Transcript Slide 1

Shihabudheen M
M E S Kalladi College
Mannarkkad
 Statistical distributions Maxwell-Boltzmann statistics –
 Distribution of molecular energies in an ideal
gas-Average molecular energy Equipartition theorem
 Maxwell-Boltzmann speed distribution law Expressions for
 rms speed,
 most probable
 speed and mean speed
 The branch of physics called statistical mechanics
considers how the overall behavior of a system of
many particles is related to the properties of the
particles themselves.
 The statistics obeyed by any system generally
belongs to one of the following two categories:
 Classical Statistics
 Quantum Statistics.
 Classical statistics treats particles as
distinguishable whereas quantum statistics treats
them as indistinguishable.
Statistical Distributions
 Statistical mechanics deals with the behavior of
systems of a large number of particles.
 We give up trying to keep track of individual particles.
 If we can’t solve Schrödinger’s equation in closed form
for helium (4 particles) what hope do we have of
solving it for the gas molecules in this room (10really big
number particles).
 Statistical mechanics handles many particles by
calculating the most probable behavior of the system
as a whole, rather than by being concerned with the
behavior of individual particles.
 Four balls and two boxes
1
2
 statistical mechanics determines the
most probable way in which a certain
total amount of energy E is distributed
among the N members of a system of
particles in thermal equilibrium at the
absolute temperature T.
 Thus we can establish how many
particles are likely to have the energy 1,
how many to have the energy 2, and so
on.
E
N
4
n4
3
n3
2
n2
1
n1
 The particles are assumed to interact with one another
and with the walls of their container to an extent
sufficient to establish thermal equilibrium
 More than one particle state may correspond to a
certain energy .
 If the particles are not subject to the exclusion
principle, more than one particle may be in a certain
state.
 Let W be the number of different ways in
which the particles can be arranged among
the available states to yield a particular
distribution of energies
 The greater the number W, the more
probable is the distribution.
 It is assumed that each state of a certain
energy is equally likely to be occupied. (equal
a priori probability )
 The program of statistical mechanics begins by
finding a general formula for W for the kind of
particles being considered.
 The most probable distribution, which corresponds to
the system being in thermal equilibrium, is the one for
which W is a maximum, subject to the condition that
the system consists of a fixed number N of particles
whose total energy is some fixed amount E.
 (The No. of particles in a state of energy E) = (The No.
of states having energy E) x (probability that a particle
occupies the state of energy E).
 If we know the distribution function, the (probability
that a particle occupies a state of energy E), we can
make a number of useful calculations.
 Mathematically, the equation is written
nε  = gε  f ε 
 g( ) = number of states of energy (statistical weight
corresponding to energy )
 f( ) = distribution function
= average number of particles in each state of energy 
= probability of occupancy of each state of energy 
 In systems such as atoms, only discrete energy levels
are occupied, and the distribution g() of energies is
not continuous.
 On the other hand, it may be that the distribution of
energies is continuous, or at least can be approximated
as being continuous. In that case, we replace g(ε) by
g(ε)dε, the number of states between ε and ε+dε.
 We will find that there are several possible
distributions f(ε) which depend on whether particles
are distinguishable, and what their spins are.
Three different kinds of particles:
 Identical particles that are distinguishable.
 Two particles can be considered distinguishable if
their separation is large compared to their de Broglie
wavelength.
 Example: ideal gas molecules.
 In quantum terms, the wave functions of the particles
overlap to a negligible extent.
 The Maxwell-Boltzmann distribution function
holds for such particles.
Three different kinds of particles
 Identical particles of zero or integral spin that cannot
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be distinguished one from another because their wave
functions overlap.
Such particles, called bosons
They do not obey the exclusion principle
The Bose-Einstein distribution function holds for
them.
Photons are bosons
We shall use Bose-Einstein statistics to account for
the spectrum of radiation from a blackbody.
Three different kinds of particles
 Identical particles with odd half-integral spin that also
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cannot be distinguished one from another.
Such particles, called fermions,
They obey the exclusion principle,
The Fermi-Dirac distribution function holds for
them.
Electrons are fermions
We shall use Fermi-Dirac statistics to study the
behavior of the free electrons in a metal that are
responsible for its ability to conduct electric current.
Statistics
Classical
statistics
MaxwellBoltzmann
Statistics
Quantum
statistics
Bose-Einstein
statistics
Fermi-Dirac
statistics
 The Maxwell-Boltzmann distribution:
 We know that classical particles which are identical
but far enough apart to be distinguishable obey
Maxwell-Boltzmann statistics.
 Or Two particles can be considered distinguishable if
their separation is large compared to their de Broglie
wavelength.
 The Maxwell-Boltzmann distribution function states
that the average number of particles fMB(ε) in a state of
energy ε in a system of particles at the absolute
temperature T is
f  ε  = A e-ε/kT .
 The is the distribution function for ‘n’ no. of particles in
‘g’ states.
 The number of particles having energy ε at temperature
T is
n  ε  = A g  ε  e-ε/kT .
 A is like a normalization constant; we integrate n(ε) over
all energies to get N, the total number of particles.
 ε is the particle energy, k is Boltzmann's constant
Molecular Energies in an ideal Gas
We assume a continuous distribution of energies so that, the
no. of particles having energies between ϵ and ϵ+dϵ is given
by
n ε  dε = A g ε  e-ε/kT dε .
where g(ϵ) dϵ is the no of states in the energy range ϵ and
ϵ+dϵ. g(ϵ) is called density of states.
It turns out to be easier to find the number of momentum
states corresponding to a momentum p, and transform back
to energy states.
Corresponding to every value of momentum there is a
value of energy.
Momentum is a 3-dimensional
vector quantity. Every (px,py,pz)
corresponds to some energy.
We count how many
momentum states are there in a
region of space and then
transform to the density of
energy states.
We need to find how many momentum states are in this
spherical shell.
The Maxwell-Boltzmann distribution is for classical
particles, so we write
p =
2mε =
p2x + p y2 + p 2z .
The number of momentum states in a spherical shell
from p to p+dp is proportional to 4πp2dp (the volume of
the shell).
Thus, we can write the number of states having
momentum between p and p+dp as
gp dp = B p2dp
where B is a proportionality constant, which we will
worry about later.
Because each p corresponds to a single ε,
g ε  dε = gp dp = B p2dp .
Now,
2
p = 2mε
p = 2mε
1 -1/2
dp = 2m ε dε ,
2
so that
p2 dp  ε ε-1/2 dε ,
and
g ε  dε  ε ε-1/2 dε .
n  ε  dε = C ε e-ε/kT dε .
The constant C contains B and all the other
proportionality constants lumped together.
n  ε  dε = C ε e-ε/kT dε
To find the constant C, we evaluate


0
0
N =  n  ε  dε = C 
ε e-ε/kT dε
where N is the total number of particles in the system.
Now

1 
0 x e dx  2a a
The result is
1
2  ax
C
N =
2
  kT 
a  1 \ kT
3/2
,
so that
n  ε  dε =
2N
 kT 
3/2
ε e-ε/kT dε .
This is the number of molecules having energy between
ε and ε+dε in a sample containing N molecules at
temperature T.
“It forms the basis of the kinetic theory of gases, which
accurately explains many fundamental gas properties,
including pressure and diffusion.”
“The Maxwell-Boltzmann distribution also finds
important applications in electron transport and other
phenomena.”
Here’s a plot of the
distribution:
Continuing with the article, the total energy of the


system is
2πN
3/2
-ε/kT
E=
 ε n  ε  dε
0
=
ε

 kT 
Evaluation of the integral gives
3/2
e
dε .
0
E=
3NkT
.
2
This is the total energy for the N molecules, so the
average energy per molecule is
3
ε = kT ,
2
exactly the result you get from elementary kinetic theory
of gases.
Calculation:

x
0
3
2
e
 ax
3
dx 
4a 2

a


2N   3

2



E
kT

kT
3


  4
2
{

kT
}


3NkT
E=
.
2
Results of Kinetic Theory
 KE of individual particles is related to the temperature
of the gas:
½ mv2 = 3/2 kT
Where v is the average velocity.
Boltzmann
Distribution
 Demonstrated that
there is a wide range
of speeds that varies
with temperature.
Equipartition of Energy
 Each degree of translational freedom takes ½ kT.
KEx +KEy+KEz = ½ kT + ½ kT + ½ kT
KEtotal = 3/2 kT
This is true for single point masses that possess no
structure.
 Each new DOF requires ½ kT of energy.
 Each new DOF contributes ½ kT to the total internal
energy of the gas. This is the Equipartition
Theorem.
Equipartition of Energy
For molecules, i.e. multi-atom particles, there are added
degrees of freedom.
Internal Energy of Di-Atom
 Three translational DOF + 2 rotational DOF = 5 DOF.
Each DOF contributes ½ kt, so the internal energy of a
diatomic gas is,
U = 5/2 NkT,
For a gas of N molecules.
Because ε = mv2/2, we can also calculate the number of
molecules having speeds between v and v + dv.
The result is
2  N m3/2 2 -mv /2kT
n  v  dv =
 kT 
3/2
v e
2
dv .
“We” (Beiser) call this n(v).
Here’s a plot (number
having a given speed
vs. speed):
The speed of a molecule having the average energy
comes from solving
mv 2
3
ε=
= kT
2
2
for v. The result is
vrms = v 2 =
3kT
.
m
vrms is the speed of a molecule having the average
energy ε.
It is an rms speed because we took the square root of the
square of an average quantity.
The average speed v can be calculated from

v=
 v n(v) dv
 n(v) dV
0

.
0
The result is
v=
8kT
.
m
Comparing this with vrms, we find that
vrms = 1.09 v .
You can also set dn(v) / dv = 0 to find the most probable
speed. The result is
vp =
2kT
.
m
The subscript “p” means “most probable.”
Summarizing the different velocity results:
v rms =
3kT
m
v=
vrms = 1.09 v
8kT
m
vp =
2kT
m
n(v)
Plot of velocity distribution again:
Example 9.4 Find the rms speed of oxygen molecules at
0 ºC.
0 m/s? 10 m/s? 100 m/s? 1,000 m/s? 10,000 m/s?
You need to know that an oxygen molecule is O2. The
atomic mass of O is 16 u (1 u = 1 atomic mass unit =
1.66x10-27 kg).
 1.66×10-27 kg 
-26
mass of O2 =  2 16 u  
=
5.31×10
kg

u


v rms =
3kT
m
vrms =
3 1.38×10-23 J/K    0+273 K 
5.31×10
-26
kg
vrms = 461 m/s
vrms =
 461
m/s mile / 1610 m3600 s/h
vrms = 1031 miles / hour .