canonical ideal

8/11/2019 Canonical Ideal

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Table of Contents

1.Introduction2. The Ideal Gas

2.1. The Partition Function for the Ideal Gas

Detailed Calculation for a two particle ideal gas Gen-eralization to N molecules Relationship Between the N-particle and single particle Partition Function2.2. Evaluation of the Partition Function

3. The translational, single-particle partition function

3.1. Density of States

3.2. Use of density of states in the calculation of the

translational partition function3.3. Evaluation of the Integral

3.4. Use ofI2 to evaluate Z13.5. The Partition Function for N particles

4. Calculating the Properties of Ideal Gases from the Par-

tition Function


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Table of Contents (cont.) 3

4.1. The Equation of State

5. The Classical Regime

5.1. The Classical Regime in terms of the de Brogliewavelength

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Section 1: Introduction 4

1. Introduction

We are going to analyze an ideal gas on the canonical ensemble, we will

not use quantum mechanics, however, we will need to take account ofsome quantum effects, and as a result the treatment is a semi-classicaltreatment. Im going to take the calculation several steps farther thanour author, in particular, Ill derive the ideal gas law.

2. The Ideal GasWe have already noted that an ideal gas is an idealization in which wecan ignore the potential energy terms. That is the interaction energyof the molecules is negligible. This means that we can write the energy

of the ideal gas in terms of a sum of energies for each molecule.The energies

1 2 3 . . . r . . .correspond to the complete set of (discrete) quantum states 1, 2, . . . , r , . . . ,

in which a single molecule can exist. To analyze this system on the

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Section 2: The Ideal Gas 5

canonical ensemble, we need to impose the conditions of the canonicalensemble. Accordingly, considerNparticles of an ideal gas contained

in a volume V in contact with a heat bath at temperature T. Wecan specify the state of the gas by counting the number of moleculesin each quantum state. If we denote the occupancy of the ith statebyNi, then Ni is the occupancy of the i

th level. As in our previousexample, the energy of the gas is given by

E(N1, N2, . . . , N r, . . .) =

r

Nrr,

the sum being taken over all single particles states. Since we have Nparticles, we must have

N= r Nr

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2.1. The Partition Function for the Ideal Gas

There are some points where we need to be careful in this calculation.

We can easily calculate the partition function for a single molecule

Z(T , V , 1) =Z1(T, V) =

r

er .

At this point it is tempting to write

ZN1 =Z(T , V , N ).

Unfortunately, the answer is wrong! Why? We can see why it iswrong by considering two particles.

Detailed Calculation for a two particle ideal gas

Assuming we can write the partition function for an ideal gas as aproduct of single particle partition functions, for two particles we

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would have

s e

st e

t= s e2s + s t (s=t)

e(s+t)

The first term on the r.h.s corresponds to all terms on the l.h.s forwhich both molecules are in the same state, the second term corre-

sponds to the molecules being in different states. We can now see theproblem. When the particles are in different state, we have countedeach state twice. The state with one molecule in state 1, the otherin state 2, could be written as s = 1, t = 2 or as s = 2, t = 1. Nowexcept for the fact that weve labelled these states as 1 and 2, thetwo states of the gas are the same. But the molecules are identical,we cannot justify counting the states twice. It only the occupationnumbers that matter - we cant distinguish between the states whereparticle labels are exchanged experimentally. Now, we already knowhow to delete useless rearrangements think golfing foursomes we

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divide by the factorial of the number of objects, in this case 2!. Thus,the correct expression for partition function of the two particle ideal

gas isZ(T , V , 2) =

s

e2s + 1

2!

s

t

(s=t)

e(s+t).

Generalization to N molecules

For more particles, we would get lots of terms, the first where allparticles were in the same state, the last where all particles are in dif-ferent states, where the last term must be divided by N! to eliminate

overcounts. Thus, we should write

Z(T , V , N ) =

s

eNs +. . .+ 1

N!

s1

. . .sN

all sN differente(s1+...+sN)

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In the terms represented by the ellipsis, some particles are in thesame states, some particles are in different states and they need to

be appropriately weighted. However, it isnt necessary for us to writethese terms. In the classical regime, the probability of a single particlestate being occupied by more than one particle is vanishingly small. Ifa few states are doubly or triply occupied, they contribute little to thepartition function and can be safely omitted. In fact, we can safely

approximate the partition function by the last term in the expressionfor the partition function.

Relationship Between the N-particle and single particle

Partition Function

Thus,

Z(T , V , N ) = 1

N!

r

er

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and the relationship between Z(T , V , N ) and Z1 is

Z=Z(T , V , N ) = 1

N![Z

1(T, V)]N

The division by N! makes this calculation semi-classical. It is oftenascribed to the particles being non-localized. If a particle is localizedto a site on a crystal lattice, this serves to distinguish that particleand division by N! is not needed.

2.2. Evaluation of the Partition Function

To find the partition function for the ideal gas, we need to evaluatea single particle partition function. To evaluate Z1, we need to re-

member that energy of a molecule can be broken down into internaland external components. The external components are the transla-tional energies, the internal components are rotations, vibrations, andelectronic excitations. To represent this we write

r =sj =trs +

intj .

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S i 2 Th Id l G 11


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In this expression trs is the translational energy, intj is the internal

energy. We use this to rewrite Z1 as

Z1=

r

er =

s

j

e(trs +intj )

or

Z1 = s

etrs

j

etrj = Ztr1 Zint

where

Ztr1 =Ztr(T , V , 1) =

s

etrs

andZint=Z

int(T) =

j

eintj .

Again, we see a partition function factorizing, this enables us to evalu-ate the pieces separately. Z1is the same for all gases, only the internal

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S ti 3 Th t l ti l i l ti l titi f ti 12


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Section 3: The translational, single-particle partition function 12

parts change. For now well assume that all internal energies are inthe ground state, so Zint = 1. So lets evaluate Z1.

3. The translational, single-particle parti-

tion function

Well calculateZtr1 once and for all, the result will apply to all molecules

in the classical regime. We know that we can writetr =p2/2m, clas-sicallyE, pcan take on any value, quantum mechanically we can stillwrite tr =p2/2m, but tr and p are restricted to discrete values. Toaddress this, we need to introduce the density of states.

3.1. Density of States

Consider a single, spinless particle in a cubical box of side L. We canwrite

= p2

2m

= 1

2m

(p2x+p2y+p

2z)

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S ti 3 Th t l ti l i l ti l titi f ti 13


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We (from our knowledge of quantum mechanics) expect it to be rep-resented by a standing wave, in 3 dimensions this is

n1,n2,n3 = (Constant)sinn1x

L

sin

n2yL

sin

n3ZL

n1, n2, n3 = 1, 2, 3 . . . . The solutions vanish at x, y, z= 0, L

We then define

k2 = 2

L2 (n21+n22+n23)

wherek is the wave vector with components

k=

n1

L ,

n2L

,n3

L We can plot these vectors in k-space the points at the tips of thevectors fill k-space. These points form a cubic lattice with spacing

/L, the volume per point ofk-space is

L

3.

How many of the allowed modes have wave vectors between k and

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Section 3: The translational single particle partition function 14


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k+ dk. Since theni are zero,k1, k2, k3 are greater than zero. Thus ifwe imagine the set of points at the tips of the wave vectors centered

at the origin forming a sphere, we should only consider the positiveoctant.

Let number of modes with a wave vectors between k and k +dkbe

f(k)dkthen

f(k)dk=1

8

volume of shell

4k2dk

(/L3) volume/point

and since V =L3

f(k)dk=V k2dk

22 .

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NowE=h, so since p= E

c, we have

p=E

c =h

c = h

2

2

c =

c = k

Thus p = k and recognizing that particles can be treated as waves,the number of particles with momentum betweenp and p+dp is

f(p)dp = V(p/)2

22

dp

= V p2dp4

h3

3.2. Use of density of states in the calculation of

the translational partition functionZtr1 is the sum over all translational states. Thus we can rewrite Z

tr1

as an integral using the density of states function. Then

Ztr1 = s etrs

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=

0V4p2dpep

2/2m

h3

3.3. Evaluation of the Integral

We have an integral of the form

In(a) =

0

xneax2

dx a >0

where

I0a=

0

eax2

dx

To evaluate this, let ax2 =u2 so

xx= u and du=

adxThus

I0(a) =

01

aeu

2

du

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Section 3: The translational single-particle partition function 17


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= 1

2

a

eu

2

du

Now

eu2

du

2

=

ex2

dx

ey2

dy

=

e(x2+y2)dxdy

To complete the evaluation, we transform to polar coordinates using

x = r cos , and y = r sin , x2 +y2 = r2, and dA = rdrd, andintegrate over the entire plane,

eu

2

du

2

=

0 2

0

er2

drd

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Section 3: The translational single-particle partition function 18


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Section 3: The translational, single particle partition function 18

= 2

0er

2

dr

now let r2 = and d= 2rdrand

eu2

du2

= 2

0

ed

2

=

0

ed

= [e]0

=

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Section 3: The translational, single particle partition function 19

and

0

eu2

du=

and finally

I0(a) =

2

a=

1

2

a.

Similarly

I1(a) =

0

xeax2

dx

= 12a

0

eudu

= 1

2a

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S s , s g p p

All other values can be found from these with a recursion relationshiparrived at by differentiating In(a) w.r.t a. The values of interest to

us are:

I2(a) = 1

4a

a

I3(a) = 1

2a2

3.4. Use of I2 to evaluate Z1

I2(a) = 1

4a

aso

Ztr1 =

0

V4p2dpep2/2m

h3 =V

2mkT

h2

32

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Section 4: Calculating the Properties of Ideal Gases from the Partition Function


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g p

3.5. The Partition Function for N particles

Using our calculations up to this point

Z=Z(T , V , N ) = 1N!

[Z1(T, V)]N

so

Z=VN

N! 2mkT

h2 3N2

(Zint(T))N.

Now from F =kTln Zwe can find all the important properties ofan ideal gas.

4. Calculating the Properties of Ideal Gases

from the Partition Function

F = kTln Z

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g p

F = N kt ln

eV

N

2mkT

h2

32

Zint(T)

where I have used

N! =

N

e

N.

This result can easily be demonstrated:

ln N! = lnN

e

N=N(ln N ln e) =Nln NN.

4.1. The Equation of State

We have characterized an ideal gas as a gas in which pV =N kT andE=E(T). The term Zint(T) in

F = N kTln

eV

N

2mkT

h2

32

Zint(T)

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refers to a single molecule, it does not depend on V. Thus, we canwrite

F =Ftr+Fintwhere

Ftr= N kTln

eV

N

2mkT

h2

32

and

Fint= N kTln Zint(T) = N kTln

j

eintj

.

NowF =E

T S, dF =

SdT

pdV +dN so

p= F

V

T,N

and

Ftr=

N kTlnVA

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Section 5: The Classical Regime 24


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where

A= e

N2mkT

h2

3

2

.

Using this, we easily recover the ideal gas equation of state

p=

F

V

T,N

= (N kT) AV

d(V /A)

dV =

N kT

V

and pV =NkT.

5. The Classical Regime

As previously noted, the classical regime is the state where most singleparticle states are unoccupied, a few contain one molecule, and aninsignificant number have higher populations.

The probability of a particle being in a translational state s with

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energy trs is given by the Boltzmann distribution

Ps = 1

Ztr1e

trs .

ForNmolecules, the mean number of molecules in the states,is given by

=N Ps

for each state s.For each translational states, the molecule can be in many differ-ent states of internal motion. A sufficient condition for the classicalregime to hold is

1, for all s.We can put this in terms of quantities that pertain to the ideal gas.=N Ps where

Ps = 1

Ztr1e

trs .

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and

Ztr1 =V 2mkT

h2

3

2

.

So

=

N

V

h2

2mkT

32

e

trs 1.

This expression is certainly true as h

0, this is in fact the Maxwell-

Boltzmann limit. We have developed classical statistical mechanics.Examining the equation shows that it is easier to meet at high temper-atures and low particle concentrations. To clarify things even more,lets rewrite the expression in more familiar terms.

5.1. The Classical Regime in terms of the de Brogliewavelength

We know that we can write

dB =

h

p =

h

2mtr .Toc Back Doc Doc



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We already know that the mean kinetic energy of a gas molecule inthe classical gas is

tr

=

3

2 kT .So using this, we have

dB = h

3mkT=

2

3

12

h2

2mkT

12

.

So N

V

h2

2mkT

32

1.

can be rewritten as

323

2

NV3dB 1.

If we let = (V /N)1/3 be the mean distance between molecules in ourgas, and omit the numerical factors, the condition for the classicalregime to hold becomes

3dB

3

.

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In other words, the classical regime is when the de Broglie wavelengthis small compared to the mean molecular separation. If we take T =273 K, for Helium with 1020 molecules/cm3, = 2

107 cm, and

dB = 0.8 108 cm. Under these conditions quantum mechanicaleffects are negligible.

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canonical ideal

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