# Lecture 8. Thermodynamic Identities (Ch. 3)

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27 Οκτ 2013 (πριν από 4 χρόνια και 6 μήνες)

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Chapter 3: Interactions and Implications.

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

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

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

(

degrees of freedom!), may be inverted to give

compare with

pressure

chemical potential

This holds for
quasi
-
static processes

(
T
,
P
,

a牥e
-
deieth牯ght thesystem⤮

shs

h

mch

the

system

s

energy

changes

when

one

particle

is

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:

shs

h

mch

the

system

s

energy

changes

when

one

particle

is

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
-
deieth牯ght thesystem)

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

to

its

kinetic

energy
.

Assuming

that

the

atmosphere

is

isothermal

(not

quite

right),

find

ad

-
de物e

the

ba牯met物c

eqati
.

Ieqiib物m, thechemicatetiasbeteeaytheightsmstbeeqa

tethatthe
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

isentropic

non
-

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
-

S

might be non
-
zero!!!

An example of a non
-

Direct approach:

ise瑲ic

-
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

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

(

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

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

N

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

-

N

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

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

to

the

system,

ad

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

(

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/

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/

B

T

S/Nk
B

ln2

T

0

S/Nk
B

0

ln2

Problem (cont.)