1. Boltzmann factors and the Maxwell-Boltzmann distribution
Thermal & Kinetic
Lecture 5
Boltzmann factors and the Maxwell-Boltzmann distribution
LECTURE 5 OVERVIEW
Recap….
Stability, instability, and metastability
Boltzmann factors and probability
Distributions of velocities and speeds in an ideal gas

2. Last time….
Free expansion.
The ideal gas law.
Boltzmann factors.

3. Equation 1.1 has a bracket missing. Should be:
Questions and errata
Equation 1.1 has a bracket missing. Should be:
Small – though important – typographical error in Eqn – should have:
dP = dn’kT

4. Boltzmann factors and distributions
Yes, but so what?
Remember that at the start of our derivation we said any force was appropriate – thus, this is a general expression.
Assuming conservative force – no equilibrium otherwise.
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?
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.

5. Stability, metastability and instability
It is possible to have a time invariant unstable state – can you sketch the potential energy curve associated with this state? ?
Unstable
(transient)
Potential energy DE
Probability of surmounting barrier proportional to:
Metastable
Stable
(ground state)
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)

6. Boltzmann factors and probability
The probability of finding a system in a state with energy DE above the ground state is proportional to: E1 E2 E3
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? ?
(a) exp (E3/kT)
(b) exp (E2/kT)
(c) exp ((-E3+E2)/kT), or
(d) exp ((-E2-E3 )/kT) ?
ANS: c

7. Boltzmann factors and probability!
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 eV.
Molecular vibrations and rotations are also quantised.

8. 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 molecules
Monatomic - 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.

9. The distribution of velocities in a gasvx vy vz
Consider components of velocity vector.
Velocity components lie within ranges:
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…..? ?
ANS: a exp (- mv2/2kT)
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) dvxdvydvz  exp (-mv2/2kT) dvxdvydvz
…… but v2 = vx2 + vy2 + vz2 
constant
f(vx, vy, vz) dvxdvydvz = Aexp (-mvx2/2kT) exp (-mvy2/2kT) exp (-mvz2/2kT) dvxdvydvz

10. 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: 1
Hence:
In addition, if we’re interested in the probability distribution of only one velocity component (e.g. vx), we integrate over vy and vz:

11. The distribution of velocities in a gas
…and we end up with:
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 of the Feynman Lectures, Vol. 1 to see how to do the integration.
which means:
NB Note misprint in G & P on p. 440 – there’s a p missing

12. Gaussian distributions
f(vx)dvx vx
FWHM
Expression for f(vx)dvx represents a Gaussian (or normal) distribution
where  is the standard deviation,
(s = FWHM/√(8ln2)) and m is the mean.

13. Distribution of molecular speeds
We need to do a few more steps to get a formula for the distribution of molecular speeds.
First, we can combine the expressions for molecular velocities to get:
Now, this expression is written in Cartesian co-ordinates (x,y,z) – this makes deriving an expression for speed tricky.
Switch to spherical polar coordinates.
Spherical
polar
coords q vz vy vx f

14. Molecular speeds: Polar coordinatesq vz vy vx f
Surface element, dS, shown – need to consider volume element, dV
Considering all directions, the tip of the velocity vector ‘sweeps’ out a spherical volume (only one ‘octant’ shown above). How many velocity states within v and v + dv ? ?
What is the volume of this thin shell?
Consider thin shell of sphere
whose radius changes from
v to v + dv.
ANS: 4pv2dv

15. Maxwell-Boltzmann distribution We had:
Taking into account discussion of spherical polar coordinates:
Maxwell-Boltzmann distribution of speeds of molecules in a gas at thermal equilibrium.
Note that G&P also leave the p ‘s out on p.441!

16. ! Maxwell-Boltzmann distribution
Maxwell-Boltzmann distribution for N2 molecules
293 K
600 K !
Maximum not at v=0;
most probable speed is less
than mean speed; (CW 3)
curve broadens as T increases

17. Maxwell-Boltzmann distribution 293 K 600 K
Shaded part of graph gives fraction of molecules with speeds between 500 and 1000 ms-1. (Integrate under curve with appropriate limits).
Coursework Set 3 includes a number of questions on this distribution function.