PFIG-1 The partition function and ideal gases

From the previous section we found that the partition function of an entire non-interacting system, Q(N,V,T), could be written in terms of individual atomic or molecular partition functions, q(V,T):

€

Q(N,V ,T) =q(V ,T)N

N!

N: number of particles

V: volume

T: temperature

We now apply this to the ideal gas where:

1) The molecules are independent

2) The number of states greatly exceeds the number of molecules (assumption of low pressure).

PFIG-2 The ideal monatomic gas

For an ideal monatomic gas the only degrees of freedom are translational and electronic, a large simplification:

electransatomic εεε +=

total energy translational energy

electronic energy

The Energy: (sum)

We can divide the problem up by considering each degree of freedom separately:

PFIG-3 The ideal monatomic q

The Partition Function:

€

q(V ,T) = qtrans(V ,T) qelec(T)

total atomic partition function translational atomic

partition function

electronic atomic partition function

(product)

The atomic partition function is the product of the partition functions from each degree of freedom:

Next, consider and … ),(trans TVq

€

qelec(T)

PFIG-4 Translations of an ideal gas: qtrans

€

qtrans = e−βε transstates∑We start with the general form

of the partition function:

From our introduction to QM, we know that for a cubic container with sides of length a,

€

ε trans =h2

8ma2 nx2 + ny

2 + nz2( ) with nx,ny,nz =1,2,…

With this transε

€

qtrans = e−βε nx ,ny ,nznx ,ny ,nz =1

∞

∑ =nx =1

∞

∑ny =1

∞

∑ exp − βh2

8ma2nx2 + ny

2 + nz2( )

⎡

⎣ ⎢

⎤

⎦ ⎥

nz =1

∞

∑

a a

a

PFIG-5 Translations of an ideal gas: qtrans

€

qtrans = e−βε nx ,ny ,nznx ,ny ,nz =1

∞

∑ =nx =1

∞

∑ny =1

∞

∑ exp − βh2

8ma2nx2 + ny

2 + nz2( )

⎡

⎣ ⎢

⎤

⎦ ⎥

nz =1

∞

∑

cbacba eeee =++Use the rule that

∑∑∑∞

=

∞

=

∞

=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

12

22

12

22

12

22

trans 8exp

8exp

8exp

zyx n

z

n

y

n

x

manh

manh

manhq βββ

+++=⎟⎟⎠

⎞⎜⎜⎝

⎛− −−−

∞

=∑

222222 8/98/48/

12

22

8exp mahmahmah

neee

manh ββββ

The three sums are all the same (only the name of the index is different), i.e.,

( )3

12

22

trans 8exp,

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∑

∞

=n manh

TVqβ

Leaving us with

PFIG-6 qtrans is nearly continuous

( )3

12

22

trans 8exp,

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−= ∑

∞

=n manh

TVqβUnfortunately, there is no simple

analytical expression for this sum,

However, since translational energy levels are spaced very close together the sum is nearly a continuous function and we can approximate the sum as an integral,

EX-PFIG1

( )3

0

8/trans

222

, ⎟⎟⎠

⎞⎜⎜⎝

⎛= ∫

∞− manhednTVq β ( ) V

hTmkTVq

2/3

2B

trans2, ⎟

⎠

⎞⎜⎝

⎛=π

work the integral

Va =3Note limit change

PFIG-7 Ideal monatomic gas , <εtrans>

With q we can calculate any thermodynamic quantity we want, such as the average energy:

In the last section we demonstrated that VT

qTk ⎟⎠

⎞⎜⎝

⎛∂

∂= trans2

Btransln

ε

Using our expression for qtrans, ( ) VhTmkTVq

2/3

2B

trans2, ⎟

⎠

⎞⎜⎝

⎛=π

€

εtrans = kBT2 ∂

∂Tln T 3 / 2 2πmkB

h2⎛

⎝ ⎜

⎞

⎠ ⎟ 3 / 2

V⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ V

= kBT2 ∂

∂TlnT 3 / 2

⎛

⎝ ⎜

⎞

⎠ ⎟ = kBT

2 ∂

∂T32lnT

⎛

⎝ ⎜

⎞

⎠ ⎟

not a function of T (so partial derivative is zero)

TkBtrans 23

=ε As we found previously. (note that this is per atom)

EX-PFIG2

PFIG-8 Ideal monatomic gas: qelec

Consider the electronic contribution to q: elecq

€

qelec = gie−βε i

(levels), i∑

Again start from the general expression for q, but this time we will sum over levels rather than states:

degeneracy of level i

energy of level i

We can choose to set the lowest electronic energy state at zero, (because in thermo, zero is always arbitrary)

€

ε1 = 0

€

qelec(T) = g1 + g2e−βε 2 + g3e

−βε 3 +

Note: only a function of T energy of level 2 relative to the ground state energy,

€

ε1 = 0( )

PFIG-9 qelec : The sum converges rapidly

The electronic energy levels are spaced far apart, and therefore we only need to consider the first term or two in the series,

€

qelec(T) = g1 + g2e−βε 2 + g3e

−βε 3 +

At 300 K, you only need to keep terms where

€

ε j <103 cm−1

General rule of thumb:

€

e−βε j > 0.008( )

terms are getting small rapidly

EX-PFIG3

PFIG-10 Electronic energy levels Look up electronic energy levels:

Note general trends,

Nobel gas atoms:

€

ε2 ≈105 cm−1

Alkali metal atoms:

€

ε2 ≈104 cm−1

Halogen atoms:

€

ε2 ≈102 cm−1

(at 300 K only keep 1st term)

(at 300 K only keep 1st term)

(at 300 K keep 1st and 2nd terms)

€

qelec(T) = g1 + g2e−βε 2 + g3e

−βε 3 +

EX-PFIG4

PFIG-11 qelec (T)

In general, for non-metals, it is sufficient to keep only the first two terms for qelec(T) ,

€

qelec(T) ≈ g1 + g2e−βε 2

However, you should always keep in mind that for very high temperatures (like on the sun), or smaller values of εj (like in metals), additional terms may contribute. If you find that the second term is of reasonable magnitude (>1% of the first term), then you should check to see that the third term can be neglected. Levels are known from experiment or quantum chemical computation.

PFIG-12 Ideal monatomic gas, Q

€

Q(N,V ,T ) =qtrans (V ,T )qelec (T )( )N

N!

For a monatomic ideal gas we have:

( ) VhTmkTVq

2/3

2B

trans2, ⎟

⎠

⎞⎜⎝

⎛=π

€

qelec(T) ≈ g1 + g2e−βε 2

with

PFIG-13 Ideal monatomic gas: U We can calculate the average energy, U=<E>

( )VVV T

qqTNkTqTNk

TQTkU ⎟

⎠

⎞⎜⎝

⎛∂

∂=⎟

⎠

⎞⎜⎝

⎛∂

∂=⎟

⎠

⎞⎜⎝

⎛∂

∂= electrans2

B2

B2

Blnlnln

qelec contribution typically small

TNkU B23

≈(dominated by the translational contribution… think about fraction in excited states at typical temperatures)

monatomic gas

€

U = NkBT2 ∂∂Tln 2πmkBT

h2⎛

⎝ ⎜

⎞

⎠ ⎟ 3 / 2

V g1 + g2e−βε 2 +( )

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ V

=32NkBT + Ng2ε 2e

−βε 2 +

qtrans qelec

!),(),,(

NTVqTVNQ

N

= )(),(),( electrans TqTVqTVq =

PFIG-14 Ideal monatomic gas: CV

RdT

RTd

dTUdC

VN

VNV 2

323

,

,

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

Molar heat capacity, CV

€

U → U 32

NkBT →32

RT

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

RCV 23

=

PFIG-15 Ideal monatomic gas: P

Pressure, P

( )TTTN V

qqTNkVqTNk

VQTkP ⎟

⎠

⎞⎜⎝

⎛∂

∂=⎟

⎠

⎞⎜⎝

⎛∂

∂=⎟

⎠

⎞⎜⎝

⎛∂

∂= electrans

BB,

Blnlnln

( ) VhTmkTVq

2/3

3B

trans2, ⎟

⎠

⎞⎜⎝

⎛=π

€

qelec(T) ≈ g1 + g2e−βε 2

!),(),,(

NTVqTVNQ

N

= )(),(),( electrans TqTVqTVq =

€

P = NkBT∂∂Vln 2πmkBT

h2⎛

⎝ ⎜

⎞

⎠ ⎟ 3 / 2

V g1 + g2e−βε 2( )

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ T

= NkBT∂∂VlnV

⎛

⎝ ⎜

⎞

⎠ ⎟ T

=NkBTV

Plug in and

the only function of V

monatomic gas

VTNkP B= Ideal Gas Law (PV=nRT)

PFIG-16 Summary: Ideal monatomic gas

€

Q(N,V ,T) =qtrans(V ,T)qelec(T)( )N

N!

( ) VhTmkTVq

2/3

2B

trans2, ⎟

⎠

⎞⎜⎝

⎛=π

€

qelec(T) ≈ g1 + g2e−βε 2

Partition function:

TNkU B23

=

B23 NkCV =

VTNkP B=

RTU23

=

RCV 23

=

VRTP =

From Q

get

Energy

Heat Capacity

Pressure

(molar)

PFIG-17 Energy of a diatomic

elecvibrottransdiatomic εεεεε +++=

total energy translational energy

electronic energy

The Energy: (sum)

rotational energy

vibrational energy m1 m2 (( ((

rigid rotor harmonic oscillator

In addition to translational and electronic degrees of freedom, a diatomic will also have rotations and vibrations. We will treat these within the rigid rotator and harmonic oscillator approximations, respectively (exactly solvable QM problems),

m1 m2

PFIG-18 Diatomic ideal gas partition function

!),(),,(

NTVqTVNQ

N

=Start from the general expression:

Use the diatomic molecular partition function,

elecvibrottrans),( qqqqTVq =

€

Q(N,V ,T) =qtransqrotqvibqelec( )N

N!

The translational partition function. This is the same as in the monatomic case with the mass now the total, m1+m2 m1 m2

€

qtrans V ,T( ) =2π m1 + m2( )kBT

h2⎛

⎝ ⎜

⎞

⎠ ⎟

3 / 2

V

Consider each degree of freedom:

PFIG-19 Diatomic: Where is E0?

m1 m2

R

V(R)

0=ν

Re

0DeD

Ene

rgy

0

ee D−=1ε R

εv=0 =hν/2

2eε

( ) … ,2,1,012

2

=+= JJJIJε

For rotations set zero at J=0, since εJ=0 at J=0:

(note: positive)

For electronic, we take the zero to be the infinitely separated atoms in their ground electronic states (note positions of ε1 and ε2). We have ε1 = –De, and,

For vibrations, we set zero at the bottom of the internuclear well, with the result that εv=0=hν/2.

⎟⎠

⎞⎜⎝

⎛ +=21vhv νε …,2,1,0=v

electronic ground state

electronic excited state

€

qelec(T) ≈ g1eDe / kBT +g2e

−ε 2 / kBT( )(Note: This is a different choice of zero than previously on PFIG-8)

vib zero point energy

PFIG-20 Diatomic vibrations, qvib

€

qvib(T) = e−βε vib

states, v∑ = e

−β v+12

⎛

⎝ ⎜

⎞

⎠ ⎟ hν

v= 0

∞

∑ = e−βhν / 2 e−βhνvv= 0

∞

∑

We use the harmonic oscillator approximation,

⎟⎠

⎞⎜⎝

⎛ +=21

vib vhνε …,2,1,0=v and the levels are non-degenerate (gv=1 for all v)

Careful: Quantum number, v (vee) Vibrational frequency, ν (nu)

recognize the series xx

n

n

−=∑

∞

= 11

0with 1<= − νβhex

€

qvib(T) = e−βhν / 2 e−βhν( )v

v= 0

∞

∑ =e−βhν / 2

1− e−βhν νβ

νβ

h

h

eeTq

−

−

−=1

)(2/

vib

m1 m2 (( ((

PFIG-21 The vibrational temperature, Θvib

It is common to define a vibrational temperature,

€

Θvib =hνkB

note units:

€

J ⋅ s( ) s−1( )J ⋅K−1( )

=K

And we can write qvib in terms of Θvib:

€

qvib(T) =e−βhν / 2

1− e−βhνΘvib =hν / kB

⎯ → ⎯ ⎯ ⎯ qvib(T) =e−Θvib / 2T

1− e−Θvib /T

ν / cm-1 Θ

vib /

K For example:

K302,cm210 vib1 =Θ= −ν

At room temp 1B cm200 −≈Tk

PFIG-22 <Evib> and CV,vib

From the partition function, we can calculate the vibrational contribution to the average energy,

€

Evib = NkBT2 ∂lnqvib

∂T= NkBT

2 ∂∂Tln e−Θ vib / 2T

1− e−Θ vib /T

⎛

⎝ ⎜

⎞

⎠ ⎟

⎟⎠

⎞⎜⎝

⎛−

Θ+

Θ= Θ 12 /

vibvibBvib vib Te

NkE

and to the molar heat capacity,

€

C V,vib =∂ E vib∂T

= R Θvib

T⎛

⎝ ⎜

⎞

⎠ ⎟ 2 e−Θvib /T

1− e−Θvib /T( )2

PFIG-23 CV,vib

€

C V ,vibR

=Θvib

T⎛

⎝ ⎜

⎞

⎠ ⎟ 2 e−Θ vib /T

1− e−Θ vib /T( )2

CV,

vib /

R

T / Θvib

CV,

vib /

R

T / K

Classical limit, CV,vib = R

EX-PFIG5

(an OK approximation for T>>Θvib) T=Θ

vib

PFIG-24 Fraction in state v, fv

€

fv =e−βhν v+

12

⎛

⎝ ⎜

⎞

⎠ ⎟

qvib= 1− e−βhν( )e−βhνv = 1− e−Θ vib /T( )e−vΘ vib /T

fraction in state v

€

fv>0 =1− fv= 0 =1− 1− e−Θ vib /T( ) = e−Θ vib /T

f v f v

f v f v

f v f v

v v

PFIG-25 Diatomic rotations

We use the rigid-rotator approximation,

( ) … ,2,1,012

2

=+= JJJIJε

degeneracy of the Jth level, 12 += JgJ

€

qrot (T) = gJ e−βε rot

levels, J∑ = 2J +1( )e−β

2J J +1( ) / 2I

J= 0

∞

∑

Define a rotational temperature: B

2

rot 2Ik

=Θ

€

qrot (T) = 2J +1( )e−Θ rot J J +1( ) /T

J= 0

∞

∑

PFIG-26 Diatomic rotations, qrot

( ) ( )∑∞

=

+Θ−+=0

/1rot

rot12)(J

TJJeJTqThis series, ,does not have a simple closed form.

However, for Θrot<<T the levels are closely spaced and we can approximate the sum as an integral,

€

qrot = dJ 2J +1( )0

∞

∫ e−Θ rot J J +1( ) /T

integrate

ThTIkTIkTq <<Θ==

Θ= rot2

B2

2B

rotrot ,82 π

important

?

EX-PFIG6

PFIG-27 <Erot> and CV,rot

At room temperature Θrot<<T is a good approximation for most rotations, so we will stay in this high temperature limit,

ThTIk

TNkTqTNkE

∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂

=∂

∂=

2B

2

2B

rot2Brot

8lnln

π

TNkE Brot =

dTEd

C rotrotV, =

Rotational energy:

Molar heat capacity:

RC =rotV,

A diatomic has 2 degrees of rotational freedom, each contributes R/2 to CV.

PFIG-28 Fraction in level J, fJ

Fraction of molecules in rotational level J,

€

fJ =2J +1( )e−Θ rot J J +1( ) /T

qrot=Θrot

T⎛

⎝ ⎜

⎞

⎠ ⎟ 2J +1( )e−Θ rot J J +1( ) /T

f J CO

Degeneracy moves maximum away from J=0

Example: CO at 300 K

PFIG-29 Rotational symmetry number, σ

Due to considerations beyond the scope of this course, we need to add a symmetry number to the rotational partition function,

diatomicr homonuclea2diatomicear heteronucl1

,rot

rot =

=

Θ=

σ

σ

σTq

symmetry number

Note that up to this point our derivation was actually for heteronuclear diatomic rotation, σ = 1.

PFIG-30 Diatomic: Put it all together

elecvibrottrans),( qqqqTVq =

€

q(V ,T) =2π m1 +m2( )kBT

h2⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

3 / 2

V ⋅ TσΘrot

⋅e−Θvib / 2T

1− e−Θvib /T⋅ g1e

De / kBT

rigid rotator, Θrot <<T

EX-PFIG7

PFIG-31 Summary: Diatomic gas

!),(),,(

NTVqTVNQ

N

=

€

q(V ,T) =2π m1 +m2( )kBT

h2⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

3 / 2

V ⋅ TσΘrot

⋅e−Θvib / 2T

1− e−Θvib /T⋅ g1e

De / kBTwith

From Q

get

Energy

Heat Capacity

eAT

T

DNeeRRRTRTU −

−

Θ+

Θ++= Θ−

Θ−

/

/vibvib

vib

vib

1223

€

C V =52

R + R Θvib

T⎛

⎝ ⎜

⎞

⎠ ⎟ 2 e−Θ vib /T

1− e−Θ vib /T( )2

translations

rotations

vib zero point energy

vib energy above zero point

electronic

Electronic heat capacity necessarily zero since assumption of all molecules in ground state

PFIG-32 Polyatomic vibrations Remember that for polyatomic vibrations we divided the motions into normal coordinates (modes) and expressed these coordinates as independent harmonic oscillators,

∑=

⎟⎠

⎞⎜⎝

⎛ +=α

νε1

vib 21

jjj vh …,2,1,0=jv

Energy is the sum over modes,

α is the number of vibrational modes: α=3n-5, linear molecule α=3n-6, nonlinear molecule

Partition function is the product,

€

qvib(T) =e−Θ vib, j / 2T

1− e−Θ vib, j /T( )j=1

α

∏

∑=

Θ−

Θ−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

Θ+

Θ=

α

1/

/vib,vib,

Bvib vib,

vib,

12jT

Tjj

j

j

ee

NkE

€

CV,vib = NkBΘvib, j

T⎛

⎝ ⎜

⎞

⎠ ⎟

2e−Θ vib, j /T

1− e−Θ vib, j /T( )2

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

j=1

α

∑

EX-PFIG8

PFIG-33 Contribution to CV,vib in CO2

asymmetric stretch

symmetric stretch

Vertical bend Horizontal bend

3n-5 = 4 normal modes degenerate, Θvib= 954 K

Θvib= 1890 K

Θvib= 3360 K

€

CV,vib = NkBΘvib, j

T⎛

⎝ ⎜

⎞

⎠ ⎟

2e−Θ vib, j /T

1− e−Θ vib, j /T( )2

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

j=1

4

∑

Bvib k

hν=Θ

PFIG-34 Polyatomic rotations Linear polyatomics: The result is the same as that for the diatomic rigid rotator.

Nonlinear polyatomics: For each of the 3 degrees of rotational freedom, (labeled A,B,C) we have a separate moment of inertia and a separate rotational temperature,

The possible cases:

Crot,Brot,Arot,CBA , Θ=Θ=Θ== III spherical top

Crot,Brot,Arot,CBA , Θ≠Θ=Θ≠= III symmetric top

Crot,Brot,Arot,CBA , Θ≠Θ≠Θ≠≠ III asymmetric top

rotrot Θ=σTq σ  = 1 for COS

σ = 2 for CO2

PFIG-35 Nonlinear polyatomic rotations, qrot

1. From quantum get energy levels, for example,

€

εJ =J J +1( )2

2IgJ = 2J +1( )2 J = 0,1,2,…for the spherical top:

2. Plug into ∑∞

=

−=0

rot )(J

JJegTq βε

3. Assume Θrot<<T and approximate the sum as an integral,

JegdJq Jβε−

∞

∫=0

rot

4. Work the integral,

( )2/3

rot

2/1

rot ⎟⎟⎠

⎞⎜⎜⎝

⎛

Θ=

TTqσπ

( )2/1

Crot,Arot,

2/1

rot ⎟⎟⎠

⎞⎜⎜⎝

⎛

Θ⎟⎟⎠

⎞⎜⎜⎝

⎛

Θ=

TTTqσπ

( )2/1

Crot,Brot,Arot,

32/1

rot ⎟⎟⎠

⎞⎜⎜⎝

⎛

ΘΘΘ=

TTqσπ

spherical top

symmetric top

asymmetric top

PFIG-36 Nonlinear rotations: U and CV

( )2/3

rot

2/1

rot ⎟⎟⎠

⎞⎜⎜⎝

⎛

Θ=

TTqσπ

( )2/1

Crot,Arot,

2/1

rot ⎟⎟⎠

⎞⎜⎜⎝

⎛

Θ⎟⎟⎠

⎞⎜⎜⎝

⎛

Θ=

TTTqσπ

( )2/1

Crot,Brot,Arot,

32/1

rot ⎟⎟⎠

⎞⎜⎜⎝

⎛

ΘΘΘ=

TTqσπ

spherical top

symmetric top

asymmetric top

from q

calc Urot

For all 3 you find, 23,

23

rot,rotRCRTU V ==

This is due to the assumption we made that the distribution of levels is continuous, (Θrot<<T).

1/2 RT for each rotational degree of freedom

PFIG-37 Summary: Polyatomic gas

TkDe

n

jT

Teeg

eeTV

hTMkTVq B

jvib,

jvib,/

1

53

1/

2/

rot

2/3

2B

12),( ⋅⎟

⎟⎠

⎞⎜⎜⎝

⎛

−⋅

Θ⋅⎟

⎠

⎞⎜⎝

⎛= ∏−

=Θ−

Θ−

σπ

€

U = 52

RT + RΘvib, j

2+

Θvib, j

e−Θ vib,j /T −1⎛

⎝ ⎜

⎞

⎠ ⎟

j=1

3n−5

∑ − NA De

€

C V =52

R + RΘvib,j

T⎛

⎝ ⎜

⎞

⎠ ⎟

2e−Θ vib,j /T

1− e−Θ vib,j /T( )2

j=1

3n−5

∑

€

q(V ,T) =2πMkBTh2

⎛

⎝ ⎜

⎞

⎠ ⎟ 3 / 2

V ⋅π1/ 2

σT 3

Θrot,AΘrot,BΘrot,C

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

⋅e−Θ vib,j / 2T

1− e−Θ vib,j /Tj=1

3n−6

∏⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ⋅ ge1e

De / kBT

eA

n

jT DN

eRRTU −⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

Θ+

Θ+= ∑

−

=Θ−

63

1/jvib,jvib,

123

jvib,

€

C V = 3R + RΘvib, j

T⎛

⎝ ⎜

⎞

⎠ ⎟

2e−Θ vib, j /T

1− e−Θ vib, j /T( )2

j=1

3n−6

∑

Linear polyatomic

Nonlinear polyatomic

PFIG-38 Compare with experiment For H2O,

K,5360K,5160K,2290vib =Θ

3n-6=3 vibrational modes:

€

C V = 3R + RΘvib, j

T⎛

⎝ ⎜

⎞

⎠ ⎟

2e−Θ vib, j /T

1− e−Θ vib, j /T( )2

j=1

3

∑

nonlinear polyatomic:

The points are experimental data and the line is calculated!

PFIG-39 What did we learn?

By considering the molecules that make up a macroscopic material we can construct q, from q we can construct Q, and from Q we can calculate any thermodynamics quantity we wish. For example, U and CV are not just collections of numbers in tables. You can now understand the molecular basis of thermodynamic quantities, and perhaps you have a little insight into why different materials have different thermodynamics properties.

Next we will concern ourselves with the laws that govern the macroscopic thermodynamic quantities. From an historical perspective, we will leave chemistry and devolve to physics…