1
‫ أحمد سعيد شلبي قسم الكيمياء كليه العلوم جامعه بنها امتحان ثالثه كيمياء خاصه‬.‫د‬.‫أ‬
‫ميكانيكا احصائيه رابعه تخلف ثالثة‬
Prof.Dr. Ahmad S. Shalabi
___________________________________________________________
Benha University
3rd year Students
Date : 62-26-2011
Faculty of Science
Special Chemistry
Time : 3 hours
Chemistry Department
Staistical Mechanics
___________________________________________________________
Statistcal Thermodynamics
___________________________________________________________
1- Based on the principle of equipartition of energy, write the values of
the translational Et , rotational Er ,and vibrational Ev energies in terms
of the gas constant (R).
2- Compare between the third law of thermodynamics and statistical
entropies clarifying the different types of discrepant behavior.
3- Calculate the translational partition function for 1 mole of oxygen
at 1 atm. pressure at 25 oC , assuming the gas to behave ideally.
[Avogadro number : N=6.023 x 1023 - π= 3.2422 - k= 1.38 x 10 -16 R= 82.06 - h= 6.624 x 10 -27] .
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GOOD LUCK
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‫ أحمد سعيد شلبي قسم الكيمياء كليه العلوم جامعه بنها امتحان ثالثه كيمياء خاصه‬.‫د‬.‫أ‬
‫ميكانيكا احصائيه رابعه تخلف ثالثة‬
Prof.Dr. Ahmad S. Shalabi
____________________________________________________________________
Benha University
3rd year Students
Date : 26-12-2011
Faculty of Science
Special Chemistry
Time : 3 hours
Chemistry Department
Staistical Mechanics
_____________________________________________________________________
Model Answers
___________________________________________________________________
1.
According to the principle of equipartition of energy of energy,
each kind of energy of a molecule that can be expressed in the
general form ax2 contribute an amount of RT to the average
energy of one mole. The translational kinetic energy of a molecule
is
m/2 (x2 + y2 + z2)
x , y, z : the components of velocity in the directions at right
angles.
momentum Px=mx. , and so the energy is given by
1/2m ( P2x + P2y + P2z )
thus, consisting of three quadratic terms.
Consequently,
. The translational energy is
3 x (1/2) x RT =
RT
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. The rotational energy of a molecule, assumed to be of constant
dimensions, is proportional to the square of the angular
momentum, so that rotation is represented by one quadratic or
square terms and contributes
1 x (1/2) x RT =
RT per mol
The rotational energy of a diatomic or linear molecule, that
exhibits rotations about two axes contributes
2 x (1/2) x RT =
RT per mol
The rotational energy of a non linear molecule, that exhibits
rotations about three axes contributes
3 x (1/2) x RT =
RT per mol
. A non linear molecule containing (n) atoms have (3n-6) modes
of vibrations, so that the vibrational energy contributes
(3n-6)RT per mol
A linear molecule containing (n) atoms have (3n-5) modes of
vibrations, so that the vibrational energy contributes
(3n-5)RT per mol
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2.
. Third Law ( thermal ) entropy
Obtained from heat capacity measurements.
It is zero at absolute temperature i.e perfect solid.
. Statistical (standard) entropy
Obtained from Boltzman-Planck equation
i.e the combined rotational and vibrational partition functions
Qr and Qv .
Both types of entropies are in agreement. However some
discrepancies occur for : CO, NO, NNO, H, D, H2O, and
CH3-CH3 ( internal rotation).
The various types of discrepant behavior will now be discussed:
o Random Orientation in the Solid
For CO, NO, and NNO
. The third law considers one arrangement for solids
CO CO
NO NO
NNO NNO
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so that there is similarity between the two ends, and the
probability of state is 1 so,
S = R ln 1 = 0
This is not the case for randomly oriented solids, i.e it is not
correct to take the entropy as zero.
. Statistical mechanics considers two arrangements for solids
CO OC
NO ON
NNO ONN
so that there is no similarity between the two ends, and the
probability of state is 2 so,
S = R ln 2 ≠ 0
This is the case for randomly oriented solids, i.e it is correct to
take the entropy as non zero.
The same situation occurs for H2O :
The third law considers ice as a perfect solid with zero entropy.
While, in fact, there is a distribution of H bonds in ice crystal and
the solid is not perfect so that entropy is not zero as predicted by
statistical mechanics
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3.
Qt = (2πmKT)3/2 . V
h3
= (2πmKT)3/2 . RT
h3
P
since Qt is dimensionless, the numerator and denominator must be
expressed in the same units m,k,h in c.g.s, p in atm., R in
cc.atm.deg.-1 mol.-1
m = M.W =
32_____ = 5.313 x 10-23 g
N
6.023 x 1023
Qt=[2x 3.1414 x(5.313x10-23)x(1.38x10-16)x 298.2]3/2 . 82.06x298.2
[6.624x10-27]3
1
= 4.28x1030