d.
variable heat capacity by integration of cp(T) or cv(T) equation
T2
P
dT
cp
R ln 2
T
P1
T1
T2
c
v
T1
v
dT
R ln 2
T
v1
Cycles and efficiencies
In an engine cycle a certain amount of heat, |QH|, is added to the cyclic device at a high temperatur

For indistinguishable particles,
1
1
dN i N i2 2 dN i 0
N
Ni
i
i
i
dN dN i 0
d ln ln
i
dU Ei dN i 0
i
Well apply the method of undetermined multipliers to determine the equilibrium distribution
of particles among the states, that is, the Ni.
1
i ln N

2.
Ideal Gas Sec 6.7
a. One molecule
Z1 e
i
Ei
kT
e
translational
states, i
Ei
kT
e
Ej
kT
rotational
states, j
e
En
kT
Z transZ rot Z int
internal
states, n
b. Two or more molecules
If the molecules are not interacting, then as before, the partition func

The pressure at which the phase transition occurs, at temperature T, is that value of P where
the two shaded areas cancel. So, tracking along an isotherm from right to left, the gas is
compressed and pressure rises until that horizontal section is reached

So, each quadratic degree of freedom, at equilibrium, will have the same amount of energy.
But, this equipartition of energy theorem is valid only in the classical limit and high
temperature limits. That is, when the energy level spacing is small compared

E
E i Z
Z
i 2 e kT
Z
i kT
T V
Z
2 Ei N i
kT i
ZU
2
kT
ln Z
U kT 2
T V
Entropy:
S k ln k N i ln N i kN
i
N E
k N i ln i kN
i
Z kT
Z U
Nk ln Nk
N T
Pressure:
TdS dU PdV
P
dU
dS
U TS NkT ln Z NkT
T
dV
dV
N
V
T
T
V
ln Z
P NkT

d ln ln
i
1
dN i
Ni
a bEi dN i
i
adN bdU
bdU
b
d ln
1 d k ln
1 S
dU
k dU
k U
1
kT
V , N
The equilibrium numbers of particles in each energy state finally are
E
N kTi
N i e
Z
This is the Boltzmann probability distribution.
1
P( E ) e
Z
E
kT
. The

IV.
Statistical Mechanics
A.
Partition Function
1.
Boltzmann Sec 6.1
a. Multiplicity
Consider a system, in contact with an energy reservoir, consisting of N weakly interacting
identical particles. The energy of each particle is quantized, the energy level

3
m 2
Dv
4v 2e
2kT
mv 2
kT
B. Adding up the States Sec 1.2, 1.7, 2.5, 6.6, 6.7
1.
Two-State Paramagnet Sec 6.6
The specifics of computing the partition function for a system depend on the nature of the
systemthe specifics of its energy levels. For in

If the number of particles is allowed to change, then we have to sum also over all possible
numbers of particles, as well as all possible energy states, giving the grand canonical
ensemble.
b. Average values
The probability that a particle will be observe

D. Eisenberg & W. Kauzmann, The Structure and Properties of Water, Oxford Univ. Press, 1969.
The phase diagram for water shown in the text figure 5.11 is a teensy strip along the T axis
near P = 0 on this figure. [One bar is about one atmosphere of air pr

introduce a correction to the pressure that is proportional to the density
and to the
N
V
number of atoms in the system, N. That is,
N 2
P P 2
V
.
With the new V and P, the gas law becomes the van der Waals equation of state:
or
2
NkT
N 2
N
P
P 2 V N

dGliquid dGgas
SliquiddT VliquiddP liquiddN S gasdT VgasdP gasdN
dP S gas Sliquid
dT Vgas Vliquid
Weve assumed that dN = 0. This result is the slope of the phase boundary curve on the PT
diagram.
Commonly, we express the change in entropy in terms of the

This result assumes that all N oscillators have the same frequency. The expression
approaches the Delong-Petit limit for high temperatures, but decreases exponentially at low
temperatures. Experiment shows that CV decreases like T3 at low temperature. The

NkT ln N 1 ln vQ ln V
F
N
N T ,V
ln N
kT ln N 1 ln vQ ln V NkT
N
V
kT ln
NvQ
4.
T ,V
Solids Sec 2.2, 3.3, 7.5
In a solid, the atoms/molecules are each confined to specific locations. Therefore, we regard
them as distinguishable particle

Summary of Important Equations in Thermodynamics
Basic notation and definition of terms
Fundamental dimensions of mass, length, time, temperature, and amount of substance (mol) are
denoted, respectively as M, L, T. , and in the abbreviations below. For ex

m
:
:
V2
2
the mass flow rate (MT-1)
the kinetic energy per unit mass (L2T-2)
gz:
the potential energy per unit mass (L2T-2)
Etot:
the total energy = m(u +
Q
:
:
2 -2
2 + gz) (ML T )
V
2
the heat transfer rate (ML2T-3)
the rate of change of energy for th

Q = U + W
Closed-system first law:
Mass balance equation:
dmcv
m i
dt
inlet
q = u + w
m o
outlet
Relation of mass flow to velocity, specific volume and area:
Steady-flow assumptions:
Steady-flow first law:
=0
and
VA
m VA
v
=0
2
2
V
V
W u Q m o ho o g

For ideal gases only: Cp Cv = mR = n
cp cv = R
R
c P c v R
Ideal gas entropy changes
a.
constant heat capacity
T
P
T
v
s2 s1 c p ln 2 R ln 2 cv ln 2 R ln 2
T1
P1
T1
v1
b.
variable heat capacity by integration
T2
T
P 2 dT
v
dT
s2 s1 c p
R l

Lets look at the high and low temperature limits.
For
.
3
1
T D , d D , so that Cv 3R
3 T
For
T D , D , and
d
4
3
4
.
T
4
12
T 3
, so that Cv
R
15
5
D
Evidently, the Debye version matches the experimental temperature dependence of the
specific h

hcn
En
2L
, whence
and
2L
dn dE
hc
2L
n 3
En
hc
U
3
.
hc
n3
8L3
E3
dn
dE
hcn
3
E
2 L
hc
0 2 LkT
0 kT
e
1
e 1
U U
8
E3
dE
E
L3 V hc 3
0 kT
e 1
U 8 5 kT
3
V
15 hc
4
We can now compute also the heat capacity and entropy of the photon gas.
U
8

photons must be the same as the photon gas in the oven, since all photons travel at the same
speed, c. By a similar token, the energy emitted through the hole is proportional to T4.
Finally, we might consider a perfectly absorbing material object exchangi

b.
Debye theory of specific heat
The oscillators do not vibrate independently. Rather, there are collective modes of vibration
in the crystal lattice. well treat the situation as elastic waves propagating in the solid. We
consider that the energy residing

5.
Photons Sec 7.4
a. Photon gas
Consider electromagnetic radiation inside a box. We may regard the electromagnetic field as
a superposition of standing waves that fit between the walls of the box. The system, then,
consists of these standing waves, rathe

whether the molecule was turning, then that axis does not count. Thats why there are no
states for an axis that runs through the carbon and oxygen atoms of carbon monoxide.
Therefore, a rotational partition function will look something like this for three

[Actually, in the classical form of the partition function, we are integrating over the possible
(continuous) values of particle momentum and position.
Lx
Z tr , x e
p x2
2 mkT
dp x dx Lx 2mkT
0 0
The classical partition functions differ from the classic

The probability peaks at about
[I did the calculation of
q A 3
, rather than
q A 4
, that is, at
NA
q A q
3.4
N A NB
.
using the COMBIN function in Excel.]
As we increase the numbers, the
s become very large very quickly, as illustrated by the text
examp

dS R
1
P
1
dU R R dVR R dN R dU R
TR
TR
TR
TR
dStotal dS
1
1
1
1
dU R dS dU dU TdS dF
TR
T
T
T
The increase in total entropy under conditions of constant T, V, and N is equivalent to a
decrease in the Helmholtz free energy of the system.
In a similar ve

( q )
P (q )
(all )
Notice that the total multiplicity is
(all ) 2 N
.
because each dipole has only two possible
states.
Here is a microstate for a system of N = 10 dipoles, with q = 6 (6 dipoles point up).
The probability function, P(q), for this system

2N
e q
max
N 2
2N
. Thats a very large number. How about the width of the curve? In the
text, the author shows that the curve is a Gaussian:
max e
The origin has been shifted to the location of
when
axis
q
x
2 N
q A 0 to q
q
qA
2
N 2 x q
2
, where

III.
Processes
A.
Cyclic Processes
1.
Heat Engines & Heat Pumps Sec 4.1, 4.2
A heat engine is a device that absorbs heat from a reservoir and converts part of it to work.
The engine carries a working substance through a PVT cycle, returning to the state a

b. Helmholtz
Lets say the system is in contact with a heat bath, so that the temperature is constant. The
pressure may not be constant. To create the system, some of its total energy can be taken
from the environment in the form of heat. So the total work

c. Rankine cycle
In some ways the steam engine is a more nearly exact example of a heat engine than is the
Otto engine. No chemical reaction or combustion takes place within the working fluid, and at
least in principle the working fluid is not replaced at

The temperatures also change from step to step. The efficiency is given by
V
e 1 2
V1
The quotient
V1
V2
1
1
T1
T
1 4
T2
T3
is the compression ratio. The greater the compression ratio, the greater is
the efficiency of the engine. However, so is T3 g

The First Law says
together, we obtain
Qh Qc W
. The Second Law says
1
T
COP
c
Th
1 Th Tc
Tc
Qh Qc
Th Tc
. Putting these
. A Carnot cycle running in reverse will give the
maximum COP.
2.
Otto, Diesel, & Rankine Sec 4.3
Real heat engines need to produce