diatomic and polyatomic ideal gas: vibrations,...
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gueDiatomic and polyatomic ideal
gas: vibrations, rotations
Peter Košovanpeter.kosovan@natur.cuni.cz
Dept. of Physical and Macromolecular Chemistry
Lecture 4, Statistical Thermodynamics, MC260P105, 3.11.2014
If you find a mistake, kindly report it to the author :-)
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Rigid rotor – harmonic oscillator approximation
Diatomic ideal gas:
µ =m1m2
m1 + m2
Rigorous separation:
H = Htr + Hinternal
ε = εtr + εint
q = qtr qint
Q(N,V ,T ) =1
N!(qtr qint)
N
where
qtr = V(
2π(m1 + m2)kBTh2
)3/2
=VΛ3
Approximate separation
Hint = Hvib + Hrot
εint = εrot + εvib
qint = qrotqvib
Q(N,V ,T ) =1
N!(qtr qrot qvib)N
P. Košovan Lecture 4: Polyatomic ideal gas 1/31
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Rotation and vibrationHarmonic approximation:
u(r) = u(re) + (r − re)
(dudr
)r=re
+12
(r − re)2(
d2udr2
)r=re
+ · · ·
u(r) = u(re) + k(r − re)2 + · · ·
Vibrational energy levels(harmonic oscillator):
εn = hν(n +12
)
ν =1
2π
(kµ
)1/2
ωn = 1 for all n
Image source: wikimedia commons
Rotational energy levels(rigid rotor):
εJ =h2J(J + 1)
2I
ν =h
4π2I(J + 1)
ωJ = 2J + 1P. Košovan Lecture 4: Polyatomic ideal gas 2/31
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Electronic and nuclear degrees of freedom
∗
Further approximate separation ofelectronic and nuclear DOF:
H = Htr + Hrot + Hvib + Hel + Hnucl
ε = εtr + εrot + εvib + εel + εnucl
q = qtr qrot qvib qel qnucl
Q(N,V ,T ) =1
N!
(qtr qrot qvib qel qnucl
)N
Electronic partition function:
qel = ωe,1eDe/kBT + ωe,2e−ε2/kBT + · · · where D0 = De −12
hν
∗Image source: wikimedia commons.P. Košovan Lecture 4: Polyatomic ideal gas 3/31
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Vibrational partition function
Vibrational energy levels and degeneracy:
εn = hν(
n +12
), ωn = 1 for all n
qvib(T ) =∑
n
e−βεn = e−βhν/2∞∑
n=0
e−βhνn =e−βhν/2
1− e−βhν
qvib(T ) is a geometric series and can be summed directly (very rare!).
High temperature limit for later comparison:
qvib(T ) = e−βhν/2∫ ∞
n=0e−βhνdn =
kBThν
=TΘv
for (hν � kBT )
Vibrational temperature: Θv = hν/kB
P. Košovan Lecture 4: Polyatomic ideal gas 4/31
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Vibrational contribution to thermodynamic functionsVibrational temperature: Θv = hν/kB.
Evib = NkBT 2(∂ ln qvib
∂T
)= NkB
(Θv
2+
Θv
eΘv/T − 1
)Cvib
v =
(∂Evib
∂T
)N
= NkB
(Θv
T
)2 Θv
(eΘv/T − 1)2
Note that for T →∞:
Evib → NkBT , Cv → NkB
• Universal form forheteronuclear diatomics
• Values of Θv are tabulated 0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2
Cvvib
(T
) /
Nk
B
T / ΘvP. Košovan Lecture 4: Polyatomic ideal gas 5/31
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Properties of some common diatomic molecules
Table from McQuarrie, Statistical Mechanics, University Science Books (2000)
P. Košovan Lecture 4: Polyatomic ideal gas 6/31
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Population of vibrational states
fn =e−βhν(n+1/2)
qvib, fn>0 =
∞∑n=1
e−βhν(n+1/2)
qvib= 1− f0 = e−Θv/T
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5
f n
n
Population of vibrational states in H2
Θv = 6215 K
H2, 300 KH2, 700 K
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5f n
n
Population of vibrational states in Br2
Θv = 463 K
Br2, 300 KBr2, 700 K
Population of states of H2 and Br2 at various temperatures.
P. Košovan Lecture 4: Polyatomic ideal gas 7/31
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Rotational partition function (heteronuclear diatomic)Rotational energy levels and degeneracy:
εJ =h2J(J + 1)
2I= BJ(J + 1), ωJ = 2J + 1
where we defined the rotational constant B = h2/(8π2I)
qrot(T ) =∑
J
(2J + 1)e−βBJ(J+1)
• qrot(T ) cannot be summed directly.• Analogous with Θv we define rotational temperature Θr = B/kB
• The high temperature limit (T � Θr):
qrot(T ) =
∫ ∞0
(2J + 1)e−J(J+1)Θr/T dJ
P. Košovan Lecture 4: Polyatomic ideal gas 8/31
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Approximations to qrot(T ) (heteronuclear diatomic)
• The high temperature limit (T � Θr):
qrot(T ) =
∫ ∞0
(2J + 1)e−J(J+1)Θr/T dJ =
∫ ∞0
e−J(J+1)Θr/T d{J(J + 1)}
=TΘr
=8π2IkBT
h2
• Low temperatures (T . 1.4Θr) – first few terms suffice:
qrot(T ) = 1 + 3e−2Θr/T + 5e−6Θr/T + 7e−12Θr/T + · · ·
• Intermediate temperatures: none of the above, but Euler-MacLaurin:
qrot(T ) =TΘr
(1 +
13
(Θr
T
)+
115
(Θr
T
)2
+4
315
(Θr
T
)3
+ · · ·
)
P. Košovan Lecture 4: Polyatomic ideal gas 9/31
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Euler MacLaurin expansion in more detail
b∑n=a
f (n) =
∫ b
af (n) dn +
12
(f (b) + f (a)
)+∞∑
j=1
(−1)j Bj
(2j)!
(f (2j−1)(a)− f (2j−1)(b)
)f k (a) is the k -th derivative of f (a).{Bj} are Bernoulli numbers:
B1 =16, B2 =
130, B3 =
142, · · ·
Example:
11− eα
=∞∑
j=0
e−αj =1α
+12− α
12+
α3
720+ · · ·
P. Košovan Lecture 4: Polyatomic ideal gas 10/31
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In most cases Θr � boiling temperature
Table from McQuarrie, Statistical Mechanics, University Science Books (2000)
P. Košovan Lecture 4: Polyatomic ideal gas 11/31
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Various approximations to qrot for HCl (Θr = 15.02 K)
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
0 2 4 6 8 10 12
qro
t(n
)
number of terms
T = 10 K
High T
Euler-MacLaurin
Direct sum
0
1
2
3
4
5
6
7
8
0 2 4 6 8 10 12
qro
t(n
)
number of terms
T = 100 K
High T
Euler-MacLaurin
Direct sum
0
5
10
15
20
0 2 4 6 8 10 12
qro
t(n
)
number of terms
T = 300 K
High T
Euler-MacLaurin
Direct sum
0
5
10
15
20
25
30
0 2 4 6 8 10 12q
rot(n
)
number of terms
T = 400 K
High T
Euler-MacLaurin
Direct sum
P. Košovan Lecture 4: Polyatomic ideal gas 12/31
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Rotational contribution to thermodynamic functions
0
0.1
0.2
0.3
0.4
0.5
0 2 4 6 8 10 12
f J
J
Population of rotational states in HCl at 300 K
Θr = 15.02 K
Euler-MacLaurin
High-T limit
Direct sum (4 terms)
Erot = NkBT 2(∂ ln qrot
∂T
)= NkBT + · · ·
Crotv =
(∂Erot
∂T
)N
= NkB + · · ·
Note that for T →∞:
Erot → NkBT , Cv → NkB
Population of vibrational states:
fJ =(2J + 1)e−ΘrJ(J+1)/T
qrot(T )
Jmax =
(kBT2B
)1/2
− 12≈(
kBT2B
)1/2
=
(T
2Θr
)1/2
P. Košovan Lecture 4: Polyatomic ideal gas 13/31
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Homonuclear diatomics: symmetry of wave functionSymmetry of the total wave function upon interchange of nuclei:• Bosons (integral spin): symmetric• Fermiions (half-integral spin): antisymmetric
Interchange of electrons:1. Inversion of all particles2. Inversion of just the electrons back
ψ′tot = ψtrans ψvib ψrot ψelec exclusive the nuclear part
• ψtrans, ψvib symmetric with respect to inversion• ψelec depends on symmetry of the ground state• The most common Σ+
g ground state is symmetric wrt inversion
• Only ψrot can control the symmetry of ψ′tot.
P. Košovan Lecture 4: Polyatomic ideal gas 14/31
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Implications of inversion for H2
H2 atom nuclei with spins 1/2:• 3 symmetric spin functions:αα, ββ, (αβ + βα)/
√2
• 1 anti-symmetric spin function:(αβ − βα)/
√2
• Nuclei with s = 1/2 arefermions⇒ ψtot anti-symmetricwrt interchange of nuclei
• Symmetric spin functionscouple with odd J
• Anti-symmetric spin functionscouple with even J
Rotational eigenstates:• J even: symmetric• J odd: anti-symmetric
Rigid rotor eigenfunctions for J ≤ 2. Imagesource: Wikimedia commons
• Statistical weights 1/3 for even/odd J• Parallel (ortho) and anti-parallel (para) nuclear spins in H2
P. Košovan Lecture 4: Polyatomic ideal gas 15/31
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Generaliztion of the effect of inversion on ψtot
For a nucleus with spin I there are (2I + 1) spin states.• α1, α2, · · ·α(2I+1)
• There are (2I + 1)2 nuclear spin functions in ψtot
• Anti-symmetric spin functions: αi(1)αj(2)− αi(2)αj(1)
• Total (2I + 1)(2I)/2 anti-symmetric functions• Total (2I + 1)2 − (2I + 1)(2I)/2 = (I + 1)(2I + 1) symmetric functions
General rule for symmetric Σ+g electronic states:
• Integral nuclear spin:• I(2I + 1) anti-symmetric spin functions couple with odd J• (I + 1)(2I + 1) symmetric spin functions couple with even J
• Half-integral nuclear spin:• I(2I + 1) anti-symmetric spin functions couple with even J• (I + 1)(2I + 1) symmetric spin functions couple with odd J
P. Košovan Lecture 4: Polyatomic ideal gas 16/31
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Example: acetylene H–C=C–H (linear polyatomic)
Figure from McQuarrie, Statistical Mechanics, University Science Books (2000)P. Košovan Lecture 4: Polyatomic ideal gas 17/31
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Homonuclear diatomic molecule
Integral spins:
qrot,nuc(T ) = (I + 1)(2I + 1)∑
J even
(2J + 1)e−ΘrJ(J+1)/T
+ I(2I + 1)∑
J odd
(2J + 1)e−ΘrJ(J+1)/T
Half-integral spins:
qrot,nuc(T ) = I(2I + 1)∑
J even
(2J + 1)e−ΘrJ(J+1)/T
+ (I + 1)(2I + 1)∑
J odd
(2J + 1)e−ΘrJ(J+1)/T
Rotational and nuclear partition function cannot be separated!
P. Košovan Lecture 4: Polyatomic ideal gas 18/31
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High temperature limit (T � Θr)
Applicable for (T & 5Θr):∑J even
≈∑
J odd
≈ 12
∑J all
≈ 12
∫ ∞0
(2J + 1)e−ΘrJ(J+1)/T dJ =T
2Θr
Equations for integral and half-integral spin both yield
qrot,nuc(T ) =(2I + 1)2T
2Θrcf. heteronucl. diatomic: qrot(T ) =
TΘr
which we can separate to
qrot(T ) =T
2Θr, qnuc = (2I + 1)2
We can combine both homo- and hetero-nuclear into one form:
qrot(T ) ≈ TσΘr
≈ 1σ
∞∑J=0
(2J + 1)e−ΘrJ(J+1)/T symmetry number: σ
P. Košovan Lecture 4: Polyatomic ideal gas 19/31
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In most cases Θr � boiling temperature
Table from McQuarrie, Statistical Mechanics, University Science Books (2000)
P. Košovan Lecture 4: Polyatomic ideal gas 20/31
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Hydrogen at low T is an exception
qrot,nuc(T ) =∑
J even
(2J + 1)e−ΘrJ(J+1)/T
+ 3∑
J odd
(2J + 1)e−ΘrJ(J+1)/T
Northo
Npara=
3∑
J odd(2J + 1)e−ΘrJ(J+1)/T∑J even(2J + 1)e−ΘrJ(J+1)/T
Cv(300 K) =34
Cv(ortho) +14
Cv(para)
P. Košovan Lecture 4: Polyatomic ideal gas 21/31
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Overall partition function of diatomics
q(V ,T ) = V(
2πmkBTh2
)3/2 (8π2IkBTσh2
) (e−βhν/2
1− e−βhν
)ωe,1eβDe
ENkBT
= kBT 2(∂ ln q∂T
)N,V
=52
+βhν
2+
βhνeβhν−1 − βDe
Possible further extensions:• Anharmonic vibrations• Vibration-rotation coupling• Centrifugal distorsion• Molecules with low electronic states: inclusion of more states in qelec
• Molecules with other than Σ ground state: coupling betweenelectronic and rotational angular momenta
P. Košovan Lecture 4: Polyatomic ideal gas 22/31
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Thermodynamic functions of diatomics (T � Θr)
Cv
NkB=
52
+
(hν
kBT
)2 eβhν
(eβhν − 1)2
SNkB
= ln
(2π(m1 + m2)kBT
h2
)3/2Ve5/2
N+ ln
8π2IkBT eσh2
+βhν
eβhν − 1− ln(1− e−βhν) + lnωe1
pV = VkBT(∂ ln Q∂V
)N,T
= NkBT
µ0(T )
kBT=− kBT ln
(2π(m1 + m2)kBT
h2
)3/2
− ln8π2IkBTσh2
+hν
2kBT+ ln(1− e−βhν)− De
kBT− lnωe1
P. Košovan Lecture 4: Polyatomic ideal gas 23/31
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Polyatomic ideal gas
q(V ,T ) = qtrans qrot qvib qelec qnuc
Q(N,V ,T ) =(qtrans qrot qvib qelec qnuc)N
N!
• Separation of individual degrees of freedom: rigid rotor, harmonicoscillator.
• Analogy with diatomics• Three degrees of freedom per rotation• (3n − 5) or (3n − 6) vibrational degrees of freedom
P. Košovan Lecture 4: Polyatomic ideal gas 24/31
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Vibrations in polyatomics
• Normal coordinates, normal modes• (3n − 5) or (3n − 6) independent harmonic oscillators
ε =α∑
j=1
(nj +
12
)hνJ
νj =1
2π
(kj
µj
)1/2
Θv,j =hνj
kB
qvib =α∏
j=1
e−Θv,j/2T
(1− e−Θv,j/T )
Evib = NkB
α∑j=1
(Θv,j
2+
Θv,je−Θv,j/T
1− e−Θv,j/T )
)
CvibV = NkB
α∑j=1
((Θv,j
2
)2
+e−Θv,j/T
1− e−Θv,j/T
)P. Košovan Lecture 4: Polyatomic ideal gas 25/31
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Rotations in polyatomicsLinear polyatomics (analogy with diatomics):
qrot =8π2IkBTσh2 =
TσΘr
Non-linear polyatomics – principal moments of inertia: IA, IB, IC .
A =h2
8π2IA, B =
h2
8π2IB, C =
h2
8π2IC
ΘA =8π2IAkB
h2 , ΘB =8π2IBkB
h2 , ΘC =8π2ICkB
h2 ,
Special cases:• Spherical top: IA = IB = IC• Symmetric top: IA = IB 6= IC• Asymmetric top: IA 6= IB 6= IC
P. Košovan Lecture 4: Polyatomic ideal gas 26/31
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Special casesSpherical top (ΘA = ΘB = ΘC):
εJ =J(J + 1)~2
2I
ωJ = (2J + 1)2
High T limit:
qrot =1σ
∫ ∞0
(2J + 1)2e−J(J+1)~2/2IkBT dJ
≈ 1σ
∫ ∞0
4J2e−J2~2/2IkBT dJ =π1/2
σ
(8π2IkBT
h2
)3/2
(1)
qrot =π1/2
σ
(T 3
ΘAΘBΘC
)1/2
P. Košovan Lecture 4: Polyatomic ideal gas 27/31
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Special cases
Symmetric top (ΘA = ΘB 6= ΘC):
εJK =~2
2
(J(J + 1)
IA+ K 2
(1IC− 1
IA
))J = 0,1,2, · · · ; K = J, J − 1, · · · ,−J
ωJK = (2J + 1)
High T limit:
qrot =1σ
∞∑J=0
(2J + 1)2e−αAJ(J+1)+J∑
K =−J
e−αCK 2, αj =
~2
2IjkBT, j = A,C;
qrot =π1/2
σ
(8π2IAkBT
h2
)(8π2ICkBT
h2
)1/2
=π1/2
σ
(T 3
ΘAΘBΘC
)1/2
P. Košovan Lecture 4: Polyatomic ideal gas 28/31
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gue
Special casesAsymmetric top (ΘA 6= ΘB 6= ΘC):• Very involved at quantum level.• Can be solved numerically.• Analytically solvable in the classical limit.High T limit:
qrot =π1/2
σ
(8π2IAkBT
h2
)1/2(8π2IBkBTh2
)1/2(8π2ICkBTh2
)1/2
Common formulation for all cases using rotational temperatures:
qrot =π1/2
σ
(T 3
ΘAΘBΘC
)1/2
Erot =32
NkBT , CrotV =
32
NkB, Srot = NkB ln
(π1/2
σ
(T 3e3
ΘAΘBΘC
)1/2)
P. Košovan Lecture 4: Polyatomic ideal gas 29/31
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Uni
vers
ityin
Pra
gue
Thermodynamic functions of a non-linear polyatomic
q = V(
2πMkBTh2
)3/2 π1/2
σ
(T 3
ΘAΘBΘC
)1/2(
3n−6∏j=1
e−Θv,j/2T
1− e−Θv,j/T
)ωe,1eβDe
ENkBT
=32
+32
+3n−6∑j=1
(Θv,je−Θv,j/T
1− e−Θv,j/T
)− De
kBT
− ANkBT
= ln(
2πMkBTh2
)3/2 VeN
+ lnπ1/2
σ
(T 3
ΘAΘBΘC
)1/2
−3n−6∑j=1
(Θv,j
2T+ ln(1− e−Θv,j/T )
)+
De
kBT+ lnωe,1
P. Košovan Lecture 4: Polyatomic ideal gas 30/31
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ityin
Pra
gue
Thermodynamic functions of a non-linear polyatomic
− ANkBT
= ln(
2πMkBTh2
)3/2 VeN
+ lnπ1/2
σ
(T 3
ΘAΘBΘC
)1/2
−3n−6∑j=1
(Θv,j
2T+ ln(1− e−Θv,j/T )
)+
De
kBT+ lnωe,1
SNkB
= ln(
2πMkBTh2
)3/2 Ve5/2
N+ ln
π1/2e3/2
σ
(T 3
ΘAΘBΘC
)1/2
−3n−6∑j=1
(Θv,j/T
eΘv,j/T − 1− ln(1− e−Θv,j/T )
)+ lnωe,1
Next lecture(s):• Statistical thermodynamics in the classical limit• Chemical equilibrium in dilute gases
P. Košovan Lecture 4: Polyatomic ideal gas 31/31
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