Zemansky Statistical Thermodynamics

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250

Heat

and

Thermodynamics

(6)

L\\

=

KA/Ax, where A'

is

the

thermal

conductivity

and A

and,

Ax

arc

the

area

and length, respectively, of the wire.

(c)

L12

=

T/R', where

A is the

electric resistance

of

the

wire.

9-25 Show

that,

in the

case of

irreversible

coupled flows

of heat

and

electricity,

(b)

(0

d

\

dr/iT

=

2/

and

d AT

rf)

=2,,.

Show that, with

AT fixed,

the equilibrium

state obtained

when

1=0 involves

a

minimum rate of

entropy

production.

(d)

Show that,

with

AS fixed, the equilibrium

state obtained when

I

s

=

involves

a

minimum rate of entropy production.

9-26

Three

identical finite

bodies of constant

heat capacity

are

at

temperatures

300, 300,

and 100°K.

If

no work

or

heat

is

supplied from

the

outside,

what is

the highest temperature

to which

any one of

the

bodies

can

be

raised by

the

operation of

heat engines

or

refrigerators?

10.

STATISTICAL

MECHANICS

10-1

Fundamental

Principles

In the

treatment of kinetic theory

given in

Chap.

6,

the molecules

of an

ideal

gas

could not be regarded as

completely

independent of one

another,

for

then they

could

not arrive at an

equilibrium

distribution of

velocities.

It was

therefore

assumed that

interaction did

take

place, but

only during

collisions

with other molecules

and

with

the

walls.

To

describe

this

limited

form of interaction we refer to

the molecules

as weakly

interacting

or

quasi-independent. The

treatment

of

strongly interacting

particles

is

beyond

the

scope

of

the

present

discussion.

The

molecules

of

an

ideal

gas

have another characteristic

besides

their

quasi-independence.

They

are

indistinguishable,

because they are

not localized

in

space. It

was

emphasized

in

Chap.

6

that the

molecules

have neither a

preferred location nor

a

preferred

velocity.

The particles occupying

regular

lattice sites in a crystal

are

distinguishable,

however,

because they are

con-

strained to

oscillate

about fixed

positions;

therefore one

particle

can be dis-

tinguished

from

its

neighbors

by its

location. The

statistical

treatment

of

an

ideal crystal as a

number

of

distinguishable,

quasi-independent

particles

will

be given

in

the

next

chapter.

In

this

chapter

we

shall

confine

our

atten-

tion to

the

indistinguishable,

quasi-independent particles

of

an ideal gas.

Suppose

that a monatomic

ideal

gas

consists of N particles,

where

N,

as

usual,

is

an

enormous

number

—say, about 10

20

. Let

the gas

be

contained in

a

cubical enclosure whose

edge has a length

L, and

let

the

energy

e

of any

particle, as

a

first step,

be entirely kinetic

energy of translation.

In the x

direction,

(mxY

m

J|

2m

=

i

2

*mx

=

where

p

x

is the x component of the

momentum.

If

the

particle is

assumed

to



252

Ileal

and

Thermodynamics

10-1

Statistical Mechanics 253

move

freely

back and

forth

between

two

planes a distance

/,

apart,

the

sim-

plest

form of

quantum mechanics

provides that, in

a

complete cycle

(from

one

wall to the

other

and

back

again, or a total distance of

2L), the

constant

momentum

p

x

multiplied by

the total path 2L is an integer

n

x

times

Planck's

constant h.

Thus,

p

x

2L

=

n

x

h.

Substituting

this

result into the previous

equation,

we

get

,

A

2

n

x

—

j

v8me

x

.

The allowed

values

of

kinetic energy t

x

are discrete, corresponding

to

integer

values of

n

x

\

but when n

x

changes

by

unity, the corresponding change

in

e

x

is

very

small,

because

n

x

itself is exceedingly large.

To

see

that a

typical value

of n

x

is

very large,

consider a cubical box containing

gaseous helium at

300°K,

whose

edge is,

say,

10

cm.

It

was shown

in Chap.

6

that the

average

energy

associated

with each

translational

degree

of

freedom

is

\kT.

Then,

e*

=

UT<=

i

X

1.4

X

10- |3I

x

300 deg

-

deg

and

n,

=

=

2.1

X

10-'

erg,

10 cm

6.6

X

10-

erg

s

10

X

10.5

X

10~

19

6.6

X

10-*'

—

s/8

X

6.6

X

10-

g

X

2.1

X

10

H

erg

«

10

9

.

Therefore,

the

change

of

energy

when

nx

changes

by

unity

is

so

small

that,

for

most

practical purposes,

the energy may

be

assumed

to vary

continuously.

This

will

be of advantage later

on, when it

will be

found useful to replace a

sum

by

an

integral.

Taking

into account the three components

of

momentum,

we

get for

the

total

kinetic

energy

of a

particle

=

Pi

+

Pi

+

Pi

_

(10-1)

2m

a«tf

w

+ S+<»-

The

specification

of

an integer

for each n

x

,

n„,

and n

z

is

a specification

of a

quantum

state

of a particle. All states characterized

by

values

of

the ri's such

that n

x

+

»„

+

nz

=

const,

will

have

the

same

energy.

To

use

an example

given

by

Guggenheim, the

states

corresponding to the

values of

n

x ,

n

y

,

and

n

z

in

Table

10-1

all have

the energy

e

=

66A

2

,

8mL

2

. There

are

twelve

quan-

Table

10-1

ni +

n;

+

n;

=

66

1

2

3

4

5

6

7

8 9

10

11 12

n

x

8

1 1

7

1 4

7 4 1

5 5

4

y

1

8

1

4

7

1 1

7

4

5

4

5

>h

1 1

8

1

4 7

4

1 7 4

5 5

turn states

associated with the same

energy

level,

and we

therefore refer to

this

energy level

as

having

a

degeneracy

of

12.

In

any

actual

case, n\

+

n

y

+

n\

is

an

enormous

number,

so

that

the degeneracy

of an actual energy

level

is

extremely

large.

However

close

they

may

be,

there

is still

only

a

discrete

number

of

energy

levels

for

the

molecules

of an ideal

gas.

It

is the fundamental

problem

of

statistical

mechanics

to determine, at

equilibrium,

the

populations

of

these

energy levels—

that

is,

the

number

of

particles

Ni

having

energy

ei,

the

number

A

;

2

having energy

e

s,

and

so on. It is

a

simple

matter

to show that

(see Prob.

10-2) the

number

of quantum states

g

;

corresponding

to

an energy

level

i

(the degeneracy of

the

level)

is very

much

larger than

the

number

of

particles

occupying

that level. Thus,

Mi

»

N

u

(10-2)

It

is very

unlikely,

therefore,

that

more

than

one particle

will

occupy the

same

quantum

state at

any

one

time.

At

any one moment,

some

particles

are

moving

rapidly

and some slowly,

so that

the

particles

arc distributed

among

a

large

number ofdifferent quan-

tum states.

As

time

goes

on, the

particles collide

with

one another and

with

the

walls,

or

emit

and

absorb

photons,

so that

each

particle undergoes many

changes

from

one

quantum state

to

another.

The fundamental

assumption of

statistical

mechanics

is that

all quantum

stales have equal likelihood

of

being

occupied. The probability

that

a

particle

may

find

itself in

a

given

quantum

state

is

the same

for all

states.

t

E.

A. Guggenheim,

Boltzmann's

Distribution

Law,

Interscience

Publishers,

Inc.,

New York,

1955.



254

I It-at

and

Thermodynamics

1

2

3

4

5

6

7

8

9 10 i 1

12 13 14 15 16

A

B

c

'

A

C

B

B

A

C

B

C

A

C

A

B

C

B

A

Fig.

10-1

There

are

six

ways in

which

three,

distinguishable

particles

A,

B, and C can

occupy

three given

quantum

slates.

Now

consider the

N

f

particles

in any of

the

g,

quantum states

associated

with the energy

«,-.

Any

one particle

would

have

$

choices

in

occupying

g<

different

quantum

states.

A

second particle

would have

the

same

gt

choices,

and

so

on. The

total

number of

ways

in which A',-

distinguishable

particles

could

be

distributed

among

g<

quantum

states would

therefore bcg

t

•'.

But the

quan-

tity

g*

1

'

is

much

too

large,

since

it

holds

for

distinguishable

particles

such

as

A, B,

and C in Fig.

10-1. This

figure

shows

six

different

ways in

which three

distinguishable

particles

can occupy

quantum

states

2, 7,

and

13.

If the

particles

had no

identity,

there would

be

only one

way to occupy

these

particular quantum states.

That

is,

one must

divide

by

6,

which

is

3

The

number

of

permutations of

A ,

distinguishable

objects is

A'; If

the

quantity

gfi

is

divided

by

this factor, the

resulting

expression will

then hold

for

indis-

tinguishable

particles. Therefore,

No.

of

ways

that

Ni

indistinguishable

}

particles

can

be

distributed

among

g.

quantum

states

Nil

(10-3)

\

It

should be pointed out

that

the

A

indistinguishable,

quasi-independent

particles were

assumed

to be

contained

within

a

cubical

box

only for

the

sake of

simplicity.

A

rectangular box

with

three

different

dimensions

could

easily

have been

chosen,

in

which case

Eq. (10-3)

would

be

unchanged.

10-2

Equilibrium Distribution

We

have seen that,

in the

case of

an ideal gas, there

are many

quantum

states

corresponding to the same energy

level

and

that

the

degeneracy

of

each

level

is

much larger than the

number of particles

which

would be

found

in

10-2

Statistical

Mechanics

255

any

one level

at any one

time. The

specification,

at any one

moment,

that

there are

A'i

particles

in energy

level

ei

with degeneracy

gi

Ni

particles

in

energy

level

et

with

degeneracy

£2

Ni

particles

in

energy

level

e,

with

degeneracy

gi

in

a

container of

volume V

when

the

gas

has

a

total

number

of

particles

N

and

an energy

V

is

a

description of

a macrostate

of the gas. The number of

ways Q

in

which

this macrostate

may

be

achieved is

given

by

a

product

of

terms of

the

type

of Eq.

(10-3),

or

=

Nx'.

N

t

l

(10-4)

The quantity

V.

is

called the thermodynamic

probability of

the particular macro-

state.

Other

names for this

quantity are the number

of

microstates

and the num-

ber

of

complexions. Whatever

its

name, the larger <i is, the greater

the proba-

bility of finding the

system

of N

particles in this state.

It is

assumed that, if

V,

N,

and

U

are

kept constant, the equilibrium state

of

the

gas

will

correspond to

that macrostate in

which Q

is

a maximum.

To

find

the

equilibrium

populations

of

the

energy levels, therefore,

we look

for the

values of

the

individual

A's that

render

52

—or

more simply, In fi—

a maximum.

Since

In

contains

factorials

of

large

numbers,

it is

convenient

to

use

Stirling's

approximation,

which

may

be derived

in the

following

way: the

natural

logarithm

of

factorial

x

is

In (x )

=

In

2

+

In 3

+

In x.

If

we

draw

a series

of

steps

on a

diagram,

as shown

in Fig.

10-2, where the

integers arc plotted

along

the x

axis and In

x

along thc^

axis, the

area

under

each step is

exactly

equal

to

the natural logarithm,

since the

width of

each

step equals unity.

The

area

under

the

steps from

x

=

1 to x

=

x

is

therefore

In

(xl).

When

x

is

large,

we may replace

the

steps by

a

smooth

curve,

shown



256

Heat

and

Thermodynamics

In

x

In

7

In 6

In 5

In 4

In

3

In

2

In

1

8

X

Fig.

10-2

The

area under

the dashed

curve

approximates

the area

under

the

steps

{In

2

+

In

3

+

In

4

+

•

•

+

In x)

when

x

is

large.

as

a

dashed

curve

in Fig.

10-2;

therefore,

approximately,

when

x

is

large,

In (xl)

~

/ In

x dx.

Integrating by

parts,

we get

In (.v )

~

x

In

x

—

x

+

1.

Then,

if we neglect 1

compared with

x,

In

(x\)

«

x

In x

—

x.

(10-5)

This

formula

is

Stirling's approximation.

Using

Stirling's approximation in

Eq.

(10-4),

we

get

In 9.

=

A'i

In

g

x

—

Ni

In h\

4-

A'i

+

Nt

In

g

2

—

A

2

In N

2

=

V

Ni

In

gi

-

Y A ,- In

A',-

-•-

V

.V,-,

or

In Q

=

y

A',- In

#

+

A',

Zw

A;

+

#8

+

(10-6)

10-2

Statistical Mechanics

257

where

we

have

used the fact

that

2A'f

=

N.

Our problem now

is to render

In V. a

maximum

subject

to

the

conditions

that

2A

r

,-

=

A*

=

const.,

SA',«,-

=

II

=

const.

(10-7)

(10-8)

Before we proceed

to

solve this

problem

by

the

method

of

Lagrange

multi-

pliers, it is important

to

bear

in

mind that the

e's

and

g's are

constants.

The

only variables are the populations

of the

energy levels,

and their

sum A'

is

constant.

Since dN

=

0,

the

differential

of In Q

is

d In

Q

L>

=

y

d

Niln

iL\

=

N

t

I

B|î.

(10-9)

Setting

the

differential

of

In U equal to zero

and taking the

differential of

Eqs.

(10-7) and

(10-8),

we

get

gi

g-i

gi

In

|f

dNi

+

In

U-

dN

s

+••••+

hi

#£

Mi

+

=

0,

A

i

JV

2

A,-

dNx

+

dNi

+

•

+

dNi

+

•

=

0,

and

«i

dNi

+

e

2

dN

t

+••+

u

dNi

+

=

0.

Multiplying

the

second

equation

by In A and

the

third by

—

/?,

where

In A

and

—0

arc

Lagrange

multipliers

(sec Art.

6-3), we get

In

&-

dNi

+

In

|r

«f#s

+

Ai

A'2

In

A

dNi

+

In

A

dN2

+

and —

/3<i dN%

—

fie* «W»

—

+

In

§

dN

t +

A;

+

In

A

dN +

-

0e;

dN

-

=

o,

=

o,

=

0.

If

we add these

equations,

the coefficient

of

each

dN may be

set equal

to zero.

Taking

the

ith

term,

In

4

?-

+

In

A

-

0a

=

0.

Ni

or Inî-

ln

A

=

-0

%

N

t

= Agie-e'<.

(10-10)



258

Heat

and Thermodynamics

The

population

of

any

energy

level

at equilibrium

is therefore

seen to

be

proportional to

the

degeneracy

of the

level and to vary

exponentially

with

the energy

of

the level.

The

next step

is

to determine the physical

significance of

the Lagrange

multipliers A

and

/?.

1

0-3

Significance of

A

and

j3

The

population A , of the z'th

energy

level

is given

by

Ni

=

Agie-f':

Summing over all the energy

levels,

we

get

and

N

A

=

H*

r»

(10-11)

The sum in

the denominator

plays a fundamental

role

in statistical mechanics.

It

was

first

introduced by

Boltzmann,

who called

it the Zuslandsumme,

or

sum

over

states. We retain

the

first

letter

of

Zuslandsumme

as

a

mathematical

symbol,

but the

accepted

English

expression

for

this

sum

is

the

partition

junction.

Thus.

?»^nr*i

(10-12)

and

-f-

(10-13)

Substituting

this result

into

Eq.

(10-10),

we

get

Ni = A'

,gm

(10-14)

It will

be

shown

later

that

Z

is

proportional

to the

volume of the

container.

Since

the properties

of a gas

depend

on temperature

as

well as

on

volume,

one

would

expect

a relation

between

/3

and the temperature.

To introduce

the

concept

of

temperature

into

statistical

mechanics,

we

must

go back to

the

fundamental

idea

of

thermal equilibrium

between

two

systems,

just

like

the

procedure

in

Chap.

8 for

relating

the

quantity X

to

temperature.

Conse-

quently,

let

us consider

an

isolated composite

system

consisting

of

two samples

of

ideal

gas

separated

by

a

diathermic wall,

as shown in

Fig.

10-3.

For

the

10-3 Statistical

Mechanics 259

1

N

t

,N

s

Mt

SAT,

=

N

>

A

A

=

11

«J

*}i

.

. .

A

A A

N„

»„...

Nj,

. . .

A A

Wj

=

iV

//-/V/y/

W///Y/////y///z/^^^^^

^

Fig. 1 0-3

An

isolated

composite

system

of

two

samples of

ideal

gas

separated

by

a

diathermic

wall.

The

total

energy

is constant.

second sample of

gas, the

symbols

expressing

energy levels,

populations,

etc.,

are

distinguished

with

a circumflex.

The

thermodynamic

probability

of

the

composite

system

Q is

the product

of

the separate

thermodynamic

proba-

bilities,

so

that

the

logarithm is

In ii

=

Y

N

In

%

+

N

+ y^

y

In

Jf

+

N.

U Ni

L,

fy

Each sample

has

a

constant number

of molecules,

so

that

and

2A

7

,- =

A'

=

const.

2A'j

=

Jv

=

const.,

but

the

energy of

each

sample

is not

constant.

Only

the

total energy

of

the

composite

system

is constant;

thus,

=

U

=

const.

JV**

+

^%

=

v

Vo

find

equilibrium

conditions,

we proceed as

before

and

get

In

A

y

dNi

=

In

A

V dNj

=

-££«4ft-lT%d$-

0.

Adding

and

setting

each

coefficient

of dN

and

dN

equal to zero, we

get two



260

Hrat

and

Thermodynamics

sets

of

equations,

and

Nj

=

Ag,-e-^,

where

all

quantities

are

different, except

0.

When two systems

separated

by

a

diathermic

wall

come

to

equilibrium, the

temperatures

are

the same

and

the

0's

are

the

same.

The

conclusion

that

is

connected

with

the

temperature

is

inescapable.

It

was

shown in

Chap.

9

that the

entropy

of an isolated

system increases

when the

system undergoes

a

spontaneous, irreversible

process. At

the con-

clusion

of

such

a

process,

when

equilibrium

is

reached,

the

entropy has the

maximum value consistent

with

its

energy and volume.

The thermodynamic

probability also increases and

approaches

a

maximum

as equilibrium is

approached.

We therefore look for some correlation between

.9

and £2. Con-

sider two

similar

systems

A and B in

thermal contact

—one with

entropy

Sa

and

thermodynamic

probability

il

A

,

the other

with

values

Sb

and

Ob-

Since entropy

is an

extensive variable, the total entropy

of the

composite

system is

S

=

Sa

+

Ss,

The

thermodynamic probability,

however,

is the product, or

£2

=

Q

A

Q

B

.

If we

let

then.

s

=

m),

f(QjiB

)

=/(a

A

)

+f(ih<).

The

only function that satisfies this

relation is the

logarithm. Introducing

an

arbitrary

constant k', we may

write

S

=

k' In S2

(10-15)

for

the

relation

between

entropy and

thermodynamic

probability.

The

first law

of thermodynamics applied to any

infinitesimal process

of

any

hydrostatic

system is

dQ

=

dl

+

P

dV.

If

the

process takes

place

between two neighboring

equilibrium states,

it may

be

performed

reversibly,

in

which case dQ

=

T

dS,

and

dU

=

T dS

-

P

dV.

10-4

Statistical

Mechanics

261

If we now

specify that

the reversible

process take place at constant I

,

we

have

the

important

link

between

thermodynamics and

statistical

mechanics:

1

= (?£

T

\dU

(10-16)

Since

both

S

and

U

may

be

calculated

by

statistical

mechanics,

the

derivative

(dS/(ll/)v gives

the reciprocal

of

the

Kelvin

temperature. This

is

the way

in which

the

macroscopic

concept

of

temperature is injected

into

statistical mechanics.

In

employing the

Lagrange

method to find the equilibrium values

of

the

energy

level

populations

[Eqs.

(10-9) and

(10-10)],

wc found that

and

Therefore,

'

In il

=

y

In

-0-

dN,:

In

|i

=

/3<e,-

-

In

A.

'

In

Q

=

7

/3e,:

dNi

-

In A^ dN

t

where

U

is the total energy

of the

system. Therefore,

d

In

2

1

d

0-

dU

=

k>du

k

'

lnn

=

k' \dUj,

Since

(dS/dlf)v

=

V^'>

we

S

ct

tnc beautiful

result

fi

~

Yt

(10-17)

W

;

hen

the

actual

values

of

the

e's

appropriate to an ideal

gas

are

introduced,

it will

be

seen

that k'

is

none other

than

Boltzmann's

constant

k.

10-4

Partition Function

Wc have

seen that

the population

jV,- of

the ;'th energy level is

Ni

=

Agie-H'i.



262

Heat

and

Thermodynamics

Substituting

\/k'T

for

/3

and

N/Z for

A, we

get

N

/V- =

—

(T

p->ilVT

vhere

Z

=

2^-'<'*T

(10-18)

(10-19)

The

partition

function Z contains

the heart of the statistical

information

about

the

particles

of

the

system,

so

that it is

worth

while to express other

properties

of the

system,

such as U, S,

and P, in terms ofZ. If

wc

differentiate

Z

with

respect to T,

holding V

constant,

wc get

(&-Z-

fri™ *-*

=

_- . V

f-a-p-'tili'T

=

It

z,

€(

'

v

''

ze/

/V/fcT

2

It follows that

V-«T>( jA.

(10-20)

and

E7

may

be

calculated

once

In

Z

is

known as a function

of

T and V. Also,

,9

=

k'

In

0,

where, according

to

Eq.

(10-6),

Hence,

.9

=

-£'

Y

-V,' In

—

+

k'N.

U

gi

Substituting

for

/Vj/gs

the value given in

Eq.

(10-18),

we

get

10-5

Statistical

Mechanics 263

and,

finally,

7 rr

a

=

Nk

In

N

+

r

-

Nk',

(10-21)

which

provides

us with

a

method of

calculating

.S

once In

Z

is known.

One more

equation

that is of

value is

the relation between

pressure and

the partition function.

Since

TdS

=

dU

+

PdV,

~ Apr-H)

From

the relation

between

S

and the partition function,

given in

Eq.

(10-21),

we

get

U

-

TS

=

-Nk'T In

4

-

Nk'T;

TV

therefore,

P

=

Nk'T

d

In z\

(10-22)

so

that

again

the

pressure may

be

calculated

once In

Z is known

as

a

function

of T and I '.

This

is the advantage

of statistical

mechanics. It

provides

us

with

a

simple

set of

rules

for

obtaining

the

properties

of a

system:

10-5

Use

quantum

mechanics

to

find the

e values of the quantum states.

Find

the partition function Z

in

terms of T and V.

Calculate

the

energy by differentiating In Z with respect to

7 .

Calculate

the

pressure

by

differentiating

In

Z

with

respect

to

V.

Calculate

the

entropy

from Z and

U.

Partition

Function of an Ideal Monatomic

Gas

To

apply

the rules laid down in the preceding article to

an

ideal gas, wc

must first calculate

the appropriate

partition function.

This

was defined

to be

levels



264

Heal

and

Thermodynamics

where

the

summation was

over all the energy levels.

Exactly

the

same

result

is

obtained

if we sum the

expression

over all

quantum states.

We at

first take

into

account

only

the

kinetic

energy

of

translation

of

the particles

confined to

a rectangular box whose

x,

y,

and

z

sides

are,

respectively,

a,

l>, and c. The energy

of

any

quantum

state; is

given

by Eq.

(10-1)

as

*'

8m

\a*

+

b*-

+

?)'

where n

t

,

n,„

and

n

z

arc quantum

numbers specifying

the various

quantum

states.

The partition function is therefore

a

threefold sum: thus,

Z

=

V V V

gHh*t«mk

f

THnJt<l.*+n,

,

l&*+n

t

*{c

i

)

71x=

1

«y=

1 tli

=

1

or

2

=

y

('

C

|

.'8»'*'n( ^/ )

y

e

-(h

,

ISmk'T)(n,

,

H

t

)

V

e

-(li

,

;S»a-'r)(ri,

I

..'c

>_

Since

the

values of

n

r

,

n,„

and n

z

that give

rise to appreciable

values

of

the

energy

are

very

large and

since

a change of

n

x

or

n

y

or

n.

by unity produces

a change

of

energy that is exceedingly small, no

error is introduced by

replacing

each sum with an integral and

by

writing

z

=

r

f

x

r

tfis.*'rit,,w

d

„

i

r

/*

r

a>n«w)(«,w

rf

„

i

[ J°°

g-oto**?nwt*>

&t\

Each

integral

is

of the

type

listed in

Table

6-2:

f

g-«*

*

«

i„

Therefore,

Z

=

and

since

abc

=

V,

2\a

<?

ISirrnk'T

2

\

A

1

4

l&irm/c'T

2\

A

2

8jrm*'r

.2 \

A

Z=

F

2wmk'

T\i

h

2

(10-23)

10-5

Statistical

Mechanics 265

and In Z

=

In

f

In r

fin

(

2Trmk'\

(10-24)

Pressure

of

an

ideal monalornic

gas

P

=

Nk'T

dlnZ\

bV

)

T

-««(;)

=

_

i- T

(10-25)

Comparing

this

result with

the

expression

for

the

pressure

P

=

NkT/V

obtained

with

the kinetic

theory

of

gases, given in

Eq.

(6-10),

where

k is

Boltzmann's

constant, we

sec that the arbitrary constant

k' introduced

in

the

equation

S

=

k' In

Q,

is

none other than the

Boltzmann

constant,

or

k'

=

k

=

R

(10-26)

2

Energy

of

an ideal monatomic

gas

U

=

NkT

2

(

d

In

Z\

\

w

h

=

INkT.

(10-27)

This

is

exactly

the same

result

which was

obtained by the kinetic

theory

of

gases for

a

monatomic ideal gas and

shows

that,

when particles each

having

three translational

degrees of

freedom

come

to

statistical

equilibrium, the

energy

per

particle equals |-

k

T.

3 Entropy

of

an

ideal

monatomic

gas

^

=

Nk In

£

+

¥

+

m

In

Z

+ |

la

T

In

(2irmk

+

$M

+ Nk

=

Nk

||

In T

*5

A

. /2armk\i



266

Heat

and Thermodynamics

If

we

take

1 mole of gas, N

=

A

T

A

and A'\k

=

R.

Therefore,

s

=

c

v

In

T + R

In v

-

R

In

^

2lrm

'

:

/

h

^

i

+

s

R

(10-28)

This

expression

is to

he

compared with

Eq.

(9-5),

namely,

s

=

cv

In

T

-\-

R\n

v

-\-

s

u

,

and

we

sec

that not

only

were we able

to

arrive at

this equation

by

the

methods of statistical mechanics

but also we

were

able

to calculate

the

con-

stant

s<,.

Equation

(10-28),

which

was

first

obtained

by

Sackur

and

Tetrode,

usually

bears

their

names.

10-6

Equipartition

of

Energy

Both

kinetic

theory and

statistical

mechanics,

when applied

to the mole-

cules of an ideal gas

(each

having three

translational

degrees

of

freedom),

yield

the

result

that

at

equilibrium the

energy

per

particle

associated

with

each degree

of translational

freedom

is

%kT.

The methods

of

kinetic

theory

could not

be applied

to

rotational

and

vibrational

degrees

of

freedom,

but

the simple

statistical

method

just

developed is

capable

of

dealing

with all

types of

molecular

energy,

not

just translational

kinetic

energy.

The

property

of

the

partition

function

which makes it

so useful

is

that,

whenever

the energy

of

a molecule is

expressed as a

sum of independent

terms

each referring

to

a

different

degree

of

freedom,

then.

6

=

t + e + t

+•;

Z

= \

e

~'

lkT

=

V

£-(«'+« +« '+•

••)/*r

i-i

Li

=

y

e

-'''

kT

y

e-< i

kT

y

r****

=

Z'Z Z'

(10-29;

If the

various

types of energy are calculated

with

classical

physics,

it is

a

simple

matter to

derive

the

classical principle

of

the

equipartition

of

energy.

We

take

Eq.

(10-20),

namely,

U

=

NkT*

d

In Z

dT

10-6

Statistical

Mechanics 267

and

rewrite

it thus:

l\

-

U

-

din Z

W

N

d(\/kT)

<«>

=

d

In Z

dp

(10-30)

Suppose e

to

consist

of terms

representing

translational

kinetic

energy

of

the

type

fymw*,

those representing

rotational kinetic energy of

the

type

i/co

2

,

those

representing vibrational

energy

gJKj

8

+ $&£*,

etc. All these

forms

of

energy

are

expressed

as squared terms of the

type

bip\.

Let there

be

/

such

terms, or

e

=

hp\

+ b

2p\

+ +

b,p).

Then,

since

the

partition

function

is

the product

of

the

separate partition

functions,

Z

=

J

*-**«»>

dp

x

J

e-1*** dpi

J

rw dp,.

yt

=

fflpi

and d

yi

=

& dpf,

J*

f+rt

dp,

=

p

g-»ji.

f

ie

d

yi

=

(S-i

J

r*an

dyi

where

Ki

does

not

contain

p.

The partition function now becomes

Let

then

Z

=

0-iKi

/3-»/T,

• •

0-lK

f

,

=

p-z^JCff,

• •

Kf,

where none

of

the

K's

contains

0.

Since

(e)

=

—

3(ln Z)/dp,

to -

-

35

(-

2

ln

+

ln

Kl

+

ln

K

*

+

'

'

'

2/3'

and

since

/?

=

1/kT,

<<>

=

{hT.

In K

(10-31)



268 Heat and

Thermodynamics

It

has therefore

been

proved

thai,

when

a

large

number

of

nondistinguishable,

quasi-

independent

particles whose

energy

is

expressed

as the

sum

of

f

squared

terms

come

to

equilibrium,

the

average energy per particle

is

f

times

^k

T.

This

is

the

famous

principle

of

the equipartition

of energy

that was

men-

tioned

but not proved in Art. 6-5.

It was

stated

that the

principle

broke

down

badly

when applied to polyatomic

molecules

which

have

many

vibrational

degrees

of

freedom.

It

is

a

simple

matter

to introduce

quantum

mechanical

expressions

for

the

energy

of rotation

and

vibration

in the

partition

function

and to calculate

the resulting

thermodynamic

properties.

For discussion

of

this

the

student is referred

to any of

the

books

on

statistical

mechanics

listed

in

the appendix.

10-7

Statistical

Interpretation

of

Work and

Heat

We have been considering

the

statistical

equilibrium

of a large

number

N

of nondistinguishable,

quasi-independent

particles

in

a

cubical

container

of

volume

V. The

energy

levels

e,- of

individual

particles

undergoing

trans-

lational

motion

only were given

by

A

2

8mL

I

(»i

+K+

nl).

Since

L

3

=

V,

then

l.

1

=

V',

and letting

B

t

be

the

sum

of the

squares

of the

quantum

numbers

appropriate

to

the

z'th energy

level,

we get

h

2

U-£-

BiV-i.

8m

Given

the

set

of

quantum

numbers

that

determines

B

f,

we

may

say that

the

corresponding

energy

e

t

depends

on volume

only. Taking

the logarithm

of

e,-,

It

2

In

a

=

In

—

+

In

B

t

-

£

In

V.

8m

3

The

effect

of a small

change

of

V

on e is given

by taking

the

differential

of

this

equation;

whence

dti

=

2dV

ti

3

V'

Therefore,

and

Kde^-^dV,

lN

idu=-\pV.

(10-32)

10-8

Statistical

Mechanics 269

Now, it

has

been

shown both

by

kinetic

theory and

by statistical mechanics

that

the pressure of

an

ideal

gas

is given by

P

=

NkT

Since

the

energy

per particle

is

translational kinetic

energy only, with

three

degrees

of

freedom,

U

=

$MsT.

It

follows that

p m

2

U

Substituting this result into Eq.

(10-32),

we get

S»<fe

=

-PdV.

(10-33)

(10-34)

A

change

of volume,

therefore,

causes

changes in

the

energy values

of

the

energy

levels,

without producing changes in the

populations of

the

levels.

When

the

JV;

change

and

the

u

remain

constant,

we

have from page

261,

d

In it

=

S«

;

dNi.

Since

kd In

=

dS,

k 2e,- dNi

=

dS,

and

setting

1:0

equal

to

l/T, we

get

finally,

2

£i

dNi

=

T

dS.

We

see

that a

reversible

heat transfer

produces changes in

the

populations

of the

energy levels without

changes in

the energy values

of

the

levels them-

selves. Thus, the equation

dU

=

2e,- dN

{

-\-

2JV<

de,- expresses

the first

and

sec-

ond

laws

of

thermodynamics,

with

2e,

dNi

= T

dS and

2A';

de{

=

—PdV.

10-8

Disorder,

Entropy, and Information

Whenever work

or kinetic energy

is

dissipated

within

a

system because

of

friction,

viscosity,

inelasticity,

electric

resistance,

or

magnetic

hysteresis,

the

disorderly

motions

of

molecules

arc

increased.

Whenever

different substances

arc

mixed

or

dissolved

or

diffused with one

another,

the

spatial

positions

of

the

molecules

constitute

a

more disorderly arrangement.

Rocks

crumble,

iron

rusts,

some

metals

corrode,

wood rots,

leather

disintegrates,

paint

peels, and

people

age. All

these

processes

involve

the

transition

from

some sort

of



270

Heat

and

Thermodynamics

orderliness to

a

greater disorder. This

transition is

expressed

in the

language

of

classical

thermodynamics

by

the statement

that

the entropy

of

the

universe

increases. Molecular

disorder

and

entropy

go

together, and

if

we

measure disorder by the number of

ways a particular

macrostate

may

be

achieved, the

thermodynamic

probability S2

is a measure

of disorder.

Then

the

equation

$

=

k

In

U is the simple relation

between

entropy and

disorder.

The number

of

ways

in

which

a

particular

macrostate

may

be

achieved

can be

given another

interpretation.

Suppose

that

you

are called upon

to

guess

a

person's

first

name.

The number

of

choices of

names

of

men and

women is staggeringly large.

With

no hint

or

clue,

the number

of

ways

in

which

one can arrive at

a

name

is very

large,

and

the

information

at one's

disposal

is

small.

Suppose,

now,

that

we are

told

the

person

is

a

man. Imme-

diately the number

of choices of

names

is reduced,

whereas

the information is

increased. Information

is increased

further if

we

arc

told that

the man's name

starts

with

H,

for then

the number

of choices

(or ways

of

picking

a

man's

name) is

reduced

very greatly.

It is

clear

that

the

fewer

the number of

ways

a particular situation or a

particular state

of

a

system

may

be

achieved,

the

greater is the

information.

A

convenient

measure

of

the

information

conveyed

when

the

number

of

choices is reduced from

Q

n

to il y i s given

by

r ,

,

ô

/

=

tln

n7

The

bigger

the

reduction, the bigger

the information.

Since

k

In

9. is the

entropy

S,

then

I

-

&

-

S

u

or

Si

=

S -I,

which

can be

interpreted

to mean that the

entropy

of a

system is reduced

by

the

amount

of

information

about the

state

of a

system.

In

the words

of

Biillouin,

Entropy

measures

the

lack

of

information

about

the

exact

state

of

a

system.

As

an

example of

the

connection

between

entropy and

information,

con-

sider

the isothermal

compression

of

an ideal

gas

(N molecules)

from a volume

''ci

to

a

volume

Vj,

We

know

that

the

reduction

of entropy is

equal to

&

-

Si

m

Nk In

&

But,

when

we decrease

the

volume

of the

gas,

we

decrease the number

of

10-8

Statistical

Mechanics 271

ways

of

achieving this state,

because

there arc

fewer microstatcs

with

position

coordinates

in the

smaller volume.

Before the

compression, each molecule

is

known to

be in

the volume

V

u

. The number

of locations each molecule

could occupy is

Vo/AV,

where AV is some

arbitrary small

volume. After

the

compression,

each

molecule is to be

found

in volume

V\,

with a

smaller

number of possible

locations

Fj/AF.

It follows that

,

0,

V

/AV

,

V

I

=

kln-

r

kln

w

^

=

kln

Yi

,

and

for

the

entire gas of N

molecules,

I

=

Nk

In

\y

,

v

\

in

agreement

with the

result of classical

thermodynamics.

The increase of

information

as

a

result of the compression is

seen

to be identical with

the

corresponding

entropy

reduction.

The connection

between

entropy

and information can be applied

to

the

problem

of

MaxwelPs

demon.

Maxwell

imagined a

small creature

stationed

near

a

trap door separating two

compartments of

a vessel containing

a

gas.

Suppose

that the demon

opened the trap

door only

when

fast

molecules

approached,

thereby

allowing the fast

molecules

to

collect

in

one

compart-

ment and slow ones in the

other. This

would obviously result

in

a

transition

from disorder

to

order

—

thus

violating the second

law. According

to Biillouin,

the

demon could not

tell the difference

between

one kind

of

molecule and

another because he and the

molecules are

in an enclosure at a uniform

temperature

and

all

are bathed

in isotropic blackbody radiation. The demon

could not sec

the

individual

atoms. However,

suppose

that we allow

the

demon, according to the analysis of Rodd, to use

a

flashlight

whose radiation

is

not in

equilibrium with the

enclosure.

Then

the

demon can

get

information

about the

molecules

and

thereby

decrease the entropy of the system.

But

other

phenomena

come

into

the

discussion:

(1)

the

filament

of the lamp in

the flashlight undergoes an increase

of entropy;

(2)

a photon scattered by

a

molecule

is

absorbed

by the demon and serves to increase his entropy;

(3)

the action

of the

demon in opening the

trap

door reduces

the

number of

microstatcs available

to

the molecules. (The

entropy

change of the battery

ofthe

flashlight

can

be

ignored.)

If all

these processes

are

taken into

account

and the corresponding entropy changes

are

calculated

from

the

standpoint

of the

increase

or

decrease

of

information,

then

Rodd

was

able to

demon-

strate

that

the total entropy change

is

positive. The second

law is not

violated.

Much

has

been

written

about reversibility and

irreversibility, order

and



272

Heat

and Thermodynamics

Statistical

Mechanics

273

disorder,

and supposed violations

of the

second

law.

It would

be hard

to

find

anything

in any language

to

compare

with

the

exposition

given in

Feynman's Lectures

on

Physics

(Chap.

46). This

chapter

is

recommended

wholeheartedly

to all students

whether

naive

or

sophisticated

in their

level

of

attainment

—

for

its brilliance,

its depth,

and its human

warmth.

PROBLEMS

(Values

of constants:

k

=

1.38

X

lO

16

erg/deg

and h

=

6.63

X

10'

erg

•

s.)

10-1

A mercury

atom

moves in

a

cubical

box

whose

edge

is

1

m long.

Its kinetic energy

is equal

to the average

kinetic

energy

of a

molecule

of

an

ideal

gas

at

1000°K. If

the

quantum

numbers

n

x ,

n

y

,

and n.

arc

all equal

to n, calculate

n.

10-2

The

quantum

states available

for

gas

molecules

of energy

e

in

a cubical

box

of

length

L correspond

to

integer

values

for

each

n

x,

«„,

and

n

z

according

to

Eq.

(10-1).

In

a

three-dimensional

Euclidean

space with

coordinates

n

x

,

n

v

,

and

n

z,

each

unit

volume will

contain

one

quantum

state.

The

total

number

of

quantum

states

g'

with

energy

less than

«'

is

equal to

the volume

of the

positive octant

of a sphere

of

radius

r

=

Li^mtîh.

(a)

Show

that

S

3A»

In

a

volume

of

1

cm'

of helium

gas at

300°K and 1

atm pressure,

e' is about

10~

12

erg.

Calculate:

(b)

g'

and

(c) the number

A'

of helium

atoms,

(d)

Show

that

g'

»

N,

10-3

Consider

a function

/

defined

by

the relation

/(0.,fi

;( ) =f(ii

A

) +/(Q ll

).

First

differentiate

partially

with

respect

to

Q,

;

, and

then

with

respect

to il

A

.

Integrate

twice

to show

that

/(12)

=

const.

In

Q

+

const.

10-4

Take

the

expression

for

the

kinetic

energy

of

a

particle in

a

cubical

box and

imagine

a

space

defined

by

the

cartesian

coordinates

n

x

,

n

u

,

and

n

z

.

Note

that

a

single

quantum

state

occupies

unit

volume in

this

space.

(a)

Setting

n

2

= n

x

+

n\

+

n\,

show

that the number

of

quantum

states

in

the small

interval

dn

is

(4tt«

2

dn).

(b)

Prove that

the

number

of

quantum states

dg, in

the

energy interval

dek(2ir//i

i

)V(2m)h

i

d6.

(c)

Show that

the

number

of

ideal-gas

molecules

dN

(

occupying these

quantum states is

given

by

d

^

=

2

Sjm

iie

~'

lkTdi

-

(d) Derive the

Maxwellian

law

of the

distribution

of

speeds,

that

is,

Eq. (6-25).

10-5

In

the

case

of

N

distinguishable

particles,

the

number

of

ways

in

which

a

macrostate

defined by

A i

particles in

gi

quantum states

with

energy

«jj

A

2

particles in

gi

quantum states

with

energy

62;

and so

on, may-

be achieved

is

given

by

Q

=

AM

Si

Sl_>

A

r

i A'

2

• •

when g,- » A

7

,-.

(«)

Using

the

Stirling

approximation, calculate In 9..

(b) Render

In

il

a maximum subject to

the equations

of constraint

2A

r

j

=

N

=

const,

and

2A

r

,e;

=

U

=

const.,

and

explain

why

U

and

P

should

be

the

same

as

for

indistinguishable

particles but

S

should

be

different

10-6

Given

A' indistinguishable, quasi-independent

particles capable

of

existing

in

energy

levels

e%, (2,

• •

,

with

degeneracies

gi,

gz,

. .

.

,

respectively.

In

any given macrostate in which there

are

A'i

particles

in

energy level

ei;

A'2

particles with

energy

etl

and so on,

assume the

thermo-

dynamic

probability

to be given by

the Bose-Einstein expression,

o

=

(gi+Ni)\(g2+N

2

)\

liE

giWiig*m\

Using

Stirling's

approximation

and

the

Lagrange

method,

render

In

f2

U

E

a

maximum subject

to

2A

r

,-

=

A'

=

const, and 2A

r

,e,-

=

U

=

const.,

and

show

that

Si

Nt

=

Ae-?»

-

1

10-7

Given

the same

system

as in Prob. 10-6,

except

that the thermo-

dynamic

probability is.

given by the

Fermi-Dirac expression

Bra

=

gi\g*i

Nil(gt-

N{)W

t

l(g

t

-

N

2

)\



274

Heat and

Thermodynamics

Using

Stirling's approximation and

the Lagrange

method,

render

In

Qfd

a

maximum

subject to 2iV,-

=

A

r

=

const, and 2A ,e;

=

U

=

const., and

show

that

Ni

=

gi

AeS«

+

1

10-8

Given

a

gaseous

system

of

jV

a

indistinguishable,

weakly

inter-

acting

diatomic

molecules:

(a)

Each molecule may

vibrate

with the same

frequency

v

but

with

an

energy e,-,

given

by

U

=

(k

+

t}hv

(i

=

0,l,2, . .

.).

Show that the vibrational partition

function Z„

is

e

-lh,lkT

z

t

=

1

-

«-*'/«

(b)

Each molecule may

rotate, and

the

rotational partition

function

Z

T

has

the

same

form

as that

for

translation,

except

that

the

volume

V is

replaced by the total solid

angle Air, the

mass is replaced

by

the moment

of

inertia

/,

and the

exponent

§

(referring

to

three

translational

degrees of

freedom) is replaced

by

-§-,

since there

are

only

two rotational degrees

of

freedom. Write

the

rotational partition

function.

(V) Taking

into account translation,

vibration,

and

rotation,

calculate

the Helmholtz

function.

(d) Calculate

the

pressure.

(e)

Calculate the

energy.

(/)

Calculate the

molar

heat capacity

at constant volume.

11.

PURE

SUBSTANCES

11-1

Enthalpy

The

laws

of

thermodynamics were

stated and their consequences

were

developed in a

sufficiently general

manner

to

apply

to systems

of

any

number

of coordinates. When there

are

three

or

more independent coordinates, one

speaks of

isothermal

surfaces and isentropic (adiabatic reversible)

surfaces.

If, as

is

often

the case,

there

are

only two

independent coordinates,

these sur-

faces reduce to

simple

plane curves. The most important

system

of two

inde-

pendent coordinates is a hydrostatic one, consisting of a single

pure substance

of constant

mass.

Once

the

thermodynamic equations arc developed

for

this

system,

we shall see how simple

it

is

to

write down the analogous equations

for any

other

two-coordinate

system.

In

discussing

some

of

the

properties of

gases

in Chap.

4,

the sum

of

U

and

PV

appeared several

times

(see

Probs.

4-8, 4-9,

and

4-11). It

has

been

found

very useful

to define

a

new function

H,

called

the

enthalpy,

1

by

the relation

H

=

U +

PV.

(11-1)

In

order to study the properties of this function, consider

the change

in

enthalpy

that

takes

place when a

system

undergoes an

infinitesimal

process

from

an

initial

equilibrium

state

to a final

equilibrium

state. We have

dH

=

dU+

P dV

+

V dP;

but dQ

=

dli

+

P dV.

Therefore, dH

=

dQ +

V

dP.

t

Pronounced en-thal'-pi.

(11-2)

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Zemansky Statistical Thermodynamics