thermal & kinetic lecture 6 the maxwell-boltzmann distribution
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Recap…. Kinetic energy and temperature. Thermal & Kinetic Lecture 6 The Maxwell-Boltzmann distribution. LECTURE 6 OVERVIEW. Distributions of velocities and speeds in an ideal gas. Last time…. Free expansion. The ideal gas law. Boltzmann factors. - PowerPoint PPT PresentationTRANSCRIPT
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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
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Last time….
Free expansion.
The ideal gas law.
Boltzmann factors.
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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)
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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
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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?
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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:
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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
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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)
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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.
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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.
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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…..???
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…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)
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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….?”
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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
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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
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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
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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