Lecture 23: Kinetic Theory of Gases (review) Maxwell ...

PHYS 342 Fall 2011

Lecture 23: Kinetic Theory of Gases (review) Maxwell-Boltzmann Distribution

Ron ReifenbergerBirck Nanotechnology Center

Purdue Universityy

no particles 1 particle 2 particles ……….. N particles

Lecture 23

Historically, (Boyle, 1662; Charles, 1787; Gay-y, ( y , , yLussac, 1802) it was known that for many common

gasses

PV/T = constant

Thi ti lit b t d t This proportionality can be converted to an equation (Clapeyeron, 1834):

PV=NkBT (N=number of gas atoms)or

PV=nRT (n=number of moles)PV=nRT (n=number of moles)

This is an empirical law; there was no microscopic d t di f h it h ld b t understanding of why it should be true.

kB= 1.38 x 10-23 J/K; R =8.314 J/K = 0.082 liter atm mol-1 K-1

Realize that N is a really large number, like 1023. It is remarkable that science learned to deal

i h h i f b f with the properties of enormous numbers of particles BEFORE it discovered how to deal with

individual atoms! The reason is that the thermodynamic properties of a system like P and T are AVERAGE values over statistically

large assemblies of particles large assemblies of particles.

As a common example, it’s always easier to deal ith th t d t th it i t d l ith with the average student than it is to deal with

any one individual student in this class.

So let’s study one gas molecule. Confine one air molecule into very narrow tube with moveable piston

Moveable i t ith

(fi l) (i iti l)

For elastic collision, magnitude of velocity does not change, so . . . .

ONLY one air l l

piston with area A

pmolecule = py(final) – py(initial)

= -m|vy| - (m|vy|)

molecule, mass m, velocity +vy,

momentum L= -2m|vy|

ppiston = -pmolecule

py=mvy

x

yL

pp p

t tFAt = 2L|v |

roundtrip transit time t

Fav

F y|vy| trans t t me

p tWhat is Fav?

Impulsive ForceF =

pt

How to find average force value of N pulses in a time T?t

tFo

. . . . .1 2 3 4 5 6 N

0 time

time =T

Sample F at k discrete values and calculate an average:

1

k

ii o o o

F p t tN F t F t F t pF 1i o o oa v

p pFk T T N t t t

p Fav =

tp

Evaluating the Average Force and Pressure

Fav = = t

ppiston 2m|vy|2L/|v |

= + m|vy|2 / L

F m|v |2 m|v |2

t 2L/|vy|

P=Pressure (one molecule) = = = Fav

Am|vy|

ALm|vy|

V

PV = 2*KE

Generalization to 3D: ONE gas molecule in a cylinder with a frictionless piston with area A

h l l hi ll l i lli i

y m

• when molecule hits wall, elastic collision

• collisions with top and bottom walls produces no change in magnitude or direction of vx e g

x Adirection of vx, e.g.

lli i i h l f i h ll

b e fo r e x y z

a fte r x y z

v v i v j v k

v v i v j v k

L • collisions with left or right wall reverses direction of vx

befo re x y zv v i v j v k

• no forces acting on molecule between collisions

a fter x y zv v i v j v k

For N molecules in the cylinder

For N non-interacting molecules, with identical mass, each with different velocity vi,x How to

l l t ?2, 2

,1 1 1

( )N N N

i xi i x

i i i

N

mv mP total P P vV V

calculate?

2 2,

1

1 N

i x x avi

N NN

m mv vV V

PV = Nm[v2x]av = 2N(½ m [v2

x]av )

KEx

[ x]av (½ [ x]av )x-direction only

Molecular Interpretation of Temperature

• Comparing to Ideal Gas Law

22 12B x av

PV Nk T N m v

• In general, 22 2 2x y zav av avav v v vv

2

21 12 x x avav

m v KE kT

2 22

23

x y zav av av

x

but v v v

v

(there’s nothing special about x-direction)

x av

2 2

3 2 31 12 22

aB B

v avv vPV Nk T N m k T m

21 12 23 B

avm k Tv

From this result we infer that 21 ( )

232av av av BKE K m k T for one mv olecule

equipartition theorem

Estimating the average speed in a gas of N particlesin a gas of N particles

2 31 ( )2 2tot av BK N m v Nk T

2 2

2 33 3( ) A Bav

A

N k TkT RTvm N m M

Am N m M

2 3 3( ) Brms av

k T RTv vm M

m M

2 2rms[Note that v ( ) ]avv

Two ChoicesHow would you

statistically characterize a gas of N atoms in

A. List of numbersAt some t= t

B. Probability Distribution

equilibrium at temperature T?

Particle 1: r1 v1 E1

Particle 2: r2 v2 E2

At some t= t1

r1

v1

Particle 3: r3 v3 E3

r1

..

P ti l N E

.

Particle N: rN vN EN

Need a different list for each time!

<Kav> = 3/2 kBT

from equipartition theorem we know for each time! from equipartition theorem, we know that each molecule has (on average) 3/2 kBT of KE

Maxwell-Boltzmann Distribution

The Probability Distribution function for an ideal (classical) gas is called the Maxwellideal (classical) gas is called the Maxwell-

Boltzmann distribution function. It specifies the probability that a single gas particle from a large distribution of particles in equilibrium at

some temperature T has a speed v or a momentum p or a kinetic energy E.momentum p or a kinetic energy E.

There are many ways to derive the Maxwell-Boltzmann (M-B) Distribution Law:

Calculate the number of particles with velocities

Maxwell Boltzmann (M B) Distribution Law:

pbetween v and v+dv as a function of energy E. Maxwellvelocity distribution results.

Enumerate all possible ways that N particles can bedistributed in energy. Use variational calculus to find thedi t ib ti l ( b f ti l ) th tdistribution law (number of particles vs. energy) thatmaximizes the system weight W. The M-B Distributionresults (see Appendix GG).

Maximize the Entropy S=kBlnW. The M-B Law results.

Use microscopic reversibility arguments.

The Distribution of Molecular Speeds in a Gasfirst derived by J.C. Maxwell in 1852

23 2

2 24 ( )( )2

Bmv k Tm n vP v dv v e dv dvk T N

2 Bk T N • Normalized

2 2 2•

• Probability that molecule has speed v e-mv2/2kB

T

N b f l l s ith s d 2

2 2 2x y zv v v v

• Number of molecules with speed v v2

• Note that P(v) is a continuous probability distribution function

• Note that only v2 appears not the components of velocity• Note that only v appears, not the components of velocity

• Connects macroscopic Thermodynamic properties with microscopic models 14

Maxwell Speed Distribution Function

Note asymmetry – there are more ways to get a large

d th ll

( )P v

speed than a small one.

The inherent asymmetry The inherent asymmetry gives rise to different

values for vmax, vav, and vrms.vv

Check out: http://www.chm.davidson.edu/ChemistryApplets/KineticMolecularTheory/Maxwell htmlets/KineticMolecularTheory/Maxwell.html

Shaded area represents probability that molecule will have a speed v ± dv/2

15

The Average Speed

23

2234( )

2B

mvk T

avgmv v P v dv v e dvk T

2

0 0

2 1

( )2

! 1

avgB

n ax

k T

n md

2 11

0

! ; 1;2 2

n axn

B

n mx e dx n aa k T

8 Bavg

k Tvm

(mean speed)

On earth, T=300K, atmosphere is mostly N2. Mass of N2 molecule is 4.7x10-26 kg. What is vavg?

16

23

26

8 8(1.38 10 )(300) 474 /3.14(4.7 10 )

Bavg

k Tv m sm

Most Probable Speed

( ) 0most probabledP vv

d

2 2

2 22 22 ( ) 0B B

most probable

mv mvk T k T

dvmvve v e

2

2 ( ) 02

2B

B

ve v ek T

k T

2 2

2

Bk Tvm

k T

2 Bmost probable

k Tvm

17

Square Root of the Mean Squared Speed

232

22 2 4

0 0

4( ) ( )2

Bmv

k Tavg

B

mv v P v dv v e dvk T

2

0 0

21

2

1 3 5 (2 1) ; 2;2 2

B

n axn n

k T

n mx e dx n ak T

1

0

2

2 23( )

n nB

B

a a k Tk Tv

2

( )

3( )

avg

B

vm

k Tv v

2

( )

31 1 3

rms avg

B

v vm

k Tmv m k T

(equipartition

182 2 2rms Bmv m k T

m ( q p

theorem)

Maxwell Boltzmann Speed Distribution for Maxwell Boltzmann Speed Distribution for different T

N2 molecule

19

What is the probability that a gas atom with mass m has a momentum between p and p+dp at temperature T?a momentum between p and p+dp at temperature T?

2 23 2 3 2

2 22 24( ) 4B Bmv mvk T k Tm mP v dv v e dv v e dv

22

( ) 42 2

1 12 2

B B

P v dv v e dv v ek T k T

p pdp pdpmv mvdv

dv

dpdv

23 222

22

2 2

42

Bp mk T

m m mv m

v m ek T

m

mm

dpm

2 23 2 3 22

23 2

2 2

2

4 142 2

B B

B

p pmk T mk T

k T

p m me dp p e dpm k T m k T

m m

2 2B Bm k T m k T

22

24 ( )( ) Bp mk Tp n pP d d d

20 3 2

2 ( )( )2

B

B

p mk Tp pP p dp e dp dpNmk T

What is the probability that a gas atom with mass m has an energy between E and E+dE at temperature T?

Change Variables:

an energy between E and E dE at temperature T?

2

2

2pEm

g

2

2

222

2p d

p mEp dp mdE

p pdp mE m Ep d

2

3 2

22( ) 4

2B

B

p mk Tn p dp eN mk T

p dp

3 2 1 21 22

2p d

m E d

p pdp mE m Ep d

E

2

3

3 22

2 1 2

242

4 1

B

B

p mk T

E k T

emk T

dp p

3

32 1 2

2

1 2

3

42

( ) 2( )

1 22

B

B

B

E k T

E k T

emk T

n E EP E dE dE e d

m E dE

E

32

( )B

d d e dN k T

21

What is the Average Energy of Gas Atom at Temperature T?

1 2

320 0

2( ) Bavg

B

E k TEE P E dE e dEk T

E E

3 2

03

2

2 1 B

B

E k TE e dEk T

3 2

0 0

1 ; 3 / 2B n

B

E k T aEa E e dE E e dE nk T

5

52

21

( 1) 3 14 1

34n B

na

k T

25

1

6 332 1Bk T

E k T Tk kT

22

32 4 24avg

B

B B BE k Tk T

Tk kT

Probability of finding gas atom with energy between E and E+dEgy

0.6 200K

0.4

0.5

)

300K

500K

0 2

0.3

0.4

P(E)

0 0

0.1

0.2

0.0

0.00 0.03 0.06 0.09 0.12 0.15E (in eV)E (in eV)

23

Experimental Verification of Maxwellian Distribution Function

APPLICATION

The Maxwell speed distribution serves as the basic input for computer calculations of molecular

24

dynamic (MD) simulations of gas flow, gas cooling, gas heating, flames, etc.

Summary

23 2

2 2( ) 4( ) Bmv k Tn v mP d d d

Velocity distribution

2 2( )( )2

Bmv

B

k TP v dv dv v e dvN k T

22

3 22( ) 4( ) Bp mk Tn p pP p dp dp e dp

Momentum distribution

3 2( )2 B

p p p pN mk T

Energy distribution

1 2

32

( ) 2( ) BE k Tn E EP E dE dE e dEN k T

Energy distribution

25

2B

N k T