The Heat Capacity of a Diatomic Gas

15.1 Introduction• Statistical thermodynamics provides deep

insight into the classical description of a MONATOMIC ideal gas.

• In classical thermodynamics, the principle of equipartition of energy fails to give the observed value of the specific heat capacity for diatomic gases.

• The explanation of the above discrepancy was considered to be the most important challenge in statistical theory.

15.1 The quantized linear oscillator• A linear oscillator is a particle constrained to move

along a straight line & acted on by a restoring force

F= -Kx = ma =

If displaced from its equilibrium position and released, the particle oscillates with simple harmonic motion of frequency , given by

Note that the frequency depends on K and m, and is

independent of the amplitude X.

m

Kv

2

1

2

2

td

xdm

• Consider an assembly of N one-dimensional harmonic oscillators, in which the oscillators are loosely coupled so that the energy exchange among them is small.

• In classical mechanics, a particle can oscillate with any amplitude and energy.

• From quantum mechanics, the single particle energy levels are given by

EJ = (J + ½)h , J = 0, 1, 2, ….. • The energies are equally spaced and the ground state

has non-zero energy.

• The internal degrees of freedom include vibrations, rotations, and electronic excitations.

• For internal degrees of freedom, Boltzmann Statistics applies. The distinguishable property arises from the fact that those diatomic molecules have different translational energy.

• The states are nondegenerate, i.e. gj = 1

• The partition function of an oscillator

• Introducing the characteristic temperature θ = h/k

• The solution for the above eq. is (in class derivation)

......132

2 TTTT eeee

• The internal energy is:

U = NkT2

since

For T → 0

For

thus

1

1

2

1

12

1

TT

T

e

Nk

e

eNkU

0

1

1

Te

15.3Vibrational Modes of Diatomic Molecules

• The most important application of the above result is to the molecules of a diatomic gas

• From classical thermodynamics

for a reversible process!

Since

Or

At high temperatures

At low temperature limit

On has

So

approaching zero faster than the growth of (θ/T)2

as T → 0

Therefore Cv 0 as T 0

The total energy of a diatomic molecule is made up of four contributions that can be separately treated:

1. The kinetic energy associated with the translational motion

2. The vibrational motion

3. Rotation motion (To be discussed later)

Example: 15.1 a) Calculate the fractional number of oscillators in the three lowest quantum states (j=0, 1, 2,) for

Sol:

forTandT 4

J = 0

• 15.2) a) For a system of localized distinguishable oscillators, Boltzmann statistics applies. Show that the entropy S is given by

• Solution: according to Boltzmann statistics

So

N

NNS J

JJ ln

J

NJ

n

J N

gNW

J

1!

1ln wkS

!ln!lnln

!ln1ln!lnln

01ln1

lnln!lnln

NNw

NNNw

g

NgNNw

JJ

J

JJJ

JJ

JJ

JJJ

J

JJ

JJ

w

w

w

w

w

lnln

lnlnln

lnlnln

lnlnln

1lnlnln

n

J

jj

jj

S

S

1

ln

ln

15.4 Rotational modes of diatomic molecules

The moment of inertia,where μ is the reduced mass

r0 is the equilibrium value of the distance between the nuclei.

From quantum mechanics, the allowed angular momentum states are

where l = 0, 1, 2, 3… …

From classical mechanics, the rotational energy equals with w is angular velocity.

The angular momentum

therefore, the energy

Define a characteristic temperature for rotation

θrot can be found from infrared spectroscopy experiments, in which the energies required to excite the molecules to higher rotational states are measured.

Different from vibrational motion, the energy levels of the above equation are degenerate.

for level

Now, one can get the partition function

For , virtually all the molecules are in the few lowest rotational states.

As a result, the series of can be truncated with negligible errors after the first two or three terms!

For all diatomic gases, except hydrogen, the rotational characteristics temperature is of the order of 10 k (Kelvin degree).

At ordinary temperature,

Therefore, many closely spaced energy states are excited. The sum of may be replaced by an integral.

Define:

Note that the above result is too large for homonuclear molecules such as H2, O2 and N2 by a factor of 2… why?

The slight modification has no effect on the thermodynamics properties of the system such as the internal energy and the heat capacity!

Using

(Note: )

again

At low temperature

Keeping the first two terms

VT

ZNkTU

ln2

NkT

UC

vrotV

,

Using the relationship

(for )

And

(for )

2

222 23

ln

TeNkT

T

ZNkTU rotT

V

rot

2

2

, 26T

eNkT

UC rotT

rotV

rotV

rot

Characteristic TemperaturesCharacteristic temperature of rotation of diatomic molecules

Substance θrot(K)

H2 85.4

O2 2.1

N2 2.9

HCl 15.2

CO 2.8

NO 2.4

Cl2 0.36

Characteristic temperature of vibration of diatomic molecules

Substance θvib(K)

H2 6140

O2 2239

N2 3352

HCl 4150

CO 3080

NO 2690

Cl2 810

15.5 Electronic Excitation

The electronic partition function is

where g0 and g1, are, respectively, the degeneracies of the ground state and the first excited state.

E1 is the energy separation of the two lowest states.

Introducing

For most gases, the higher electronic states are not excited (θe ~ 120, 000k for hydrogen).

therefore,

At practical temperature, electronic excitation makes no contribution to the external energy or heat capacity!

0ln2

VT

ZNkTU

0

V

V T

UC

15.6 The total heat capacity

For a diatomic molecule system

Since

Discussing the relationship of T and Cv (p. 288-289)

Heat capacity for diatomic molecules

• Example I (problem 15.7) Consider a diatomic gas near room temperature. Show that the entropy is

• Solution: For diatomic molecules

,2

2ln

2

7 25

23

2

rot

T

h

mkNkS

)!(0

1ln1

121

0

contributenotdoes

e

eNk

Te

Nk

S

S

etemperaturroomAt

SSSSS

T

TT

vib

excit

excitrotvibtrans

• For translational motion, the molecules are treated as non-distinguishable assemblies

23

2

23

2

2ln

2

5

1lnln

2

freedomofdegreesthree2

3

h

mkT

N

VNkNkS

NZNkT

US

h

mkTVZ

NkTU

tr

tr

• For rotational motion (they are distinguishable in terms of kinetic energy)

rotsystem

rotsystem

rot

rot

T

h

mk

N

VNkNkS

TNkNk

h

mkT

N

VNkNkS

TNkNk

ZNkT

US

TZNkTU

2

2ln

2

7

2ln

2ln

2

5

2ln

assemblyhabledistinguisforln

moleculerhomonucleatodue2

1

2

252/3

2

23

2

• Example II (problem 15-8) For a kilomole of nitrogen (N2) at standard temperature and pressure, compute (a) the internal energy U; (b) the Helmholtz function F; and (c) the entropy S.

• Solution:

calculate the characteristic temperature first!

29.23352 Nforkandk

rvib

J

kkmolJkmol

eNk

eNk

eNkNkTU

Tbecause

NkTe

NkNkTU

UUUU

vib

r

Tvib

rotvibtrans

7

113

25.11

25.11

2983352

1089.1

227210314.80.1

1

33522272

1

33523352

2

12985.2

1

1

2

1

2

5

1

1

2

1

2

3

4.16

11015.5ln11002.6

10195.3ln

10195.3

103.274.22

1086.2734.22

10626.6

29810381.11065.424.22

2

1lnln

&ln

1lnln

626

33

33

23

2023

23

19

23

682

123263

23

2

NkT

NkTNkTF

Jkm

h

mkTVZ

NZNkTF

rotationvibassuchparticleshableDistinguisZNkTF

particleshabledistinguisnonNZNkTF

t

t

tt

J

NkT

NkTNkTNkT

FFFF

NkTF

ZZ

TforT

ZZNkTF

eZ

e

e

e

eZ

ZNkTF

tvibrot

rot

rotrot

rotrot

rotrotrot

vib

vib

vibvib

72326

24.11

24.11

62.5

2983352

29823352

10817.32981038.11002.641.15

41.15

4.1662.563.4

63.4

63.4ln72.1029.2

298

2ln

62.51

1ln62.5ln

11

ln