Beyond Going U and CV (17.6, 7, 8)
So far we have considered the energy and heat capacities of two systems: perfect gases
(monatomic & diatomic) and perfect atomic crystal.
Again, looking ahead a couple of weeks, we note
E j
th
Pj ( N , V ) =
= Pressure in the j energy state
V N
By our definition of ensemble averaging:
our definition of ensemble averaging:
P = Pj ( N , V ) p j ( N ,V , ) =
E j E j ( N ,V )
e
N
Q ( N ,V , )
1 ln Q
ln Q
so : P = k BT
=
V N ,T V N ,
1
1
=
( N ln q ln N !) N , =
( N ln V + terms with no V ) N ,
V
V
1 1 Nk BT nRT
=N=
=
( ideal gas eqn.)
V
V
V
V
j
j
Math stuff:
recall Q = e
E j ( N ,V )
j
1 Q
=
Q V N ,
E j
Q
E j ( N ,V )
so
=
e
V N ,
j
V N ,
E j
E j ( N ,V )
e
N ,
=P
Q Lecture 7
V
j
Note that for any ideal gas (monatomic, diatomic, and polyatomic),
q(
q(N,V) is proportional to V, so the ideal gas equation applies to any
ideal gas.
Lecture 7
2
1
Lets look at things just a bit more deeply. In statistical mechanics the
energies, Ej(N,V), are eigenvalues of the N-particle Hamiltonian.
Remember, we couldnt find exact eigenvalues even for a 3-body system
(except H2+ in the B.O. approximation) so what can we do for N-bodies?
1st we can assume that the N-particles are independent.
i =1
2nd: If the particles are distinguishable, we know which is which
(examples would be our perfect crystal of atoms at specific
(examples would be our perfect crystal of N atoms at specific
coordinates).
b
c
Then E j ( N , V ) = la (V ) + m (V ) + n (V ) + ...
14444244443
4
4
N terms, each with its own energy value
j
E j ( N ,V )
=
e
(
b
c
la (V ) + m (V ) + n (V ) +...
l , m , n ,...
qs are individual particle partition functions (we can
are individual particle partition functions (we can
distinguish each particle).
Lecture 7
As in the last slide, if the particles are identical and have the same range
N
of states, then: Q ( N , V , T ) = q (V , T ) where q ( V,T ) = e i / kBT
i
But wait. In general, particles (identical ones in a gas) are
indistinguishable So we can say any one particle has given energy
indistinguishable. So we cant say any one particle has a given energy.
What to do? 1st lets be nave (and get the right answer!)
N
Then E j ( N , V ) = i (V )
So Q = e
1 ln Q
ln Q
= k BT
V N ,
V N ,T
And we equate the ensemble average <P> to the experimentally
observed pressure P.
N
3/ 2
q (V , ) ; q V , = 2 m V
For our ideal monatomic gas: Q ( N,V, ) =
( )2
N!
h
So, we have shown P =
) = e
l
a
l
e e ...
m
b
m
n
qa (V , T ) qb (V , T ) qc (V , T )
3
c
n
Lets say we have a 3 atom system (identical and indistinguishable),
where each atom could be in one of many states.
b
c
Eabc = 1a + 2 + 3
E =1 +2+3
abc
These states are all identical since we cannot
cba
distinguish between identical atoms (a,b,c)
bac
We have over-counted this energy by 6=3!
3
acb
so Q = q (V , T ) / 3!
bca
Lecture 7
4
cab
If we have N indistinguishable systems then Q= q ( V,T ) / N !
N
Indistinguishable
But is this strictly correct?
In the case above, all three energy states were different, but what
happens if they are not?
Consider what happens with fermions protons,
neutrons, (electrons, and particle with half-integer spin).
Two identical particles cant have the same state (violates the Pauli
exclusion principle) because the wavefunction vanishes
(wavefunction must be antisymmetric with respect to interchange).
How does this impact state counting?
On the next slide we consider the number of allowed microstates
On the next slide, we consider the number of allowed microstates
allowed for two particles with three energy levels (1, 2, 3). We
will look at both indistinguishable and distinguishable fermions
and bosons.
Lecture 7
5
How can we get around these problems?
Notice that if every particle (Boson or Fermion) is in a different
that if every particle (Boson or Fermion) is in different
energy state, then our counting problems (double counting or Pauli
exclusion forbidden states) disappear.
What is this condition?
q (V , T ) = e i / kBT # of possible energy eigenstates that can be populated
i
2 mk BT
q (V , T ) =
2
h
3/ 2
V
2 mk BT
so we want:
2
h
3/ 2
for translational states
N h2
V >> N
V 2 mk BT
3/ 2
<< 1
3/ 2
N h2
or as given in the text:
as given in the text:
<< 1
V 8mk BT
This condition is valid for almost all systems except liquid He,H2
(because low Temp) and electrons (because low mass).
(b
Lecture 7
1 + 1
1 + 2
2 + 1
1 + 3
3 + 1
2 + 2
2 + 3
3 + 2
3 + 3
Fermion
Boson
Fermion
x
x
x
3 states
states
a
1
a
1
a
2
a
1
a
3
a
2
a
2
a
3
a
3
6 states
states
+
+
+
+
+
+
+
+
+
b
1
b
2
b
1
b
3
b
1
b
2
b
3
b
2
b
3
Boson
x
x
x
6 states
states
9 states
states
Note that when we have indistinguishable particles (either Bosons or
Fermions) in the same state this causes some problems in counting
Fermions) in the same state this causes some problems in counting.
Lecture 7
6
When # states >> # particles Boltzmann statistics hold.
N
q (V , )
for identical, indistinguishable particles
Q ( N ,V , ) =
N!
Separating the degrees of freedom
the degrees of freedom
With our new knowledge of independent, distinguishable and
indistinguishable entities, we can separate the degrees of freedom.
ln Q
Start with the basic definition: E =
N ,V
N
q (V , ) For identical, indistinguishable
E =
ln
N!
atoms/molecules
V , N
i
e
( q (V , ) )
i
E =
( N ln q (V , ) ) N ,V = N q (V , ) = N q (V , ) N ,V
e
= +N
E
7
Distinguishable
i
i
q (V , )
=N
Where is average energy for
Lecture 7
one particle (atom/molecule)
8
Now, the energies within the particle (atom/molecule) are independent:
= itrans + rot + kvib + lelect .
j
qrot = ( 2 J + 1) e
J
They are also distinguishable, so:
q = qtrans qrot qvib qelec
And we can determine average contributions to each degree of freedom,
rot
e.g:
rot e j
rot =
=
j
j
j=sum over states
over states
qrot
g
rot
J
Jrot e
J
qrot
h 2 J ( J +1) / 2 Ik BT
so
rot
2
h2
2 I J ( J + 1) ( 2 J + 1) e h J ( J +1) / 2 IkBT
=J
qrot
Since we worked so hard at Q.M. last semester, we will
define and use molecular partition functions based on
sums over energy levels.
rot
J
J=sum over rotational energy levels
rot
recall: g J = 2 J + 1
Jrot = h 2 J ( J + 1) / 2 I
Lecture 7
9
Lecture 7
10