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A Brief Introduction to
Statistical Thermodynamics
A Web-Book by Rituraj Kalita, Dept. of
Chemistry, Cotton College,
Guwahati-781001 (Assam, India)
Preface
Ch. 1
Ch. 2 Ch. 3
Ch. 4
Ch. 5
Ch. 6
Ch.
7
Bibliog.
Background topics/ vocabulary
General topics
Advanced
(avoidable) topics
© 2006.
Copyright reserved. The book or any portion of it
can't be
reproduced/
re-published/
circulated.
Ch. 3: Molecular Partition
Function for
Gases
and their Thermodynamic Functions
3.1 Towards the Evaluation of the Molecular Partition Function q for Gases:
The molecular energy e discussed earlier is actually the complete (overall) molecular energy that is expressible (approximately) as a sum of several specific-mode molecular energy components such as translational, electronic etc. energies. Because of this fact, the (complete or overall) molecular p. f. (denoted as q) may be expressed as a product (we'll show that soon) of several such specific-mode p. f. factors (qtr, qel etc.). The basic idea is that a molecule, say an O2 molecule, at the same time, undergoes several types of molecular motions such as translational, electronic, intra-nuclear, rotational and vibrational, with at least some approximate degree of independence among these motions (talking in irritating quantum-mechanical jargon, we say that the molecular Hamiltonian h may be separated into specific-mode Hamiltonians, at least as a reasonable approximation), and so such specific-mode (translational, electronic etc.) energies and corresponding specific-mode energy-states (e.g., a translational molecular state, an electronic molecular state) can be thought of. When you've recently discussed the electronic motion and the associated electronic energy of H-atom in your quantum mechanics course (and derived that electronic energy to be of the form Z2E0/n2), you were made to overlook the fact that the H-atom may also have other (i.e., non-electronic) form of energy, particularly the translational energy (the translational energy is associated with translational motion, i.e., with the movement of the atom/molecule as a whole from one place to another). At this point, I would also like to recall an interesting incident of 'ragging' I faced as a fresher student then, during the mid-eighties, at the college where I now teach: I was asked to move my head, hand, feet, fingers and toes. I could do each of that, naturally, but then I was asked to move all these body-parts together! This came out to be totally impossible for me; but, strangely, the gas molecules always keep undergoing all the specific-mode motions simultaneously. An O2 molecule moving from one place to another (i.e., undergoing translation) is simultaneously also undergoing haphazard electron-motions, a 'to and fro' vibration in the internuclear bond and a revolving rotation of the nuclear framework!
For the non-monatomic molecules (i.e.,
those having two or more nuclei within it, e.g., O2, NO, NH3)
the molecular energy is approximated (via
Born-Oppenheimer Approximation and Rigid-Rotor-Harmonic-Oscillator
Approximation etc.) to be a sum (as
e = etr +
eel +
enuc +
evib +
erot) of a translational energy
etr, an electronic energy
eel, an (intra-)nuclear energy
enuc, a vibrational energy
evib and a rotational energy
erot, each of these specific-mode energies
being associated with a corresponding specific-mode energy state (so, the
overall molecular state f is obtained as a product of
specific-mode states as:
f = ftrfelfnucfvibfrot).
Thus, a complete (overall) molecular state for such a molecule is a combination
of a translational state, an electronic state, an (intra-)nuclear state, a
vibrational state and a rotational state. For system of such gas molecules,
q (expressed as a sum over states)
is given by q =
Ss
exp{-es
/(kT)}, and so we have:
q = Ss
exp{-e
/(kT)}= Ss
exp{-(etr+eel+enuc+evib+erot)/(kT)}
= Ss
exp{-etr/(kT)}.exp{-eel/(kT)}.exp{-enuc/(kT)}.exp{-evib/(kT)}.exp{-erot/(kT)}
As an overall state is composed of a combination of the specific-mode states, so the sum over all
the overall molecular states is same as the sum over all translational states,
all electronic states, all
(intra-)nuclear states, all vibrational states and all rotational states. So:
q
= Str-states
Sel-states
Snuc-states
Svib-states
Srot-states exp{-etr/(kT)}.exp{-eel/(kT)}.
exp{-enuc/(kT)}. exp{-evib/(kT)}. exp{-erot/(kT)}
As the specific-mode states are independent of one another, so the above five
summations could be performed independently of one another. This means
q = [Str-states
exp{-etr/(kT)}].
[Sel-states
exp{-eel/(kT)}].
[Snuc-states
exp{-enuc/(kT)}].
[Svib-states exp{-evib/(kT)}].
[Srot-states exp{-erot/(kT)}]
The sums within the square-brackets in this expression are called respectively
as translational p. f. qtr, electronic p. f. qel,
(intra-)nuclear p. f. qnuc, vibrational p. f. qvib,
rotational p.f. qrot.
So we have
q = qtr qel qnuc
qvib qrot
Notes:
(i)
A complete (overall) molecular energy-level for such a
molecule is similarly a combination of a translational level, an electronic level, an (inra-)nuclear
level, a vibrational level and a rotational level, with the degeneracy g of the
overall level equaling the product of the degeneracies of the specific-mode
energy levels, i.e., g = gtr
gel gnuc gvib grot. The
specific-mode degeneracies get multiplied thus, because every translational
state (within the overall level) may coexist with every electronic state
(therein), and so on.
(ii) The intra-nuclear (molecular) state (clearly different from the
roto-vibrational inter-nuclear state) is popularly also called just
as the nuclear state, so we're now calling it '(intra-)nuclear state'.
Soon, we'll call it just as nuclear state, and the corresponding
energy as nuclear energy!
For the monatomic molecules (i.e., those having only one nuclei within it, e.g., He, Ne, Ar, H) the molecular (i.e., atomic) energy is approximated to be a sum (as e = etr + eel + enuc) of a translational energy etr, an electronic energy eel and an (intra-)nuclear energy enuc, and so for them we get q = qtr qel qnuc (for them, rotational and vibrational motions do not exist!).
While evaluating the partition functions, it is also important to note the
convention in which the zero-values of the molecular energies are expressed.
Just as the gravitational potential energy (expressed
as Vgrav = mgh) of a piece of chalk in the teacher's hand may be measured with a
zero-value reference (i.e., h = 0, Vgrav = 0) to the classroom ground, to the nearby street, or to
the world-wide sea-level reference,
the energy of a molecule also may be expressed with
several different zero-value references. However, the widely
popular convention is that any specific-mode (i.e., translational etc.)
energy of a molecule is considered to be zero at the ground energy level (of
that mode). In contrast to this popular
convention (which is analogous to the classroom-ground reference for the piece
of chalk), another
convention particularly useful for treating chemical reactions and equilibriums
(i.e., molecule-interconversions) considers instead the (overall) ground
energy level for all neutral gaseous atoms to be at zero energy (see Ch. 4,
this convention is obviously analogous to the world-wide sea-level reference
one). In
a few coming sections, we'll try to find simple expressions for all the
specific-mode p. f. 's, using the above popular convention.
3.2 Derivation of the Expression for the Translational Partition Function (qtr):
From a quantum mechanics outlook, the
translational energy of a gas molecule enclosed within a system is nothing but
the energy of a particle in a 3-dimensional box model (with the mass of
the particle equaling the mass m of the molecule).
[Why? This is because the energy of the particle
in the 3-D box is clearly by virtue of only its translational motion (i.e.,
motion from one place to another).] Similarly,
every energy-state of the particle in a 3-D box problem implies a translational
energy state of the gas molecule. For simplicity of derivation, we choose a
cubical 3-D box (of equal sides, say L).
Note: Mass m of a gas molecule
equals M/NA, where M is the molar mass. Thus, the mass of an ordinary
(16O2) oxygen molecule
(with M = 0.032 kg mol1) is
5.314 x 1026 kg.
From quantum mechanics, we know that
the translational energy
etr of the molecule in such a cubical box
is
etr = (nx2 + ny2
+ nz2) h2 / (8 m L2), where (nx,
ny, nz) are all positive integers ranging from 1, 2, 3
onwards. Furthermore, a specific set of (nx, ny, nz)
[say the set (nx = 235, ny =
641, nz = 3726)] implies a particular
translational state.
So we have, qtr = Σtr-st
exp [etr
/ (k T)] = Σtr-st exp
[ (nx2 + ny2
+ nz2) h2 / (8 m L2 k T)]
where tr-st implies translational states, and the above sum is made over all
translational states.
Note: The above defining expression for
translational energy,
etr = (nx2 + ny2
+ nz2) h2 / (8 m L2) doesn't exactly
obey the aforesaid popular convention, as per which the ground level (with nx
= ny = nz = 1) should has energy zero. However,
as h2 / (8 m L2) is extremely small, this inconsistency is
negligible.
As every possible (positive integral)
set of (nx, ny, nz) implies one translational
state, and every translational state is indicated by one possible set of (nx,
ny, nz), so the above sum over all translational states is
identical with the similar sum over all the positive integral values of the
three numbers (nx, ny, nz). This means (in
expressions below, nx,ny,nz varies from 1 to
+a):
qtr = Σnx,ny,nz exp [
(nx2 + ny2 + nz2)
h2 / (8mL2 kT)]
= Σnx,ny,nz [exp{
nx2h2/(8mL2 kT)}]. [exp{
ny2h2/(8mL2 kT)}]. [exp{
nz2h2/(8mL2 kT)]
As each of the above three exponential
functions within the square brackets involve only one out of the three variables
(nx, ny, nz), so each of the summations over
each of the variables (nx, ny, nz) can also be
performed independently. This gives qtr as a product of three sums
as:
qtr = [nx=1Σ+a
exp{ nx2h2
/ (8mL2kT)}]. [ny=1Σ+a
exp{ ny2h2
/ (8mL2 kT)}].
[nz=1Σ+a
exp{ nz2h2
/ (8mL2 kT)}]
Expanding the three sums within the
square brackets above, it becomes obvious that all the three sums are actually
identical, differing only in the name (nx, ny, nz)
of the dummy variable. So we now have a simple expression: qtr = [nx=1Σ+a
exp{ nx2h2
/ (8mL2kT)}]3
To evaluate the sum nx=1Σ+a
exp{ nx2h2
/ (8mL2kT)}, we note that for any macroscopic system (with L at
least 0.5 mm, say) the quantity a = h2/(8mL2kT) is
extremely small, and so it can be shown that the difference between two
consecutive terms (say, the two terms with nx
= 342 and nx = 343) in the series is
much smaller than the two terms themselves. This is the condition for
approximating a sum by an integral, and so the sum nx=1Σ+a
exp{ nx2
a} can be replaced by the integral I = nx=1∫+a
exp{ nx2
a} dnx. Further, as this integral value is quite large, so it makes
little error if the lower limit is replaced by nx = 0 instead of nx
= 1. So we have: I = nx=1∫+a
exp( nx2
a) dnx = nx=0∫+a
exp( nx2
a) dnx = 0∫+a
exp( a s2) ds.
Note: The sum [nx=1Σ+a
exp{ nx2
a}] = [nx=1Σ+a
exp{ nx2
a} Δnx], where
Δnx = 1. So wasn't there a
Δnx hidden in the sum (which got
converted to dnx)? (If not interesting, don't bother about this
math-insight!)
We know that value of the standard
integral 0∫+a
exp(as2)ds =
(1/2) (p/a)1/2. So we have:
qtr = [0∫+a
exp( a s2) ds]3
= [(1/2)(p/a)1/2]3 = (1/8)(p/a)3/2
= (1/8) [(8pmL2kT)/h2]3/2
= 81/2[(pmkT)3/2 /
h3] L3 = 23/2[(pmkT)3/2
/ h3] L3 = [(2pmkT)3/2
/ h3] L3
As L3 is nothing but the volume V of the system, so we have qtr
= [(2pmkT)3/2 / h3]
V
Though derived for the system within a cubical-box shaped enclosure only, this
is the generally valid expression for qtr. So we have:
qtr = (2pmkT)3/2
V / h3
Notes: (i) Making a substitution v =
a s2, we could get I = (1/2) a1/2
v=0∫+a
v1/21 exp(v)
dv (please verify!). The integral 0∫+a
vn1 exp(v)
dv (where n is a positive real number) is known in mathematics as
the gamma function, G (n).
We know that G (2)
= G (1) = 1 and
G (1/2) = p1/2.
[For n > 1, this function obeys the recursion
relation G (n) = (n1)
G (n1) which gives, for n = 2, 3, 4
etc. natural-number values of n,
G (n) = (n1)!].
So we have, I = (1/2) a1/2
w=0∫+a
w1/21 exp(w)
dw = (1/2) a1/2 p1/2
= (1/2) (p/a)1/2.
(ii) Take care to express V in m3 (SI) or cm3 (CGS) unit
(not dm3), and m, k & h also in the SI or CGS unit. (iii) In a
problem, the volume V might have to be found from the P-T data, using the ideal
gas equation. Thus, for 2 moles of He gas (obviously here m = {(4.002/6.022) x 1026}
kg at P = 1 atm and T = 298 K, V = (2 mol) x (0.08206 atm L K1
mol1) x (298 K) / (1 atm) =
48.91 L = 4.891 x 102 m3.
So, qtr = (2 x 3.142 x 6.646 x 1027
kg x 1.381 x 1023 J K1
x 298 K)3/2. (4.891 x 102
m3) / (6.626 x 1034
J s)3 = 3.788 x 1029.
To perform this sort of complicated calculations, some software-package
calculators such as the free package Assam-Calcu
are the most suitable tools. (iv) The expression for qtr may also be written as qtr
=
V/Λ3 where
Λ = h / (2pmkT)1/2
is called the thermal de Broglie wavelength for the gas system (Λ
obviously has the dimension
of length: in the example in (iii), its value is
5.054 x 1011
m = 0.5054 A0). [Please check!]
3.3 Derivation of the Expression for the Electronic Partition Function (qel):
The electronic energy levels eel
and electronic degeneracies gel for any molecule may be
obtained (from quantum mechanics) via solution of its molecular electronic
Schrφdinger equation. Having obtained these energy levels and degeneracies,
the electronic partition function qel is expressed (and calculated)
simply from the following, basic, 'sum over levels' form:
qel = gel,0 exp{
eel,0
/(kT)} + gel,1 exp{
eel,1
/(kT)} + gel,2 exp{
eel,2
/(kT)} + .........
(where
eel,0 ,
eel,1 ,
eel,2 are respectively the ground, 1st-excited, 2nd-excited
energy-level energies and gel,0 ,
gel,1 , gel,2 are respectively
their corresponding degeneracies.)
Using the above-mentioned popular convention, the
eel are expressed with the ground-level
electronic energy as zero (i.e., eel,0
= 0), so the actual expression looks as follows:
qel = gel,0 + gel,1 exp{
eel,1
/(kT)} + gel,2 exp{
eel,2
/(kT)} + .........
In most cases, there arises a great simplification to this
expression. For most of the stable gas molecules (e.g., Ne, Ar, H2, N2,
O2, HCl, HI, CO etc.) the excited electronic energy levels lie quite high
(i.e., compared to the ground one), so that at usual, not-too-high temperatures
(say, 300 K or even 1000 K), the exp{eel
/(kT)} values for any of the excited levels is practically zero (as
eel >> kT), and so the excited-level
contribution-terms to qel are negligible. So in such cases qel
usually equals almost gel,0 (the degeneracy of the ground electronic level):
qel = gel,0
(An exception to this generalization is NO, for which eel,1
= 0.015 eV only: however, here also eel,2
, eel,3 etc. lie quite high.) Furthermore,
for many of those stable gases (e.g., Ne, Ar, H2, N2, HCl,
HI
etc.) the ground electronic level (generally a singlet 1S+
one) is non-degenerate (i.e., gel,0 = 1) so that qel
equals simply 1 (a notable exception is O2, for which gel,0
= 3 this level is a triplet 3S-g
one). The excited levels lie similarly high even for the atomic alkali metal
gases (e.g. Na, Cs etc. atomic gases) and for atomic hydrogen (i.e., atomic H),
though for them gel,0 = 2 (a doublet term): so for them even at
1000-2000 K temperatures, qel equals practically 2. Similarly, also
for electron gas [e (g),
available within ionized gases] qel = 2.
Note: For NO at 300 K, eel,0 = 0, eel,1
= 0.015 eV and gel,0 = gel,1 = 2 while the 2nd excited and
still upper levels hardly contribute (as for them
eel >> kT), so we get:
qel = 2 + 2 exp{(0.015 x 1.609 x
1019 J)
/(1.381 x 1023 J K1
x 300 K)} = 3.117
3.4 Derivation of the Expression for the Nuclear Partition Function (qnuc):
The (intra-)nuclear energy levels for a molecule are
always such that the upper (i.e., the excited) nuclear levels lie extremely
above the ground level, so that at any temperature of chemical interest the
excited nuclear levels are hardly occupied, and thus the contributions of the
excited nuclear levels to the nuclear p. f. is negligible. Considering the
universally accepted convention of the ground level nuclear energy being taken
as zero, we so get qnuc = gnuc,0
(where gnuc,0 is the degeneracy of the ground nuclear level).
Note: For an atom (i.e., monatomic molecule) with
nuclear-spin number I (I = ½ for H, 1 for 14N),
gnuc,0 = 2I+1. For a diatomic molecule with atoms a & b, gnuc,0
= (2Ia + 1)(2Ib + 1).
However we have an even more simplifying, but nevertheless
useful*, expression for qnuc. We note that the ground nuclear
degeneracy gnuc,0 of an atom never actually changes in any physical
or chemical process (but simply combining into their product during formation of
a molecule). So in
any calculation about change of a thermodynamic property for any physical or chemical process (say that of
entropy change for a reaction, ΔSreac) it makes no difference to the
calculation-result
if nuclear p. f. of all the atoms and molecules are taken simply as 1. This idea
gives rise to the very popular, simplifying convention among chemists, according
to which qnuc
= 1 (for any atom/ molecule/ ion).
* Notes: (1) This simplifying convention is
obviously not useful for calculating the absolute value of entropy (the
so-called third-law entropy) of a system (in contrast to calculating an
entropy-change). (2) With this simplifying
convention, the relation
q = qtr qel qnuc
qvib qrot
(for non-monatomic molecules) obviously becomes simply q = qtr qel
qvib qrot.
3.5 Derivation of the Expression for the Vibrational Partition Function (qvib):
At this point it should be repeated that qvib
and qrot exists only for non-monatomic molecules. Let us now evaluate
qvib for the simplest case of diatomic molecules, using the
rigid-rotor
harmonic-oscillator model. According to this simple model (please note that
this model is rather an approximate one), a diatomic
molecule (e.g., CO)
has a characteristic frequency of vibration ν,
equaling* (2p)1(K/m)½.
From quantum mechanics of harmonic oscillators, we further know that the
vibrational energy levels
evib are given by
(v + ½)hν
(where v = 0, 1, 2, ... etc.), each of these levels being non-degenerate
(i.e., gvib = 1). However, considering the popular convention of the
ground (v = 0) vibrational level having the zero energy, the vibrational energy
levels are to be re-assigned as:
evib = vhν
(with v = 0, 1, 2, 3, ...... etc.)
* Note: To clarify further, let us note that here K is the
(oscillator) force constant
[equaling the second derivative (d2U/dR2)Re
of molecular electronic energy U w.r.t. internuclear distance R at R = Re,
the equilibrium internuclear distance] and
m is the reduced mass equaling
(mamb)/(ma+mb), where ma & mb are the masses of
the constituent atoms a & b respectively. Each of ν,
K and m is a molecular constant for a given
(diatomic) molecule (i.e., say, for the CO molecule). Frequency ν
is sometimes also expressed in terms of wavenumber
ύ (ύ = 1/λ, λ being the wavelength,
ύ being in cm1
or m1 unit).
So the vibrational partition function is qvib =
v=0Σa
gvib,v exp {
vhν /( kT )}
= v=0Σa
[exp {
hν /( kT )}]v
[as
degeneracy gvib,v
= 1]
= 1 + y + y2 + y3 + ......
[where we take y = exp {
hν /( kT )}]
= 1/(1 y)
[using the well-known resulting expression for this series for |y| < 1]
= 1/ [1 exp{
hν /( kT )}]
[putting the value of y]
Here ( hν / k ) is another characteristic
molecular constant with the dimension of temperature. This quantity is called
the characteristic vibrational temperature, denoted as
θvib.
Thus we get the following simple expression for
vibrational p. f. of diatomic molecules:
qvib
= 1/ [1 exp{
hν /( kT )}]
= 1/ [1 exp{
θvib / T}]
For example, let
us consider H2 gas at 2000 K temperature. If it is given that here
the force
constant K is 554.4 N m1 [we may
easily get the reduced mass m for
H2 as:
m =
{(1.6678 x 1027
x 1.6678 x 1027)/(1.6678 x 1027
+ 1.6678 x 1027)}
kg
= 8.369 x 1028
kg, please check it yourself], then ν =
(2p)1(K/m)½
= 1.295 x 1014
s1 and θvib = hν /
k = 6215 K.
Thus we have qvib =
1/ [1 exp{
6215 K / 2000 K}] = 1.0468.
For polyatomic molecules (atomicity a > 2), there exists
several characteristic frequencies of vibration
νi and several corresponding
characteristic vibrational temperatures
θvib,i
corresponding to the several normal modes of vibration i [the number of
these normal modes equal (3a5) for
linear molecules and (3a6) for non-linear
molecules, where the atomicity a is the number of atoms in the molecule].
The vibrational partition function here is a product of the factors from each of
the normal modes. So for polyatomic molecules we get:
qvib
= 1/ Πi [1
exp{
hνi /( kT )}]
= 1/ Πi [1
exp{
θvib,i / T}]
Note: We see that for most molecules, qvib values are
at most marginally more than 1 (in other cases practically equal to 1), and so
is generally immaterial in contributing to q and to thermodynamic functions.
3.6 Derivation of the Expression for the Rotational Partition Function (qrot):
Let us start with evaluation of qrot for the
simplest case of heteronuclear diatomic molecules (e.g., HCl, CO, HI, and even
HD or O16O18 but not
simple H2 which is homonuclear). Considering the aforesaid
rigid-rotor
harmonic-oscillator model, we get that the rotational energy levels are:
erot = J(J+1) h2
/ (8π2I ) with degeneracy grot =
(2 J+1) where J = 0, 1, 2, 3, .....
As the above expression obeys the
criterion that the ground rotational level (J = 0 level) has zero energy, so it is
already consistent with the aforesaid popular convention (i.e., it can be safely
used to find qrot).
Note: Here I is the moment of inertia (a
characteristic molecular constant) of the
diatomic molecule equal to
m Re2 , where
m is the aforesaid reduced mass
(mamb)/(ma+mb)
and Re is the equilibrium internuclear distance (popularly known as
the bond-length).
So the rotational partition function is qrot =
J=0Σa
(2J+1)
exp {
J(J+1)h2 / (8π2I kT )}
Here h2/(8π2Ik) is another characteristic
molecular constant with the dimension of temperature. This quantity is called
the characteristic rotational temperature, denoted as
θrot.
Also, h/(8π2Ic) is another
molecular constant with the dimension of wavenumber, known as the spectroscopic
rotational constant, B [so that h2/(8π2I)
= Bhc and
θrot
= Bhc/k]. Thus we get:
qrot =
J=0Σa
(2J+1)
exp {
J(J+1)θrot
/ T }
There are several approximate approaches to get simpler
expressions for this sum. The simplest approach, valid for
(θrot
/ T) << 1 (i.e., for T >>
θrot)
uses the idea that under the condition T >>
θrot,
the above sum could be transformed to the integral I stated below:
I = J=0∫+a
(2J+1) exp {
J(J+1)θrot
/ T } dJ. Using the transformation v = J(J+1)θrot
/ T we get dv = (2J+1)(θrot
/ T ) dJ or (2J+1) dJ = (T/θrot)
dv. As at J = 0, v = 0 and at
J = + a, v = +a, we
finally get: qrot = (T/θrot)
v=0∫+a
exp(v) dv = (T/θrot)
[as, clearly, 0∫+a
exp(v) dv = 1].
So, we get the following simplest, most used expression for this qrot:
qrot =
T/θrot =
(8π2I kT / h2)
= kT/(Bhc)
[valid for T >> θrot]
Above expression is better than 1%
accurate for T > 33 θrot.
For T not so extremely larger than
θrot
but
(θrot
/ T) still
smaller than 0.7, the four-term
Mulholland expression
[qrot = (T/θrot)
{1 + (1/3).(θrot/T)
+
(1/15).(θrot/T)2
+
(4/315).(θrot/T)3}] is
good enough.
However for
(θrot
/ T) ≥ 0.7,
the following direct 4-term summation is sufficiently accurate:
qrot = 1 + 3 exp (
2 θrot
/ T )
+ 5
exp (
6 θrot
/ T )
+ 7
exp (
12 θrot
/ T )
(Please
check this expression: i.e., do find out wherefrom the above numbers 3, 2, 5, 6, 7 &
12 appear!)
It
will be shown [from a detailed discussion of the
symmetry-interaction of
their intra-nuclear and rotational states] that for the homonuclear diatomic
molecules (e.g., H2, O2, N2) qrot
is:
qrot =
T/(2θrot)
= 8π2I kT / (2h2)
= kT/(2Bhc)
[valid for T >> θrot]
Introducing the symmetry factor
s
the value of which is 2 for homonuclear diatomic molecules and 1 for heteronuclear
ones, we get the following general relation for
T >> θrot:
qrot =
T/(sθrot)
= 8π2I kT / (sh2)
= kT/(sBhc)
[valid for T >> θrot]
The above
relations are valid even for linear polyatomic (i.e., with atomicity a > 2)
molecules (for CO2,
s is 2, not 1 can you tell why?).
For non-linear polyatomic molecules, the analogue
of the above relation (obviously valid only for some sufficiently large values of T) is:
qrot = (p0.5/s).{83
π6 (Ixx Iyy Izz)
k3 T3
/ h6}0.5
= (p0.5/s).{T
3 / (θrot-x θrot-y θrot-z
)}0.5
(here Ixx, Iyy and Izz are the 3 moments of
inertia along the 3 perpendicular directions,
and
s
= 3 for NH3, 12 for CH4
etc.
Guess defining expressions for
θrot-x
, θrot-y and θrot-z
yourself!)
As a simple
example, it is known that for ordinary O2 (i.e.,
for O2 containing only the O16 isotope)
θrot
= 2.07 K. So at 298 K, the condition T
>> θrot is obeyed, and we
get qrot = T/(sθrot)
where s
= 2 (as
ordinary O2
is homonuclear). So,
here,
qrot =
71.98. We may also get that for O2, the
spectroscopic rotational constant B = kθrot/(hc)
[from relation
θrot
= Bhc/k] is obviously 143.9 m1 = 1.439 cm1. Also, using
h2/(8π2Ik)
=
θrot
we get that here the moment of inertia I =
mRe2
equals I = h2/(8π2kθrot)
= 1.945 x 1046 kg m2, so that (we know that for O2,
m is obviously
32*32/(32+32) a.m.u. = 2.657 x 1026 kg) the bond-length
Re = √( I/m)
= 0.8556 x 1010 m = 0.8556 A0. Thus, from knowledge of
θrot
or of B, even the bond-length value of a diatomic molecule may be calculated
(and vice-versa)!
To calculate or to compare
populations in two rotational levels (irrespective
of the molecular occupation of other-mode levels)
the (mode-specific form of the) Boltzmann distribution law (or its
corollary about ratio of populations) comes handy. Thus for HCl gas for which
θrot
= 15.02 K, the ratio of populations at J = 1 (the 1st excited) and J = 3 (the
3rd excited) rotational levels at a temperature T = 300 K are (see section 2.2
for the background theory):
population-ratio NJ=3/NJ=1 = (gJ=3/gJ=1). exp{ (εJ=3 εJ=1)/(kT)}
= {(2x3+1)/(2x1+1)}. exp{ (3x4 θrot/
T 1x2
θrot
/ T)}
= {7/3}. exp{ (12 θrot/T
2
θrot
/ T)}
= (7/3). exp(
10 θrot
/ T ) = (7/3). exp (
150.2
K / 300 K) = 2.333 x 0.6061 = 1.414
Note: Have you noted that a upper rotational level is here more
(1.414 times more) occupied than a lower one! This can happen because the
degeneracy of the upper level is more (i.e., gJ=3 is 7 while gJ=1
is 3).
3.7 Expressions
for the Thermodynamic Functions (U, P and S) for Gases:
Let us now find what sort of expressions for the thermodynamic functions U, P & S of gases arise from the above knowledge about qtr, qel, qnuc, qvib, and qrot (as q = qtrqelqnuc for monatomic gases and q = qtrqelqnucqvibqrot for non-monatomic gases). In doing this, we'll first find generalized expressions for them in terms of the specific-mode partition functions and then apply them for specific examples of the commonly occurring gases. As q is a function of V & T, and the said thermodynamic functions are expressed in terms of N, V, T & q, so this procedure ultimately gives us these said thermodynamic quantities as just functions of N, V & T (while N, V & T are treated as independent variables).
3.7.1 Expression for energy U (with
corollary-expressions for molar heat capacities)
For non-monatomic gases, q =
qtrqelqnucqvibqrot and so
ln q = ln qtr + ln qel + ln qnuc + ln qvib
+ ln qrot. Using U = NkT2(∂ lnq /∂T)V, we get
U = NkT2(∂ lnqtr/∂T)V + NkT2(d lnqel/dT)
+ NkT2(d lnqnuc/dT) + NkT2(d lnqvib/dT)
+ NkT2(d lnqrot/dT)
[as qel, qnuc, qvib, and qrot
are independent of V, so the criterion about constancy of V is omitted in
these four corresponding
differential expressions]
So, naturally, the translational, electronic, nuclear, vibrational and rotational
contributions to U are defined
as:
(i) translational energy Utr = NkT2(∂ lnqtr/∂T)V
(ii) electronic energy
Uel = NkT2 d lnqel/dT
(iii) nuclear energy Unuc = NkT2 d lnqnuc/dT
(iv) vibrational energy
Uvib = NkT2 d lnqvib/dT
(v) rotational energy Urot = NkT2 d lnqrot/dT
(where, obviously, U = Utr+Uel+Unuc+Uvib+Urot)
As
qtr
= [(2pmkT)3/2 / h3]
V, so ln qtr
= ln [V(2pmk)3/2/
h3]
+ ln T3/2
[where (2pmk)3/2/
h3 is obviously a constant]
So (at constant V),
(∂ lnqtr/∂T)V
= 0 + d
ln T3/2
/dT =
(3/2)
d
ln T
/dT =
(3/2).(1/T)
so that Utr = NkT2 (3/2).(1/T) = (3/2)
NkT i.e.,
Utr = (3/2) NkT
As generally qel = gel,0 (i.e., a constant), so,
generally,
Uel
(= NkT2 d lnqel/dT)
= 0.
Similarly using either qnuc = gnuc,0 or the conventional qnuc =
1, we (always) get
Unuc = 0
[This
zero value of Unuc is consistent with the convention of avoiding Unuc
via use of
q =
qtrqelqvibqrot]
For usual diatomic gases, applying relation
qvib = 1/ {1
exp( θvib / T )}, we get
ln qvib =
ln {1 exp( θvib
/ T )}, and so [remember that
Uvib = NkT2 d lnqvib/dT]
d lnqvib/dT
=
[1/ {1
exp( θvib / T )}].
{ exp( θvib
/ T )}. ( θvib). (
1 / T2 )
=
( θvib / T2
).
[exp(
θvib / T ) / {1 exp(
θvib / T )}]
=
( θvib / T2
).
[1
/ {exp( θvib / T )
1}]
So for
such
diatomic gases
Uvib =
Nkθvib
/{exp( θvib/ T )
1}
(≈ 0 for T <<
θvib)
For polyatomic gases also,
Uvib
≈ 0
at a not-so-high temperature T.
For common diatomic (or linear-molecule) gases
with T >> θrot,
qrot = T/(sθrot)
so that
ln qrot = lnT
ln(sθrot)
This gives
d lnqrot/dT
= 1/ T ,
ln (sθrot)
being a constant. So, Urot = NkT. However, for non-linear gases
T >>
θrot, we get
qrot = (constant). T3/2
which gives Urot = (3/2) NkT
So we get: For linear-molecule
gases Urot = NkT but for non-linear ones Urot =
(3/2) NkT
For monatomic gases, q = qtrqelqnuc and U = Utr+Uel+Unuc. [Here Uvib & Urot (as well as qvib & qrot) simply do not exist.]
So for Uel ≈ 0 & Uvib ≈ 0 (Uvib ≈ 0 for not-so-high T), U ≈ (5/2) NkT for linear-molecule non-monatomic gases but U ≈ 3 NkT for non-linear ones. However, for monatomic gases with Uel ≈ 0, U ≈ (3/2) NkT. Talking in a per-mole basis, we thus see that the molar energy Umol ≈ (3/2) RT for monatomic gases, Umol ≈ (5/2) RT for linear-molecule non-monatomic gases but Umol ≈ 3 RT for non-linear ones (provided Uel ≈ 0 & Uvib ≈ 0). [To get the per-mole expressions, remember the fact that NAk = R , where NA is the Avogadro number.]
The molar heat
capacity at constant volume, Cv,mol is nothing but (dUmol/dT)V.
So we may easily get that
Cv,mol ≈ 1.5 R
for monatomic gases, Cv,mol ≈ 2.5 R for
linear-molecule non-monatomic gases but Cv,mol ≈ 3 R
for non-linear ones (provided
Uel ≈ 0 & Uvib
≈ 0).
The values of the molar heat capacity at constant pressure, Cp,mol
and the ratio γ may be
obviously obtained from the relations
Cp,mol
= Cv,mol
+ R &
γ
= Cp,mol
/ Cv,mol.
Note: Please try to show that for monatomic gases with Uel
≈ 0,
γ
= 1.67
3.7.2 Expression for pressure P (with
corollary-expression for enthalpy H)
We see above that
qtr = (2pmkT)3/2 V/ h3
is obviously of the form V. ftr (T) while
each of qel, qnuc, qvib
and qrot are either constants
or functions of T only, so q is a function of only T & V in the specific
form V. f
(T) where f (T) is a function
of T only (as mentioned earlier, in Sec 2.3). This
gives that ln q = ln V + ln f (T). As P = NkT (d lnq /dV)T, we get:
P = NkT (1/V + 0) = NkT/V. So, i.e., for all gases we get
P = NkT / V or
PV = NkT.
It can be easily
realized that P is wholly translational in the exact sense (unlike U or
S) i.e., P = Ptr or in other words
there's no non-translational contribution to P (i.e., Pel, Pnuc,
Pvib, Prot = 0). This may be understood as follows: Using q = qtrqelqnucqvibqrot,
P =
NkT (d lnq /dV)T
=
NkT (d lnqtr /dV)T +NkT (d lnqel /dV)T + NkT (d lnqnuc /dV)T
+ NkT (d lnqvib /dV)T + NkT (d lnqrot /dV)T.
we have only Ptr = NkT (d lnqtr /dV)T = NkT / V
while the other four terms above are zero. So P = Ptr.
The relation PV = NkT found above immediately gives enthalpy H (= U + PV) = U + NkT (with, obviously, molar enthalpy Hmol = Umol + RT).
This relation PV = NkT which is identical with the experimentally observed ideal gas relation PV = nRT = NkT [as n = N/NA so that Nk = nNAk = nR] also means the constant k found in the defining relation q = Sl gl exp {-el /(kT)} for q is nothing other than the Boltzmann constant k (= 1.381 x 1023 J K1) found in the ideal gas relation PV = NkT.
3.7.3 Expression for entropy S (with
corollary-expression for Helmholtz free energy A)
Starting with the relation q =
qtrqelqnucqvibqrot
for non-monatomic gases and using
S = Nk ln(q/N)
+ U/T
+ Nk
, we get:
S
= Nk ln qtr + Nk ln qel + Nk ln qnuc + Nk ln qvib
+ Nk ln qrot Nk ln N
+ (Utr+Uel+Unuc+Uvib+Urot)
/ T + Nk
In this expression, we see that if only the translation motion would have
existed, i.e., if q would have equaled just qtr, then S would have
been (Nk ln qtr Nk ln N + Utr
/ T + Nk). [Here Utr/T = 1.5 Nk] So we call this quantity as Str, the
translational contribution to entropy (i.e., translational entropy). Definition
of the other four contributions are rather obvious:
(i)
electronic entropy Sel =
Nk ln qel + Uel
/ T (ii) nuclear entropy
Snuc =
Nk ln qnuc + Unuc
/ T (iii) vibrational entropy Svib
=
Nk ln qvib + Uvib
/ T
(iv) rotational entropy Srot =
Nk ln qrot + Urot
/ T
[where, obviously, S = Str+Sel+Snuc+Svib+Srot
with Str = Nk ln (qtr / N) + Utr
/ T + Nk]
The above general
expression for Str attracts some special interest, a simplified
per-mole form of it being called by a special name. To arrive
at that per-mole relation, we take [noting that Utr
/ T = (3/2) Nk ]
Str = Nk ln (qtr / N) + (3/2) Nk + Nk =
Nk ln (qtr
/ N) + (5/2) Nk
For one mole of the gas, N = NA (with NAk = R) and transl.-entropy
is Str,mol. So we have:
Str,mol
=
2.5 R + R ln (qtr
/ NA). Putting
qtr = (2pmkT)3/2 V/ h3
where m = M/NA we get
(here M is the molar mass of the gas):
Str,mol
=
R [2.5 + ln {(2pMkT)3/2 V/
(NA5/2
h3)}]
This means,
Str,mol
= R
[2.5 + ln {(2pk)3/2
/ (NA5/2
h3)} +
1.5 ln M + ln V + 1.5 ln T]. This is
the Sackur-Tetrode equation, where
ln {(2pk)3/2
/ (NA5/2
h3)}
is a constant (its value in SI
unit being
the number 16.016).
The Sackur-Tetrode
equation could be written also in an alternative form, involving the pressure P
instead of the volume V (via use of PV = NkT), and involving the relative
molecular mass Mr instead of the molar mass M. To get this form,
we note that for one mole V = NAkT/P and that m = Mr mo
(where mo is the atomic mass unit - a.m.u., equal to 1.6606 x 1027
kg, i.e., 1/12 -th of the mass of a C-12 atom). Thus the
relation
Str,mol =
R [2.5 + ln {(2pmkT)3/2 V/
(NA h3)}]
gives:
Str,mol =
R [2.5 + ln {(2pMr
mo kT)3/2 NAkT/
(PNA h3)}]
=
R [2.5 + ln {(2pmo)3/2
( kT)5/2 /
h3}
ln P +
ln Mr]. Putting P = Po (P/Po) where Po
is the standard reference value of 1 atm (i.e., 1.013 x 105 N m2),
we get ln P = ln Po + ln (P/Po). So,
Str,mol =
R [2.5 + ln {(2pmo)3/2
( kT)5/2 /
h3}
ln
Po
ln (P/Po) +
ln Mr]
=
R [2.5 + ln {(2pmo)3/2
k5/2 / (Poh3)}
+ (5/2) ln T
ln (P/Po) +
ln Mr]
or, Str,mol
=
R [2.5 + ln {(2pmo)3/2
k5/2 / (Poh3)} + 1.5 ln Mr
ln (P/Po) + 2.5 ln T ]
Here also ln {(2pmo)3/2
k5/2 / (Poh3)}
is a constant (its value in SI unit being 3.6639).
From Sel =
Nk ln qel + Uel
/ T we find that for the usual case of qel = gel,0
(along with
Uel = 0), the electronic entropy Sel = Nk ln gel,0
(so, if gel,0 = 1, Sel = 0). On the per-mole basis, Sel,mol
= R ln gel,0.
Similarly, as always Unuc = 0, so Snuc = Nk gnuc,0
in the exact sense; however, if we are interested only in the entropy change
in the usual physical/ chemical processes, the convention qnuc = 1
giving simply Snuc = 0 is perfectly good enough.
Svib
and Srot (as you must have understood by now) arise only for
non-monatomic gases. Using the diatomic-gas Uvib expression Uvib
= Nk θvib
[1
/ {exp( θvib / T )
1}],
we get that for diatomic gases
Svib =
Nk
ln [1/{1 exp(
ζvib)}] + Nk ζvib.
[1
/ {exp(ζvib)
1}]
where
ζvib = θvib
/ T. However for the usual values of temperature with T << θvib
, ζvib is very large, giving Svib
≈ 0
so there's hardly much need to
calculate Svib.
Similarly for the polyatomic gases also, in the usual cases,
Svib ≈ 0.
The rotational entropy Srot, however, evokes much more interest
because of its non-negligible value. For diatomic and other linear
molecules that obey qrot = T/(sθrot)
and thus Urot = NkT, we get: Srot
= Nk
ln{ T/(sθrot)
} + Nk = Nk
[ ln{ T/(sθrot)
} + 1],
the molar value being Srot,mol
= R.[ln{ T/(sθrot)
} +
1]. It can be easily shown that (try it yourself) that for
non-linear molecules at sufficiently enough temperatures,
Srot =
Nk [ ln qrot
+ 1.5].
The
mode-specific contribution to A are defined with the help of those for U and S.
Thus
Atr
= Utr
TStr =
NkT ln (qtr / N)
NkT. The other contributions to A are as follows:
Ael
=
NkT ln qel, Anuc
=
NkT ln qnuc, Avib
=
NkT ln qvib, Arot
=
NkT ln qrot.
Let us now calculate the molar
translational and the molar rotational entropy of HCl at 298.15 K temperature
and 1 atm pressure, given that
for HCl, θrot = 15.3 K (you may note that for this case Sel
and Svib are practically zero). Using PV (= NkT) = nRT, we get
V = (1mol x 0.08206 L atm K1mol1
x 298.15 K/ 1atm) = 24.466 L = 2.4466 x 102
m3
Using the Sackur-Tetrode equation, the molar
translational entropy is (M
= 36.5 x
103
kg):
Str,mol
=
R [2.5 + ln {(2pk)3/2
/ (NA5/2
h3)} +
1.5 ln M + ln V + 1.5 ln T]
=
R [2.5 + 16.016 + 1.5 ln
(36.5 x
103)
+ ln (2.4466
x 102)
+ 1.5 {ln (298.15)}]
= 8.314 J K1mol1.
[2.5
+ 16.016 4.9657 3.7105 + 8.5464]
= 152.9 J K1mol1
Similarly, using Srot,mol = R [ln
qrot + 1] =
R [ln{ T/(sθrot)
} +
1] for diatomic gases, we get (for HCl being heteronuclear,
s = 1):
Srot,mol = R. [ln
(298.15 K / 15.3 K) + 1]
= 8.314 J K1mol1.
[ln 19.487 + 1] = 33.00
J K1mol1
Note: Please cross-check the
above value (~153 J K1mol1)
of Str,mol
using
the alternative Sackur-Tetrode
form [in
doing so use Mr = 36.5, (P/Po) = 1 and T =
298.15].
3.8 Statistical-Thermodynamic Calculation of Entropy of Gases in Contrast to its
Calorimetric Measurement -- The Concept of Residual Entropy of Solids at 0 K:
Similar to the Sackur-Tetrode calculation of molar translational entropy, the absolute (actual) entropy per mole of a (pure) nearly-ideal gas may be calculated (Subsection 3.7.3 above) by calculating its various components, taking care to include the (generally ignored) nuclear component (~ R.gnuc,0). On the other hand, the difference DSmol = Smol(T)Smol(0) between the molar entropy Smol(T) of a pure gas at the given temperature T (Kelvin), and the entropy Smol(0) of its pure crystalline solid form at 0 K temperature (0 K in only a practical sense, as exactly absolute zero temperature can't be attained), may be determined calorimetrically via measurements of the molar heat capacity (at constant pressure) Cp at different temperatures in its various (solid, liquid, gas etc.) phases, and of the molar enthalpies DHmol of phase transitions: the molar entropy change DSmol is a sum of several integrals of (Cp/T).dT, throughout the complicated transition of the solid at 0 K into the nearly-ideal gas at T K, plus the familiar phase-transition contributions (DHmol/Tph-tr) to this entropy. However, the third law of thermodynamics states that the entropy of a pure perfectly crystalline solid at absolute zero temperature is zero, implying thereby that the aforesaid Smol(0) = 0 so that DSmol = Smol(T), and thus meaning that the aforesaid calorimetric determination of DSmol also determines the molar entropy Smol(T) of the gas at the given temperature. Thus there is a provision for comparison of the gas-phase entropy predicted by statistical thermodynamics on one hand, and that determined by calorimetric measurements on the other.
However for some substances, such as carbon monoxide, it is seen that the statistical-thermodynamically predicted and the calorimetrically measured values of the gas-phase entropy do not exactly match. In such cases, the discrepancy that arises invariably involves a somewhat lower value for the calorimetrically measured entropy. The obvious explanation for this is that the entropy of the crystalline solid phase of that substance at the absolute zero temperature {i.e., Smol(0) above} is not exactly zero (thus apparently going against the third law). Such a non-zero, positive entropy at the absolute zero temperature for the crystalline solid phase of a substance is called its residual entropy. Actually, the residual entropy arises because of a disorder or randomness of the molecular arrangement within the crystal lattice of some substances, even at temperatures approaching absolute zero, and implying that such crystals are not exactly perfect ones. Thus the third law of thermodynamics is not actually being disobeyed in such cases!
As
an example, carbon monoxide is found to crystallize with the CO molecules
linearly oriented in its crystal lattice. But during crystallization, many CO
molecules were found to align themselves in the direction exactly opposite to
that for the other CO molecules (figure 3.1).
Figure 3.1: The random linear alignments of molecules in crystalline CO
Thus there is a disorder or randomness in the crystal lattice -- as there is freedom about which of the CO molecules would be oppositely oriented. A theoretical estimate about the residual entropy of the crystal in such situations concerns the thermodynamic probability or the number of distinct ways W in which the crystal structure may be attained, involving the Boltzmann equation S = k.lnW.
As an example, for CO a reasonable assumption would be to consider half of the N
molecules of the crystal to be aligned in an opposite way. Assuming this, the
number of ways in which the crystal
structure may be attained equals W =
N!/{(N/2)!.(N/2)!}, i.e., the number of ways in which the N/2 number of
oppositely aligned molecules may be chosen out of the total N molecules. [It
may be noted here that as the molecules in the crystal are localized in space
they are distinguishable, unlike the molecules in gaseous carbon monoxide, and
so we get this value for W.]
Thus here we have, S = k.lnW
= k ln
[N!/{(N/2)!.(N/2)!}]
= k [ln(N!)
ln{(N/2)!} ln{(N/2)!}] = k
[ln(N!) 2ln{(N/2)!}]
Applying Stirling's approximation to ln(N!) & ln{(N/2)!},
we get:
S = k [N lnN N
2{(N/2) ln(N/2) (N/2)}]
= k [N lnN
N N ln(N/2) + 2(N/2)]
= k [N lnN
N lnN + N ln2] = Nk ln2
Thus the molar residual entropy here is NAk ln2 = R ln2 =
11.758 J K1mol1.
This predicted estimate for residual entropy of CO is of the same order to its
reported experimental values.
3.9 Looking back towards a Verification for the Relation
b = 1/(kT):
In Section 2.2 we assumed that the Lagrange undetermined multiplier b (in the Boltzmann distribution law and in the definition of q) equals 1/(kT) [where k is the Boltzmann constant and T is the thermodynamic temperature]. Let us now verify this relation, as we are now in a position to do so!
Let us begin with the relation q =
Sl
gl exp (-bel)
from Section 2.2, and differentiate it w.r.t.
b
at constant volume V. This gives:
(dq/db)V
= -Sl
gl el
exp (-bel),
or, Sl
gl el
exp (-bel)
= - (dq/db)V
Using Nl
= (N/q)
gl exp(-bel)
from Section 2.2, we get:
U
=
Sl
Nl el =
(N/q) Sl
gl el
exp (-bel
) =
-
(N/q)
(dq/db)V
=
-N (d ln q /db)V
So, the translational system-energy is definable as: Utr
= -N (d ln qtr /db)V.
Avoiding the replacement b
= 1/(kT) in the definition of qtr, we may easily derive that
qtr = (2pm/b)3/2 V/ h3
, so that ln qtr = ln
{(2pm)3/2 V/ h3}
-
(3/2) ln b.
This gives:
(d ln qtr /db)V
= 0
-
(3/2).(1/b) =
-(3/2).(1/b).
This gives:
Utr
= -N.{-(3/2).(1/b)}
= (3/2). (N/b)
As we know (from experimental
knowledge and from kinetic theory of gases etc.) that the translational
(kinetic) energy for gases is
Utr
= (3/2).NkT, so obviously,
(3/2)(N/b)
= (3/2)NkT implying
b
= 1/(kT).
Note: Using b = 1/(kT) we arrived at the approximate Boltzmann equation S = k lnW* (where k is the Boltzmann constant). As lnW* ≈ lnW, we get that the constant k in the Boltzmann equation S = k lnW is also nothing other than the Boltzmann constant!
3.10 Looking back at the rotational p. f. for
homonuclear diatomic molecules:
In any discussion about the rotational p. f. of homonuclear diatomic molecules, we need to note that all the rotational states of such molecules can't coexist with all the nuclear states, because of certain symmetry consideration of the complete molecular wavefunction with respect to exchange of the pair of identical nuclei therein. Just as any many-electron electronic wavefunction (of an atom or molecule) must be antisymmetric with respect to exchange of any pair of electrons therein (electrons being a set of identical Fermions), the complete molecular wavefunction of a molecule (e.g., CH4) with two or more identical nuclei (e.g., the four 1H nuclei in CH4) must also either be antisymmetric (for Fermion kind of nuclei, e.g., for 1H as in CH4) or be symmetric (for Boson kind of nuclei, e.g., for 14N in N2 or 16O in O2) with respect to exchange of any pair of identical nuclei therein. Now, the complete molecular wavefunction f of a molecule is a product of five specific-mode wavefunctions: a translational, an electronic, a nuclear, a rotational and a vibrational wavefunction, i.e., f = ftr fel fnuc frot fvib. Out of these, each of ftr, fel & fvib are always symmetric with respect to exchange of a pair of identical nuclei, and so these three never determines the overall symmetry of f. It is the rotational wavefunction frot and nuclear wavefunction fnuc that are, with respect to exchange of a pair of identical nuclei in the molecule, symmetric in some cases and antisymmetric in other ones: thus frot is symmetric for even-J states (i.e., for J = 0, 2, 4, 6 etc.) and anti-symmetric for odd-J states (i.e., for J = 1, 3, 5 etc.); whereas fnuc is antisymmetric if the nuclear-spin function part of fnuc is of antisymmetric type [e.g., of the form w(a,b) = {an(a) bn(b) an(b) bn(a)} for H2 that satisfies w(b,a) = w(a,b), this w(a,b) being built from the two one-nuclear spin functions an (corresponding to mI = ½) & bn (for mI = ½) found in the 1H nuclei], otherwise fnuc is symmetric. Furthermore, we may note that a pair of identical nuclei are Fermions if the nuclear-spin number I is half-integral (e.g., 1H with I = ½) and Bosons if the nuclear-spin number I is integral (e.g., 14N with I = 1, or 16O with I = 0).
Thus, for homonuclear diatomic molecules with
half-integral nuclear-spin number I (e.g., H2) the nuclei are
Fermions, and so the complete molecular wavefunction f
must be antisymmetric. Thus there are two kinds of H2 molecules: in
the ortho variety (~75% for H2 molecules are found to be
ortho-hydrogen at room-temperature) the nuclear wavefunction fnuc is
symmetric, coexisting with only the odd-J rotational states so that frot
is antisymmetric, making f (= ftr
fel fnuc
frot fvib)
antisymmetric. On the other hand, in the para variety (at
room-temperature, the rest ~25% of H2 molecules are found to be para-hydrogen)
the nuclear wavefunction fnuc is
anti-symmetric, coexisting with only the even-J rotational states so that frot
is symmetric, again making f antisymmetric as a
whole. Now there are three symmetric nuclear- spin functions w(a,b)
for H2: w(a,b) = an(a)
an(b), bn(a)
bn(b) & {an(a)
bn(b) + an(b)
bn(a)}, in contrast to the only one
antisymmetric w(a,b) mentioned above, and so there
are 3 symmetric nuclear ground-level wavefunctions in H2 in
contrast to only 1 antisymmetric one. Thus we get that the combined
rotational-nuclear partition function for the homonuclear diatomic molecule
H2 is:
qrot-nuc = 1{1 +
5
exp (
6 θrot
/ T ) + 9
exp ( 20 θrot
/ T ) + ..... }
+
3{3 exp (
2 θrot
/ T )
+ 7
exp (
12 θrot
/ T ) + 11
exp ( 30 θrot
/ T ) + ....}
dividing this by nuclear p. f. expression qnuc = (2I + 1)2
= 4 for the homonuclear diatomic molecule H2 to get
qrot (as qrot-nuc
must, obviously, equal qrot qnuc), we finally get:
qrot = (1/4){1 + 5
exp (
6 θrot
/ T ) + 9
exp ( 20 θrot
/ T ) + ..... }
+
(3/4){3 exp (
2 θrot
/ T )
+ 7
exp (
12 θrot
/ T ) + 11
exp ( 30 θrot
/ T ) + ....}
where the upper curly-bracketed sum is over the even values (0, 2, 4 etc.) of J,
whereas the lower curly-bracketed sum is over the odd values (1, 3, 5 etc.) of J
(please verify that the coefficients 1, 5, 9 etc. and
the exponents 0, 6, 20 etc. are indeed so). In
general for homonuclear diatomic gases consisting of Fermion-nuclei we
have
qrot = {I/(2I+1)}{1 + 5
exp (
6 θrot
/ T ) + 9
exp ( 20 θrot
/ T ) + ..... } +
{(I+1)/(2I+1)}{3 exp (
2 θrot
/ T )
+ 7
exp (
12 θrot
/ T ) + 11
exp ( 30 θrot
/ T ) + ...}
Now for temperatures T >> θrot
both the curly-bracketed sum are naturally equal, and as their total sum (i.e.,
the unrestricted-J sum giving
qrot in the heteronuclear case) was earlier shown to be (T/θrot)
as in Section 3.6, so each curly-bracketed sum above is (T/θrot)/2,
and so qrot here equals [(1/4)(T/ θrot)/2
+ (3/4)(T/ θrot)/2] i.e.,
(T/ θrot)/2. This is the
basis behind the formula qrot =
T/(2θrot) for homonuclear diatomic molecules,
valid under the condition T >> θrot, and
already mentioned in Section 3.6 earlier.
The
ortho and para forms of a homonuclear diatomic molecule are rather distinct and
can even be separated. The ratio of the numbers of ortho- and para- molecules
are naturally given by their respective contributions to the combined
rotational-nuclear p.f. and thus to the complete molecular partition function.
So for H2 gas at equilibrium at a temperature T, the ratio ro/p
= Northo/Npara of ortho-hydrogen and para-hydrogen
molecules is:
Northo/Npara = 3{3 exp (
2 θrot
/ T )
+ 7
exp (
12 θrot
/ T ) + 11
exp ( 30 θrot
/ T ) + ....} / {1 + 5
exp (
6 θrot
/ T ) + 9
exp ( 20 θrot
/ T ) + 13
exp ( 42 θrot
/ T ) + ..... }
For H2, θrot is
rather high (85.3 K) so that at the room temperatures the condition T >> θrot
is not exactly valid, and so for H2 gas at room temperature, ro/p
only approximately equals its high temperature limit 3.0. At very low
temperatures, ro/p for H2 approaches 0.
Problem: Show that at 300 K, ro/p for H2 is 2.990, but at 20
K it is only 0.002! Also show that even at 300 K, qrot for H2 is
1.934, a lot different from the T/(2θrot)
value 1.758!
On
the other hand for homonuclear diatomic molecules with integral nuclear-spin number I (e.g., 14N2, 16O2) the nuclei are
Bosons, and so the complete molecular wavefunction f
must be symmetric. If, however, I = 0 (e.g., 16O2) then only
one spin state (symmetric) is there, so that there are only ortho variety
existing, with
only the even-J rotational states allowed. But for integral yet non-zero I (e.g.,
for 14N2 with I = 1) there are both symmetric and
anti-symmetric spin states. So for such molecules even-J rotational states
coexist with symmetric nuclear states giving the ortho form (67% of N2 at
room-temperature are ortho-nitrogen), while odd-J rotational states coexist with
anti-symmetric nuclear states giving the para form (the rest 33%). At any
temperature T,
qrot and equilibrium ro/p are given here by the relations:
qrot = {(I+1)/(2I+1)}{1 + 5
exp (
6 θrot
/ T ) + 9
exp ( 20 θrot
/ T ) + ..... } +
{I/(2I+1)}{3 exp (
2 θrot
/ T )
+ 7
exp (
12 θrot
/ T ) + 11
exp ( 30 θrot
/ T ) + ...}
and
ro/p = {(I+1)/I}
{1 + 5
exp (
6 θrot
/ T ) + 9
exp ( 20 θrot
/ T ) + 13
exp ( 42 θrot
/ T ) + ..... } / {3 exp (
2 θrot
/ T )
+ 7
exp (
12 θrot
/ T ) + 11
exp ( 30 θrot
/ T ) + ....}