Kinetic Theory of GasesKinetic Theory of Gases

Everything up to this point has been empiricalThere is clearly a great deal of truth in the equationsHowever, there is no physical pictureHowever, there is no physical picture

To gain insight, one must develop and validate models We look at a molecular (kinetic theory of gases) modelto begin that processto begin that process.We begin that process today

Reading Test #1 Reading Test #1

HCl and NH3 gases at the same temperature and pressure are introduced at opposite ends of a glass tube. A ring of solid NH4Cl forms wherethe two gases meet inside the tube.the two gases meet inside the tube.HCl + NH3 NH4ClWhere will the ring form?g

A At the centerA. At the centerB. Closer to the HCl endC Closer to the NH3 endC. Closer to the NH3 end

Three Postulates ofKinetic Theory of Gases

1. Gas composed of hard spherical particles that are small relative to the mean distances between them.

2. Particles are in constant motion and have

3 Neither attractive nor repulsive forces exist

kinetic energy.

3. Neither attractive nor repulsive forces exist between the particles except on contact (collision).( )

ConcepTest #2

Which of the following gases deviates most from the postulates of kinetic molecular theory?p y(i.e., deviates most from ideal gas behavior)

A. He

B. H2

C NC. N2

D. NH3

E. CH4

Kinetic-Molecular TheoryKinetic-Molecular Theory

Consistent with Ideal Gas Law Dalton’s LawConsistent with Ideal Gas Law, Dalton’s Law,Graham’s Law and many other observations

Consider a molec leConsider a moleculecolliding elastically with a wallg y

velocity, vx

l

x-coordinateWal

l

velocity, -vx

R l ti hi b t FRelationship between Force and Change of Momentumg

Momentum, p = mv, pF = ma = m dv/dt = d(mv)dt = dp/dt

What is change of velocity on each collision?

v = vfi l – vi i i l = v – (– v ) = 2v

What is change of momentum on each collision?

v vfinal vinitial vx ( vx) 2vx

What is change of momentum on each collision?

p = 2vxm

Collision rate for one moleculeCollision rate for one molecule with a wall

vxBox with dimensionsx by y by zx

-vxWal

l x by y by z

x (1ength of 1 side)

W

Number of collisions in time t = vxt/2x

N b f lli i it ti /2Number of collisions per unit time = vx /2x

Force (momentum change)(m m m g )per Unit Time for One Particle

F= ma=dp/dt = (Change of Momentum per Collision) x

(Number of Collisions per Unit Time)

dp/dt = Fx = (2vxm) (vx/2x)

= mvx2/xx

This is the force exerted on the wall by one particle.

Pressure is Forceper Unit Area

yz

/ /

xz

/ 2/Px = Fw/A = Fx/yz

Px = mvx2/xyz

Fx = (2vxm) (vx/2x) = mvx2/x

Px = mvx2/V

What changes are required when N l l d di t ib ti f l iti molecules and a distribution of velocities

are considered?

M l i l b N Px = mvx

2/VMultiply by N Replace vx

2 with 2vxy

xy zPx = Py = Pz = P implies

22 2 2

v V= xyz2 2 2

3x y zv v v

and

V= xyz

2 2xP m / V P Nm v /3Vxv

Fundamental Equation from qSimple Kinetic Theory of Gases

2 /2PV Nm v /3

Parallels the ideal gas law:

PV nRTg w

2Nm 2Nm vnRT

3

2 2rms

3RT 3RTv v vM M

M Mwith M (molar mass) =mN/n

C T #3ConcepTest #3

At a given temperature, which of the following gases has the smallest v ?following gases has the smallest vrms?

A. HeB HB. H2

C. N2

D. NH3E CH4E. CH4

Connection between Kinetic Energy Connection between Kinetic Energy & Temperature

Average Kinetic Energy per Molecule

1 2m1 v2

Substitute this expression into the pressure equation to Substitute this expression into the pressure equation to obtain the relationship between kinetic energy and T:

k3E RT (per mole) or2

3

k B3k T (per molecule)2

ConcepTest #4pFlask A contains 8 g H2, and Flask B contains 8 g He. The fl k h th l d t t C flasks have the same volume and temperature. Compare the gas density (g/cm3), the kinetic energy per mole, and the total kinetic energy of the gases in the two flasks.gy g

Density Ek /n Total KEA A>B A>B A>BA. A>B A>B A>BB. A>B A=B A>BC A=B A=B A=BC. A=B A=B A=BD. A=B A=B A>BE A B A B A BE. A=B A>B A>B

Collisions Between MoleculesMolecule A travels with velocity vA in a

Collisions Between MoleculesMolecule A travels with velocity vA in a

container of B molecules

Collision will occur each time the distance Collision will occur each time the distance between the center of molecule A and

molecule B is dABm cu AB

where dAB = (dA + dB)/2

dA = molecule A diameter

dB = molecule B diameter B

dAB = collision diameter for AB collisions

A(vA) colliding with stationary B(0)

The collision frequency ZA for one A molecule with NBB isThe collision frequency ZA for one A molecule with NBB is

ZA = d2AB vA NB/V, s-1

Going from one A molecule to NA A molecules/cm3to NA A molecules/cm

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

2 1d v N N s

Z i th C lli i D it

2 3AB A A B

ABd v N N sZ

V cm

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.

1

So 1

2 2 2rel A Bv v v

Getting the correct relativecollision speedscollision speeds

Using this information we go back to correct the expression g f g pfor the number of collisions per sec of one molecule of A with B molecules with density NB/V.

1

2 2 22AB A Bd v v N B AB A B

AZV

B

1

2 2 22d v v N N

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

2

2

AB A A B

AB

d v v N NZ

V

B

Consider collisions only between A molecules

We found2

1 AB A BA

d v NZ s

V

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

V

If the B molecules become A s and are allowed to move,then

12 2 2 2A A A Av v v v and d2

AB becomes d2A

f 2

12 A A A

A

d v NZ s

So for a one component gas,

Collision frequency

2 23 12

A A Ad v NZ m s

AZ sV

f q y

and the collision density 2

2AAZ m sV

and the collision density

More on a gas of A molecules

Consider a gas of A molecules with average speed vAand number density nA = NA/V. One can show (but we will

) h h fl f l l 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

This is a very useful relation to remember but hereThis is a very useful relation to remember, but herewe wish to focus on the average velocity, and seeif we can understand the basis for the molecularspeed distribution that seems ubiquitous.

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

in an unbiased manner from the allowed outcomes.

Mother Nature’s choicesOne example of this is that a gas will uniformly fill a box that is free of any external forces. If there are

t l f ( h it ) th external forces (such as gravity), the gas assumes an exponential distribution of the form

( )mgzmgzN

where mgzo is a sort of average gravitational potentialand z is a “scale height” of the gas

( ) omgzN z e

and zo is a scale height of the gas.

We see this characteristic form of an energy relation gyvirtually everywhere: rates of reaction , energy distributions of gas particles . . . How does it arise? How does it arise?

Possible arrangements of particleswith fixed total energywith fixed total energy

While the real systems of interest typically have ~NAparticles we illustrate the concept on an embarrassinglyparticles, we illustrate the concept on an embarrassinglysmall, but simple, system. Consider a system comprised of 4 particles. Any particle can h v nl discr t n r i s: E 0 1 2 3 4 (qu ntiz d)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?What is the average energy per particle, E?

A. 3

B 4B. 4

C. 0.75

D. 12

E. Not enough information given

Possible arrangements of particleswith fixed total energywith fixed total energy

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

I have chosen two different distributions of energies, withg ,both satisfying energy conservation.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 energywith fixed total energy

Consider a system comprised of 4 particles. Any particle can have only discrete energies: E = 0 1 2 3 4 (quantized)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 thep g ,ergodic 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 0I. 3 0 0 1 0II. 2 1 1 0 0III. 1 3 0 0 0

T l /4 1 5 1 0 25 0 25 0

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

Possible arrangements of particleswith fixed total energyf gy

Energy 0 1 2 3 4I. 3 0 0 1 0II. 2 1 1 0 0III. 1 3 0 0 0

Totals/4 1.5 1 0.25 0.25 0

2.0

2.5

3.0

Plot n(E) vs. EE

n(E

)

1.0

1.5

2.00.75( ) 3f E e

0 1 2 3 4 50.0

0.5

Energy

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

Maxwell distribution of molecular speedsThe probability that a system with fixed total energy has particular energy, E, is now given by

P(E) a exp [-E/kBT]

Probability that molecule has velocity along the x-axis between vxy y g xand vx + dvx

dPx = B exp [-mvx2/2kBT]

Similar expressions for velocities along the y-axis and z-axis

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 2/ k d d dexp[-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

x y z x y z

But we must convert to spherical polar coordinatesto get the volume element in v!to 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,x r sin cos,y = r sin sin

z = r cosVolume element transform:

dvxdvydvz 4v2dv2

23 24 B

mvk TdP B v e dv

4 BdP B v e dv

For a normalized distribution, 0

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

0

2 2

3 32 222 2 24 4B

mv Mvk T RTm MP v dv v e dv v e dv

4 4

2 2B

P v dv v e dv v e dvk T RT

Characteristics of P(v)

2

32

2 24MvRTMP v dv v e dv

4

2P v dv v e dv

RT

Characteristics of P(v)

2

32

2 242

MvRTMP v dv v e dv

RT

2 RT

ConcepTest

(I) If the curves represent 3 different gases at the same T which

Maxwell speed distributions are shown for three situations.

(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,

xxxxxxxxxxxxxx

( ) f p m g ff ,which curve shows the highest T?

bluexxxxxxxxxxxxxx-

xxxxxxxxxxxxxxx

blue

red

green

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

red

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

Helium velocity distribution as f(T)