Chapter 3: Interactions and Implications.
Start with
Thermodynamic Identities
Diffusive Equilibrium
and Chemical Potential
Sign
“

”
:
out
of
equilibrium,
the
system
with
the
larger
S
/
N
will
get
more
particles
.
In
other
words,
particles
will
flow
from
from
a
high
/
T
to
a
low
/
T
.
Let
’
s
fix
V
A
and
V
B
(the
membrane
’
s
position
is
fixed),
but
assume
that
the
membrane
becomes
permeable
for
gas
molecules
(exchange
of
both
U
and
N
between
the
sub

systems,
the
molecules
in
A
and
B
are
the
same
)
.
U
A
, V
A
, N
A
U
B
, V
B
, N
B
For sub

systems in
diffusive equilibrium:
In equilibrium,

the chemical
potential
Chemical Potential: examples
Einstein
solid
:
consider
a
small
one,
with
N
=
3
and
q
=
3
.
let
’
s
add
one
more
oscillator
:
To keep
dS
= 0, we need to
decrease
the
energy, by subtracting one energy quantum.
Thus, for this system
Monatomic ideal gas:
At normal
T
and
P
, ln(...) > 0, and
< 0 (e.g., for He,
~

5
∙10

20
J ~

0.3 eV.
Sign
“

”
:
usually,
by
adding
particles
to
the
system,
we
increase
its
entropy
.
To
keep
dS
=
0
,
we
need
to
subtract
some
energy,
thus
U
is
negative
.
The Quantum Concentration
At
T
=300K,
P
=10
5
Pa ,
n << n
Q
. When
n
n
Q
, the quantum statistics comes into play.

the
so

called
quantum
concentration
(one
particle
per
cube
of
side
equal
to
the
thermal
de
Broglie
wavelength)
.
When
n
Q
>>
n,
the
gas
is
in
the
classical
regime
.
n=N/V
–
the concentration of molecules
The
chemical
potential
increases
with
the
density
of
the
gas
or
with
its
pressure
.
Thus,
the
molecules
will
flow
from
regions
of
high
density
to
regions
of
lower
density
or
from
regions
of
high
pressure
to
those
of
low
pressure
.
0
when
n
n
Q
,
0
when
n
increases
Entropy Change for Different Processes
Type of
interaction
Exchanged
quantity
Governing
variable
Formula
thermal
energy
temperature
mechanical
volume
pressure
diffusive
particles
chemical
potential
The last column provides the connection between statistical physics and
thermodynamics.
The partial derivatives of
S
play very important roles because they determine
how much the entropy is affected when
U, V
and
N
change:
Thermodynamic Identity for
dU
(
S,V,N
)
if monotonic as a function of
U
(
“
quadratic
”
degrees of freedom!), may be inverted to give
compare with
pressure
chemical potential
This holds for
quasi

static processes
(
T
,
P
,
†
a牥e

deieth牯ght thesystem⤮
†
shs
h
mch
the
system
’
s
energy
changes
when
one
particle
is
added
to
the
system
at
fixed
S
and
V
.
The
chemical
potential
units
–
J
.

the so

called
thermodynamic
identity
for
U
Thermodynamic Identities
is
an
intensive
variable,
independent
of
the
size
of
the
system
(like
P
,
T,
density)
.
Extensive
variables
(
U,
N,
S,
V
..
.
)
have
a
magnitude
proportional
to
the
size
of
the
system
.
If
two
identical
systems
are
combined
into
one,
each
extensive
variable
is
doubled
in
value
.
With these abbreviations:
†
shs
h
mch
the
system
’
s
energy
changes
when
one
particle
is
added
to
the
system
at
fixed
S
and
V
.
The
chemical
potential
units
–
J
.
The coefficients may
be identified as:
This identity holds for small changes
S
provided
T
and
P
are well defined.

the so

called
thermodynamic identity
The 1
st
Law for quasi

static processes (
N
= const
)
:
The thermodynamic identity holds for the
quasi

static processes
(
T
,
P
,
†
a牥e

deieth牯ght thesystem)
The Equation(s) of State for an Ideal Gas
Ideal gas:
(
fN
degrees of freedom)
The
“
energy
”
equation of state (
U
T
)
:
The
“
pressure
”
equation of state (
P
T
)
:

we have finally derived the equation of state of an ideal gas from first principles!
In other words, we can calculate the thermodynamic information for an isolated
system by counting all the accessible microstates as a function of
N
,
V
, and
U
.
Ideal Gas in a Gravitational Field
Pr
.
3
.
37
.
Consider
a
monatomic
ideal
gas
at
a
height
z
above
sea
level,
so
each
molecule
has
potential
energy
mgz
in
addition
to
its
kinetic
energy
.
Assuming
that
the
atmosphere
is
isothermal
(not
quite
right),
find
ad
牥

de物e
the
ba牯met物c
eqati
.
Ieqiib物m, thechemicatetiasbeteeaytheightsmstbeeqa
tethatthe
U
that appears in the Sackur

Tetrode
equation represents only the kinetic energy
Pr
.
3
.
32
.
A
non

quasistatic
compression
.
A
cylinder
with
air
(
V
=
10

3
m
3
,
T
=
300
K,
P
=
10
5
Pa)
is
compressed
(very
fast,
non

quasistatic)
by
a
piston
(
A
=
0
.
01
m
2
,
F
=
2000
N,
x
=
10

3
m)
.
Calculate
W
,
Q
,
U
,
and
S
.
holds for
all processes
,
energy conservation
quasistatic
,
T
and
P
are well

defined for any intermediate state
quasistatic adiabatic
isentropic
non

quasistatic adiabatic
The non

quasistatic process results
in a higher
T
and a greater entropy
of the final state.
S =
const
along the
isentropic
line
P
V
V
i
V
f
1
2
2
*
Q =
0
for both
Caution
: for non

quasistatic adiabatic processes,
S
might be non

zero!!!
An example of a non

quasistatic adiabatic process
Direct approach:
adiabatic quasistatic
ise瑲ic
adiabatic non

quasistatic
The entropy is created because it is an
irreversible
,
non

quasistatic
compression.
P
V
V
i
V
f
To
calculate
S
,
w
e
can
consider
any
quasistatic
process
that
would
bring
the
gas
into
the
final
state
(
S
is
a
state
function)
.
For
example,
along
the
red
line
that
coincides
with
the
adiabata
and
then
shoots
straight
up
.
Let
’
s
neglect
small
variations
of
T
along
this
path
(
U
<<
U
,
so
it
won
’
t
be
a
big
mistake
to
assume
T
const)
:
U = Q = 1J
For
any
quasi

static
path
from
1
to
2
,
we
must
have
the
same
S
.
Let
’
s
take
another
path
–
along
the
isotherm
and
then
straight
up
:
1
2
P
V
V
i
V
f
1
2
U = Q = 2J
isotherm:
“
s瑲aigh琠up
”
:
Total gain of entropy:
The
inverse
process,
sudden
expansion
of
an
ideal
gas
(
2
–
3
)
also
generates
entropy
(
adiabatic
but
not
quasistatic
)
.
Neither
heat
nor
work
is
transferred
:
W
=
Q
=
0
(we
assume
the
whole
process
occurs
rapidly
enough
so
that
no
heat
flows
in
through
the
walls)
.
The
work
done
on
the
gas
is
negative,
the
gas
does
positive
work
on
the
piston
in
an
amount
equal
to
the
heat
transfer
into
the
system
P
V
V
i
V
f
1
2
3
Because
U
is
unchanged,
T
of
the
ideal
gas
is
unchanged
.
The
final
state
is
identical
with
the
state
that
results
from
a
reversible
isothermal
expansion
with
the
gas
in
thermal
equilibrium
with
a
reservoir
.
The
work
done
on
the
gas
in
the
reversible
expansion
from
volume
V
f
to
V
i
:
Thus, by going 1
2
3 , we will increase the gas entropy by
Systems with a
“
Limited
”
Energy Spectrum
The
definition
of
T
in
statistical
mechanics
is
consistent
with
our
intuitive
idea
of
the
temperature
(the
more
energy
we
deliver
to
a
system,
the
higher
its
temperature)
for
many,
but
not
all
systems
.
“
Unlimited
”
Energy Spectrum
T
> 0
the multiplicity increase
monotonically
with
U
:
U
f N/2
U
S
C
T
U
U
S
U
T
U
C
U
Pr. 1.55
ideal gas in thermal equilibrium
self

gravitating ideal
gas
(not in thermal
equilibrium)
T
> 0
Pr
.
3
.
29
.
Sketch
a
graph
of
the
entropy
of
H
2
0
as
a
function
of
T
at
P
=
const,
assuming
that
C
P
is
almost
const
at
high
T
.
At
T
0
,
the
graph
goes
to
0
with
zero
slope
.
At
high
T
,
the
rate
of
the
S
increase
slows
down
(
C
P
const
)
.
When
solid
melts,
there
is
a
large
S
at
T
=
const,
another
jump
–
at
liquid
–
gas
phase
transformation
.
S
T
ice
water
vapor
“
Limited
”
Energy Spectrum: two

level systems
e
.
g
.
,
a
system
of
non

interacting
spin

1
/
2
particles
in
external
magnetic
field
.
No
“
quadratic
”
degrees
of
freedom
(unlike
in
an
ideal
gas,
where
the
kinetic
energies
of
molecules
are
unlimited),
the
energy
spectrum
of
the
particles
is
confined
within
a
finite
interval
of
E
(just
two
allowed
energy
levels)
.
S
U
E
the multiplicity and entropy
decrease
for some range of
U
in this regime, the system is
described by a
negative
T
S
U
T
U
2
N
B
Systems
with
T
<
0
are
“
hotter
”
than
the
systems
at
any
positive
temperature

when
such
systems
interact,
the
energy
flows
from
a
system
with
T
<
0
to
the
one
with
T
>
0
.
½ Spins in Magnetic Field
The total energy of the system:
N

the number of
“
up
”
spins
N

the number of
“
down
”
spins
Our plan:
to arrive at the equation of state for a two

state paramagnet
U=U
(
N,T,B
)
using the multiplicity as our starting point.

the magnetic moment of an individual spin
E
E
1
=

B
E
2
= +
B
0
an arbitrary choice
of zero energy
The magnetization:
(
N
,
N
)
S
(
N
,
N
)
=
k
B
ln
(
N
,
N
)
†
U
=
U
(
N
,
T
,
B
)
From Multiplicity
–
to
S
(
N
)
and
S
(
U
)
The multiplicity of any macrostate
with a specified
N
:
Max.
S
at
N
=
N
(
N
=
N
/2):
S
=
Nk
B
ln2
ln2
0⸶93
From
S
(
U,N
)
–
to
T
(
U,N
)
Energy
E
E
The same in terms of
N
and
N
:
Boltzmann factor!
The Temperature of a
Two

State Paramagnet
0
E
1
E
2
E
1
E
2
Energy
E
1
E
2
T
Energy
E
1
E
2

N
䈠
N
䈠
U
E
1
E
2
T
=
+
and
T
=

are
(physically)
identical
–
they
correspond
to
the
same
values
of
thermodynamic
parameters
.
Systems
with
neg
.
T
are
“
warmer
”
than
the
systems
with
pos
.
T
:
in
a
thermal
contact,
the
energy
will
flow
from
the
system
with
neg
.
T
to
the
systems
with
pos
.
T
.
The Temperature of a Spin Gas
The
system
of
spins
in
an
external
magnetic
field
.
The
internal
energy
in
this
model
is
entirely
potential,
in
contrast
to
the
ideal
gas
model,
where
the
energy
is
entirely
kinetic
.
B
E
6
E
5
E
4
E
3
E
2
E
1
B
At
fixed
T
,
the
number
of
spins
n
i
of
energy
E
i
decreases
exponentially
as
energy
increases
.
spin 5/2
(six levels)

ln
n
i
E
i

ln
n
i
E
i

ln
n
i
E
i
T
=

ln
n
i
E
i
T
= 0
For
a
two

state
system,
one
can
always
introduce
T

one
can
always
fit
an
exponential
to
two
points
.
For
a
multi

state
system
with
random
population,
the
system
is
out
of
equilibrium,
and
we
cannot
introduce
T
.
Boltzmann distribution
no T
the slope
T
The Energy of a Two

State Paramagnet
U
approaches the lower limit
(

N
B
)
as
T
decreases or, alternatively,
B
increases (the effective
“
gap
”
gets bigger).
U
B/k
B
T
N
B

N
B
U
T

N
B
1
The equation of state of a two

state paramagnet:
(
N
,
N
)
S
(
N
,
N
)
=
k
B
ln
(
N
,
N
)
†
U
=
U
(
N
,
T
,
B
)
S
(
B/T
) for a Two

State Paramagnet
S
U
0
N
䈠

N
䈠
Problem 3.23
Express the entropy of a two

state
paramagnet as a function of
B/T .
Nk
B
ln
2
B/T
0,
S = Nk
B
ln2
B/T
,
S =
0
S
(
B/T
) for a Two

State Paramagnet (cont.)
S/Nk
B
ln2
0.693
k
B
T/
B㴠x

1
high

T
(low

B
) limit
low

T
(high

B
)
limit
ln2
0.693
Low

T limit
Which x can be considered large (small)?
ln2
0.693
The Heat Capacity of a Paramagnet
The
low

T
behavior
:
the
heat
capacity
is
small
because
when
k
B
T
<<
2
B
,
the
thermal
fluctuations
which
flip
spins
are
rare
and
it
is
hard
for
the
system
to
absorb
thermal
energy
(the
degrees
of
freedom
are
frozen
out)
.
This
behavior
is
universal
for
systems
with
energy
quantization
.
The
high

T
behavior
:
N
~
N
and
again,
it
is
hard
for
the
system
to
absorb
thermal
energy
.
This
behavior
is
not
universal
,
it
occurs
if
the
system
’
s
energy
spectrum
occupies
a
finite
interval
of
energies
.
k
B
T
2
B
k
B
T
2
B
compare with
Einstein solid
E
per particle
C
equipartition theorem
(works for
quadratic
degrees
of freedom
only!)
Nk
B
/2
The Magnetization, Curie
’
s Law
B/k
B
T
N
M
1
The high

T
behavior for all
paramagnets (
Curie
’
s Law
)
B/k
B
T
N

N
The magnetization:
Negative
T
in a nuclear spin system
NMR
MRI
Fist observation
–
E. Purcell and R. Pound (1951)
Pacific Northwest National Laboratory
By
doing
some
tricks,
sometimes
it
is
possible
to
create
a
metastable
non

equilibrium
state
with
the
population
of
the
top
(excited)
level
greater
than
that
for
the
bottom
(ground)
level

population
inversion
.
Note
that
one
cannot
produce
a
population
inversion
by
just
increasing
the
system
’
s
temperature
.
The
state
of
population
inversion
in
a
two

level
system
can
be
characterized
with
negative
temperatures

a
s
more
energy
is
added
to
the
system,
ad
S
actually
decrease
.
An animated gif of MRI images of a human head.

Dwayne Reed
Metastable Systems without Temperature (Lasers)
For
a
system
with
more
than
two
energy
levels
,
for
an
arbitrary
population
of
the
levels
we
cannot
introduce
T
at
all

that's
because
you
can't
curve

fit
an
exponential
to
three
arbitrary
values
of
#
,
e
.
g
.
if
occ
.
#
=
f
(
䔩
is
not
monotonic
(
population
inversion
)
.
The
latter
case
–
an
optically
active
medium
in
lasers
.
E
1
E
2
E
3
E
4
Population inversion
between
E
2
and
E
1
Sometimes,
different
temperatures
can
be
introduced
for
different
parts
of
the
spectrum
.
Problem
A two

state paramagnet consists of 1x10
22
spin

1/2 electrons. The component of the
electron’s magnetic moment along
B
is
B
=
9.3x10

24
J/T.
The paramagnet is
placed in an external magnetic field
B =
1T
which points up.
(a)
Using Boltzmann distribution, calculate the temperature at which
N
=
N
/e.
(b)
Calculate the entropy of the paramagnet at this temperature.
(c)
What is the maximum entropy possible for the paramagnet? Explain your reasoning.
(a)
k
B
T
E
1
=

B
B
E
2
= +
B
B
B
spin 1/2
(two levels)
B

B
Problem (cont.)
If your calculator cannot handle cosh’s and sinh’s:
k
B
T/
䈠
匯Nk
B
0.09
(b)
the maximum entropy corresponds to the limit of
T
(
N
=
N
)
:
S/Nk
B
ln2
For example, at
T
=300K:
E
1
E
2
k
B
T
k
B
T/
䈠
匯Nk
B
T
S/Nk
B
ln2
T
0
S/Nk
B
0
ln2
Problem (cont.)
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