The Ideal Monatomic Gas. Canonical ensemble: N, V, T 2

The Ideal Monatomic Gas

Canonical ensemble: N, V, T

2

Canonical partition function for ideal monatomic gas

A system of N non-interacting identical atoms:

!

NqQ

N

states i energy levelsof a single atom

jijq e e

2

2 2 22 3

, ,8x y z x y z

hl l l l l l

mV

From quantum mechanics:Atom of ideal gas in cubic box of volume V

3


Translational quantum numbers, , :x y zl l l

2 2 2 2

2 3

2 22 2 2 2

2 3 2 3 2 3

8

states i

8 8 8

x y z

i

x y z

yx z

x y z

h l l l

mV

l l l

h lh l h l

mV mV mVx y z

l l l

q e e

e e e q q q

4

With suitable assumptions:

3 2

2

2mq V

h

• Approximate each sum and solve as an integral

• Then compute q

• Then compute Q

5

The calculated Q is the canonical partition function for ideal monatomic gas

We have considered the contribution of atomic translation to the partition function

However, we have not included other effects, such as contribution of the electronic and nuclear energy states.

We will get back to this matter later in this slide set.

6


Then:

!

NqQ

N

3 2

2

2

! !

N

NN

mV

hqQ

N N

2

3 2ln ln ln ln ln !

! 2

Nq mQ N N V N

N h

7


The internal energy of a N-particle monatomic ideal gas is:

,

ln 3

2V N

Q NU

But U of a monatomic ideal gas is known to be:

,

ln 3

2V N

QU NkT

8


Then:

1

kT

Note that is only a function of the thermal reservoir, regardless of the system details. Therefore, this identification is valid for any system.

9

Relationship of Q to thermodynamic properties

Then:

We showed that:

,

ln , ,

V N

Q V NU E

2

,

ln

V N

QU kT

T

10


Multiplying by kT2:

For a system with fixed number of particles N:

, ,

ln lnln

V N T N

Q Qd Q dT dV

T V

2 2 2

, ,

ln lnln

V N T N

Q QkT d Q kT dT kT dV

T V

2 2

,

lnln

T N

QkT d Q UdT kT dV

V

11


Combining:

But: 22

1U U Ud dU dT UdT T d TdU

T T T T

dVV

QkTQdkT

T

UdTTdU

dVV

QkTQdkT

T

UdTTdU

dVV

QkTQdkTUdT

T

UdTdT

T

QkTTdU

T

UdTUdTTdU

NT

NT

NT

NV

,

222

,

222

,

22

2

,

2

2

lnln

lnln

lnln

ln

12

Therefore:

,

, ,

lnln

ln lnln

T N

V N T N

U QdU kTd Q kT dV

kT V

Q QkTd Q T kT dV

T V

13


We identify that:

Comparing this result with:

dU TdS PdV

,

,

ln

lnln

T N

V N

QP kT

V

QdS kd Q T

T

14


The entropy:

,

lnln

V N

QS Q T

T

The Helmholtz free energy:

, , ln , ,A T V N kT Q T V N

2

, ,

ln lnln

V N V N

Q QA U TS kT kT Q T

T T

15


The Helmholtz energy:

, , ln , ,A T V N kT Q T V N

This is a very important equation – knowing the canonical partition function is equivalent to having an expression for the Helmholtz energy as function of temperature, volume and number of molecules (or moles). From such expression, it is possible to derive any thermodynamic property

16


The chemical potential of a pure substance:

The enthalpy:

, ,

, , ln , ,

T V T V

A T V N Q T V NkT

N N

2

, ,

ln ln

V N T N

Q QH U PV kT kTV

T V

17


Heat capacity at constant volume:

2

,

,

,

22

2, ,

ln

ln ln2

V N

VV N

V N

V N V N

QkT

TUC

T T

Q QkT kT

T T

18


The previous expression can also be written as:

22 2 2

2 2, ,,

2V

V N V NV N

kT Q kT Q kT QC

Q T Q T Q T

This form will be useful when discussing fluctuations, later in this slide set

19

Thermodynamic properties of monatomic ideal gasesThe previous slides showed how to evaluate thermodynamic properties given Q

It is time to discuss the effect of the electronic and nuclear energy states to the single atom partition function before proceeding with additional derivations

We will assume the Born-Oppenheimer approximation: translational energy states are independent of the electronic and nuclear states

Besides, we will assume the electronic and nuclear states are independent of each other

20

Thermodynamic properties of monatomic ideal gases

With these assumptions:

trans elec nuc

The single atom partition function is:

states of the atom

trans elec nuc

kTq e

As discussed in the previous class for independent energy modes:

trans elec nucq q q q21

Thermodynamic properties of monoatomic ideal gases

In which:

,

translational states i

trans i

kTtransq e

,

electronic states i

elec i

kTelecq e

,

nuclear states i

nuc i

kTnucq e

22


3 2 3 2

2 2

2 2trans

m m kTq V V

h h

3trans

Vq

2

2

h

m kT

De Broglie wavelength: based on dual wave-particle nature of matter

23


The electronic partition function:

,,

,1 ,2 ,3

,2 ,1 ,3,1

,electronic states i electronic levels j

,1 ,2 ,3

,1 ,2 ,3

...

elec jelec i

elec elec elec

elec elec elecelec

kT kTelec elec j

kT kT kTelec elec elec

kT kTelec elec elec

q e e

e e e

e e e

,1

,1 ,2 ,3

,1 ,2 ,3

...

...

elec

elec elec elec

kT

kT kT kTelec elec elece e e

24


The electronic partition function:

,1 ,2 ,3

,1 ,2 ,3 ...elec elec elec

kT kT kTelec elec elec elecq e e e

Additional information and approximations:-The degeneracy of the ground energy level is equal to 1 in noble gases, 2 in alkali metals, 3 in Oxygen;-The ground energy level is the reference for the calculations – it is conventional to set it to zero;-The differences in electronic levels are high. For example, argon:

,2 1521elec

kJ

mol At room

T:

,2

450 0elec

kTe e

25


Then, the electronic partition function is approximated as:

,1elec elecq

26


The nuclear partition function:

The analysis is similar to that of the electronic partition function, only that the energy levels are even farther apart.

It results:

,1nuc nucq

Also, in situations of common interest to chemical engineers, the atomic nucleus remains largely undisturbed. The nuclear partition function becomes only a multiplicative factor that will cancel out in calculations 27

Thermodynamic properties of monatomic ideal gasesCompiling all these intermediate results:

trans elec nucq q q q

3 2

,1 ,1 2

2elec nuc

m kTq V

h

3 2

,1 ,1 2

2

! !

N

elec nucN

m kTV

hqQ

N N

28


For the monoatomic ideal gas, the logarithm of the canonical partition function is:

3 2

,1 ,1 2

,1 ,1 2

2

ln ln ln! !

3 2ln ln ln ln ln !

2

N

elec nucN

elec nuc

m kTV

hqQ

N N

m kTN N N N V N

h

29

Thermodynamic properties of monatomic ideal gasesLet us now use this expression to compute several properties, beginning with the pressure:

,

ln

T N

Q NkTP kT

V V

Av

Rk

N

Av Av

NkT NRT NRTP PV

V N V N

where N is the number of molecules and Nav is Avogadro’s number.

30


PV nRT

31


PV nRTWe derived this very famous equation from very fundamental principles – an amazing result

32


Internal energy and heat capacity at constant volume:

2

,

ln

V N

QU kT

T

,1 ,1 2

3 2ln ln ln ln ln ln !

2elec nuc

m kTQ N N N N V N

h

2 3 3

2 2

NU kT NkT

T

,

3

2VV N

UC Nk

T

These expressions are more complicated if excited energy levels are taken into account – see eq. 3.4.12 and Problem 3.2 33

Thermodynamic properties of monatomic ideal gasesHelmholtz energy:

Before obtaining its expression, let us introduce Stirling’s approximation:

ln ! ln ln lnN N N N N N N e

This approximation is increasingly accurate the larger N is. Since N here represents the number of atoms, it is typically a very large number and this approximation is excellent.

34


Helmholtz energy:

3

2 ,1 ,1

2

2ln ln elec nucVem kT

A kT Q NkTh N

Ignoring the nuclear partition function by setting it equal to 1:

3

2 ,1

2

2ln ln elecVem kT

A kT Q NkTh N

35


Entropy:

U AA U TS S

T

5322 ,1

2

2ln elecVem kT

S Nkh N

known as Sackur-Tetrode equation 36


Chemical potential:

,

ln , ,

T V

Q T V NkT

N

37


,1 ,1 2

,

,1 2

3 3

2 2

,1 ,12 2

3 2ln ln ln ln ln

2

3 2ln ln ln ln

2

2 2ln ln

elec nuc

T V

elec

elec elec

m kTN N N N V N N N

hkT

N

m kTkT V N

h

m kT V m kT kTkT kT

h N h P

3

2

,1 20 0

3

2

,1 20 0

2ln ln ln

2ln ln

elec

elec

m kT kT kT kTkT kT kT

h P P P

m kT kT PkT kT

h P P

38


Now compare this formula and the formula well-known to chemical engineers of the chemical potential of a pure ideal gas:

00

, , lnP

T P T P kTP

3

2

,1 20 0

2ln lnelec

m kT kT PkT kT

h P P

39

Energy fluctuations in the canonical ensemble

In the canonical ensemble, the temperature, volume, and number of molecules are fixed.

The energy may fluctuate. Assume its fluctuations follow a Gaussian distribution:

2

1

21

2

x x

f x e

x Mean of the distribution Standard deviation

x f xVariable Probability density

40

Energy fluctuations in the canonical ensembleGiven this distribution, the average value of any function G(x) is calculated as follows:

2

1

21

2

x x

G f x G x dx e G x dx

The variance (standard deviation to power 2) is:

41

2 2 2 22 2 2 22 2x x x xx x x xx x x x 2x 2x 2x 2x


Let us apply this formalism to the average energy and its fluctuation:

3322

EE U NkT x

NkT

32

Ex

NkT

2 22 2 22

1

32

x x E E

NkT

42

2E 2E


2 22 2 22

1

32

x x E E

NkT

2

2

states i2 states i2 2

1

32

iiEEkTkT

iiE eE e

Q QNkT

43

44

NVNV

j

kTEj

j

kTEj

NVNVNVNV

NVNVNV

NVNV

j

kTEj

NVNV

j

kTEj

NVNV

j

kTEj

j

kTE

T

Q

Q

kT

T

QkT

Q

kT

Q

eE

and

eEQkTT

Q

Q

E

T

Q

Q

kT

T

Q

Q

kT

T

Q

Q

kT

Then

T

Q

Q

kT

T

Q

Q

kT

T

Q

Q

kT

T

E

T

Q

Q

kT

T

QkTE

Since

eEQkTT

Q

Q

E

T

E

ekT

E

T

QEQ

T

E

eEeE

j

j

j

j

jj

,

2

222

,

2

/2

/22

,

2

,2

2

,

2

22

,

2

,2

2

,

2

22

,

,

2

,

2

/22

,,

/

2

2

,,

//

)(2

12

2

ln

1


2 222 2 222

2 2 2, ,,

12

32

V N V NV N

kT kTkT TQ Q Q

Q T Q T Q TNkT

22 2 2

2 2, ,,

2V

V N V NV N

kT Q kT Q kT QC

Q T Q T Q T

We previously found that:

45


The 2/3 factor is a particularity of using monoatomic ideal gases as example. However, the factor is common and shows that relative fluctuations decrease as the number of molecules increases.

Comparing these two expressions:

22

22 2

32 1233 3

2 2

V

kT NkkT C

NNkT NkT

2 1

3 N

1 N

46


47

= [E/(3NkT/2)]-1

Gibbs entropy equation

But, using relationships developed in previous slides:

,

lnln

V N

QS Q T

T

2, ,

states j

2states j

,

ln 1

1 1

j

j

V N V N

E

kTE

jkT

V N

U Q Q

kT T Q T

eE

eQ T Q kT

48


states j

states j

states j

, , ln , ,

ln ln , ,

, ,

1ln , ,

, ,

ln , ,

j j

j

j j

E E

kT kT

E

kTj

p N V E p N V E

e e Q N V T

Q N V T

E eQ N V T

kT Q N V T

UQ N V T

kT

49

QT

QT ln

ln


Combining the expressions developed in the two previous slides (algebra omitted here):

50

),,(ln),,(

lnln

,

jstatesj

j

NV

EVNpEVNpk

T

QTQkS


If there is only one possible state:

states j

, , ln , , 1 ln1 0j jS k p N V E p N V E k

If there are only two possible states, assumed to have equal probability:

states j

1 1 1 1, , ln , , ln ln 0.693

2 2 2 2j jS k p N V E p N V E k k

If there are only three possible states, assumed to have equal probability:

1 1 1 1 1 1ln ln ln 1.0986

3 3 3 3 3 3S k k

51

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The Ideal Monatomic Gas. Canonical ensemble: N, V, T 2