Problem: A simple model for a crystal is a gas of harmonic oscillators. Determine A, S , and U from the partition function for this model.
Solution: For this model the crystal is modelled as a collection of harmonic
oscillators so we need the partition fu
The work done on the left is
wL = P1 4V = P1 (0 V1 ) = P1 V1 .
(42.46)
The work done on the right is
wR = P2 4V = P2 (V2 0) = P2 V2 .
(42.47)
4U = U2 U1 = wL + wR = P1 V1 P2 V2
(42.48)
U2 + P2 V2 = U1 + P1 V1 H2 = H1
(42.49)
Now,
Thus
For JouleThomson ex
for an isothermal process.
Recall the denition of Helmholtz free energythe energy of the system available
to do work.
We learned previously that the maximum amount of work one can extract from
the system is the work done during a reversible process. Hence
The ux of material through a plane depends on the concentration dierence
J = D
dC
1 dn
dC
=
= D
dx
A dt
dx
where D is the diusion constant
dC
1 dn
= D
A dt
dx
(45.2)
This is Ficks rst law of diusion (in one dimension).
The change in concentration in a lam
so the useful relation becomes
nRT
nRT
n2 a
nR
U
P =
+2
=T
V T
V nb
V nb V nb
V
2
na
=+2
V
(17.11)
The equation of state for U : Express U in terms of T, V, and P.
Start with the total dierential of U
U
U
dT +
dV
dU =
T V
V T
U
but U V = CV and V T = T
2. Propagation: The radical formed in the initiation step reacts with some so
molecule M0 to form another molecule M00 and another radical R0 . This step
repeats an indenite number of times.
R+M0 M00 + R0 .
3. Termination: The radicals interact with each
36. Electronic Spectroscopy of
Molecules
The electronic spectra of molecules are quite dierent than that of atoms.
Atomic spectra consist of single sharp lines due to transitions between energy
levels.
Molecular spectra, on the other hand, have numerous l
The Laplacian
2
=
2
2
2
+ 2+ 2
x2 y
z
.
Normalized wavefunctions for the 3D particle in a box,
ny y
nz z
nx x
22
sin
sin
.
sin
n (x) =
a
b
c
abc
(31.19)
(31.20)
The energy levels for the 3D particle in a box,
Enx ,ny ,nz
Orthonormality:
Z
n2 h2
n2 h2
12. Rudiments of Statistical
Mechanics
When we study simple systems like a single molecule, we use a very detailed
theory, quantum mechanics.
However, most of the time in the real world we are dealing with macroscopic
systems, say, at least 100 million mo
16.3. Equations of State
The macroscopic properties of matter are related to one another via a phenomenological equation of state.
The state of a pure, homogeneous material (in the absence of external elds) is
given by the values of any two intensive prop
Among these functions are x, y, and z.
So we can see immediately that the IR active modes of any molecule having this
point group will be A1 , B1 , and B2 .
The A2 mode is IR forbidden and any vibrations having this symmetry will not
appear in the IR spec
Also recall that the mathematical denition of activity ai of some species i is
implicitly stated as
ai
lim
=1
(46.9)
g ( )
where g( ) is any reference function (e.g., pressure, mole fraction, concentration
etc.), and is the value of at the reference stat
2. The Postulates of Quantum
Mechanics
2.1. Postulate I
Postulate I: The state of a system is dened by a wavefunction, , which contains all the information that can be known about the system.
We will normally take to be a complex valued function of time a
13. The Boltzmann Distribution
Consider a isolated system of N molecules that has the set cfw_ i energy levels
associated with it.
Since the system is isolated the total energy, E, and the total number of particles
will be constant.
The total energy is g
Substituting the expression for salt into this gives
ln asalt
v+ + v + v+ v
+
=
RT
v+ + v+ v v
+
+
=
RT
RT

cfw_z

cfw_z
v+ ln a+
So,
(23.5)
v ln a
ln asalt = v+ ln a+ + v ln a
(23.6)
asalt = av+ av
(23.7)
or, alternatively,
It is the case that 1 mol
The above considerations leave four vectors.
In fact, there will always be the same number of vectors as symmetry elements.
Altogether, the vectors represent what is call an irreducible representation of the
group.
These vectors make up the :
C2v
E
C2
A1