Thermal & Kinetic Lecture 6

The Maxwell-Boltzmann distribution

Recap….

Kinetic energy and temperature

Distributions of velocities and speeds in an ideal gas

LECTURE 6 OVERVIEW

Last time….

Free expansion.

The ideal gas law.

Boltzmann factors.

Boltzmann factors “Available energy is the main object at stake in the struggle for existence and the evolution of the world.” Ludwig Boltzmann

“At thermal equilibrium all microscopic constituents of a system have the same average energy” (Grant & Phillips, p. 421)

…….now let’s consider the distribution of energy in the system.

For a system in thermal equilibrium at a certain temperature, the components are distributed over available energy states to give a total internal energy U.

…but what is the probability of finding a particle in a given energy state?

The probability of finding a component of the system (eg. an

atom) in an energy state is proportional to the Boltzmann factor:

)exp(kT

(NB: T is in Kelvin)

Boltzmann’s Law

NB. You will be expected to be able to derive Eqn. 2.21 (i.e. Boltzmann’s law) – see derivation under Section 2.3 in the notes. You have also seen this derivation in the Mathematical Modelling course.

Erratum

Small – though important – typographical error in Eqn. 2.17 – should have: dP = dn’kT

Boltzmann factors and distributions

Yes, but so what? )/exp(' kTn

At the start of the derivation (Section 2.3) we said that any force was appropriate – thus, this is a general expression.

Assuming conservative force.

Boltzmann factors appear everywhere in physics (and chemistry and biology and materials science and…….)

Why? Because the expression above underlies the population of energy states and thus controls the rate of a process.

Eg. Diamond is less thermodynamically stable than graphite (can burn diamond in air at ~ 1300 K). Given a few million years diamonds might begin to appear a little more ‘grubby’ then they do now. Why?

Stability, metastability and instability

Potential energy

Metastable

Stable(ground state)

It is possible to have a time invariant unstable state – can you sketch the potential energy curve associated with this state?

??Unstable(transient)

The hill represents a kinetic barrier. The ball will only surmount the hill when it gains enough energy. (Diamond is metastable with respect to graphite)

)/exp( kTE

Probability of surmounting barrier proportional to:

Boltzmann factors and probability

)/exp( kTEThe probability of finding a system in a state with energy E above the ground state is proportional to:

PROBLEM: You know that electrons in atoms are restricted to certain quantised energy values. The hydrogen atom can exist in its ground state (E1) or in an excited state (E2, E3, E4 etc….). At a temperature T, what is the relative probability of finding the atom in the E3 state as compared to finding it in the E2 state?E1

E2

E3

You know that electrons in atoms are restricted to certain quantised energy values. The hydrogen atom can exist in its ground state (E1) or in an excited state (E2, E3, E4 etc….). At a temperature T, what is the relative probability of finding the atom in the E3 state as compared to finding it in the E2 state?

exp

(E3/

kT)

exp

(E2/

kT)

exp

((-E

3+E2)

/kT)

exp

((-E

2-E3)

/kT)

9%

22%

64%

5%

a) exp (E3/kT)

b) exp (E2/kT)

c) exp ((-E3+E2)/kT)

d) exp ((-E2-E3)/kT)

This is a very important value to memorise as it gives us a ‘handle’ on what processes are likely to occur at room temperature.

The value of kT at room temperature (300 K) is 0.025 eV.

Boltzmann factors and probability

!!

Molecular vibrations and rotations are also quantised.

The distribution of velocities in a gas

Let’s now return to the question of molecular speeds in a gas.

For the time being we’re concerned only with ideal gases – so:

No interactions between moleculesMonatomic - no ‘internal energies’ – no vibrations or rotations

For an ideal gas the total energy is determined solely by the kinetic energies of the molecules.

In common with a considerable number of textbooks I use the terms ‘atom’ and ‘molecule’ interchangeably for the constituents of an ideal gas.

Consider the distribution of kinetic energies – i.e. the distribution of speeds – when the gas is in thermal equilibrium at temperature T.

Speed v is continuously distributed and is independent of a molecule’s position.

The distribution of velocities in a gasConsider components of velocity vector.

Velocity components lie within ranges:

vxvy

vz

vx → vx + dvx vy → vy + dvy

vz → vz + dvz

What is the kinetic energy of a molecule whose velocity components lie within these ranges??? ANS: ½ mv2

..which means that the probability of a molecule occupying a state with this energy is…..???

…which means that the probability of a molecule having this energy is

proportional to:

exp

(-m

v2/2

KT)

exp

(-mvx

/kT)

exp

(-mv/

kT)

exp

(-mv2

/kT)

83%

13%

0%3%

a) exp (-mv2/2KT)

b) exp(-mvx/kT)

c) exp(-mv/kT)

d) exp(-mv2/kT)

f(vx, vy, vz) = Aexp (-mvx2/2kT) exp (-mvy

2/2kT) exp (-mvz2/2kT)

Therefore the probability, f(vx, vy, vz) dvxdvydvz , that a molecule has velocity components within the ranges vx → vx + dvx etc.. obeys the following relation:

f(vx, vy, vz) exp (-mv2/2kT)

…… but v2 = vx2 + vy

2 + vz2 constant

The distribution of velocities in a gas

“OK, but how do we work out what the constant A should be….?”

What is the probability that a molecule has velocity components within the range -∞ to +∞?

0

exp

(-mv3

/kT) 1

None

of the

se

3% 2%

95%

0%

a) 0

b) exp(-mv3/kT)

c) 1

d) None of these

The distribution of velocities in a gas

“OK, but how do we work out what the constant A should be….?”

What is the probability that a molecule has velocity components within the range - to +??? ANS:

Hence:

z

zy

yx

x dvkT

mvdv

kT

mvdv

kT

mv

A)

2exp()

2exp()

2exp(

1 222

In addition, if we’re interested in the probability distribution of only one velocity component (e.g. vx), we integrate over vy and vz:

zz

yy

xx

zz

yy

xx

xx

dvkTmv

dvkT

mvdv

kTmv

dvkTmv

dvkT

mvdv

kTmv

dvvf

)2

exp()2

exp()2

exp(

)2

exp()2

exp()2

exp()(

222

222

The distribution of velocities in a gas

…and we end up with:

xx

xx

xx

dvkTmv

dvkTmv

dvvf

)2

exp(

)2

exp()( 2

2

We can look up the integral in a table (if required, you’ll be given the values of integrals of this type in the exam) or (better) consult p. 40-6 of the Feynman Lectures, Vol. 1 to see how to do the integration.

m

kTdv

kT

mvx

x 2)

2exp(

2

which means:

xx

xx dvkT

mv

kT

mdvvf )

2exp(

2)(

2

Gaussian distributions

Expression for f(vx)dvx represents a Gaussian (or normal) distribution

)2

1exp(

2

1)(

2

xxg

where is the standard deviation, = FWHM/√(8ln2)) and is the mean.

f(vx)dvx

vx0

FWHM