昆明理工大学：大学物理（一）：Chapter6(1)

Part Two Thermodynamics
第 二 篇 热学
第六章 气体动理论
Chapter 6 Kinetic Theory of Gases
第七章 热力学基础
Chapter 7 Thermodynamics
Chapter 6 The Kinetic Theory of Gases
气体动理论
§ 6-1 Essential Concept of the Kinetic Theory of
Gases 气体动理论的基本概念
§ 6-3 Representation of Pressure for Ideal Gas
气体动理论的压强公式
§ 6-4 Average Translational Kinetic Energy and
Temperature 理想气体的 平均平动动能和 温度
公式
§ 6-2State Parameters Equilibrium State Ideal Gas
Law 状态参量 平衡态 理想气体状态方程
§ 6-6 Maxwell Speed Distribution
麦克斯韦速率分布律
§ 6-7 Mean Free Path & Average Collision Rate
分子的平均碰撞次数及平均自由程
§ 6-8 Boltzmann Distribution
玻耳兹曼分布律
§ 6-5 Equipartition Theory of Energy Internal
energy 能量均分定理, 理想气体的内能
1,对分子无规则热运动有一个清晰的图景；
2,掌握气体分子运动论的两个基本公式 —理想气
体的压强公式及平均平动动能与温度的关系式，
理解压强和温度的微观解释；
3.掌握能均分原则和理想气体内能公式；
4.理解麦克斯韦速率分布律，明确分布曲线的物
理意义；
5,理解分子的平均自由程和平均碰撞次数的规律。
教学基本要求
Introduction
一、重要人物：
焦耳,J.P.Joule 1818-1889，英
国物理学家，职业是酿酒商。
是发现能量守恒与转换定律的
代表人物，英国皇家学会会员，
法国科学院院士。
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 。
克劳修斯,Clausius,1822~1888,
德国物理学家, 提出热力学第
二定理 （ 1850）, 提出熵的概
念 （ 1865）, 给出理想气体压
强公式 。
麦克斯韦,Maxwell,1831~1879,
英国物理学家, 数学家, 主要
贡献,（ 1） 电磁理论； （ 2）
热学：麦克斯韦速度分布；各
态经历假说 。
玻 耳 兹 曼, Boltzmann,
1844~1906,奥地利物理学家,
主要贡献：能量分布定律和
热力学第二定理的统计解释 。
吉布斯,Gibbs,1839~1903,美
国物理学家, 化学家, 现代化
学热力学和统计物理学的奠基
人 。
热能的应用带来了第一次工业革命！
温
饱
基本问题
二、热的本质
人类对热的认识：
（ 1）古代：水、木、金、火、土；
（ 2）热质说：是没有质量的流质；
（ 3）热的分子学说：是物质
运动的一种表现，即分子运
动的表现。
三、研究方法
The studied object（对象） is,
containing a vast（ 大量 ）
number of particles:
a system
molecules
The adopted methods are:
(1) Thermodynamics （ 热力学 ）, study heat
phenomena in the view of energy transformation
based on some experimental laws.
(2) Statistical Mechanics or physics（ 统计力学或统计
物理 ）, based on the mechanics law and the
statistical（ 统计 ） theory.
§ 6-1 Essential Concepts of the Kinetic Theory of
Gases 气体动理论的基本概念
1,All matters consist of myriad （ 无数 ） of tiny
微小 molecules which are are separated
????? ???????? m o l.N
A mole of any pure substance contains a definite
number of identical molecules,which is called
Avegadro’s constant N0,denoted by
太多！！地球村,60亿！
2,The molecules are moving forever and
random(无序 ）
Famous experiment—Brownian Motion.
Liquid
无规运动
Pollen grain花粉
3,There is interaction between
molecules.
The force is attractive when
their distance is less than r0;
otherwise,it is repulsive(排斥的 ）
m10 100 ??r
The interaction force between
molecules varies with the
distance between molecules,as
shown in Fig.6-1.
f
r
r00
Fig.6-1
（ Attention,
boy~girl,18.7cm)
§ 6-2 State Parameters Equilibrium State Ideal
Gas law 状态参量 平衡态 理想气体状态方程
1,State Parameters of Gas 状态参量
For a gas,in order to
describe its properties,three
parameters （ 参数 ） are
needed:
(1) volume V,geometric parameter,cubic meters m3
(2) pressure P,mechanical parameter,Pascal,N/m2
(3) temperature T,thermal parameter,K or C°
33m10?Note,1litter （ 1公升） =
),,( TVP
These parameters are macroscopic quantities(宏观
量 ）,which are called as state parameter of the system.
),,( TVP
(V,P,T)
2,Equilibrium state（平衡） and Equilibrium
process
The gas,mass of m,is contained
in a vessel（ 容器 ） and has a
volume of V,If the gas has a same
pressure and temperature
everywhere,the gas is said to be
in an equilibrium state.
力学平衡,P
化学平衡,m
热学平衡,T
对气体的平衡态：（ P,V,T）
A system(gas) in equilibrium state can be
described by its state parameters(P,V,T) among
which there is a relationship(关系 ),In some cases,it is
simple enough that it can be expressed（ 表示成 ） in
the form of an equation
0?),,( TVPf
Hence,there are two variables
to be independent.
(V,P,T)
A equilibrium state can be represented by a dot
（ 点 ） on the pressure—volume diagram (briefly
p-V diagram),shown in the following Figure
(V,P,T)P
V
a
o
),,( aaa TVP
b
),,( bbb TVP
状态变化
P
T
a
o
b
),,( aaa TVP
),,( bbb TVP
T
a
o
b
),,( aaa TVP
),,( bbb TVPV
T-P图
V-T图
Thermodynamic process( 热力学过程 ):the
operation(操作 ） of changing the system from its
initial state(a) to its final state(b) is called a
thermodynamic process as shown in above figure.
),,( aaa TVP ),,( bbb TVP热力学过程
P
V
a
o
b
),,( aaa TVP
),,( bbb TVP
),,( ccc TVP
V
P
o
a
c
),,( aaa TVP
),,( TVP
),,( aaa TVP
Equilibrium process 平 衡 过 程, In an equilibrium
process,the system remains approximately in
thermodynamic equilibrium at all stages,represented by
a smooth curve on p-V diagram.
Is a real process considered as Equilibrium process?
),,( ccc TVP
),,( TVP
3,The ideal gas law 理想气体状态方程
RTMPV
?
?
(6-1)
which is the equation of state of ideal
gas,and holds for equilibrium state.
Here,R is a constant and has the same
value for all gases,called ideal gas
constant(理想气体普适常数 ):
? is mole mass.
For the ideal gas of mass M,its state parameters have a
simple relationship
KJ /m ol318 ??,R in SI units.
(V,P,T)
Eq.(6-1) can be rewritten as
n R TPV ?
or
RTP ???
?
Mn ? the number of mole
where ? is density（密度） of gas.
RT
P
V
M ?? ??
(V,P,T)
We now define an ideal gas as one for
which Eq.(6-1) holds for all pressure
and temperature.
For a fixed mass of an ideal gas,we have
tc o n sTPV t a n? or
2
22
1
11
T
VP
T
VP ?
(V,P,T)
RTMPV
?
?
(6-1)
§ 6-3 Representation of Pressure for Ideal
Gas 理想气体的压强公式
1,The microscopic model of an ideal gas(理想气体模型 ）
⑴ A gas consists of a very number of rapidly moving
molecules which are so small that the volume of the
molecules is negligible(忽略的 ） compared with(与 …,.
相比 ) the volume of gas,That is the size of molecules is
more smaller than the distance between molecules so
that the volume of the molecules is negligible.(与分子间
的距离相比较, 分子的大小可以忽略不计 )。
(V,P,T)
thin稀薄
⑵ Molecules exert no force to each others except for
the instantaneous impulsive force during the
collision with the wall of the container and the
collision with each others(除碰撞的瞬间外, 分子间
没有相互作用力, 分子自由运动 ).
(V,P,T)
Free自由除碰撞
⑶ Molecules are in constantly random（ 无规则的 ）
motion,colliding with each other and with the wall of
the container; all collisions are perfect elastic.（ 分子与
分子间, 分子与器壁间的碰撞都是完全弹性的 ） 。
This implies that there is not any loss of energy.
(V,P,T)
完全弹性碰撞
⑷, The motion of an individual（ 单个 ） molecule
obeys （ 遵守 ） Newton’s law.（ 单个分子的运动遵从
牛顿力学 ） 。
(V,P,T)
服从牛顿定律
The statistical hypothesis(假设 ）,
(V,P,T)
分子
When the gas is in the equilibrium state,the
probability(概率 ） at which molecule move in all the
direction is same.
（ 1）分子向各方向运动机会相等；
（ 2）分子数密度分布均匀；
（ 3）分子速度在各方向分量平均值相等，等于零,
In detail（ 详细 ）,it includes
0??? zyx vvv
(V,P,T)
（ 4） 分子速度在各方向分量平方的平均值相等
（ 6-2a）
222
zyx vvv ??
And it can be proved that
2222
3
1
vvvv zyx ???
（ 6-2b）
Question,How are the statistical average made?
(V,P,T)
For a single particle,its motion and state is
described by the following,position,displacement,
velocity,acceleration,momentum,kinetic energy,
et.al,which are called as micro-quantity(微观量 ） 。
For a system of gas,its
state is decided by macro-
quantity (P,V,T).
Question,what relation is there between macro-
quantities and micro-quantities?
(V,P,T)
? ?.,,,,,,,kEvzyx ?
2,Derivation of pressure equation of ideal 压强公式
的推导
推导依据：理气微观模型和统计假设。
?x
?y
?z
x
y
z
Studied wallA压强P
立方容器，三边长分别为 ?x,?y,?z，分子数为 N。
研究对象
?x
?y
?z
x
y
z
A
个别分子碰撞器壁 A是间歇的；大量分子
碰撞器壁将产生持续的压力 。 或说压强
是大量分子碰撞器壁, 由于每个分子都
给予器壁冲量而产生的宏观效果 。
统计的观点,
压强P
研究方法：
对单个分子 用牛顿运动定律求出一次碰
撞给面的冲量, 再对所有分子给面的冲量
求和, 用统计的方法求出 P 与 v2 的关
系 。
?x
?y
?z
x
y
z
A
压强P
ixixixi mvmvmvI 2?????
⑴ The impulse exerted by one molecule during each
collision（ 先考虑一个分子 i撞击一次施于器壁 A的冲
量）,
A
ix
x
v
t ???
⑵ The times n of collision by a single molecule in one
second（ 平均碰撞一次所用时间 ）
x
ixvnum
??
?
A
x?
x
ix
x
ix
ix
mvv
mvf
??
2
2
2 ???
⑶ The average force exerted by a
single molecule to the wall in one
second（ 单位时间内平均冲力 ）
t
I
f
?
?
1??t
A
f
f?
⑷ The force on the wall by the gas(N molecules):
N 个分子单位时间内对 A面的平均冲力,
?
??
?
?
N
i x
ixmvF
?
(5) The pressure on the wall:
（ 统计的方法 ）
zyA
F
S
FP
??
??
A
??
??
?
??
? ?
?
?
?
N
i
ix
zyx
N
i x
ix
zy
mv
mv
P
??????
?
??
?
?????? ? x
N
i
ix
zyx
vmN
VN
vmN
???
by using statistical hypothesis
??
?
?
?
?
??
?
v
N
v
v
N
ix
x
22
3
1
3
1 vnmvm
V
N ??
A
A
理想气体压强公式,
2
3
1 vnmP ? （ 6-4）
knvmnP ?3
2
2
1
3
2 2 ?? )( （ 6-5）
or
average translational
kinetic energy
平均平动动能
NN
mv
N
v
mvmε
N
i
ik
N
i
i
N
i
i
k
??
?
????
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
Average translational kinetic energy平均平动动能
（ 微观量的统计平均值）(6-6)
A
大量分子
无数次
‘无情’
打击的结果。
A
The physical meaning of the pressure
equation:
knvmnP ?3
2
2
1
3
2 2 ?? )( （ 6-5）
单位体积
的分子数
分子的平均
平动动能
In a word:
The pressure,as a macroscopic parameter of a gas,
is a statistic average quantity because the average
translational kinetic energy and n in Eq.(6-5) have
definite meaning only for a great number of molecules.
Therefore pressure has definite meaning only for a
system that consists a vast number of molecules.
压 强 公 式 的 物 理 意 义,
压强是大数分子 碰撞器壁的统计平均效果, 只
对大数分子组成的系统才具有确定的意义 。
knP ?3
2?
（ 6-5）
Example 6.1:理想气体分子模型的主要内容是什么？
Example 6.2:试从分子运动论的观点解释为什么当气
体的温度升高时, 只要适当地增大容器的容积就可以
使气体的压强保持不变 。
§ 6-4 Average Translational Kinetic Energy and
Temperature平均平动动能和 温度的关系
Taking an ideal gas as example,we try to find the
relationship with the microscopic quantities of
molecules,
(V,P,T)What is the temperature of
a system associated with?
V ?the size of container in which the particles move,
P ? the average translatoinal kinetic energy
For an ideal gas of mass M with mole mass ?,we have
RTMPV ??
Considering
0N
MN
??
the pressure P can be given by
where the constant k is defined as
KJNRk
o
/,2310381 ????
which is called Boltzmann constant（玻耳兹曼常量）,
理想气体
n k TTN RVNRTVMP ???
0?
(6-6)
重要公式
On the other hand,the pressure equation is given by
knP ??
??
we can obtain
Thus the average translational kinetic energy per
molecule depends only on the temperature,not on the
pressure,volume,or molecular species.
kTk 23??
(6-7)
n k TP ?
In other word,that temperature is the
measurement of the average translational kinetic
energy,This implies that the higher the temperature
of a system,the greater the average translational
kinetic energy so that,热运动越剧烈, 。
炙热
大小的量度,表征大量气体分子热运动的剧烈程度,是
一统计平均量,正如压强一样, 温度只对大数分子组成
的系统才有确定的值, 温度对个别分子无意义 。
kT v m k
?
? ?
?
? ?? ?
温度的物理意义
从统计的观点看，温度 T 是分子的 平均平动动能
静悄悄
Dancing is to start
Very hot!!!
Temperature is a quantity which cannot be defined
in terms of mass,length,and time,It is the fourth
fundamental quantity.
Remember that T is always absolute
temperature in Kelvins in this chapter.
第四个基本量，单位（ SI)：
开尔文，简称开,K。
Example 6.3,What is the average translational kinetic
energy of a molecule of a gas at a temperature of 300K?
Solution,From Eq.(6-7),
)J(.
.kTk
???
???
???????
???????????
?
?
?
?
?
?? 温度：
300K
Example 6.4,What is the total translational kinetic
energy of the molecules in one mole of a gas at a
temperature of 300K?
Solution,From Eq.(6-7),we have
)J(.
.kTk
???
???
???????
???????????
?
?
?
?
?
??
)J(NU k ?????? ? ?
Hence,the total kinetic energy of
this gas is
温度：
300K
1 mole
Matter enlargeConsist of enlargeConsist of分子
atoms
enlarge
Consist of
atoms
核
电子
§ 6—5 Equipartition Theorem of Energy
Internal Energy of Ideal Gas
能量均分定理 理想气体的内能
1,Matter:
In the thermodynamics
分子
or
Translational and rotational motions + vibration
The effects of vibration of atoms are ignored in
the following.
Let’s consider the size of the molecules in the gas to
modify the model of ideal gas to approach to the real
gas as possible as.
All the molecules are divided into monatomic（ 单
原子 ）,diatomic（ 双原子 ）,and polyatomic（ 多原
子 ） molecules.
理想气体 计入分子大小的效应
For simplicity,we assume that the atoms in a
molecule are,rigid balls（ 刚性球 ）, connected by
the rigid rods（ 刚性棒 ） at low temperature.
Monatomic molecule:
Particle model of Monatomic molecule
单原子理想气体的质点 模型, 只有平动 。
Example,He, Ar
Diatomic molecule（双原子分子）
Rigid model of diatomic molecule
( Dumbbell model哑铃模型 )
双原子理想气体的刚体 模型
可以平动和转动,
Example,O2,H2,N2
Polyatomic molecule(多原子分子）
Rigid model of polyatomic molecule
多 原子理想气体的刚体模型，可以平动和
转动。
Example,H2O,CH4,NH3
2,The Degrees of Freedom of molecule 分子的
自由度
The number of degrees of freedom is defined as
the independent coordinates introduced to
determine the position of a moving body in
space.(物体的自由度定义为为确定其在空间的位置
引入的独立座标数 )。
The number of degrees of freedom is labeled as i.
i =3
Only translational motion
(1) Monatomic molecule(单原子分子）
(x,y,z)
3 corresponding to translational motion
2 corresponding to rotation
i = 5 = 3+2
(2) Diatomic molecule（ 双原子分子 ）
(x1,y1,z1) (x2,y2,z2)
(3) Polyatomic molecule（多原子分子）
i = 6 = 3+3
3 corresponding to translational motion
3 corresponding to rotation
From below equations：
kT v m k? ? ? ? ? ?? ?
and
????
?
???? vvvv
zyx
We have
kT)vm(vmvmvm zyx
?
??
?
?
?
??
?
??
?
??
?
? ????
3,Equipartition Theorem of Energy 能均分定理
（不严格证明）
It means that on the average(平均来看 ）,the energy
per particle associated with each degree of translational
freedom at equilibrium state is
说明：平均而言, 每一个平动自由度均具有相等的
动能 。
kTVmVmVm zyx 21212121 222 ???
The molecule of a gas takes part in rotation and
vibration at the same time with the exception of
（ 除 …,.) the translational motion,How are the
contribution（ 贡献 ） of these motions to the energy
of molecule considered?
转动
The classical statistical mechanics（ 统计力学 ）
proves that,
The energy per particle associated
with each degree is
kT21
for a gas which is at equilibrium state
with the temperature of T,
This conclusion is called as the principle of the
equipartition of energy（ 能均分定理 ）,
Hence,the total average kinetic energy of a molecule is,
kTik 2??
自由度, i
kTk 23??
kTk ????
in which the average rotational energy is
kT
单原子分子,He,Ar,...
双原子分子,H2,O2,N2,……..
in which the average rotational energy is
kT
2
3
多原子分子,Co2,NH3,…...
kTk 26??
AN
3,Internal Energy of Ideal Gas理想气体的内能
For an ideal gas,from the principle of the
equipartition of energy,the internal energy of one mole
ideal gas is
(6-12) RTikTiNE
A 22 ???
??
?
??
没有势能
所有分子
动能之和
kNR A?
Therefore,the internal energy of M /μ mole ideal gas is
given by
RTiME
2?
?
(6-13)
Note,内能是状态函数,
?
Mn ?
A monatomic ideal gas has only three translationl
degrees of freedom so that its internal energy is equal
to
RTME
2
3
?
?
单原子分子
Note:只有平动 !
For the diatomic ideal gas,its internal energy without
that of vibration is
RTME r
?
?
RTME
2
5
?
?
双原子分子
and the contribution of rotation motion is
The degrees of freedom of polyatomic molecule is six,and
RTME r
2
3
?
?
RTME 3
?
?
多原子分子
and the contribution of rotation
motion is
例题：有容积不同的 A,B两容器, A中装有单原子分
子理想气体, B中装有双原子分子理想气体, 若两种
气体的压强相同, 那么, 这两种气体的单位体积的内
能 和 的关系：? ?
AV
E ? ?
BV
E
（ A） ? ? ? ?
BA V
EVE ? （ B） ? ? ? ?BA VEVE ?
（ C） ? ? ? ?
BA V
EVE ? （ D）不能确定
提示：
平?nP 3
2?
平动动能
§ 6-6 Maxwell Speed Distribution
麦克斯韦速率分布律
1,Introduction
(1) Some molecules are
moving rapidly,and
some slowly,The speed
of the molecule is
varying with the time
and varies over a wide
range of magnitude.
每个分子速度
随时间变化
同一时刻
不同分子
速度不同
(2) The method of photography （ 摄影 ）, in a
equilibrium state,there is a characteristic（ 特征 ）
distribution of molecular speeds for a given gas,which
depends,as we will see below,on the temperature.
(3) We will study the distribution of molecular
speeds by the statistics（ 统计学 ）,
Some definitions:
N,total number of molecules in a gas.
?N,the number of molecules having speeds
between v and v+ ?v.
总分子数
N
?N：
v~v+?v
the ratio of the number of molecules having speeds
between v and v+ ?v to the total number
or
probability at which the speeds of molecule lies
between v and v+ ?v.
总分子数
N
?N：
v~v+?v
N
N? is
vNN ???
where is called as distribution function（ 分布函数 ）
and represents:)v(f
vvfNN ??? )(
Obviously,we have
vN
N)v(f
?
??
N,气体中总分子数；
?N：速度位于 v~v+ ?v的分子数；
N
N?, 速 率 位于 v~v+ ?v的分子占总分子的百分比, 也
可理解为分子速率 位于 v~v+?v的几 率；
)v(f, 速 率 位于 v附近单位间隔 的分子占总分子的百
分比, 也可理解为分子速率 位于 v附近单位间隔的几 率；
非常重要 important!!!!
It is called Maxwell distribution function of speed (麦
克斯韦速率分布函数 ),
?????
?
?
?? ve)
kT
m()v(f
kT
mv
/
?
?
2,Maxwell Distribution of Speed
麦克斯韦速率分布律（函数）
Maxwell found the distribution
function,that is
vve)
kT
m(
N
N
kT
mv
/ ?
?
?? ?????
?
?
??
vve)
kT
m(NN
kT
mv
/ ?
?
?? ?????
?
?
??
Therefore,we have
Some discussions:
(1)The number of molecules between v1 and v2 is
??? 21vv dvvNfN )(
??? 2
1
v
v
dvvf
N
N )(
意义：速率位于 [v1,v2]
的分子数占总分子数
的百分比；或分子速
率位于 [v1,v2]的几率。
1v 2v
)v(f
v
(2) Normalization
（归一化）,
1
0
??
?
dv)v(f
equals to the area below the curve of,)v(f
)v(f
v
面积等于 1
)v(f
v
(3) The characters of the curve:
shape and maximum point
最大点
(4) the relation between the curve and the temperature
for a given gas
m1>m2
(5) the relation between the curve and the mass of
molecule for two kinds of gas at the same temperature
T.
m1
m2
)v(f
v
(6) The physical meaning of,v)v(f ?
)v(f
v
]vv,v[ ??
为图中阴影面积, 代表速率位于 v~v+?v的分
子与总分子的百分比 或分子速率 位于 v~v+?v的几率 。
v)v(f ?
3,Three speeds:
They are:
the most probable speed (最可几速率）
the root-mean-square speed （方均根速率）
average speed （平均速率）,
Pv
2v
v
is the speed at which the distribution function has
maximum value,Using the Maxwell speed
distribution,we have:
Pv
?
RT.
m
kTv
P 411
2 ??
which relates to the probability.
问题,最可几速率 的意义是什么？Pv
)v(f
v
Note,The average value of a quantity can be
calculated by
?
?
?
?
?
?
??
?
0
0
21
dvvfvA
N
dvvNfvA
N
vAvA
A
)()(
)()(
)()( ?
vvNfN ??? )(
is defined as the square root of the average of the
square of speed,It can be calculated as following:
2v
m
kTdvvfv
N
dvvNfv
N
dvvNfvA
v 3
0
20
2
02 ???? ??? ?
??
)(
)()()(
Hence：
which relates to the average kinetic energy of molecule.
??? ?0 4 2 dxex ax adxe ax ?210 2 ??? ?
?
RT.
m
kTv 73132 ?? 方均根速率
（平均速率） can be obtained byv
??
?
?
?
??? ????? dv)v(vf
N
dv)v(v N f
N
vvv ?
It relates to the collisions among the molecules,heat
transfer （ 热传导 ） and diffusion of mass(物质的扩散 ）,
??
RT.
m
kTv 5918 ??
Pizza
The relation among these three speeds,（ 非常重要 !!!）
? for a given gas at the temperature T.
2vvv P ??
? for two different gases with the
same temperature T if m1>m2.
?),m(v)m(v PP 21 ?
m1
m2
)v(f
v
问题：已知 f(v)为麦克斯韦速率分布函数, 则,（ 1）
速率 v>100m/s的分子数占总分子数的百分比表达式为
（ 2） 速率 v>100m/s的分子数表达式为
问题：已知 f(v)为麦克斯韦速率分布函数, vP为分子
的最可几速率, 则
（ 1） 表示 ；? Pv dv)v(f
0
（ 2） 速率 v>vP的分子的平均速率的表达式为：
§ 6—7 The Mean Free Path
分子的平均碰撞次数及平均自由程
1,The mean free path
The distance between the successive( 相继 ）
collision is called the free path,The average of free
path is called the mean free path,labeled as,?
Free path average
什么因素影响分子的
平均自由程？
points
crowded
Size:d Density,n
mean free path
Two factors influencing on the mean free path,
? The size of molecule,diameter(直径） d; if the
molecules is point,what does it happen?
? the density of molecule of gas.
The average collision rate(平均碰撞次数 ） is the
average number of collision per unit time a molecule
suffers as it moves in the gas,labeled as, ObviouslyZ
vZ ??
where is the average speed of the molecules.v
Z
v??
(1)Assuming the diameter of molecule is d and only a
molecule is moving at an average speed,others at
rest,we have
v
nvdZ 2??
d
d
v
横截面积 2d?
一秒钟走的距离。v
(2) Considering all the molecules are moving,it can
be proved that the above formula becomes：
nvdZ 22 ??
d
d
v
分子之间的相对运动！
v v2
(3) The mean free path is given by
ndZ
v
22
1
?? ??
which shows that depends only on the number of
molecule per unit volume.
?
d
d
v
分子大小 分子密度
Using,the mean free path can also be
expressed as
n k TP ?
Pd
kT
22 ?? ?
对空气分子 d ~ 3.5 ? 10 -10 m
标准状态下 Z ~ 6.5 ? 10 9s, ? ~ 6.9 ? 10 -8 m
一些数量级概念（参见 p158表及例 6-5)：
问题：一定量的理想气体, 在温度不变的条件下,
当容器增大时, 分子的平均碰撞次数 和平均自由
程 的变化情况是,Z
?
（ A） 减小而 不变；Z ?
（ C） 增大而 减小；Z ?
（ B） 减小而 增大；Z ?
（ D） 不变而 增大；Z ?
问题：在一封闭容器中盛有 1mole氦气 （ 视为理想气体 ）
这时分子无规则运动的平均自由程仅决定于：
（ A）压强 P； （ B）体积 V；
（ C）温度 T； （ D）平均碰撞次数 。Z