We want to find # collisions per unit timecoll/T = Z, so first lets get number of collisions of A

Collisions of A with Stationary B

Collision FrequencyThe collision frequency ZA for one A molecule with NBB is

ZA = d2AB vA NB/V, s-1

When we have a density NA/V of A molecules, weexpress the total rate of A-B collisions as

ZAB is the Collision Density

We have still ignored the fact that the B moleculesare moving. Thus we need to replace vA with the average relative speed between A and B, vrel.

2 1

2 3AB A A B

ABd v N N sZ

V cm

Going from one A molecule to NA A molecules/cm3

When we have a density NA/V of A molecules, weexpress the total rate of A-B collisions as

ZAB is the Collision Density

2 1

2 3AB A A B

ABd v N N sZ

V cm

We have still ignored the fact that the B moleculesare moving. Thus we need to replace vA with the average relative speed between A and B, vrel.

1

2 2 2rel A Bv v v

Average Relative Velocity

<VA>

<VB>

Replace v with relative velocity

The A-B collision density for NA/V molecules with NB/V is

1

2 2 22

2

AB A A B

AB

d v v N NZ

V

B

The Simpler Case of A Pure Gas

We found

for a single A molecule with B molecules.If the B molecules become A’s and are allowed to move,then

1

2 2 2 2A A A Av v v v

21 AB A B

A

d v NZ s

V

2 23 1

2

2

2A A A

AA

d v NZ m s

V

212

A A AA

d v NZ s

V

and d2AB becomes d2

A

So for a one component gas,

Collision frequency

and the collision density

Mean Free Path

Mean Free Path () =

Distance Traveled in unit timeNumber of Collisions in unit time

uA

2 dA2 uANA / V

V

2 dA2 NA

ConcepTest #1

In which of the following systems will the mean free path () be the shortest?

A.  Pure CCl4, 1 Pa pressure B.  Pure CCl4, 10 Pa pressure C.  Pure He, 1 Pa pressureD.  Pure He, 10 Pa pressure

ConcepTest #1

In which of the following systems will the mean free path () be the shortest?

A.  Pure CCl4, 1 Pa pressure B.  Pure CCl4, 10 Pa pressure C.  Pure He, 1 Pa pressureD.  Pure He, 10 Pa pressure

Collision Flux

Collision Flux – Number of collisions with an imaginary area in a given time interval, divided by the area and the time interval

We already know that Collision Frequency is the number of hits per second, so if we just divide this by the area, we get the flux

Consider a gas of A molecules with average speed vA and number density nA = NA/V. One can show (but we will not) that the flux, F, of molecules passing in one direction through a plane of unit area is given by

2 1 molecules cm4

A An vF s

Molecular Speed Distributions

There is a key, universal, fundamental concept:Mother Nature does not play favorites, but rather selects

in an unbiased manner from the allowed outcomes.

At any given temperature (total energy), collisions between gas molecules lead to exchange of energy between the molecules. However, when there are a large number of particles, the distribution function describing the molecular energies (or velocities) is essentially constant, at the equilibrium distribution.

General Principle!!

Energy is distributed among accessibleconfigurations in a random process.

The ergodic hypothesis

Consider fixed total energy with multiple particles and various possible energies for the particles

Determine the distribution that occupies the largest portion of the available “Phase Space.” That is the observed distribution.

Possible arrangements of particleswith fixed total energy

While the real systems of interest typically have ~NAparticles, we illustrate the concept on an embarrassinglysmall, but simple, system. Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0, 1, 2, 3, 4 . . . (quantized).Let the total energy of the system be 3.

What is the average energy per particle, E?

A. 3

B. 4

C. 0.75

D. 12

E. Not enough information given

Possible arrangements of particleswith fixed total energy

While the real systems of interest typically have ~NAparticles, we illustrate the concept on an embarrassinglysmall, but simple, system. Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0, 1, 2, 3, 4 . . . (quantized).Let the total energy of the system be 3.

What is the average energy per particle, E?

A. 3

B. 4

C. 0.75

D. 12

E. Not enough information given

Possible arrangements of particleswith fixed total energy

Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0, 1, 2, 3, 4 . . . (quantized).Let the total energy of the system be three.

What are the possible arrangements of energy?

Energy 0 1 2 3 4

I. 3 0 0 1 0

II.

III.

IV.

Plot n(E) vs. E

Possible arrangements of particleswith fixed total energy

Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0, 1, 2, 3, 4 . . . (quantized).Let the total energy of the system be three.

With all possible arrangements enumerated, We make theergodic assumption, namely that the system visits all of theavailable configurations with equal probability.

What is the average occupancy of each energy level?

Energy 0 1 2 3 4

I. 3 0 0 1 0

II. 2 1 1 0 0

III. 1 3 0 0 0

Totals/4 1.5 1 0.25 0.25 0Plot n(e)vs. E

Energy 0 1 2 3 4 5

n(E

)

0.0

0.5

1.0

1.5

2.0

2.5

3.0

Plot n(E) vs. E

Possible arrangements of particleswith fixed total energy

Energy 0 1 2 3 4

I. 3 0 0 1 0

II. 2 1 1 0 0

III. 1 3 0 0 0

Totals/4 1.5 1 0.25 0.25 0

0.75( ) 3E

f E e

Energy randomization leads to an exponential distribution of occupied energy levels at constant total energy

Maxwell distribution of molecular speedsThe probability that the molecular velocity is between

vx & vx+dvx, vy & vy+dvy, and vz & vz+dvz is given by

dPxdPydPz = B3 exp[-mvx2/2kBT] x exp[-mvy

2/2kBT] x

exp[-mvz2/2kBT] dvxdvydvz

This expression simplifies using u2 = ux2+uy

2+uz2

2

23 B

mvk T

x y z x y zdP dP dP B e dv dv dv

But we must convert to spherical polar coordinatesto get the volume element in v!

Spherical Polar Co-ordinates

Maxwell distribution of molecular speedsVolume element (dvx dvy dvz) transformation from

cartesian to spherical polar coordinates:x = r sin cos,y = r sin sin

z = r cosVolume element transform:

dvxdvydvz 4v2dv2

23 24 B

mvk TdP B v e dv

For a normalized distribution, 0

1 We use this constraint to set B, givingP v dv

2 2

3 32 222 2 24 4

2 2B

mv Mvk T RT

B

m MP v dv v e dv v e dvk T RT

Characteristics of P(v)

2

32

2 242

MvRTMP v dv v e dv

RT

Characteristics of P(v)

2

32

2 242

MvRTMP v dv v e dv

RT

ConcepTest

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400 m/s 800 m/s 1200 m/s 1600 m/s

(I) If the curves represent 3 different gases at the same T, which curve shows the gas of greatest molar mass?

(II) If the curves represent the same gas at 3 different T,which curve shows the highest T?

A. blue; blue C. red; blueB. blue; red D. red; red

blue

red

green

Maxwell speed distributions are shown for three situations.

ConcepTest

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400 m/s 800 m/s 1200 m/s 1600 m/s

(I) If the curves represent 3 different gases at the same T, which curve shows the gas of greatest molar mass?

(II) If the curves represent the same gas at 3 different T,which curve shows the highest T?

A. blue; blue C. red; blueB. blue; red D. red; red

blue

red

green

Maxwell speed distributions are shown for three situations.

Mixture of A and B

Using average speed of A, uA

Using average relative speed of A and B,

uAB uA2 uB

2 1/ 2

Number of collisions of one molecule of A with B molecules (per unit time); SI unit: s—1

ZA dAB

2 uANB

V ZA dAB

2 uA2uB

2 1

2

NB

V

Total number of A-B collisions (per unit volume per unit time); SI unit: m—3 s—1

ZAB dAB

2 uANANB

V 2 ZAB dAB

2 uA2uB

2 1

2

NANB

V 2

Pure A Using average speed of A, uA

Using average relative speed, uAA 2 uA

Number of collisions of one molecule of A with A molecules (per unit time); SI unit: s—1

ZA dA

2 uANA

V ZA 2 dA

2 uANA

V

Total number of A-A collisions (per unit volume per unit time); SI unit: m—3 s—1

ZAA dA

2 uANA2

2V 2 ZAA 2 dA

2 uANA2

2V 2

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