The Meaning of Absolute Zero
The Boltzmann distribution shows us what is special about absolute
zero
ni*  n0* exp[ ( i   0 ) / k BT ]
When T = 0,
ni*  n0* exp[ ( i   0 ) / k BT ]  0
for i ≠ 0
All molecules in the system are in their lowest energy (ground) state.
That is, n0* = N at T = 0. There can be no temperature lower than
absolute zero, because molecules cannot go into a lower energy state.
The Molecular Partition Function
The equilibrium probability distribution (Boltzmann distribution) tells us
the probability of finding a molecule in state i. This may be written as
*
n
exp[( i   0 ) / k BT ]
*
i
Pri  
N  exp[( i   0 ) / k BT ]
i
The sum in the denominator of this expression plays an important role in
the equilibrium properties of large collections of molecules. It is called
the molecular partition function, q(T)
q(T )   exp[( i   0 ) / k BT ]
i
The partition function is a function of temperature and the energy levels
of the molecule.
Properties of the Molecular Partition Function
What kind of physical interpretation can we make of q(T)?
q(T )   exp[( i   0 ) / k BT ]
i
  ni* / n0*
i
It is a weighted sum of the occupation numbers relative to the ground
state occupation number, and it increases as the occupation number of
excited states goes up. q(T) increases with increasing temperature.
If T  0, then q(T)  1 – all molecules are the same ground state.
If T  , then q(T)  the number of molecular states. That is, more
(higher-energy) states are occupied as T increases.
The molecular partition function, q(T) is, roughly, the number of
thermally accessible states per molecule.
Sample Problems: Equilibrium Configuration
1) Given a set of occupation numbers ni and molecular state energies i, how
would you determine whether or not the configuration corresponds to an
equilibrium configuration?
To answer this question you need to make use of a property of the equilibrium
distribution, i.e.
*
ni
 exp[ ( i   j ) / k BT ]
*
nj
This expression can be rearranged to give (j - i)-1 ln(ni*/nj*) = 1/kBT = b, a
constant for all possible i and j.
If you pick any two molecular states and insert their values of the occupation
numbers and energies, then this configuration is an equilibrium configuration
only if the left hand side of the equation equals a constant that does not change
when the same calculation is performed on a different pair of states.
Alternatively, you can make a graph of ln(ni*) against i. If the configuration is an
equilibrium one, then this will be a straight line with slope b.
Sample Problems: Equilibrium Configuration
2) Having established that a given configuration is an equilibrium configuration,
how can you determine the temperature?
This follows pretty easily from the previous question. Having established (or
been told) that the configuration is an equilibrium one, the temperature is
obtained from inverting the right hand side of the equation above and dividing by
kB. Make sure your units of energy are consistent with the units of the
Boltzmann constant.
1.00E+07
ln(ni*)
= ln(n0 + 0/kBT - i /kBT
*)
So, a graph of
i is a
straight line for the equilibrium
distribution, and the slope of that
line is -b, or -1/kBT.
ln(ni*)versus
number of molecules
The alternative approach gives:-
T /K
1.00E+06
50
200
1.00E+05
1.00E+04
1.00E+03
1.00E+02
1.00E+01
1.00E+00
0
2E-21
4E-21
Energy (J)
6E-21
8E-21
Temperature
…is a property of an equilibrium configuration.
What happens if the configuration isn’t an equilibrium configuration?
Can you still define a temperature for such a configuration?
The answer is no. Temperature is a property of equilibrium systems only (you
can regard it as being defined by the equilibrium occupation numbers). Away
from equilibrium, temperature, strictly, ceases to exist and even if there is a
nonequilibrium most probable configuration, such as in a pumped laser cavity,
the configuration can never be described by a single quantity like T.
Flash Quiz
• If energy levels were closer together, would
the partition function go up, down, or stay
the same?
Answer
• If energy levels were closer together, would the
partition function go up, down, or stay the same?
q(T )   exp[( i   0 ) / k BT ]
i
  ni* / n0*
i
The occupancy of each level, ni*, would go up,
because i is now closer in energy to 0. So the
overall partition function would also go up.
Degeneracy of Molecular States
So far we have always referred to molecular states. It if often more natural to
talk about molecular energy levels. Spectroscopy, for example, specifically
resolves differences in energy levels. The difference between states and energy
levels arises because a number of different states may have the same energy
and hence share the same energy level. Such states are referred to as
degenerate states.
The number of states that have the same energy is called the degeneracy of
that level. The degeneracy of the energy level  is denoted by g().
A molecule with a doublydegenerate first excited state would
be represented as shown in this
energy-level diagram.
3
2
energy
1
1'
0
The occupation probability of each of these states, considering only the most
probable configuration, depends only on the energy of the state: Pr1 = n1*/N =
n1’*/N. On the other hand, the probability that energy level 1 is occupied is
given by Pr(1) = n1*/N + n1*/N = g(1)  Pr1.
There are two equivalent ways of writing the partition function q(T), depending
on whether you sum over the individual molecular states i or molecular energy
levels i,
q(T )  exp[( i   0 ) / k BT ]
summing over states

i
q(T )   g ( i )exp[( i   0 ) / kBT ]
summing over energy levels
i
Similarly, you may write the Boltzmann distribution by summing over the
molecular energy levels i,
n* ( i ) g ( i )

exp[( i   0 ) / k BT ]
*
n ( 0 ) g ( 0 )
Homework Problem
Consider two energy levels at i = 2.05 x 10-21J and j = 2.64 x 10-21J. The
occupation numbers of these in a particular system at T = 200K are ni = 4,761
and nj = 11,535.
Energy i is singly-degenerate. (That is, there is only one state of energy i .)
kB = 1.381 x 10-23 J K-1
How many states are there with energy j?
Work and Heat
The First Law of Thermodynamics decomposes the change in energy of a
system into two parts, work and heat. We can write this as
dE = dEwork + dEheat

ni  i,
Using our definition of the energy of a system at equilibrium as E 
we can ask “How can a small change in the energy of a system come iabout?”
We write a change in E as DE as
*
DE   ni*D i   Dni* i   Dni*D i
i
i
i
We will focus on a very small change in the total energy. To represent this very
small change we shall replace D by d (the differential) and we have
dE   ni*d  i   dni* i
i
i
where we have neglected the third term on the right hand side as it will be
considerably smaller than the first two terms since it involves multiplying two
very small quantities together.
(16)
Work and Heat
Putting this expression into words - a small change in E can be accomplished by
small changes in the molecular energy levels themselves and by small changes
in the equilibrium configuration.
dE   ni*d  i   dni* i
i
Work: Small changes in the
molecular energy levels.
i
Heat: Small changes in the
occupation numbers of each
of the molecular energy
levels.
With respect to work we can now ask, “How is it possible to change the energies
of individual molecular states?”.
The Ideal Gas
We will build our ideal gas using the particle in a box model. The energy levels
of a 1-D particle in a box are
i 2h 2
i 
8mL2
m = mass of particle, L = length of 1-D box, and Planck’s constant h = 6.626 ×
10-34 J s.
Clearly we can change the energy levels by changing the size of the box, L. As
L is decreased and the particle more confined, the energies increase.
A particle in a 3-D box in the form of a cube of side-length L has similar energy
levels, but now we require three quantum numbers to specify the state.
 i ,i
x
y ,iz
h2
2
2
2

i

i

i

x
y
z 
2
8mL
As translation is the only possible degree of freedom for an atomic (or
monatomic) gas, the allowed energies are all specified. We can now calculate
the molecular partition function, energy, and other properties of the system.
The Ideal Gas
Using the known energies, we can calculate the total energy of the ideal gas
system as
E
ni*x ,i y ,iz  ix ,i y ,iz
…(18)

ix
iy
iz
The molecular partition function is similarly given by a sum over all energy
states (with =0)
q(T )   exp(  ix ,i y ,iz / k BT )
ix
iy
(19)
iz
…and the equilibrium occupation number if a state is given by
n*i ,i
x y , iz

N
exp(  ix ,i y ,iz / k BT )
q
These summations over huge numbers of states are not especially convenient,
so we’d like to replace them with an analytical expression. To do this we need
the sums to converge conveniently (they don’t) or to approximate the discrete
values with a continuous function. That is, we assume that very many states
are occupied.
In order to decide this we need the concept of a characteristic temperature.
The Characteristic Temperature
A characteristic temperature  = (1 - 0)/kB is, roughly, the lowest temperature
at which the first excited state is significantly populated.
At T>> , many states are occupied.
At T << , only the ground state is significantly occupied.
How big is trans? For He in a 1m3 box, trans ~ 10-17K.
For heavier particles, the characteristic temperature is even lower. Using the
partition function we can estimate the number of occupied states of such an
ideal gas a T = 1K to be q(1K) ~ 108.
Under the condition of T>> ,
qtrans
And the energy of the system is (from equation (15))
 2 mk BT 


2
 h

3/ 2
E  23 NkBT  23 nRT
V …(20)
(21)
…back to Work
Remember that we want to find an expression for the work, the energy change
associated with changing allowed (translational) energy levels in our ideal
atomic gas.
dEwork   ni*d  i
i
Just a little bit of calculus does this for us. Recall that  ix ,i y ,iz
h2

i 2  i y2  iz2 
2  x
8mL
We will just write this as
 i  L2  V 2 / 3
d
 d  i  i dV   23 V 5/ 3dV
dV
 d  i   23  iV 1dV
Which we now substitute into dEwork


dEwork   ni*d  i   23 V 1   ni* i  dV
i
 i

  23 V 1EdV   23 V 1 ( 23 nRT )dV
 nRT
 
 V

 dV   PdV

This derives the ideal gas equation, and the macroscopic relationship between
pressure applied to compress a system and increase its energy.
Work and The Ideal Gas - Summary
The energy of particle confined in a box of
length L goes like
i2
i  2
L
i.e. decreasing L means increasing εi
Since L = V1/3 we could change the
volume to change the energy levels.
It turns out (see online notes) that the
change in εi is related to the change in
volume by
d i  
2 i
dV
3V
and
dEwork = -PdV where P = nRT/V, the ideal gas equation
Summary
You should now
• Be able to explain the absolute (Kelvin) temperature scale, and
what is so special about absolute zero.
• Define degeneracy, molecular partition function, characteristic
temperature, heat and work
• Describe the assumptions and properties of an ideal gas as a
theoretical model.
• Do the homework problem.
Next Lecture
• Heat and heat capacity
• Separation of degrees of freedom and partition functions
Some Additional Notes – not examinable
The Total Energy of a System
We will see that the molecular partition function is a convenient route
from allowed energies to many macroscopic system properties.
The total energy of a system of molecules is defined by
E   ni i
(5)
i
For a system at equilibrium, we may write this as
E   ni* i  n0*   i exp[ b ( i   0 )] 
i
i
N
 i exp[ b ( i   0 )]

q(T ) i
dq(T )
=
However q(T) = i exp[-b(i - 0)], so we can write
 i exp[b ( i   0 )]
db i
Rearranging this gives
E
N dq
d ln q
d ln q
N
 Nk BT 2
q db
db
dT
(15)