ch2. kinetic theory of gases- model

8/13/2019 Ch2. Kinetic Theory of Gases- Model

1/42

CHAPTER 2

KINETIC THEORY OFGASES

BSK1133


2/42

Science Methodology

OBSERVATION

- from natural

phenomena

Develop MODELINFERENCE:

A possible explanation

Hypothesis / Theory

TEST theory against

experiment

TEST model with

computer simulation

Test

successful?

Simulation

successful?

Theory becomes LAW

MODEL accepeted and

applied

YesYes

NoNo


3/42

Kinetic Theory

The kinetic theory of gases is the study of themicroscopic behavior of molecules and the

interactions which lead to macroscopic

relationships like the ideal gas law.

The study of the molecules of a gas is a

good example of a physical situation where

statistical methods give precise anddependable results for macroscopic

manifestations of microscopic phenomena.


4/42


5/42

THE KINETIC MODEL OF GASES

3. The molecules do not interact, except

during infrequent, and perfectly elastic

collisions.

Elastic collision:

is a collision in which total translationalkinetic energy of the molecules is conserved


6/42

The model based on the assumptions relate the

pressure and volume of the gas as follows:

where Mis mNAmolar mass of molecule and

is its root mean square speed

2

1

1

2

n

cn

i

rmsc

2

3

1rms

nMcpV


7/42

Ideal Gas law :

nRTpV


8/42

Kinetic model

Gas law

So,

2

3

1rms

nMcpV

nRTpV

2

1

3

M

RT

crms


9/42

The root mean square speed crms

of the

molecules of gas is:

Proportional to the square root of the

temperature T1/2, and

Inversely proportional to the square root of

the molar mass M1/2


10/42

i.e. the higher the temperature, the faster

the molecule travel on average

at a given temperature, heavy

molecules travel more slowly on

average than lighter molecules


11/42

axwell distribution of speedscrms is just an expression for the root mean square

speed of the molecules

In actual gas, individual molecules move at a wide

range of speedsa distribution of speeds

After a collision the speeds of molecules will be re-

distributed

this distribution is given by Maxwell as follows:

RT

M

eRT

M

f

222

3 2

24

)(


12/42

The equation can be

used to calculate the

fraction of molecules ina range of speeds, by

integrating the equation

between the value of the

range,1

to2

,

Number of molecules

in the range,

N = 2

1

df )(


13/42

evaluate the mean(average) speed:

identify the most probablespeed:

calculate the rmsspeed

21

8

M

RTc

21

2

M

RTcmp

by using the Maxwell distribution we can:

2

1

3

M

RTcrms


14/42

calculate the relative mean speed

Where, k(Boltzmanns constant) = R/NA

and the reduced mass,

21

8

kTcrel

BA

BA

mm

mm

and,


15/42


16/42

With increasingtemperature and

decreasing molar mass.

The speedcorresponding

to peak of the distribution

increases

The distributionbecomes

broader

The peak(number of

molecules) goes down

Distribution of molecular speeds with

temperature and molar mass


17/42

Pressure and kinetics energy

The pressure of an ideal gas can becalculated by assuming that the walls ofthe container are flat, and that the

collisions of the molecules with the wallsare elastic. That means that themolecules do not lose kinetic energy intheir collisions with the walls

Pressure = FA


18/42


19/42

Pressure calculation

Consider N molecules ofgas with individual mass, mmoving in 1 cube x xwith speed x:

l

l

l

Z

X

A

y-Vx

+Vx

Momentum before strike the walls = mxMomentum after strike the walls = -mx Momentum at wall A for 1 strike = mx(- mx)

=2 mx

m= momentum

= speed


20/42

l

mv

lmv

t

mvmomentumrateForce

IILawNewton

V

lt

mvmomentum

x

x

x

x

x

2

2

22

2

,

2

2


21/42

V

mvP

V

mvP

V

mvP

l

mv

lxl

mv

Area

Force

pressure

zy

x

x

x

zy

x

22

2

3

2

2

2

,

1


22/42

N

i

zyx

zyx

xx

iv

Nv

where

vvvv

arevalueaveragesodifferentarespeedmolecular

vvvv

but

V

vNmP

moleculesNhaveweif

1

2

2222

2222

2

1

,

:,

,


23/42

2

2

22

2222

3

1

3

,

3

1

3

1

vNmPV

VvNmP

so

vv

vvvv

rootsquarespeedMean

rootsquarespeedaveragev

x

zyx


24/42

Kinetic energy

k

k

k

k

EnRT

nRTPVgasideafor

EPV

vNmPV

vNmE

moleculesNfor

moleculeforvmE

3

2

:

3

2

2

1

3

2

2

1

:

)1(21

2

2

2


25/42


26/42

2/1

1

2

2/1

2

2

21

2

22

2

11

21

21

2

1

2

1

)(

23,

23

2..

:2

21

21

21

M

M

v

v

vMvM

TsameatEE

TTat

RTERTE

EandEenergykineticswithgasesge

gasesbetweenValDif ferenti

kk

kk

kk

rms


27/42


28/42


29/42

EXAMPLE

Calculate VrmsO2at 20oC

22

1

11

,

478

)314.8(3

3

smkgJwhere

ms

molJKR

M

RT

m

kTVrms


30/42

Kinetic Theory

The average energy associated with the

molecular motion has its foundation in the

Boltzmann distribution, a statistical

distribution function. Yet the T and E of a gascan be measured precisely.

RT

M

eRT

Mf

222

32

24

)(


31/42

kTE

AeEf 1)(

With increasing energy, E, it progressively

less likely that any given particle will attainthat E, so more particles will be found withlower E. It is assumed that any unlimitednumber of particle can occupy any E state

Maxwell-Boltzmann Details

The probability that aparticle will haveenergy, E

Normalizationconstant A

The probability foroccupying a givenE statedecreaseexponentially with E

Boltzmans constant k times theabsolute T. The implication of thisterm is that for a higher T, it ismore probable that a given

particles can be found with E


32/42

Molecule Speed Distribution

Where,

Speed distribution , f(v)=distribution function

f(v) dv =fractionation from total molecules that have speed in therange v to v+ dv

Ektotal for gases is finite (limited value) so, f(v) 0 when v f(v) dv = dN

N

NdNdvvf )(


33/42

Equation for speed distribution : 3-D

Maxwell-Boltzmann equation:

k = Boltzmann constant

= 1.381 x 10-23JK-1

m = 1 mass molecule = M/NA

kT

mv

kT

m

vdv

NdN

vf 2exp24

/

)(

22/3

2


34/42

Maxwell-Boltzmann derivatives

222

22

2

2/1

2

2

(exp),(

tan2

2exp

)(

vucwherecspeednet

dvdukT

vumAN

vudN

tconsionnormalizatkT

mA

dukT

muA

N

udN

o

o

From Maxwell-Boltzmann equation:


35/42

R

M

R

Nx

N

M

k

mwhere

dvRT

MvkT

MvNdNor

ckT

mc

kT

m

dcN

cdNcf

coordinatesphericaldwdvdudcc

vuc

kT

mAwhere

dwdvdukT

wvumA

N

wvudN

o

o

o

o

o

2exp

24

2exp

24

)(

)()(

4

2

2

(exp

),,(

22/3

2

222/3

2

222

2/3

222

2


36/42


37/42

100 K

300 K

500 K

1000 K

f(v)/m-1s

0 v/ms-1

f(v)

v1 v2V

e.g Plotted f(v) versus v for O2at several temperature

dv

NdNvf

/)(


38/42


39/42

2/12/1_

0

32/2

2/3

0

_

1

_

88

24

)(

)(.

1

M

RT

m

kTv

dvvekT

m

dvvfvv

dvvfvvor

vN

v

kTmv

N

i

i

2. Average speed, v is calculate as the average of vusing the probability distribution


40/42

3. Root- mean-square speed is define as the

square root of Vrms

2/12/1

2/1

0

2

2/1

1

2

2/12_

33

)(.

1

M

RT

m

kTV

dvvfvV

v

N

vV

rms

rms

N

i

irms


41/42

EXAMPLE

Calculate the most probable speed, Vmp, themean speed, v, and the root-mean-square speed

(v2)1/2 or Vrmsfor hydrogen molecules at 0oC


42/42

Solution

13

2/1

13

11

1050.1

)10016.2()15.273)(3145.8)(2(2

msx

molkgxKmolKJ

MRTVmp

13

2/1

13

112/1_

1069.1

)10016.2)(1416.3(

)15.273)(3145.8)(8(8

msx

molkgx

KmolKJ

M

RTv

13

2/1

13

112/1

10016.2

)15.273)(3145.8(33

molkgx

KmolKJ

M

RTVrms

ch2. kinetic theory of gases- model

Documents