原理与热力学第二定律熵的概念建立，使热力学第二定律得到统一的定量的表述,
Chapter 18 Second Law of Thermodynamics
Example 18-6 Entropy change in a free expansion.
Consider the adiabatic free expansion of n moles of
an ideal gas from volume V1 to volume V2,where
V2>V1,Calculate the change in entropy (a) of the gas
and (b) of the surrounding environment,(c) Evaluate
△ S for 1.00 mole,with V2= 2.00V1.
SOLUTION
Chapter 18 Second Law of Thermodynamics
Example 18-7 Heat conduction,A red-hot 2.0kg
piece of iron at temperature T1=880K is thrown into a
huge lake whose temperature is T2=280K,Assume
the lake is so large that its temperature rise is
insignificant,Determine the change in entropy (a) of
the iron and (b) of the surrounding environment(the
lake.
SOLUTION
Chapter 18 Second Law of Thermodynamics
In each examples,the entropy of our system plus
that of the environment either stayed constant or
increased.
The second law of thermodynamics
Reversible and irreversible process (可逆过程与不可逆过程 ):
Carnot cycle(卡诺循环 )
Chapter 18,Second Law of
Thermodynamics（ 热力学第二定律 ）
Entropy(熵 )
Chapter 18 Second Law of Thermodynamics
1 开尔文说法：不可能制造出这样一种 循环 工作的热机，它只使 单一 热源冷却来做功，而不 放出热量给其他物体，或者说 不 使 外 界发生任何变化,
第二定律的提出
1 功热转换的条件第一定律无法说明,
2 热传导的方向性、气体自由膨胀的不可逆性问题第一定律无法说明,
一 热力学第二定律的两种表述
Chapter 18 Second Law of Thermodynamics
New words
Engine Efficiency 热机效率
Reversible and Irreversible Process
可逆和不可逆过程 p433
Entropy 熵 p440
Chapter 18 Second Law of Thermodynamics
§ 18-1 The second Law of Thermodynamics P429-430
1 功热转换 p429
通过摩擦而使功变热的过程是不可逆的 。
例如：飞轮转动,焦耳实验 （ 重物下落 ） 。
The initial potential energy of the rock changes to
kinetic energy as the rock falls,but you will never
see the reverse happen-a rock at rest on the
ground suddenly rise up in the air because the
internal energy is transformed into kinetic energy.
Some one-way processes P429-430
自然过程是不可逆的，是按一定的方向进行的。
Chapter 18 Second Law of Thermodynamics
Heat flows naturally from a hot object to a cold
object; heat will not flow spontaneously a cold
object to a hot object.
2 热传导现象 P429
热量由高温物体 自动地 传向低温物体的过程是不可逆的。
Second Law of Thermodynamics,p430
Heat flows naturally from a hot object to a cold
object; heat will not flow spontaneously form a
cold object to a hot object.
Chapter 18 Second Law of Thermodynamics
§ 18-3 Reversible and irreversible process (可逆过程与不可逆过程 ),p433
Reversible process(可逆过程 )
----无摩擦的准静态过程,
Reversible process is carried out infinitely
slowly,so that the process can be considered a
series of equilibrium states,and the whole
process could be done in reverse with no change
in magnitude of the work done or heat
exchanged.
Chapter 18 Second Law of Thermodynamics
准静态无摩擦过程为可逆过程可逆过程,在系统状态变化过程中,如果逆过程能重复正过程的每一状态,而不引起其他变化,这样的过程叫做可逆过程,
Chapter 18 Second Law of Thermodynamics
Irreversible process is one that can not be
reversed by means of small changes in the
environment.
在 不引起其他变化 的条件下,不能使逆过程重复正过程的每一个状态的过程称为不可逆过程,
(自然过程都是不可逆的 ) All real process are
irreversible.
非 准静态过程为不可逆过程,
Chapter 18 Second Law of Thermodynamics
§ 18-2 Heat Engines P430
The working substances absorb heat from high
temperature resource;
one part of energy
becomes work,and
the other part release
to the low
temperature resource.
After a cycle,the
working substances
resume their original
state.
Chapter 18 Second Law of Thermodynamics
历史上热力学理论最初是在研究热机工作过程的基础上发展起来的。热机是利用热来做功的机器。 1698
年萨维利和 1705年纽可门先后发明了 蒸汽机,当时蒸汽机的效率极低,1765年瓦特进行了重大改进,大大提高了效率,人们一直在为提高热机的效率而努力，从理论上研究热机效率问题,一方面指明了提高效率的方向,另一方面也推动了热学理论的发展,
各种热机的效率液体燃料火箭 柴油机汽油机 蒸汽机
%48
%8
%37
%25
1 热机发展简介
Chapter 18 Second Law of Thermodynamics
热机,持续地将热量转变为功的机器,
工作物质 （工质）：热机中被利用来吸收热量并对外做功的物质,
Chapter 18 Second Law of Thermodynamics
高温热源 T1
低温热源 T2
热机
W
放Q
吸Q
热机从高温热源吸取热量,一部分转变成功,另一部分放到低温热源 。 The working
substances absorb heat
from high temperature
resource; one part of
energy becomes work,
and the other part
release to the low
temperature resource.
2 Schematic working process of steam engine
Chapter 18 Second Law of Thermodynamics
In general,if a system
starts from a certain initial
state,through a closed
series of thermodynamic
processes,the system
returns to its initial state,
then these processes are
called a cycle.
3 循环过程 Cycle
系统经过一系列变化状态过程后，又回到原来的状态的过程叫 热力学循环过程,
p
Vo
A
B
AV
BV
c
d
Chapter 18 Second Law of Thermodynamics
4 循环过程的特点
important features of cyclic process
经过一个循环，内能不变。
循环曲线所包围的面积为系统做的净功。
循环曲线为闭合曲线。
DU = 0
p
Vo
W
A
B
AV
BV
c
d
Chapter 18 Second Law of Thermodynamics
p
Vo
W
A
B
AV
BV
c
d
从状态 A经状态 c达到状态 B，系统对外界做功 W1，
系统对外做 净 功的数值为 W=W1-W2
BA A c B VVSW?1
从状态 A经状态 d达到状态
B，外界对系统做功 W2，
BA A d B VVSW?2
A d B cSW?
Chapter 18 Second Law of Thermodynamics
热力学第一定律 WQ?
2Q
总放热 （ 取绝对值）
QQQW 21
净功
1Q
总吸热即：工质以传热的方式从 高温热库 得到能量，
有一部分仍以传热的方式放给 低温热库,二者的差额等于工质对外做的 净功 。
Chapter 18 Second Law of Thermodynamics
p
Vo
A
B
AV
BV
c
d
正循环 热机循环过程沿 顺时针 方向进行,系统对外做功,返回时,系统放热,对外做负功；
循环面积为正值,这种循环叫正循环,或热循环 。
热机高温热源低温热源
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
逆循环 制冷机循环过程沿 逆时针 方向进行，外界对系统做功，
返回时，系统放热，对外做负功；循环面积为负值。
这种循环叫逆循环，或 致冷循环 。
W
致冷 机高温热源低温热源
p
Vo
A
B
AV
BV
c
d
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
5 Efficiency of a Engine P432
对于正循环,其效率是在一次循环过程中工质对外做的净功占它从高温热库吸收的热量的比率 。 其是热机效能的一个重要标志 。
热机高温热源低温热源
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
热机高温热源低温热源
1Q
2Q
W
热机（ 正 循环） 0?W
吸Q
W
e?
The efficiency e of any engine is
defined by
.
||
||
|h e at a b s o r b e d|
| w o r kdt r an s f o r m e|
HQ
We
Chapter 18 Second Law of Thermodynamics
1
21
Q
Qe
热机高温热源低温热源
1Q
2Q
W
,|| 放吸 QQW由能量守恒吸放吸
Q
QQe ||
吸放
Q
Q ||1
1?
Using Ther.-I-law to the whole cyclic process,
LHne t QQQUU,021?
c y c l en e tn e t AWQ
LH QWQ
ne tLH WQQ
Chapter 18 Second Law of Thermodynamics
P432,Example 18-1
Chapter 18 Second Law of Thermodynamics
致冷机致冷系数
21
22
Q
W
Q
e
致冷机（ 逆 循环） 0?W
coefficient of performance
Chapter 18 Second Law of Thermodynamics
1 Kelvin-Planck statement,不可能制造出这样一种 循环 工作的热机,它只使单一 热源冷却来做功,而 不 放出热量给其他物体,或者说 不 使 外 界发生任何变化,即热全部转变为功的过程是不可能的 。
1851年开尔文总结出热力学过程进行的限度。
§ 18-3 Second Law of Thermodynamics –
further discussion P432
No device is possible whose sole effect is totransform a given amount of heat completely into
work.
Chapter 18 Second Law of Thermodynamics
高温热源 T1
热机 W
吸Q吸 放吸 Q
QQe || 1?
由于 e=1是不可能的，
第二类永动机（单热机）不能制成。
热 功?
%10 0
Chapter 18 Second Law of Thermodynamics
No device is possible whose sole effects is to
transfer heat from one system at one
temperature into a second system at a higher
temperature.
2 Clausius statement of the second law of
Thermodynamics:
不可能把热量从低温物体 自动 传到高温物体而 不 引起 外界的变化,
p438
Chapter 18 Second Law of Thermodynamics
高温热源 T1
低温热源 T2
吸Q
放Q
Chapter 18 Second Law of Thermodynamics
3 两种表述的统一性
1.从开尔文表述入手假定单热机是可以造成的，则高温源 低温源Q
高温热源 T1
低温热源 T2
1Q WQ?2
单热机 2Q
致冷机高温热源 T1
低温热源 T2
2Q
W )( 1Q
Chapter 18 Second Law of Thermodynamics
从克劳修斯表述入手反之,假定热量能自动地从低温源传到高温源,
则单热机也能造成 。
高温热源 T1
低温热源 T2
2Q 热机 2Q
W1Q
高温热源 T1
21 QQ?
W
单热机
Chapter 18 Second Law of Thermodynamics
永动机
1、第一类永动机不需要能量输入而能继续做功的机器叫第一类用动机。
第一类永动机违反了热力学第一定律，因此是不可能的。
1、第二类永动机有能量输入的单热源热机叫第二类用动机。
第二类永动机违反了热力学第二定律，因此是不可能的。
Chapter 18 Second Law of Thermodynamics
永动机的设想图
Chapter 18 Second Law of Thermodynamics
非 自发传热自发传热高温物体 低温物体? 热传导
热功转换完全功不 完全热自然界一切与热现象有关的实际宏观过程都是不可逆的,
热力学第二定律的实质无序有序 自发非均匀、非平衡 均匀、平衡自发
Chapter 18 Second Law of Thermodynamics
§ 18-5 Refrigerators,Air Conditioners,and
Heat Pumps P438-440
1 致冷循环,
如果工质做逆循环,即沿着与热机循环相反的方向进行循环，则在一次循环中，工质从低温库中吸热，向 高温热库放热,而外界必须对工质做功 W,由于工质 从低温库中吸热使它的温度降低，这种循环叫 致冷循环 。
Chapter 18 Second Law of Thermodynamics
高温热源 1T
低温热源 2T
致冷机
1Q
2Q
W 这就是说 工质 把从低温库中吸收的热和外界对它做的功一并以热量的形式传给 高温热库 。
由热力学第一定律
21 QQW WQQ 21
Chapter 18 Second Law of Thermodynamics
工质用较易液化的物质,如氨 。 氨 气 在 压 缩 机
(compressor motor)内被急速压缩,它的压强增大 (high
pressure,而且温度升高,进入冷凝器 (heat exchanger
or condenser)（ 高温热库 ） 后,由于冷却水 （ 或周围空气 ） 放热而凝结为液态氨 (cool to become liquid)。 液态氨经节流阀 (low-pressure tube)的小口通道后,降压降温,再进入蒸发器 。 此处由于压气机的抽吸作用因而压强很低 。 液态氨将从冷库 （ 低温热度 ） 中吸热,
使冷库温度降低而自身全部蒸发为蒸气 。 此氨蒸气最后被吸入压气机进行下一循环 。
第七节 致冷循环 / 四、冰箱工作原理2 冰箱工作原理 (p438) figure 18-9
Chapter 18 Second Law of Thermodynamics
冰箱循环示意图
Chapter 18 Second Law of Thermodynamics
3 Coefficient of performance (CP)p438
QL为工质从低温热库吸收的热量,W是外界对工质做的功，吸热越多，做功越少，则致冷机性能越好。
致冷机致冷系数
LH
LL
Q
W
Q
e
致冷 机高温热源低温热源
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
||||
||
LH
L
Q
W
QCP L?
O
p
V
The efficiency of any refrigerator (致冷系数
coefficient of performance) can be defined as
Based on the Ther-I-law,
|W | = |QH| - |QL|
Chapter 18 Second Law of Thermodynamics
New words and expressions
Engine Efficiency 热机效率
Reversible 可逆过程 p433
Irreversible Process 不可逆过程 p433
Entropy 熵 p440
Review
Chapter 18 Second Law of Thermodynamics
Two kinds of statement of the
second law of thermodynamics
1 R.J.E.Clausius Statement (P430)
Heat flows naturally from a hot object to a
cold object; heat will not flow spontaneously
form a cold object to a hot object.
热量由高温物体 自动地 传向低温物体的过程是不可逆的。
Chapter 18 Second Law of Thermodynamics
不可能制造出这样一种 循环 工作的热机,它只使单一 热源冷却来做功,而 不 放出热量给其他物体,或者说 不 使 外 界发生任何变化,即热全部转变为功的过程是不可能的 。
2 Kelvin-Planck statement,(P438)
No device is possible whose sole effect is to
transform a given amount of heat completely
into work.
Chapter 18 Second Law of Thermodynamics
Efficiency of a Engine
热机高温热源低温热源
1Q
2Q
W
The efficiency of any engine is
defined by
.
||
||
|h eat ab s o r b ed|
| w o r kdt r an s f o r m e|
1Q
We
1
21
Q
Q
e
Chapter 18 Second Law of Thermodynamics
21
22
Q
W
Q
e
致冷 机高温热源低温热源
1Q
2Q
W
The coefficient of performance (CP)p438
Chapter 18 Second Law of Thermodynamics
The work done in isothermal process
1
2
V
Vn R T ln
2
1ln
p
pn R T?dV
V
RTnWQ V
V
T
2
1
绝热方程
TV 1?
pV
Tp 1
常量常量常量
Equations of adiabatic process
Chapter 18 Second Law of Thermodynamics
§ 18-3 Carnot Engine and Its Efficiency P433-436
French scientist and engineer
N,L,Sadi Carnot first
proposed the ideal engine’s
concept in 1824.
1824 年法国的年青工程师卡诺提出一个工作在 两 热源之间的 理想 循环 —卡诺 循环,给出了热机效率的理论极限值 ; 他还提出了著名的卡诺定理,
Chapter 18 Second Law of Thermodynamics
萨迪,卡诺是法国青年工程师,热力学的创始人之一,是第一个把热和动力联系起来的人 。 他出色地,创造性地用,理想实验,的思维方法,提出了最简单,但有重要理论意义的热机循环 —— 卡诺循环,并假定该循环在准静态条件下是可逆的,与工质无关,创造了一部理想的热机 （ 卡诺热机 ） 。
卡诺的目标是揭示热产生动力的真正的,
独立的过程和普遍的规律 。 1824年卡诺提出了对热机设计具有普遍指导意义的卡诺定理,指出了提高热机效率的有效途径,揭示了热力学的不可逆性,被后人认为是热力学第二定律的先驱 。
Chapter 18 Second Law of Thermodynamics
1 Carnot Engine (p433)
Carnot engine is ideal engine,all processes
are reversible and no wasteful energy transfers
occur due to friction.
Carnot cycle consists
of two isothermal
processes and two
adiabatic processes.
卡诺 循环是由两个准静态 等温 过程和两个准静态 绝热 过程组成,理想气体的卡诺循环
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
abQ
cdQ
Chapter 18 Second Law of Thermodynamics
Operation of Carnot engine
低温热源 2T
高温热源 1T
卡诺热机
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
abQ
cdQ
Isothermal
Adiabatic
2 Efficiency of Carnot Engine (p433-435)
A--B expand isothermally
B--C expand adiabatically
C--D compress at constant
temperature
D--A compress adiabatically
卡诺循环
Chapter 18 Second Law of Thermodynamics
1,a?b,isothermal process T1,feature is DU = 0.
2
1
2
1
d
d
1
V
1
V
V
V
ab
V
V
n R T
VpWQQ
1
2
1 ln V
Vn R T?
1
2
11 ln V
Vn R TWQ
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
abQ Isothermal
A B,使气缸和温度为 T1 的高温库接触,气体 等温膨胀,体积由 V1增到 V2，它从高温库中 吸收热量 Q1
(1)
Chapter 18 Second Law of Thermodynamics
2,b?c the adiabatic process
(2)
V
p
2T
1T
A
B
CD
1p
2p
4p
3p
21 TT?
Adiabatic
B C,将气缸从高温库移开，使气体做绝热膨胀，
体积变为 V3，温度降到 T2.
213112 TVTV
Chapter 18 Second Law of Thermodynamics
4
3
22 ln V
Vn R TQQ
cd
3,c?d,Isothermal process T2,DU = 0
(3)
V
o
p
2T
W
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
cdQ
Isothermal
C D：使气缸和温度为 T2 的低温库接触,使气体等温压缩,体积由 V3减小到 V4,状态 D和 A位于同一条绝热线上,气体向低温库中 放出热量 Q2.
Chapter 18 Second Law of Thermodynamics
4,d?a the adiabatic process
214111 TVTV
(4)
4
3
1
2
V
V
V
V?
214111 TVTV
由 213112 TVTV
(3)
V
o
2T
1T
A
B
CD
1p
2p
4p
3p
1V 4V 2V 3V
21 TT?
Adiabatic
D A：将气缸从低温库移开，沿 绝热线压缩 气体，
直到它回到起始状态 A，体积变为 V1,完成一个循环。
p
Chapter 18 Second Law of Thermodynamics
1
2
4
3
1
2
1
2
ln
ln
11
V
V
V
V
T
T
Q
Q
e
卡诺 循环的效率
1
21
T
T
the efficiency of the Carnot engine is given by
1
21
T
T
e
Chapter 18 Second Law of Thermodynamics
3 Carnot Theorem (P435)
卡诺热机 效率与工作物质无关,只与两个热源的温度有关,两热源的温差越大，则卡诺循环的效率越高,
All reversible engines operating between the
same two constant temperatures TH and TL
have the same efficiency,Any irreversible
engine operating between the same two fixed
temperatures will have an efficiency less than
this.
Chapter 18 Second Law of Thermodynamics
1） 在 相同 高温热源和低温热源之间工作的任意工作物质的 可逆机 都具有 相同 的效率,
2） 工作在 相同 的高温热源和低温热源之间的一切 不 可逆机的效率都 不可能 大于可逆机的效率,
1
21
1
21
T
TT
Q
QQe
( 不 可逆机 )
（ 可逆 机 ）
Chapter 18 Second Law of Thermodynamics
1) The efficiency of the cycle of Carnot engine
depends only on T1,and T2.
卡诺机必须有两个热源 。 热机效率与工作物质无关,
只与两热源温度有关 。
Summary:
2) The efficiency of Carnot engine is always less
than 1,
热机效率不能大于 1 或等于 1，只能小于 1。
例如：波音飞机不用价格较贵的高标号汽油作燃料，
而采用航空煤油作燃料。
Chapter 18 Second Law of Thermodynamics
A phony claim,A engine manufacturer maker the
following claims,The heat input per second of the
engine is 9.0kJ at 475K,The heat output per second
is 4.0kJ at 325K,Do you believe these claims?
SOLUTION,The efficiency of the engine is:
56.0
0.9
0.411
1
2
Q
Qe
However,the maximum possible efficiency is given
by the Carnot efficiency,
Example 18-3 on page 435
32.0
4 7 5
3 2 511
1
2
T
Te
So the manufacturer’s claims violate the second law
of thermodynamics and cannot be believed.
Chapter 18 Second Law of Thermodynamics
Example,One mole of a monatomic ideal gas is
taken through the reversible cycle shown in Fig.
Process bc is an adiabatic expansion,with
pb=10.0atm and Vb=1.00?10-3 m3,Find (a ) the
energy added to the gas as heat,(b) the energy
leaving the gas as heat,(c) net work done by the
gas,and (d) the efficiency of the cycle.
Solution:
(a) a?b,V=const,W=0
)J(1047.1)(23 3D babVin VPPTnCUQ
Vb 8Vb V
p
idiabatic
o
c
b
a
Chapter 18 Second Law of Thermodynamics
( J)109,1 8||Q||QQW( c) 2o u tin
%4.62)(
inQ
Wed
( J )105,54)V(Vp25ΔTnCQ( b) 2cacpo u t
Chapter 18 Second Law of Thermodynamics
Carnot Refrigerator
The basic elements of an ideal refrigerator that
operates in the reverse of the Carnot engine.
4 Carnot Refrigerators (致冷机 ) and Its Efficiency (439)
V
o
p
2T
W
1T
A
B
CD
21 TT?
2Q
1Q 高温热源 1T
低温热源 2T
卡诺致冷机
1Q
2Q
W
Chapter 18 Second Law of Thermodynamics
Adiabatic
Isotherm
O
p
VV1
ap1
p2
p3
p4
V2 V3V4
b
cdQ
L
QH
Since Carnot refrigerator
is a Carnot engine
operating in reverse,
so the coefficient of
performance of Carnot’s
cycle is:
LH
L
TT
TCP
||||
||
LH
L
QCP
or
Chapter 18 Second Law of Thermodynamics
§ 18-5 &18-6 Entropy & The Second Law of
Thermodynamics (P440-445)
We have not arrived at a general statement of
the second law of thermodynamic,This general
statement can be made in terms of a quantity,
called entropy (熵 ).
1 Entropy and Change in Entropy (440-441):
In the study of the Carnot cycle,we found that
H
L
H
L
T
T
Q
Q?
||
||
H
H
L
L
T
Q
T
Q ||||?
Chapter 18 Second Law of Thermodynamics
Recalling our original convention as used before,
0
L
L
H
H
T
Q
T
Q
Now consider any reversible
cycle,as represented by the
smooth curve,Any
reversible cycle can be
approximated as a series of
Carnot cycles.
0 TQ
So we can write
Chapter 18 Second Law of Thermodynamics
0 TdQ
The symbol means take the integral around a
closed path.?
系统经历任意可逆循环一周后，
其热温比之和为零。
In the case of infinitely Carnot cycles,the eqution
becomes
Chapter 18 Second Law of Thermodynamics
Let’s divide the cycle of Fig,into two parts,then
0III abba TdQTdQ
0
III
b
a
b
a T
dQ
T
dQ
baba TdQTdQ III
the integral of dQ/T between any two
equilibrium states does not depend on the path
of the process.
Chapter 18 Second Law of Thermodynamics
We define a new quantity — entropy S
T
dQdS?
From above Eq.,for reversible cycle,we have
0dS
The change of entropy DS should be
D babaab TdQdSSSS
It is independent of the path between the two
points a and b,This is an important result.
— entropy is a state variable!
Chapter 18 Second Law of Thermodynamics
2 Entropy and the 2nd law of Thermodynamics:
热力学第二定律的统计表述 —— 熵增加原理,
The entropy of an isolated system never
decreases,It either stays constant (reversible
process) or increases (irreversible process).
孤立系统中的熵永不减少,
Since all real processes are irreversible,we can
state the second law as:
The total entropy of any system plus that of its
environment increases as a result of any natural
process:
0?D?D?D e n vs y s t SSS
Chapter 18 Second Law of Thermodynamics
在 封闭 系统中发生的任何 不可逆 过程,都导致了整个系统熵的 增加,系统的总熵只有在 可逆 过程中 才是不变的,
孤立系统 不 可逆过程 0?DS
孤立系统 可逆 过程 0?DS0?D S
孤立系统中的 可逆 过程，其熵不变；孤立系统中的不 可逆过程，其熵要增加,
平衡态 A 平衡态 B （ 熵不变）可逆 过程非平衡态 平衡态（熵增加）不可逆 过程自发过程
Chapter 18 Second Law of Thermodynamics
熵增加原理的应用,
给出自发过程进行方向的判据,
Entropy tells us the direction of process go,
Hence it has been called ―time’s arrow‖.
热力学第二定律亦可表述为,一切自发过程总是向着熵增加的方向进行,
熵增加原理与热力学第二定律熵的概念建立，使热力学第二定律得到统一的定量的表述,
Chapter 18 Second Law of Thermodynamics
Example 18-6 Entropy change in a free expansion.
Consider the adiabatic free expansion of n moles of
an ideal gas from volume V1 to volume V2,where
V2>V1,Calculate the change in entropy (a) of the gas
and (b) of the surrounding environment,(c) Evaluate
△ S for 1.00 mole,with V2= 2.00V1.
SOLUTION
Chapter 18 Second Law of Thermodynamics
Example 18-7 Heat conduction,A red-hot 2.0kg
piece of iron at temperature T1=880K is thrown into a
huge lake whose temperature is T2=280K,Assume
the lake is so large that its temperature rise is
insignificant,Determine the change in entropy (a) of
the iron and (b) of the surrounding environment(the
lake.
SOLUTION
Chapter 18 Second Law of Thermodynamics
In each examples,the entropy of our system plus
that of the environment either stayed constant or
increased.
