Download Polyatomic Gases Non-interacting, identical Find Each molecule has # atoms

Polyatomic Gases
Non-interacting, identical ⇒ Z = N1! Z1N
Find Z1
Each molecule has # atoms ⇒ 3# position coordinates
3# =
3
C.M.
+
+ (3# −3 − nr )
rotation
nv , vibration
nr
8.044 L14B1
MONATOMIC
Xe
DIATOMIC
HS
LINEAR TRI.
CO2
NON-LINEAR TRI.
H2 O
3
0
0
3
3
2
1
6
3
2
4
9
3
3
3
9
8.044 L14B2
C.M. Motion:
Particle in a box ∆E s kT ⇒ classical
Rotation:
(H2 νrot = 3.65 × 1012 Hz → 175 K ) ⇒ Q.M.
Vibration:
(H2 νvib = 1.32 × 1014 Hz → 6, 320 K ) ⇒ Q.M.
H = HCM + Hvib + Hrot ⇒ problem separates
8.044 L14B3
Vibration
Hvib
nv


1
1 Ki 2 
2
=
ȧi
Ki ai +
2
2 ωi
i=1 2

nv 1 dimensional harmonic oscillators, use Q.M.
Ĥψn = nψn
n = (n + 1
2 )h̄ω n = 0, 1, 2, · · ·
The energy levels are non-degenerate.
8.044 L14B4
−(n+ 1
hω/kT
2 )¯
p(n) = e
ε
∞
/
e−n/kT
n=0
hω
7
2
hω
1
5
2
hω
1
3
2
hω
1
1
2
hω
1
∞
−(n+ 1
hω/kT
2 )¯
e
=
−1
e 2 ¯hω/kT
n=0
n
∞ ¯ /kT
−hω
e
n=0
=
¯
p(n) = 1 − e−hω/kT
1
e− 2 ¯hω/kT / 1
e−¯hω/kT
n
¯
− e−hω/kT
= (1 − b)bn
8.044 L14B5
Geometric or Bose-Einstein
p(n)
n
b
1
<n> =
= ¯hω/kT
1−b
e
−1
→ e−h̄ω/kT
when kT h̄ω
8.044 L14B6
1
For kT h̄ω < n > →
2
h̄ω + 1 ¯
hω
1 + kT
··· − 1
2 kT
h̄ω
kT
1
kT
1
≈
1−2
=
1
¯
hω
h̄ω
kT
h̄ω 1 +
2
kT
kT
=
−1
h̄ω 2
< >= (< n > +
1
2
)h̄ω → kT
kT h̄ω (Classical)
→ 1
2 h̄ω kT h̄ω (Ground state)
8.044 L14B7
3
< ε>
<n>
2
kT
1
1
2
0
-1
1
2
3
kT/h ω
hω
T
8.044 L14B8
∂<>
d<n>
CV = N
= N h̄ω
∂T
dT
V
h̄ω
= Nk
kT
2
e¯hω/kT
e h̄ω/kT
h̄ω 2 −h̄ω/kT
→ Nk
e
kT
−1
2
→ Nk
kT h̄ω
(energy gap behavior)
kT h̄ω
8.044 L14B9a
CV /Nk
1
ENERGY
GAP
BEHAVIOR
2 LEVEL SYSTEM
SHOWING SATURATION
1
2
kT/h ω
3
8.044 L14B9b
High and low temperature behavior without solving the
complete problem Consider first the high T limit.
e - ε/ k T
∆ contains ∆
h̄ω states
kT
Z1 =
∞
hω
ε
e−n/kT
n=0
≈
∞
0
1 −E/kT
kT ∞ −y
kT
e
dE =
e dy =
∝ β −1
h̄ω
h̄ω 0
h̄ω
8.044 L14B10
Zvib = Z1N ∝ β −N

Uvib

1  ∂Z 
=−
= −β N (−N )β −N −1 = N kT
Z ∂β N
Cvib = N k
Next, consider the low T limit.
e - ε/ k T
⇒ consider only 2 states
kT
1
2
hω
3
2
hω
ε
8.044 L14B11
p(n = 1) ≈
−3
2 h̄ω/kT
1
e
hω/kT
−¯
=
≈
e
3 h̄ω/kT
−1
−
h̄ω/kT
e¯hω/kT + 1
e 2
+e 2
p(n = 0) ≈ 1 − e−h̄ω/kT
< E >= 12 N h̄ω 1 − e−¯hω/kT + 32 N ¯
hωe−¯hω/kT
= 12 N ¯
hω + N h̄ωe−¯hω/kT
2
h̄ω
¯
hω
∂<E>
−h̄ω/kT = N k
−h̄ω/kT
CV =
= N h̄ω
e
e
∂T
kT 2
kT
8.044 L14B12
Angular Momentum in 3 Dimensions
θ, φ)
CLASSICAL, 3 numbers: (Lx, Ly , Lz ); (|L|,
QUANTUM, 2 numbers: magnitude and 1 component
ˆ·L
ˆψ
2ψ
ˆ2 ψ
≡
L
=
l(l
+
1)h̄
L
l,m
l,m
l,m l = 0, 1, 2 · · ·
Lˆz ψl,m = mh̄ ψl,m
m = l,
l − 1,· · · − l
2l+1 values
8.044 L14B13
Specification: 2 numbers l & m → ψl,m
or
|l, m >
Molecular rotation
In general
1 2
1 2
1 2
L1 +
L2 +
L3
Hrot =
2I1
2I2
2I3
For a linear molecule
1
1 2
2
Hrot =
(L1 + L2) =
L·L
2I⊥
2I⊥
I3 = 0
L3 = I3 θ˙3
= 0
I1 = I2 ≡ I⊥
8.044 L14B214
Ĥrot
1 2
ˆ
=
L
2I⊥
Ĥrot |l, m > = l |l, m >
ε/k
2 0 ΘR
9
l= 4
1 2 ΘR
7
l= 3
it is 2l + 1 fold degenerate.
6 ΘR
5
l= 2
l = kΘR l(l + 1)
2 ΘR
0
3
l= 1
l= 0
h2
¯
=
l(l + 1)|l, m >
2I⊥
l depends on l only;
1
¯
h2
ΘR ≡
(rotational temp.)
2I⊥k
8.044 L14B15
1 −l(l+1)ΘR/T
p(l, m) =
e
ZR
ZR =
e−l(l+1)ΘR/T
l,m
For T ΘR
=
(2l + 1)e−l(l+1)ΘR/T
l
ZR ≈ 1 + 3e−2ΘR/T = 1 + 3e−2ΘRkβ
1 ∂Z
6ΘR k e−2ΘRkβ
−2ΘR /T
< >= −
=
≈
6Θ
k
e
R
Z ∂β
1 + 3e−2ΘRkβ
8.044 L14B16
CV |rot
∂<>
=N
= 6ΘR N k
∂T
= 3N k
2ΘR
T
2
2ΘR −2ΘR/T
e
2
T
e−2ΘR/T
(energy gap behavior)
For T ΘR, convert the sum to an integral.
ZR ≈
∞
0
(2l + 1)e−l(l+1)ΘR/T dl
8.044 L14B17
x ≡ (l2 + l)ΘR/T
dx = (2l + 1)ΘR/T dl
T ∞ −x
T
1
ZR ≈
e dx =
=
β −1
ΘR 0
ΘR
kΘR
1 ∂Z
(−1)(−1)Z/β
< >= −
=
= β −1 = kT
Z ∂β
Z
CV |rot
∂<>
=N
→ Nk
∂T
(classical result)
8.044 L14B18a
H = HCM + Hrot + Hvib
CV (T ) = C
|
VCM
all T
+
C |
V rot
appears at modest T
+
C |
V vib
only at highest T
8.044 L14B19a
Raman Scattering
BEFORE
εi
νi
AFTER
εf
νf
∆ε = εf - εi = h( νi - νf )
FREQUENCY CHANGES IN THE SCATTERED LIGHT CORRESPOND TO
ENERGY LEVEL DIFFERENCES IN THE SCATTERER.
WHICH ENERGY LEVEL CHANGES OCCUR DEPEND ON SELECTION
RULES GOVERNED BY SYMMETRY AND QUANTUM MECHANICS
8.044 L14B20
Example Rotational Raman Scattering
Selection rule: ∆l = ±2
∆νl↑ = −(kΘR/h)[(l + 2)(l + 3) − l(l + 1)]
= −(4l + 6)(kΘR/h)
⇒ uniform spacing between lines of 4(kΘR/h)
Il↑ ∝ number of molecules with angular momentum l
∝ (2l + 1)e−l(l+1)ΘR/T
8.044 L14B21
ROTATIONAL RAMAN SPECTRUM OF A DIATOMIC MOLECULE
BOLTZMANN FACTOR
I( ∆ν)
LEVEL DEGENERACY
∆ν
4(k ΘR/ h )
0
-6(k ΘR/ h )
8.044 L14B22
MIT OpenCourseWare
http://ocw.mit.edu
8.044 Statistical Physics I
Spring 2013
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