## Statistical Mechanics - 2nd Year, Trinity Term

Cover the following topics:

(i) The Boltzmann distribution and the population of energy levels in a range of circumstances. You have already done a fair amount of this in Rotations, Vibrations and Heat Capacities (First Year), and will find it helpful to revise it.

(ii) The basic assumptions, i.e. S = klnΩ and A = -kTlnQ. See summary.

(iii) The expression for the entropy of a monatomic gas (Sackur-Tetrode equation) and preferably its derivation. Application of it (question 9).

(iv) Expressions for the translational partition function of a gaseous molecule, rotational partition function of a diatomic molecule, and a vibrational partition function of a harmonic oscillator. You should be able to derive the first and third of these. Application of these expressions.

(v) Outline of how heat capacities of gases and their variation with temperature may be calculated using statistical mechanics. The heat capacities of ortho and para hydrogen.

(vi) Explanation of how the equilibrium constant of a chemical reaction may be calculated from a knowledge of the energy levels of the atoms or molecules of which it is comprised (questions 12, 13 and 14). This is most important for transition state theory, which will be done separately.

(vii) The Third Law of thermodynamics and residual entropies (question 8).

To help you understand some of the concepts,two applets are shown below (Click here for information concerning applets; the relevant filenames, if you wish to download them, are respectively tutorials/statmech/twoLevelHeatCap.jar and tutorials/statmech/twoLevelHeatCapAppletJ.html for the two level system and tutorials/statmech/rotHeatCap.jar and tutorials/statmech/rotHeatCapAppletJ.html for the rotational heat capacity). The first one is a slight advance on one in the first year work and shows how the heat capacity, internal energy and entropy of a two level system vary with temperature, energy and degeneracy of the upper of the two levels.

The applet calculates the heat capacity, energy and entropy (on the left) for the pair of energy levels shown on the right. As the temperature is varied (using the middle slider) the corresponding heat capacity is marked with a circle, the relative population of the two levels is shown in green on the right, and the relative number of excitations from the ground to the excited state is shown as the width of the blue arrow connecting the two states. As well as adjusting the energy gap between the two levels and the temperature, you can also vary the ratio of the degeneracy of the upper to that of the lower state.

Try the following:

(i) Set the energy slider at a low value, say 50 cm−1 and vary the temperature. Note how at low temperature there are almost no molecules being excited. This is the quantum effect; the available energy (kT) is far too small to excite molecules to the upper level (more accurately, almost no change in population is necessary to change the Boltzmann distribution to that appropriate for a higher temperature and hence negligible energy is needed to raise the temperature). Note that at high temperature the populations of the two levels are tending to equality. Vary the temperature slowly between the two limits and note that the temperature at which the largest number of excitations occur corresponds to the maximum in the heat capacity. The heat capacity is zero at both high and low temperatures because there are no changes in population with temperature at these limits.

(ii) Keeping the temperature constant, vary the spacing (energy) between the two levels and note that the temperature for the heat capacity to be a maximum increases directly with energy. Pick a value of the heat capacity, then change the energy to a new value. Verify that you can adjust the temperature to recover your starting value of the heat capacity.

(iii) Examine the effect of changing the degeneracy ratio. By following the changes in population as you do this, see if you can come up with a verbal explanation of why the maximum value of the heat capacity is sensitive to this degeneracy ratio.

(iv) Note that for this two level system the energy and the entropy approach constant limiting values at high temperature. What are the values of the two limits?

The second applet calculates the heat capacity associated with rotation of a diatomic molecule.

The applet calculates the heat capacity and internal energy (on the left) for the set of rotational energy levels shown on the right. As the temperature is varied (using the middle slider) the corresponding heat capacity is marked with a circle, the relative populations of the levels are shown in green on the right, and the relative changes in the number of excitations between the states are shown as the width of the blue arrow connecting the two states. You can vary the rotational constant B with the slider on the left. The heat capacity is very sensitive to the spin statistics and the slider on the right allows you to select a number of spin combinations.

Try the following:

(i) The spacing of rotational energy levels is approximately quadratically dependent on the quantum number J and the degeneracy increases with J. This leads to some important differences between the temperature dependences of rotational and vibrational heat capacities. The first is seen clearly at the starting settings of the applet. There is a small maximum in the heat capacity just before the behaviour becomes fully classical. This is because the system behaves almost like a two level system with a high degeneracy ratio. Vary the temperature in the region just below this point and watch the changes in population of the energy levels. They reflect this near-two level behaviour very clearly.

(ii) The near-two level becomes much more marked for certain spin statistics. Examine the effects for the various ortho and para hydrogen and deuterium situations. Note that the "equilibrium" curves are somewhat hypothetical in that they assume equilibrium flipping of the nuclear spins. However, the amplitude of the maximum in the curve can always be explained by the combination of the degeneracy ratio of the first two levels and the ratio of the spacing between the second two levels and that between the first two levels.

(iii) Note how the increasing degeneracy of rotational levels leads to a completely different pattern of occupation compared with that for vibrations (see first year applet of vibrational heat capacities).

### Problems

Where the question number is enclosed in a button, e.g. , you can obtain help or comments about the question by clicking it.

.
Click here for values of the fundamental constants.

#### Systems with only two or three energy levels

Calculate the mean electronic energy per atom of oxygen atoms at 300 K given that the 3P0, 3P1, and 3P2 states occur at 226.5 cm-1, 158.5 cm-1, and 0 cm-1 respectively.
To what limit do the populations of the three levels tend as the temperature is raised to a high value and what are the limiting values of (a) the energy and (b) the corresponding electronic heat capacity at high temperature?
[If you find the degeneracies difficult you should briefly revise term symbols of atoms by doing the quiz in the first year (See question 2 of Atomic Spectra).]

The population of an excited state of the oxygen atom relative to the ground state is 5.695 x 10-2 at 10000 K and 5.840 x 10-3 at 5000 K. Assuming that spin-orbit coupling may be neglected,
(i) Calculate the energy of this excited state relative to the ground state,
(ii) Calculate the ratio of the degeneracies of the ground and excited states.
(iii) Given that the ground state is 3P, use your result from (ii) to determine the degeneracy of the excited state and use this value to deduce a term symbol for the excited state.
[If you find the degeneracies difficult you should briefly revise term symbols of atoms by doing the quiz in the first year (See question 2 of Atomic Spectra).]

Write a formula for the electronic partition function, qelec, for the NO molecule, which has a 2Π1/2 ground electronic state and a low lying electronic state, 2Π3/2, at 121 cm-1 above the ground state (both states have a degeneracy of two).

Derive a formula for the electronic contribution to the internal energy, U, for NO and sketch the temperature dependence of the electronic contribution to U(T) and to the heat capacity, CV(T) from 0 to 600 K.

Why is NO paramagnetic? Sketch the temperature dependence of the paramagnetism of gaseous NO. How is this temperature dependence different from that of O2?

The energy levels of the ground state of the caesium atom are split in an external magnetic field, B, to give

ε(B) = ε0 ± gμB/2

where μ is the Bohr magneton.
(a) Show that the electronic contribution to the partition function, qelec, is given by

(b) The total magnetization, M, of the sample is given by M = Nmav where mav is the mean magnetic moment per atom in the field direction. Using the Boltzmann distribution, show that the magnetization of a sample of caesium atoms is given by:

(c) From the general result derived in (b) show that, for caesium atoms in small fields, μB<<kBT, M = χB, where the paramagnetic susceptibility χ follows Curie's Law,

Using this result calculate the mean magnetic moment per atom for a sample in a magnetic field of 0.3 T at a temperature of (i) 300 K, (ii) 30 K, and (iii) 3 K; compare your values with those obtained from the general result for mav in (b) above.

#### Continuous series of levels

The rotational energy levels of a (heteronuclear) diatomic molecule are given by EJ = BhcJ(J+1), with B the rotational constant such that θ = Bhc/kB is the characteristic rotational temperature.

(a) Explain why the fraction of molecules found in the rotational energy levels specified by the quantum number J is

where qrot is the rotational partition function.

(b) Write down the expression for qrot and, by transforming the sum to a suitable integral, obtain qrot in the high-temperature limit.

(c) To a first approximation the intensity of a rotational absorption transition from JJ' is determined by f(J). Explain briefly why this is so, and show that the line with maximum spectral intensity (J = Jmax) corresponds to

(d) The Stokes branch of the pure rotational Raman spectrum of the (heteronuclear) molecule consists of approximately equally spaced lines separated by 9.8 cm−1. Identify which rotational transition has maximum intensity at a temperature T = 298 K.

In the infrared emission spectrum of the first overtone (v = 2 → 0) of CO formed in an oxy-acetylene flame a series of transitions corresponding to v = 3 → 1, 4 → 2, etc. are observed. These lines can be resolved because they have slightly different frequencies because of anharmonicity. The intensities of the transitions depend on the populations of the v = 2, 3, 4, etc. levels and are proportional to

I = (v + 1)(v + 2)exp[-(v + 2)hcν/kBT]

where v refers to the lower level in each transition. Values of the intensity are

 I 4 5.9 5.5 4.6 3.9 3.2 v 0 1 2 3 4 5

Use a graphical method to determine the temperature of the flame (the frequency of the stretching vibration of CO is 2200 cm-1).

#### Heat capacities

The rotational heat capacity curves of para (antisymmetric nuclear spin wavefunction) and ortho-hydrogen and HD are shown below.

(a) From the relative values of T at which CV is R/2 identify which sets of rotational levels are associated with ortho and para-H2. Give your reasons for this identification.

(b) Calculate the ratio of degeneracies of the first excited and ground states for the two forms of H2. By comparison with the behaviour of the heat capacity for a two level system (see applet above), suggest reasons why there is a sharp maximum for para but not one for ortho-H2.

(c) The nuclear spin of H is 1/2. Write down the possible nuclear spin states for H2 and hence calculate the nuclear spin degeneracy of ortho and para-H2.

(d) Use the difference in in reduced masses between HD and H2 to determine the relative spacing between ground and excited states in HD and para-H2. Hence explain the appearance of the heat capacity curve for HD.

(e) Compare the heat capcity behaviour shown here for hydrogen with that shown earlier in rotation and outline the main differences.

Use the following data for O2 to plot a graph of the heat capacity, CV, as a function of temperature. Discuss the shape of this graph and the origin of the temperature variation in terms of the properties of the O2 molecule. To what extent is the form of the function applicable to other molecules?

 T/K 200 300 500 1000 1500 CV/R 2.503 2.533 2.739 3.197 3.399

The molecules O2 and Cl2 have vibrational frequencies equivalent to 1553 cm-1 and 557 cm-1 respectively. Estimate C V for Cl2 at 300 K.

Show that the following data for H2O are consistent with the presence of one vibration (a bending mode) with a frequency similar to that of O 2 and two vibrations with substantially higher frequencies.

 T/K 300 500 1500 CV/R 3.041 3.238 4.677

Estimate a mean value for the higher pair of frequencies.

#### Entropy

In terms of the canonical partition function Q = Q(N,V,T), the internal energy and Helmholtz free energy are given respectively by U = NkBT2(dlnQ/dT)N,V and A = −kBTlnQ

(a) For a gas of independent particles, show that the entropy S is given in terms of the molecular partition function q by

where lne = 1. Stirling's approximation is lnN! ≈ NlnN - N for N >> 1.

(b) By considering an atomic gas, and neglecting the electronic degrees of freedom, show that the translational contribution to S may be expressed in the form

where m is the relative atomic mass, and C is independent of the chemical identity of the gas.

(c) Account for the differences between the spectroscopic entropies of the following pairs of gases (given in J K-1 mol-1)

 S(Kr) − S(Ar) = 9 S(DI) − S(HI) = 6 S(N2O) − S(CO2) = 6 S(F2 at T = 350 K) − S(F2 at T = 250 K) = 7

Comment on the fact that the ratio of the entropy of Cl2 at 350 K to that at 250 K is larger than the corresponding ratio for F2.

(Atomic masses are Ar = 40, Kr = 84, C = 12, O = 16, N = 14, F = 19, I = 127)

Comment on the values of the entropies of the following compounds at 1 atm pressure and 298 K (values in J K-1 mol-1):

 - CO2 N2O CH4 CH3D H2S H2O Spectroscopic 213.5 219.9 185.4 200.4 205.4 188.6 Calorimetric 213.7 215.0 185.2 188.9 205.2 185.1

Indicate how the following expression for the molar electronic entropy of a gas may be derived:

Selec = Rlnq elec + Uelec/T

where qelec and Uelec are respectively the molecular electronic partition function and the electronic contribution to the molar internal energy.

A sample comprising a mole of non-interacting sodium atoms is placed in a magnetic field B.

(a) Comment on the observation that, at low temperatures, the only thermally accessible states are found to have energies given by:

ε+ = + αB and ε - = -αB

where α is a positive constant, such that αB<<kT.

(b) Give an expression for qelec.

(c) Show that the difference in populations of the two states is approximately equal to NαB/kT.

(d) Hence obtain an expression for Uelec .

(e) Calculate Selec in zero magnetic field and comment on the physical significance of your result.

(f) Hence obtain an expression for the change in Selec when the sample is transferred isothermally from zero field to a field B. Comment on the sign of this result and on any practical implications of it.

#### Equilibrium Constants

Why is the rotational partition function of a diatomic molecule reduced by a factor of two if the molecule is homonuclear?
Show that this factor of two causes the equilibrium constant for the reaction

to be 4 at 298 K.

In the chemical oxygen-iodine laser the spectroscopically long-lived 1Δg excited state of O2 is produced in the gas phase by the decomposition of hydrogen peroxide. The O2(1Δg) then reacts with an iodine atom and excites the iodine atom from its ground 2P3/2 state to its excited 2P1/2 state and itself drops down to the ground state, 3Σg−. The energy level diagram is shown below.

The exchange of energy is an equilibrium process

Calculate ΔU0 and its contribution towards the equilibrium constant at 300 K.
The equilibrium constant is also affected by the degeneracies of the four states involved. Calculate the total degeneracy on either side of the equilibrium and hence the contribution of the degeneracies to the equilibrium constant.
Calculate the overall equilibrium constant at 300 K.

Discuss how the equilibrium constant KP for the dissociation

may be calculated from a knowledge of the separation DE of the 2P1/2 and 2P3/2 levels in the atom, the rotational constant B, the vibrational wavenumber νe and the dissociation energy of the molecule.

For this system, ΔE = 402.7 cm-1 , B = 0.8895 cm-1, and ν e = 916.6 cm -1 . Discuss whether the rotational, vibrational, and electronic contributions to KP are significant at 1000 K. (kBT/hc = 695.0 cm-1 ).]