Thermal & Kinetic Lecture 7 Maxwell-Boltzmann distribution, Equipartition of energy (….and some problems with classical physics) Deriving the Maxwell-Boltzmann distribution function LECTURE 7 OVERVIEW Equipartition and degrees of freedom
Dec 16, 2015
Thermal & Kinetic Lecture 7
Maxwell-Boltzmann distribution, Equipartition of energy
(….and some problems with classical physics)
Deriving the Maxwell-Boltzmann distribution function
LECTURE 7 OVERVIEW
Equipartition and degrees of freedom
Velocity distribution – TAKE NOTES
Consider velocities of molecules first:
)2
exp()(
)/exp()(2
kT
mvAvf
kTEAEf
xx
So, our velocity distribution in one dimension is: )2
exp(2
)(2
kT
mv
kT
mvf xx
How might we determine what the constant, A, should be?[See Q7(b) of the ’04-’05 Thermal and Kinetic paper for a similar question].
Given:
??
dxx )exp( 2
Gaussian function, <vx> = 0
)21
exp(2
1)(
2
x
xg
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:
zyxzyx
zyxzyx dvdvdvkT
vvvm
kT
mdvdvdvvvvf )
2
)(exp(
2),,(
2222/3
This expression is written in Cartesian co-ordinates (x,y,z).
Switch to spherical polar coordinates.
Sphericalpolar
coords
vz
vy
vx
222zyx vvvv
Surface element, dS, shown – need to consider volume element,
dV
vz
vy
vx
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 ?
Molecular speeds: Polar coordinates
Consider thin shell of sphere whose radius changes from v to v + dv. ?? What is the volume of this
thin shell?
What is the volume of the thin shell?
4p v
3/3
2p v
dv
4p v
2 d
v
None
of thes
e
15%
28%
57%
0%
a) 4v 3/3b) 2v dvc) 4v 2 dvd) None of these
Maxwell-Boltzmann distribution
zyxzyx
zyxzyx dvdvdvkT
vvvm
kT
mdvdvdvvvvf )
2
)(exp(
2),,(
2222/3
We had:
Taking into account discussion of spherical polar coordinates:
dvkT
mv
kT
mvdvvf )
2exp(
24)(
22/32
Maxwell-Boltzmann distribution of speeds of molecules in a gas at thermal equilibrium.
Maxwell-Boltzmann distribution
To convert from velocity to speed we have carried out two steps:
1. Convert from 1D to 3D probability.2. Consider all directions.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/maxspe.html#c1
)2
exp(2
4)(22/3
2
kT
mv
kT
mvvf
Function no longer a Gaussian!
293 K
600 K
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 2)• curve broadens as T increases
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 2 includes a number of questions on this distribution function.
½ m<v2> = (3/2)kT
In our derivation of the ideal gas law, we set the constant of proportionality between mean kinetic energy and temperature as 3k/2
From the Maxwell-Boltzmann distribution function, we can now show why the mean kinetic energy is given by 3k/2
0
24
2/32
0
22/3222
)2
exp(2
4
)2
exp(2
4
dvkT
mvv
kT
mv
dvkT
mv
kT
mvvv
The integral can be found in standard integral tables (see Lecture Notes Set 2b) and leads to the result:
kTvm
m
kTv
2
3
2
1
3
2
2
Typographical errors in Section 2.6: (i) Boltzmann’s (not Planck’s) constant!; (ii) Factor of 4missing from Eqn. 2.32. Apologies.
Equipartition of Energy and Degrees of Freedom
kTvm2
3
2
1 2 This statement has very important implications for both classical and quantum theory.
We’ve found that the mean kinetic energy for the gas molecules is3kT/2. However, this expression was derived by considering motionsof the molecules in the x, y and z directions.
We say that each molecule has three degrees of freedom
Theorem of equipartition of energy
“Each quadratic term in the expression for the average total energy of a particle in thermal equilibrium with its surroundings contributes on average ½ kT to the total energy”
or “Each degree of freedom contributes an average energy of ½ kT”