Download Monatomic ideal gas: partition functions and equation of state.

Peter Košovan
[email protected]
Dept. of Physical and Macromolecular Chemistry
Statistical Thermodynamics, MC260P105, Lecture 3, 27.10.2014
If you find a mistake, kindly report it to the author :-)
Faculty of Science, Charles University in Prague
http://www.natur.cuni.cz/chemie/fyzchem
Monatomic ideal gas: partition
functions and equation of state.
http://www.natur.cuni.cz/chemie/fyzchem
For non-interacting particles a, b, c, d, . . . we can decompose the
system Hamiltonian to individual contributions:
Ĥ ≈ Ĥ a + Ĥ b + Ĥ c + Ĥ d + . . .
Consequently, the energy is a sum of individual particle energies
E ≈ εa + εb + εc + εd + . . .
• Complex multi-body problem approximated as a superposition of
simpler one- and two-body problems.
• Often encountered in Physics
• Allows for systematic corrections
• e. g. Separation of a single molecule Hamiltonian
Ĥ ≈ Ĥtranslation + Ĥrotation + Ĥvibration + Ĥelectronic + Ĥnuclear + . . .
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Systems of non-interacting particles
ε ≈ εtranslation + εrotation + εvibration + εelectronic + εnuclear + . . .
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Lecture 3: Monatomic ideal gas
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Suppose we can decompose Hamiltonian to individual contributions.
Non-interacting distinguishable particles a, b, c, . . . :
X
X
a
b
c
Q(N, V , T ) =
e−βEj =
e−β(εk +εl +εm +... )
j
=
X
e
−βεak
k ,l,m,...
X
k
e
−βεbl
X
e
−βεcm
· · · = qa qb qc . . .
m
l
where k , l, m, . . . are indices of quantum states of particles a, b, c, . . .
Single-particle partition function:
qi (V , T ) =
X
i
e−βεj
j
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Lecture 3: Monatomic ideal gas
Faculty of Science, Charles University in Prague
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Systems of non-interacting particles
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• If all particles are the same but distinguishable, qa = qb = qc = q :
Q(N, V , T ) =
a
X
b
c
e−β(εk +εl +εm +... ) = qa qb qc · · · = q N (V , T )
k ,l,m,...
• The N-body problem reduces to a one-body problem!
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Lecture 3: Monatomic ideal gas
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Distinguishable vs. indistinguishable particles
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• If all particles are the same but distinguishable, qa = qb = qc = q :
Q(N, V , T ) =
a
X
b
c
e−β(εk +εl +εm +... ) = qa qb qc · · · = q N (V , T )
k ,l,m,...
• The N-body problem reduces to a one-body problem!
• But identical particles are indistinguishable:
Q(N, V , T ) =
X
e−β(εk +εl +εm +... )
k ,l,m,...
• Permutation of k and l yields the same quantum state of the system.
• Correct for double-counting: N! permutations.
P. Košovan
Lecture 3: Monatomic ideal gas
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Distinguishable vs. indistinguishable particles
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Indistinguishable particles
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Q(N, V , T ) =
q N (V , T )
N!
where q(V , T ) =
Lecture 3: Monatomic ideal gas
X
j
e−βεj
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• For indistinguishable particles:
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Indistinguishable particles
Q(N, V , T ) =
q N (V , T )
N!
where q(V , T ) =
X
e−βεj
j
• We silently ignored multiply occupied states: εi = εj :
P. Košovan
Lecture 3: Monatomic ideal gas
Faculty of Science, Charles University in Prague
http://www.natur.cuni.cz/chemie/fyzchem
• For indistinguishable particles:
4/24
Indistinguishable particles
Q(N, V , T ) =
q N (V , T )
N!
where q(V , T ) =
X
e−βεj
j
• We silently ignored multiply occupied states: εi = εj :
• Bosons: ψ symmetric with respect to permutation
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⇒ any number of particles can occupy a given quantum state.
Lecture 3: Monatomic ideal gas
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• For indistinguishable particles:
4/24
Indistinguishable particles
Q(N, V , T ) =
q N (V , T )
N!
where q(V , T ) =
X
e−βεj
j
• We silently ignored multiply occupied states: εi = εj :
• Bosons: ψ symmetric with respect to permutation
⇒ any number of particles can occupy a given quantum state.
• Fermions: ψ antisymmetric with respect to perumtation
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⇒ two particles cannot occupy the same quantum state.
Lecture 3: Monatomic ideal gas
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• For indistinguishable particles:
4/24
Indistinguishable particles
Q(N, V , T ) =
q N (V , T )
N!
where q(V , T ) =
X
e−βεj
j
• We silently ignored multiply occupied states: εi = εj :
• Bosons: ψ symmetric with respect to permutation
⇒ any number of particles can occupy a given quantum state.
• Fermions: ψ antisymmetric with respect to perumtation
⇒ two particles cannot occupy the same quantum state.
• We will see that except for low temperatures or high densities the
number of states N N.
• Then multiply occupied states are rare and the equation above is an
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• For indistinguishable particles:
excellent approximation
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Lecture 3: Monatomic ideal gas
4/24
Ĥψ = Eψ
ny
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Time-independent Schrödinger equation:
Particle in a well of length a:
Ĥ = −
εn =
h2 ∂ 2
2m ∂x 2
h2 n2
8ma2
0 1 2 3 4 5 6 7
nx
Number of states with εnx ,ny ,nz < ε:
n = 1, 2, . . .
Particle in a 3d box of length a:
εnx ,ny ,nz =
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7
6
5
4
3
2
1
0
h2
(n2 + ny2 + nz2 )
8ma2 x
1 4πR 3
Φ(ε) =
8
3
π 8ma2 ε 3/2
... =
6
h2
Lecture 3: Monatomic ideal gas
Faculty of Science, Charles University in Prague
Some useful results from quantum mechanics (QM)
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Degeneracy of translational energy levels
π 8ma2 ε 3/2
Φ(ε) =
6
h2
Number of levels between ε and (ε + ∆ε):
π 8ma2 3/2 1/2
2
ω(ε, ∆ε) = Φ(ε + ∆ε) − Φ(ε) =
ε
∆ε
+
O
(∆ε)
4
h2
Let’s plug in some numbers:
ε = 3kB T /2, T = 300 K, m = 10−22 g, a = 10 cm, ∆ε/ε = 0.01:
ω(ε, ∆ε) ≈ O(1028 )
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Density of energy levels from previous slide:
Degeneracy of translational energy levels far exceeds number of
particles at sufficiently low density and high temperature.
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Lecture 3: Monatomic ideal gas
6/24
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• Sometimes we have to explicitly account for
multiple occupancy of quantum states
• At low T and high density
• Partition function in terms of molecular
*Constrained sum:
X
N=
nk
energy levels (occupation numbers nk ):
Q(N, V , T ) =
X
j
e−βEj =
X∗
e−β
P
i
k
εi ni
{nk }
Ej =
X
k
• Evaluation of the restricted sum is awkward.
• Convenient solution: grandcanonical ensemble (GCE).
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Lecture 3: Monatomic ideal gas
εk nk
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Fermi-Dirac and Bose-Einstein statistics
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Fermi-Dirac and Bose-Einstein statistics
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Ξ(µ, V , T ) =
∞
X
λ = eβµ
eβµN Q(N, V , T )
N=0
µ = kB T ln λ
Rearrangements (* is the restricted sum from the previous slide):
∞ X P
∞
P
P
X
X
X∗
∗
λ i ni e−β i εi ni
Ξ(µ, V , T ) =
λN
e−β i εi ni =
=
N=0
∞
X
N=0 {nk }
{nk }
X ∗ Y
N=0 {nk }
λe−βεk
nk
k
Crucial step (now each nk ranges over all possible values):
n1max
Ξ(µ, V , T ) =
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max
2
X nX
···
Y
λe−βεk
n k
n1 =0 n2 =0
k
Lecture 3: Monatomic ideal gas
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Absolute activity:
Formulation in GCE:
8/24
Fermi-Dirac and Bose-Einstein statistics
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n1max n2max
Ξ(µ, V , T ) =
X X
···
Y
n1 =0 n2 =0
=
n1max X
nk
k
max
λe
−βε1
2
n1 nX
n1 =0
=
λe−βεk
n2
n2 =0
nkmax YX
λe−βε2
λe−βεk
nk
k nk =0
Last equation is a general result. Special cases:
Fermions, nkmax = 1:
Y
ΞFD =
1 + λe−βεk
k
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Bosons, nkmax = ∞:
∞ nk
YX
ΞBE =
λe−βεk
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We continue rearranging
k nk =0
Lecture 3: Monatomic ideal gas
9/24
Fermi-Dirac and Bose-Einstein statistics
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j=0 x
ΞBE =
j
= (1 − x)−1 for x < 1 to rewrite BE formula
∞ YX
k nk =0
λe−βεk
nk
=
Y
1 − λe−βεk
−1
for λe−βεk < 1
k
and cast both FD and BE formula in the same form
±1
Y
Ξ FD =
1 ± λe−βεk
for λe−βεk < 1
BE
k
• These are the only exact distributions in statmech.
• Low T and high density: FD or BE statistics apply, q(N, V , T ) has no
meaning (symmetry requirements of the wave function)
• High T and low density: both degenerate to Boltzmann statistics
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P∞
Use
• Nobel prize in Physics 2001: experiment Bose-Einstein condensate.
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Lecture 3: Monatomic ideal gas
10/24
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Number of particles:
N=N=
X
n k = kB T
k
∂ ln Ξ
∂µ
Number of particles
in state k :
V ,T
X λe−βεk
∂ ln Ξ
=λ
=
∂λ V ,T
1 ± λe−βεk
k
Energy E and average energy per particle, ε:
E = Nε =
X
k
n k εk =
X λεk e−βεk
1 ± λe−βεk
k
pV = kB T ln Ξ = ±kB T
X ln 1 ± λe−βεk
nk =
λe−βεk
1 ± λe−βεk
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Thermodynamic quantities from FD and BE statistics
k
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Lecture 3: Monatomic ideal gas
11/24
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For λ 1 (equivalent βµ 0):
nk =
λe−βεk
1 ± λe−βεk
we can write approximate relations
X
X
nk = λe−βεk , N =
nk = λ
e−βεk
k
k
and using q(N, V , T ) =
P
k
e−βεk
nk
e−βεk
e−βεk
= P −βε =
k
N
q(N, V , T )
ke
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The limiting case of Boltzmann
i. e. both FD and BE reduce to Boltzmann statistics.
P. Košovan
Lecture 3: Monatomic ideal gas
12/24
The limiting case of Boltzmann
X λεk e−βεk
X
E=
→
λεj e−βεj ,
−βε
1 ± λe k
k
E
ε= P
j
k
P
nk
j
→ P
εj e−βεj
j
e−βεj
Using ln(1 + x) ≈ x for small x:
X
X
pV = kB T ln Ξ → ±kB T ±λ e−βεj = λkB T
e−βεj = λkB Tq
j
j
so that
βpV = ln Ξ = λq
Ξ = eλq =
∞
X
(λq)N
N=0
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N!
=
∞
X
λN Q(N, V , T ), where Q(N, V , T ) =
N=0
Lecture 3: Monatomic ideal gas
qN
N!
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Similarly, for small λ we obtain:
13/24
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At high temperature and low
pressure (Boltmzmann limit):
Q(N, V , T ) =
q N (V , T )
N!
Further decomposition of q:
q(V , T ) = qtr qel qnucl
Nuclear partition function:
qnucl = ωn1 + . . .
Except for rare cases just a
constant.
P. Košovan
Lecture 3: Monatomic ideal gas
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Ideal monatomic gas
14/24
Ideal monatomic gas
Q(N, V , T ) =
q N (V , T )
Electronic partition function:
qelec = ωe1 + ωe2 e−β∆ε1,2 + . . .
N!
Further decomposition of q:
q(V , T ) = qtr qel qnucl
• Zero energy at electronic
ground state
• Can be degenerate
Nuclear partition function:
• Higher excited states typically
several eV
qnucl = ωn1 + . . .
• At 300 K 1eV ≈ 40 kB T
Except for rare cases just a
constant.
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• First two or three terms
usually suffice
Lecture 3: Monatomic ideal gas
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At high temperature and low
pressure (Boltmzmann limit):
14/24
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Quantitative view of electronic levels
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Table from McQuarrie, Statistical Mechanics, University Science Books (2000)
Lecture 3: Monatomic ideal gas
15/24
Quantitative view of electronic levels
f2 =
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Fraction of electrons in second excited state:
ωe2 e−β∆ε1,2
ωe1 + ωe2 e−β∆ε1,2 + . . .
Table from McQuarrie, Statistical Mechanics, University Science Books (2000)
Higher electronic states need to be considered at higher temperatures.
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Lecture 3: Monatomic ideal gas
16/24
Translational partition function
h2
(n2 + ny2 + nz2 ),
8ma2 x
εnx ,ny ,nz =
qtr =
∞
X
nx , ny , nz = 1, 2, . . .
e−βεnx ,ny ,nz
nx ,ny ,nz =1
∞
X
∞
−βh2 nx2 X
−βh2 nx2
=
exp
exp
8ma2
8ma2
nx =1
nx =1
nx =1
!3
!3
Z ∞
∞
X
−βh2 n2
−βh2 n2
=
≈
exp
dn
exp
8ma2
8ma2
n=0
−βh2 nx2
exp
8ma2
X
∞
n=1
Alternative approach
qtr =
∞
X
ε=0
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ω(ε)e
−βε
Z
∞
≈
ω(ε)e
n=0
−βε
Z
π 8ma2 3/2 ∞ 1/2 −βε
dε =
ε e
dε
4
h2
n=0
Lecture 3: Monatomic ideal gas
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Translational energy levels:
17/24
We evaluate
Z
qtr =
!3 −βh2 n2
2πmkB T 3/2
V
exp
dn
=
8ma2
h2
n=0
∞
and introduce thermal de Broglie wavelength Λ
1/2
h2
V
Λ=
such that qtr = 3
2πmkB T
Λ
Translational energy and momentum per particle:
3
2 ∂ ln qtr
εtr = kB T
= kB T , |~p| = (3mkB T )1/2 ,
∂T
2
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Lecture 3: Monatomic ideal gas
Λ≈
h
|~p|
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Translational partition function
18/24
We evaluate
Z
qtr =
!3 −βh2 n2
2πmkB T 3/2
V
exp
dn
=
8ma2
h2
n=0
∞
and introduce thermal de Broglie wavelength Λ
1/2
h2
V
Λ=
such that qtr = 3
2πmkB T
Λ
Translational energy and momentum per particle:
3
2 ∂ ln qtr
εtr = kB T
= kB T , |~p| = (3mkB T )1/2 ,
∂T
2
Λ≈
h
|~p|
Boltzmann applies at Λ3 V , i. e. when quantum effects diminish.
P. Košovan
Lecture 3: Monatomic ideal gas
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Translational partition function
18/24
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q(V , T ) = qtr qel qnu
qtr =
2πmkB T
h2
3/2
V =
V
λ3
qel = ωe1 + ωe2 e−β∆ε12 + . . .
qnu = ωn1 + . . .
• The above holds for particles without inernal structure
(monatomic ideal gas)
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Summary – single-particle partitition function
• Vibrational and rotational terms arise for composite particles
(next lecture)
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Lecture 3: Monatomic ideal gas
19/24
A(N, V , T ) = −kB T ln Q = kB T ln N! − NkB T ln q(V , T )
"
#
2πmkB T 3/2 Ve
= −NkB T ln
− NkB T ln ωe,1 + ωe,2 e−β∆ε1,2
2
N
h
E = kB T
2
∂ ln Q
∂T
=
N,V
Nωe,2 ∆ε1,2
3
NkB T +
2
qel
In both above cases, the first term is dominant.
∂ ln Q
NkB T
p = kB T
=
∂V
V
N,T
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Thermodynamic functions of an ideal monatomic gas
Equation of state (EOS) of an ideal gas.
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Lecture 3: Monatomic ideal gas
20/24
Thermodynamic functions of an ideal monatomic gas
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∂ ln Q
S = kB T
∂T
+ kB ln Q
N,V
"
#
3
2πmkB T 3/2 Ve5/2
= NkB + NkB ln
2
N
h2
+ NkB ln ωe,1 + ωe,2 e
−β∆ε1,2
+ NkB
ωe,1 + ωe,2 e−β∆ε1,2
qel
Sackur-Tetrode equation:
"
#
3
2πmkB T 3/2 Ve5/2
S = NkB + NkB ln
+ Sel
2
N
h2
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Lecture 3: Monatomic ideal gas
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Entropy:
21/24
Sackur-Tetrode equation:
"
#
2πmkB T 2/2 Ve5/2
3
+ Sel
S = NkB + NkB ln
2
N
h2
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Quantitative view of entropy
Table from McQuarrie, Statistical Mechanics, University Science Books (2000)
Higher electronic states have to be considered at higher temperatures.
P. Košovan
Lecture 3: Monatomic ideal gas
22/24
Chemical potential:
∂ ln Q
q
µ = −kB T
= −kB T ln
∂N T ,V
N
"
#
2πmkB T 3/2 V
= −kB T ln
− kB T ln qel qnu
N
h2
"
#
2πmkB T 3/2 kB T
p
= −kB T ln
− kB T ln qel qnu + kB T ln 2
p
p
h
p
= µ0 (T ) + kB T ln p
where
"
µ0 (T ) = −kB T ln
P. Košovan
2πmkB T
h2
3/2
kB T
p
#
− kB T ln qel qnu
Lecture 3: Monatomic ideal gas
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Thermodynamic functions of an ideal monatomic gas
23/24
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• Separation of different contributions to Ĥ, E, Q and q
• Fermi-Dirac and Bose-Einstein statistics
• Boltzmann as high-temperature and low-density limit
• Thermal wavelength
• EOS of an ideal gas from first principles (Boltzmann limit)
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Lecture 3: Monatomic ideal gas
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Summary
24/24
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• Separation of different contributions to Ĥ, E, Q and q
• Fermi-Dirac and Bose-Einstein statistics
• Boltzmann as high-temperature and low-density limit
• Thermal wavelength
• EOS of an ideal gas from first principles (Boltzmann limit)
What comes next
• Polyatomic gas molecules – vibration and rotation
• Condensed matter – dense gases and liquids (classical limit)
• Distinguishable particles – crystals, magnets
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Lecture 3: Monatomic ideal gas
Faculty of Science, Charles University in Prague
Summary
24/24