Chapter 22 Molecular properties

of gases

J. J. Thomson discovered electrons in 1897.

22-1 The atomic nature of matter ( 物质的原子本质 )

Rutherford discovered the nature of atomic nucleus.

He was at his lab at McGill Univ. in 1905.

In 1828 Robert Brown observed through his microscope that tiny grains of pollen suspended inwater underwent ceaseless random motion. We now call this phenomenon “Brownian motion”.

The modern trail to belief in atoms can be said to have started in 1828: the observation of Brownian motion.

1. Brownian motion

See 动画库 \ 力学夹 \4-01 布朗运动

(I) The ideal gas consists of particles, which are

in random motion and obey Newton’s Laws of

motion. These particles are “atoms” or “molecules”.

(II) The total number of particles is “large”. The rate at which momentum is delivered to any area A of the container wall is essentially constant.

(III) The volume occupied by molecules is a negligibly small fraction of the volume occupied by the gas.

2. Properties of the ideal gas

(IV) No forces act on a molecule except during a

collision.

(V) All collisions are elastic and negligible duration.

Total kinetic energy of the molecules is a constant, and total potential energy is negligible.

22-2 A molecular view of pressure( 压强 )

We will take the ideal gas as our system. Consider N molecules of an ideal gas confined within a cubical box of edge length L, as

in Fig 22-2.

LL

y

x

z

Lm1A

v

2A

Fig 22-2How to relate pressure to microscopic quantities?

?,~ ... vρP

The average impulsive force exerted by themolecule on is 1A

Lmv

vLmv

F x

x

xx

2

22

(22-4)

The total force on by all the gas molecules is the sum of the quantities for all the molecules.Then the pressure on A1 is (22-5)

1ALmvx /2

)(1

...)(1

22

213

22

21

2

212

xxxx

xx

vvL

m

L

mvmv

L

FFL

P

If N is the total number of molecules in container,

the total mass is Nm. the density is . (22-6)

The quantity in parenthesis is average value of

for all the molecules in the container. (22-7)

3/ LNm

2xv

avxvP )( 2

][

22

3 N

v

N

v

L

mNP i

xii

xi

)( 22

213

xx vvL

mP

Nvvi

xiavx /)()( 22

For any molecules, , and

so Eq(22-7) becomes (22-8)

1.The result is true even when we consider collisions between molecules.2. The result is correct even with consideration of the collisions between molecules and other walls in the box.

3. The result is correct for boxes with any kinds of shape.

2222zyx vvvv

avavzavyavx vvvv )(3

1)()()( 2222

avvP )(3

1 2

avxvP )( 2 (22-7)

(22-9)

In Eq(22-8,9), we relate a macroscopic quantity ( the

pressure P) to an average value of a microscopic

quantity, that is to or .

P

vv avrms

3)( 2

rmsvavv )( 2

4. The “root-mean-square” ( 均方根 )speed of the molecules ：

Sample problem 22-1

Calculate the of at , .

Under these conditions .

Solution:

rmsv cT 0 atmP 132 /1099.8 mkgH

smmkg

PaPvrms /1840

/1099.8

)1001.1(3332

5

2H

Sample problem 22-2

In Fig 22-2 , L=10cm, P=1atm, T=300K(a) How many moles of oxygen are in the

box?(b) How many molecules ?Solution:(a)

(b)

molkmolkJ

mPa

RT

PVn 041.0

)300)(/31.8(

)10.0)(1001.1( 35

molecules

molnNN A

22

23

105.2

)1002.6)(041.0(

L

L

y

x

z

Lm1A

The mass of the H2 molecule is 3.3*10-24g. If

1.6*1023 hydrogen molecules per second strike 2.0

cm2 of wall at an angle of 600 with the normal when

moving with a speed of 1.0*105cm/s, what pressure

do they exert on the wall?

Problem

22-3 The mean free path

1. Mean free path :The average value of the straight-line distance a molecules travels between collisions.

2. Which kinds of physical quantities are related to mean free path?

(22-3)

where P and T are macroscopic quantities pressure and temperature, d is the diameter of a molecule of the gas.

Pd

KT22

See 动画库 \ 力学夹 \4-11 平均自由程

See 动画库 \ 力学夹 \4-10 碰撞频率

For air molecules at sea level, . At an altitude of 100 km, .cm16

m710

Sample problem 22-4

What are (a) the mean free path and (b) the average collision rate for nitrogen at T=300k, ? Suppose , .

Solution:

(a)

(b) The average collision rate is

mPd

KT 8

2103.9

2

mdN101015.3

2

smvav /478PaP 51001.1

ondcollisionsv

Z av sec/101.5 9

22-4 The distribution of molecular speeds1.The Maxwell speed distribution

where N is the total number of molecules; T is temperature, m is the mass of each molecule.N(v) expresses particle number in unit speed unit speed rangerange at v.

KT

mv

evKT

mNvN 222/3

2

)2

(4)(

N(v)Fig 22-6

(22-14)

(c) The number of molecules with speeds in the range from v to v+dv is N(v)dv.

(b)The total number of molecules: (22-15)N is equal to the total area under speed distribution curve in Fig 22-6.

0

)( dvvNN

Notes:

(a) Avoid the temptation to interpret N(v) as “the number of molecules having a speed v”.

Fig 22-7

V(m/s)

N(v) T=80K

T=300K

See 动画库 \ 力学夹 \4-05 麦克斯韦速率分 布律 1 兰媚尔实验

2. Consequences of the speed distribution(i) The most probable speed . It is the

speed at which N(v) has its maximum value. (22-

16)(ii) The average speed

Pv

0)(

dv

vdN

M

RT

m

KTvP

22

KT

mv

evKT

mNvN 222/3

2

)2

(4)(

p

iav

vM

RT

m

KT

dvvvNNN

vv

88

)(1

0(22-17)

(22-18)

N(v)

800

Fig 22-6

rmsvavv

pv

400 600

0

22 )(1

)( dvvNvN

vv avrms

av

rms

vM

RT

m

KTv

33

(iii) The root-mean square speed rmsv

(v) The ideal gas law

p

M

RTvrms

33

VnM /nRTPV

(iv) Average translational kinetic energy per

molecule

KTm

KTm

mvN

vvvm

vvvmN

K

rmsN

Ntrans

2

3)

3(

2

12

1)(

2

1

)(2

1

222

22

1

222

21

(22-21)

Scientists contributed to ideal gas law:

Boyle( 玻意耳 , 英国）， Charles( 查理 , 法国），Gay-Lussac( 盖吕萨克 , 法国）

Sample problem 22-5

The speeds of ten particles in m/s are 0, 1.0, 2.0, 3.0, 3.0, 3.0, 4.0, 4.0, 5.0 and 6.0. Find (a) the average speed (b) (c) of these particles

Solution:(a)

rmsv Pvavv

sm

vN

vn

nav

/1.3)0.60.50.40.30.30.30.20.10[10

1

1 10

1

(b)

(c) Of the ten particles, three have speeds of 3.0m/s, two have speeds of 4.0m/s, and

each of the other has a different speed. Then

sm

vN

vn

nrms

/5.3

)0.60.50.40.30.30.30.20.10[10

1

1

22222222

10

1

2

smvP /0.3

Sample problem 22-6

A container filled with N molecules of oxygen gas is maintained at 300K . What fraction of the molecules has speeds in the range 599-601 m/s? The molar mass M of oxygen is 32g/mol.

The fraction in that interval is ,

where , .

From

smv /2

N

vvN

N

dvvNf

)()(

0

601

599

smv /6000

3106.2 f

Solution:

KT

mv

evKT

mNvN 222/3

2

)2

(4)(

22-5 The distribution of molecular energy

1) Consider a special case that translational kinetic energy is the only only formform of energy that a molecule can have.

2) Let us consider again the situation of Sample problem 22-6.

3) The number with kinetic energies between E and E+dE is the same as the number with speed between v and v+dv.

Mathematically, we express this conclusion as (22-22) (22-23)Since the energy is only kineticthe energy is only kinetic, we must have or , thus (22-24)then

dvvNdEEN )()(

dE

dvvNEN )()(

2

2

1mvE

m

Ev

2

)2

(2 2

1

E

mdE

dv

KT

E

eEKT

NEN

2

1

2/3)(

12)(

(22-25)

KT

mv

evKT

mNvN 222/3

2

)2

(4)(

Eq(22-25) is called Maxwell-Boltzmann energy distribution.

(i) Since we have assumed that the molecules can have only Ktrans, this distribution applies only to a monatomic gas. But Boltzmann Factor is always present in N(E) no matter what the form of energy E.

(ii) The total number of molecules N is determined from (22-26)

0)( dEENN

(iii) Maxwell-Boltzmann distribution is precisely thesame for any gas at a given temperature.

KT

E

eEKT

NEN

2

1

2/3)(

12)(

(22-25)

Boltzmann Factor

Sample problem 22-8

Find (a) The average energy and (b) the most

probable energy of a gas in thermal equilibrium at

temperature T.

Solution:Solution: (a)

Substituting Eq(22-25), we obtain

0

E/KT3/2

3/2av dEeE(KT)π

2E

0

)(1

dEEENN

Eav

KT

E

eEKT

NEN

2

1

2/3)(

12)(

Using Appendix I

(b) Taking the derivative of Eq(22-25), and setting

it equal to zero, then

KTEav 2

3

2

2

1

2

1pP mvKTE

*22-6 Equations of state for real gases

The equation of state for an ideal gas holds well

enough only for real gases at sufficiently low densities.

It does not hold exactly for real gases at any density.

Two methods used for real gases:

nRT,nbVV

anP

2

2

))((

(I) The Virial expansion ( 维里展开 )(II) van der Waals equation of states (proposed in 1873):

(He received the 1910 Nobel prize for his work.)in which a and b are constants obtained by experiment.

(22-33)

1.The volume correction

In section 22-1 we assumed that the volume occupied by the molecules of an ideal gas is negligible. This is not quite true for real gases.

Let us regard each molecule of real gas as a hard sphere of diameter d. We can find an approximate value of b

)3

4(

2

1 3dNb A

2. The pressure correctionFor ideal gas, we assumed that the molecules of an ideal gas exert forces on each other only during collisions. That is not quite true for real gases.

a) If the molecule is not near the wall of the vessel, it would experience no net force due to the forces exerted on it by the surrounding molecules.

b) However, if the molecule is located near the wall of the vessel, as in Fig 22-11. the molecule c would experience a net force of attraction away from the wall.

Thus the pressure measured at the wall is somewhat less than what we may call the true pressure.

If P in Eq 22-33 is to be the measured pressure, we must increase it by to obtain the “true” pressure.

2)(V

na

F

R

R

Fig 22-11

Wall

c