canonical ideal
TRANSCRIPT
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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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Section 2: The Ideal Gas 6
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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Section 2: The Ideal Gas 7
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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Section 2: The Ideal Gas 8
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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Section 2: The Ideal Gas 9
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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Section 2: The Ideal Gas 10
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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Section 2: The Ideal Gas 11
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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Section 3: The translational, single-particle partition function 13
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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Section 3: The translational, single-particle partition function 14
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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Section 3: The translational single particle partition function 15
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Section 3: The translational, single-particle partition function 15
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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Section 3: The translational single particle partition function 16
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Section 3: The translational, single-particle partition function 16
=
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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Section 3: The translational, single-particle partition function 17
= 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
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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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Section 3: The translational, single-particle partition function 20
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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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Section 5: The Classical Regime 25
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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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Section 5: The Classical Regime 26
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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
Section 5: The Classical Regime 27
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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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Section 5: The Classical Regime 28
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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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