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Chapter 5 Gases

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Chapter 5. Gases. Early Experiments. Torricelli performed experiments that showed that air in the atmosphere exert pressure. Torricelli designed the first barometer. A torricellian barometer. Simple manometer. Unit of Pressure. 1 standard atmosphere =1 atm =760 mm Hg =760 torr - PowerPoint PPT Presentation

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Page 1: Chapter 5

Chapter 5

Gases

Page 2: Chapter 5

2

Early Experiments

Torricelli performed experiments that showed that air in the atmosphere exert pressure.

Torricelli designed the first barometer.

Page 3: Chapter 5

3

A torricellian barometer

Page 4: Chapter 5

4

Simple manometer

Page 5: Chapter 5

5

Unit of Pressure

1 standard atmosphere

=1 atm

=760 mm Hg

=760 torr

=101325 Pa

2

force F NPressure pascal

area A m

Page 6: Chapter 5

6

Boyle’s Experiment

Page 7: Chapter 5

7

Boyle’s Law : PV=k A gas that obeys Boyle’s law is called an ideal

gas

Page 8: Chapter 5

8

Charles’s Law: V=bT

The volume of a gas at constant pressure increases linearly with the temperature of the gas.

Page 9: Chapter 5

9

Plot V vs. T using kelvin scale

Page 10: Chapter 5

10

Avogadro’s Law : V=an A gas at constant temperature and pressure

the volume is directly proportional to the number of moles of gas.

Page 11: Chapter 5

11

Ideal Gas Law

Equation of state for a gas PV=nRT P: atm V: Liter n: moles R: 0.082 L atm K-1

T: K

Page 12: Chapter 5

12

Laws for Gas experiments

Boyle’s Law : PV=k Charles’s Law : V=bT Avogadro’s Law : V=an The ideal Gas Law : PV=nRT

(so called equation of state for idea gas)

Page 13: Chapter 5

13

Ideal Gas The volume of the individual particles can

be assumed to be negligible. The particles are assumed to exert no

force on each other. It expresses behavior that real gases

approach at low pressure and high temperature.

Page 14: Chapter 5

14

Gas StoichiometryStandard Temperature and Pressure (STP) T=0oC P=1 atm V=22.4 L

Natural Temperature and Pressure (NTP) T=25oC P=1 atm V=24.5L

Page 15: Chapter 5

15

Plot of PV versus P for 1 mol of ammonia.

Page 16: Chapter 5

16

Page 17: Chapter 5

17

2NaN3(s)→2Na(s)+3N2(g)

Page 18: Chapter 5

18

Dalton’s Law of Partial Pressures

total 1 2 3

31 2

1 2 3

total

P =P +P +P +......

n RTn RT n RT = + + +....

V V VRT

=(n +n +n ...)( )V

RT =n ( )

V

Page 19: Chapter 5

19

Mole Fraction and Partial Pressure

toraltotal

total

total

PxPP

P

PPP

Px

RTVPRTVPRTVP

RTVP

n

nx

RT

VPn

RT

VPn

RT

VPn

nnn

n

n

nx

111

321

11

321

111

2211

321

111

...

...)/()/()/(

)/(

..... ),( ),()(

...

Page 20: Chapter 5

20

The Model of Ideal Gas in Kinetic-Molecular Theory

The volume of the individual particles can be assumed to be negligible.

The particles are assumed to exert no force on each other.

The particles are in constant motion. The average kinetic energy of a collection of

gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

Page 21: Chapter 5

21

An ideal gas particle in a cube wholse sides are of length L (in meters).

Page 22: Chapter 5

22

The velocity u can be broken down into three perpendicular components, ux, uy, and u2.

Page 23: Chapter 5

23

Pressure of an Ideal Gas

Let the container be a rectangular box with

sides of length Lx, Ly and Lz.

Let v be the velocity of a given molecule.

2 2 2 2

2 2 21 1 1

2 2 2: translational energy

x y z

tr x y z

tr

v v v v

mv mv mv

Page 24: Chapter 5

24

In the xy plane,vx

2 + vy2 = vyx

2 by the Pythagorean theorem.

Page 25: Chapter 5

25

Page 26: Chapter 5

26

A molecule collide with wall W where W is parallel to the xz plane. Let i have the velocity components vx, vy, vz

2

1

2

1

2

1

2

2

1221

t

t xxxw

t

t xx

t

t xxx

xx

xxxx

dtFmv so Fives Fhird law gNewton's t

dtFmv

dtF)(tP)(t P to tg from tIntegratin

dtFPmomentum ddt

dP

dt

dvmmaF

Page 27: Chapter 5

27

The integration can be extended over the whole time interval t1 to t2.

x

xw

xx

x

xw

x

xxx

w

t

twx

l

vmNF

N

vv

l

mvF

v

ltttvl

F

dttFtt

FttFmv

222

2

12

1212

2 ,

Won wall exerted force average theis where

])(1

[ )(22

1

速度平方之平均值

Page 28: Chapter 5

28

The translational energy Etr= 1/2mv2

2 2 21

22

3

2

3

trtr

tr

tr tr

tr

m v vm

PV N

N is the total translation kinetic energy E of gas

PV E

V

vTherefore

vvvv

V

vmN

lll

vmN

ll

lv

mN

ll

F

A

FP

zyx

x

xyz

x

yz

x

x

yz

ww

3

mNP

2

2222

22

2

single particle

Page 29: Chapter 5

29

Temperature dependence with translation kinetic energy

0

0

2 3

3 23 3

2 2

3

2

3 ( )

2'

tr tr

tr tr

tr

PV E nRT E nRT

NE N nRT RT

N

RT

N

kT translational kinetic energy of a molecule

k Boltzmann s constant

single particle

one mole of particles

Page 30: Chapter 5

30

2

2

12 2

1 3 =

2 23

- -

3 3

:

:

tr

rms

rms

m v kT

kTv

m

v is called root mean square speed v

kT RTv

m Mm mass of one gas molecule

M molar mass of the gas

Page 31: Chapter 5

31

Distribution of Molecular Speeds in an Ideal Gas

Root mean square speed is assumed that all

molecules move at the same speed.

The motions of gas molecules should have

distribution of molecular speeds in

equilibrium.

Page 32: Chapter 5

32

( )

( )

( ) :

:

v

v

Use the distribution function G v

dNG v dv

NG v the fraction of molecules with speed

in the range v to v dv

dNthe probability that a molecule will have

Nits speed between v and v dv

Page 33: Chapter 5

33

2

2

3

222

-

( ) 42

.

mvv kT

cv

dN mG v dv e v dv

N kT

Maxwell distribution laws

The function with Ae form is called

normal or gaussian distribution

指數函數→機率

Page 34: Chapter 5

34

Plot of O2 molecules

Page 35: Chapter 5

35

Plot of N2 molecules

Page 36: Chapter 5

36

Application of The Maxwell Distribution

23

232

0 0

3 1

2 2

2

( ) 42

1 8 8 4 or

2 2( )2

8

mv

kT

average speed

mv vG v dv e v dv

kT

m kT kTv

mkT m mkT

RTv

M

Page 37: Chapter 5

37

Application of The Maxwell Distribution

1.225:1.128:1382

20

M

RT:

πM

RT:

M

RT:vv:v

M

RTs v one find

dv

dG(v)when

peedprobable smost

rmsmp

mp

Page 38: Chapter 5

38

Velocity distribution for nitrogen

Page 39: Chapter 5

39

Collisions with a Wall-2 -11

(collisions cm s )wdN

A dt,

,

,

0

0

( ) ( )

:

( ) ( )

( )

1( )

v yy y v y y y

yy

y

y yv y y y

y y

wy y y

y

wy y y

dNg v dv dN Ng v dv

Nv dt

the fraction of molecules within distance v dt of Wl

v dt vdN Ng v dv dt

l l

dN Ng v v dv dt

A l A

dN Ng v v dv

A dt V

vydt

ly

dNw:粒子撞擊牆壁的數目

Page 40: Chapter 5

40

21 1

2 22

0 0

1

2

0

0

1

20

1( )

2 2 4

1

2

1 1 1 8

4 4

ymvy kT

y y y yy

w

w

v m RTg v dv e v dv v

l kT M

dN N RT

A dt V M

PNN NPV nRT RT

N V RT

dN PNN RTv

A dt V RT M

Page 41: Chapter 5

41

3 3 8.314 2981928( / )

0.002

8 8 8.314 2981776( / )

0.002

2 2 8.314 2981574( / )

0.002

rms

mp

RTv m s

M

RTv m s

M

RTv m s

M

H2 at 25oC and 1atm

分子量單位 :Kg

Page 42: Chapter 5

42

1

20

23 11

3 1 -1

24 2 1

1 1 1 8

4 4

1 (1 )(6.02 10 )1776 100

4 (82.06 )(298 )

4.4 10

wdN PNN RTv

A dt V RT M

atm molcm s

cm atm mol K K

cm s

Page 43: Chapter 5

43

Definition of Pressure The pressure of a gas results from collisions

between the gas particles and the walls of the container.

Each time a gas particle hits the wall, it exerts a force on the wall.

An increase in the number of gas particles in the container increases the frequency of collisions with the walls and therefore the pressure of the gas.

Page 44: Chapter 5

44

The effusion of a gas into an evacuated chamber.

Page 45: Chapter 5

45

Suppose there is a tiny hole of area A in the wall

and that outside the container is a vacuum.

Escape of a gas through a tiny hole is called

effusion.

01

2

2

1

(2 )

'

1

2

A PN AdN

dtMRT

Graham law of effusion

Rate of effusion for gas M

Rate of effusion for gas M

collisions × area

Page 46: Chapter 5

46

Diffusion

Relative diffusion rates of NH3 and HCl molecules

Page 47: Chapter 5

47

Molecules Collisions and Mean Free Path

Intermolecular collisions are important in reaction kinetics.

Assume a molecule as a hard sphere. No intermolecular forces exist except at the

moment of collision.zAA: the number of collisions per unit time

that one particular A molecule makes with

other A molecule [collisions s-1]

Page 48: Chapter 5

48

Cylinder swept by gas particles

1( )

2 A B A Bd d r r

Page 49: Chapter 5

49

Calculate zAA and zAB

2

:

,

( ) .

AB

AB

cyl AB A B

v average speed of A relative to B

In time dt the moving molecule will travel a

distance v dt and will sweep out a cylinder

of volume V v dt r r

(rA+rB=d)

Page 50: Chapter 5

50A B

Page 51: Chapter 5

51

Since the stationary B molecules are uniformly distributed throughout the container volume V, the number of B molecules with centers in the cylinder equals (Vcyl/V)NB.

2

2

( ) ( ) ( )

( )

cyl B BAB AB A B AB

cyl AB A B

V N Nz z r r v

V dt V

V v dt r r

Page 52: Chapter 5

52

2 2

2

12 2 2 2

12 2

1 1 12 2 02 2 2

8 8

( ) ( )

( ) [ ] ( )

8 1 1( ) [ ( )] ( )

82 2 ( )

AB A BA B

BAB A B AB

BAB A B A B

BAB A B

A B

AAAA A A A

A

RT RTv v v

M M

Nz r r v

V

Nz r r v v

V

RT Nz r r

M M V

If B A

P NN RTz d v d

V M RT

Page 53: Chapter 5

53

Mean Free Path The average distance of a molecule travels

between collisions. In a mixture of gases A and B, A differs from

B.

:

:

: ( )

A

A

AA AB

average speed of A v

distance v t

number of collisions z z t

Page 54: Chapter 5

54

average distance traveled by an A molecule

between collisions

( )A

AAA AB

v

z z

In pure gas A, there are no A-B collisions, zAB=0

1 12 2 02 2

1 1

2 ( ) 2

A

AA

v RT

z PNNd d

V

Page 55: Chapter 5

55

Diameter d of H2 in the hard-sphere is 1.48 Å

Page 56: Chapter 5

56

1

9 1

5

1776 100

4.3 10

4.1 10 4100A

AAA

AA

v cm s

z collisions s

cm

Page 57: Chapter 5

57

Real Gas No gas exactly follows the ideal gas law. Compression factor (壓縮因子 )

Z(P,T)=PV/nRT

Z<1, P<Pid, V<Vid strong intermolecular attraction

Z=1, P→0 and T→∞

Z>1, P>Pid, V>Vid strong intermolecular repulsion

Page 58: Chapter 5

58

Page 59: Chapter 5

59

Page 60: Chapter 5

60

Van der Waals equation

2( )( ) ( )

a: the effect of intermolecular attractive forces on the gas pressure

b: nonzero volume of molecules and volume

exclusive by intermolecular repulsive force

a VP V b RT V

nV

Page 61: Chapter 5

61

Page 62: Chapter 5

62

Page 63: Chapter 5

63

Page 64: Chapter 5

64

Calculate pressure for 1 mole of CO2 at 0oC in containers with 22.4 L

P = 0.995 atm

Use idea gas equation

Use van der Waals equation

Page 65: Chapter 5

65

Calculate pressure for 1 mole of CO2 at 0oC in containers with 0.2 L

Use idea gas equation

Use van der Waals equation

P = 52.6 atm

Page 66: Chapter 5

66

Calculate pressure for 1 mole of CO2 at 0oC in containers with 0.05 L

Use idea gas equation

Use van der Waals equation

P = 1620 atm

Page 67: Chapter 5

67

Analysis of the van der Waals Constants~a constant

The a constant corrects for the force of attraction between gas particles.

attraction between particles↑ a ↑ As the force of attraction between gas particles

becomes stronger, we have to go to higher temperatures for the molecules in the liquid to form a gas.

Gases with very small values of a, such as H2 and He, must be cooled to almost absolute zero before they condense to form a liquid.

Page 68: Chapter 5

68

Analysis of the van der Waals Constants~b constant

a rough measure of the size of a gas particle the volume of a mole of Ar atoms is 0.03219 L r = 2.3 x 10-8 cm

Page 69: Chapter 5

69

Chemistry in the Atmosphere

Page 70: Chapter 5

70

The variation of temperature and pressure with altitude.

Page 71: Chapter 5

71

Concentration (in molecules per million molecules of “air”) of some smog components versus time of day.

Page 72: Chapter 5

72

Diagram of the process for scrubbing sulfur dioxide from stack gases in power plants.