Statistical Mechanics

Question 1

population of a state

Answer 1

the average number of molecules that occupy each state

there are on average ni molecules in a state of energy εi

Question 2

Boltzmann distribution

Answer 2

gives the number of molecules ni in a particular energy state εi at temperature T

k - Boltzmann constant

Question 3

using the Boltzmann distribution

Answer 3

used to predict the populations of states in a system at thermal equilibrium

yellow bars: population
black lines: energy states available to that system

shows the distribution of molecules across energy states of increasing energy at different temperatures

at T=0K - all molecules occupy the lowest energy state
as T increases - more molecules have enough energy to populate higher states
as T → ∞ - all the energy states become equally populated

Question 4

relative population of states

Answer 4

considers the ratio of the number of molecules in state i, ni with energy εi to the number of molecules in state j, nj with energy εj

removes the need for a proportionality constant

TEMPERATURE is the only parameter that governs the population of the available energy states

Question 5

the total number of molecules, N

Answer 5

N is equal to the sum of the number of molecules in each energy state (Avogradro’s number NA or L)

Question 6

equation for ni (rewrite Boltzmann distribution)

Answer 6

lowest energy state = ε0
number of molecules in that state = n0

assume ε0 = 0

rewrite Boltzmann distribution, assuming εj is the lowest energy state, ε0 and that ε0 = 0

Question 7

substitute equation for ni into equation for N

Answer 7

Question 8

partition function, q

Answer 8

an abbreviation for the sum over all energy states of e^-βεi

it measures how the total number of molecules is distributed (partitioned) over all available states
- contains the information required to calculate the bulk properties of a system of independent (non-interacting) molecules

Question 9

worked problem 1.1

Answer 9

as q increases, T increases as ε1 becomes thermally accessible
- therefore, tends to a more even distribution of ε0 and ε1

as T → ∞, the e^-βε term gets larger, (→ 1)
- therefore, q → 2

as T increases, second term increases as the higher energy states are becoming more thermally accessible, therefore populated

Question 10

worked problem 1.2

Answer 10

Question 11

worked problem 1.3

Answer 11

as before, q increases as T increases as more states become thermally accessible

at 300 K, only ground state ε0 is populated and has appreciate contribution to q

at 800 K, all states are contributing as kT&raquo_space; ΔE

as T → ∞, q → 5 which relates number of accessible states

in general:
q → total number of states as T → ∞
q = 1 at T = 0K as only ε0 is thermally available

Question 12

worked problem 1.4

Answer 12

Question 13

principle of equal a priori probabilities

Answer 13

states that all possibilities for the distribution of energy are equally probable - hence, the treatment of energy states can be used for any type of molecular energy

allows us to assume that a translational, rotational and vibrational energy level of the same energy have equal probability of being occupied

energy states concerned with molecular motion: translational, rotational and vibrational energy levels

Question 14

vibrational energy levels

Answer 14

solve Schrodinger equation for SHO:

evenly spaced non-degenerate energy levels
E = (v+1/2)ħω
energy difference between levels
Δ = ħω

vibrational levels in spectroscopy:

G(v) = ωe(v+1/2) determined in wavenumbers
spacing between levels:
ΔG(v) = ωe

Question 15

rotational energy levels

Answer 15

solve Schrodinger equation for a particle on a sphere or rigid rotor

E = (ħ^2/2I)J(J+1)
separation increases with increasing energy

rotational levels in spectroscopy

F(J) = BJ(J+1) determined in wavenumbers
difference between consecutive levels increases from 2B to 4B to 6B etc.
typical values for B range from 0.1 to 60 cm-1

unlike vibrational energy levels, rotational energy levels exhibit multiple degeneracy with (2J+1) states at each energy level

Question 16

translational energy levels

Answer 16

derived and described using the particle in a box model

energy levels can be typically treated as a continuum (translational motion of gases can be treated classically)
exceptions to this: very light molecules at low temperatures in confined spaces

Question 17

energy levels and states

Answer 17

a number of states (gi) may be available at an energy (εi)
- hence, that energy level is gi-fold degenerate

diagram shows two energy levels, ε0 and ε1, each of which is 2-fold degenerate (g0 = 2 and g1 = 2)

modify parition function, q equation to accound for the degeneracy and sum over other levels

Question 18

contributions to the molecular partition function

Answer 18

the Born-Oppenheimer approximation allows us to sum the contributions from modes of energy to obtain the total energy of an isolated molecule

T = translation
R = rotation
V = vibration

we can therefore rewrite expression for the molecular partition function

Question 19

consider q^V, q^R and q^T

Answer 19

for translational energy levels, the energy separation between them is far smaller than the thermal energy available to a molecule at room temperature

Δε^T &laquo_space;kT

hence, q^T is large

for vibrational energy levels, the energy separation between them is larger than the thermal energy available at room temperature

Δε^V&raquo_space; kT

hence, most molecules occupy the lowest energy level (vibrational ground state) at 298 K

q gives us an indication of the number of states that are thermally accessible to a molecule at the temperature of the system

Question 20

worked problem 2.1

Answer 20

Question 21

worked problems 2.2

Answer 21

Question 22

Answer 22

Question 23

total energy of molecules in a system

Answer 23

assume that we have a total of N molecules that are independent (non-interacting)

if each molecule occupies a state with energy εi

the total energy E is:

Question 24

mean energy of a molecule

Answer 24

Question 25

substitute expression for ni into equation for < ε >

Answer 25

assumptions made for this expression: 1. by convention, we have assumed that ε0 = 0 when determining q - this is NOT always the case if εgs ≠ 0 εgs is the zero-point energy for the SHO - hence, the true mean energy would be: < ε > + εgs 2. in some case, partition function might depend on the volume as well as temperature (as in the case for q^T)

Question 26

expression for < ε > as a partial derivative

Answer 26

relates the partition function to the mean energy of a molecule

Question 27

internal energy of non-interacting molecules

Answer 27

continuing with the assumption that molecules are non-interacting total energy for N molecules, E_N the equation: E_N = N< ε > is for non-interacting molecules - assumes that the total energy of the system scales by multiplying the mean energy of each molecule by N if molecules interact (electrostatic attractions/repulsions, H-bonding etc.), additional energy terms arise that do not scale linearly with N

Question 28

the canonical ensemble

Answer 28

to account for interactions between molecules: consider a closed collection of N interacting molecules at constant volume and temperature - these molecules can be distributed across a set of energy states - each molecule has a total energy Ei these energy states adjust to any intermolecular interaction this system is replicated multiple times in forming this system, we have partitioned the molecules among the available energy states, which gives rise to the canonical partition function, Q

Question 29

the canonical partition function, Q

Answer 29

(just like q) Q can be used to calculate the mean energy of an entire system composed of molecules that may be interacting with eachother the sum is over all members of the ensemble Q is more versatile than q because it accounts for molecular interactions - can be applied to condensed phases and real gases

Question 30

energy of an ensemble of interacting molecules

Answer 30

scaling factor N is not required Q already considers the partitioning of N molecules as a whole and not as the product of N separate molecular contributions

Question 31

relationship between Q and q: when molecules are independent and distinguishable

Answer 31

Question 32

relationship between Q and q: when molecules are independent and indistinguishable

Answer 32

Question 33

determining internal energy of a system

Answer 33

does not matter which relationship between Q and q is used for internal energy the mean energy of a gas composed of N indistinguishable molecules is N times the mean energy of a single molecule

Question 34

worked problem 3.1

Answer 34

Question 35

worked problem 3.2

Answer 35

Question 36

worked problem 3.3

Answer 36

Question 37

from quantum mechanics, energy level of a system containing molecule of mass m free to move in a 1-dimensional container of length X

Answer 37

Question 38

partition function can be factorised into 3-dimensions

Answer 38

Question 39

lowest energy level, ε0 is defined as zero energy therefore consider the subsequent energy levels relative to the n=1 level

Answer 39

Question 40

modify summation to find qX^T

Answer 40

two reasonable assumptions are made in order to modify the summation: 1. the translational energy levels are so close together under typical conditions that we replace the sum with an integral (the continuum approximation) 2. extending the lower limit from 1 to 0 introduces neglible error and therefore allows the solution of a standard integral

Question 41

solution to qX^T

Answer 41

Question 42

translational partition function in 3-dimensions

Answer 42

partition function increases with volume of the container and mass of the molecule - an increase in each of these results in the separation between translational energy levels becoming smaller - more levels become thermally accessible β = 1/kT - an increase in temperature increases the value of the partition function

Question 43

large values of q^T are typical for light molecules in small volumes at room temperature equation q^T can be rewritten

Answer 43

Λ - thermal wavelength of molecule: has dimensions of length continuum approximation led to this expression - therefore it is only valid if q^T is very large for this to be true, the thermal wavelength of the molecule must be far smaller than the linear dimensions of the container V>>Λ

Question 44

using expression for q^T to determine the mean molecular energy

Answer 44

consider a monatomic perfect gas this allows us to: 1. consider only translational motion (no rotational or vibrational states) 2. assume the atoms are independent and non-interacting

Question 45

using expression for q^T to determine the internal energy

Answer 45

consider a monatomic perfect gas

Question 46

using expression for q^T to determine the pressure

Answer 46

consider a monatomic perfect gas

Question 47

using expression for q^T to determine the Gibbs energy

Answer 47

consider a monatomic perfect gas this result allows us to understand Gibbs' energy on a statistical level q - the number of thermally accessible states N - number of molecules q/N = average number of thermally acessible states per molecule G(T) - G(0) is proportional to the log(ln) of the average number of thermally accessible states available to the molecules as q/N increases, - log(ln) term becomes larger - G(T) - G(0) becomes more negative the thermodynamic tendancy to lower Gibbs' energy is driven by the tendency to maximise the number of thermally accessible states

Question 48

worked problem 4.1

Answer 48

Question 49

worked problem 4.2

Answer 49

Question 50

worked problem 4.3

Answer 50

Question 51

the rotational partition function

Answer 51

q^R evaluated by summation

Question 52

if many rotational states are occupied, kT >> ε^R

Answer 52

at 298K, a considerable number of rotational states are occupied the continuum approximation can be used and the sum in the partition function is replaced with an integral

Question 53

evaluating integral for the rotational partition function to show that:

Answer 53

this equation is only valid for linear non-symmetricsl diatomic molecules (i.e. A-B) validity of using this equation (continuum approximation) for rotational states is less clear cut than for translational states - when molecules have a large value for B, they have a small reduced mass - these molecule are borderline for the assumption that kT >> ε^R - they have the form: H-X

Question 54

to assess the temperature above which continuum approximation equation is valid, use the characteristic rotational temperature of the molecule

Answer 54

Question 55

number of thermally accessible rotational states for a symmetrical linear molecule

Answer 55

half the number available to an asymmetric linear molecule a homonuclear diatomic molecule (A-A) or symmetrical linear molecule (O=C=O) has a rotation through 180° that results in an indistinguishable state of the molecule continuum approxmation for rotational partition function is no longer valid for symmetrical linear molecules expression for the rotational partition function beocmes

Question 56

symmetry number, σ

Answer 56

heteronuclear diatomic: σ =1 homonuclear diatomic/linear symmetric molecule: σ = 2

Question 57

fermions

Answer 57

nuclei with half-integer nuclear spin 1H 19F I=1/2

Question 58

bosons

Answer 58

nuclei with integer or zero spin 16O 14N I=0

Question 59

The Pauli principle states: when the labels of any two identical FERMIONS are exchanged, the sign of the total wavefunction CHANGES

Answer 59

Ψ(2,1) = - Ψ(1,2)

Question 60

The Pauli principle states: when the labels of any two identical BOSONS are exchanged, the sign of the total wavefunction STAYS THE SAME

Answer 60

Ψ(2,1) = Ψ(1,2)

Question 61

Born-Oppenherimer approximation: the total wavefunction for the molecule

Answer 61

Question 62

consider C16O2 (CO2)

Answer 62

rotation through 180° results in exchange of two identical bosons (16O, I=0) hence: Ψtot(2,1) = Ψtot(1,2) consider each contribution to the total wavefunction in turn: ELECTRONIC WAVEFUNCTION electronic ground state for CO2 is 1Σg - hence symmetric Ψele(2,1) = Ψele(1,2) VIBRATIONAL WAVEFUNCTION vibrational ground state (v=0) has a symmetric wavefunction Ψvib(2,1) = Ψvib(1,2) NUCLEAR WAVEFUNCTION I=0, so exchaging the 16O nuclear spins is symmetric Ψnuc(2,1) = Ψnuc(1,2) ROTATIONAL WAVEFUNCTION symmetries of rotational wavefunctions depend on J Ψrot for even J are symmetrical wrt rotation through 180° Ψrot(2,1) = Ψrot(1,2) Ψrot for odd J are asymmetrical wrt rotation through 180° Ψrot(2,1) = -Ψrot(1,2) to obey Pauli principle for a boson, only J values that do NOT change sign on exchange are allowed - hence for CO2, only rotational levels with even J occupied

Question 63

why are only half of the thermally available rotational states allowed to be occupied for symmetrical linear molecules

Answer 63

the rotational partition function is only half of the value that is obtained for asymmetrical linear molecules

Question 64

worked problem 5.1

Answer 64

Question 65

worked problem 5.2

Answer 65

Question 66

the electronic partition function, q^E

Answer 66

q^E has been neglected from the molecular partition function as it is not associated with molecular (or atomic) motion

Question 67

calculating q^E

Answer 67

summation over electronic levels i typical energy gaps between ground and excited electronic states are about 10^-19 to 10^-18 J more than 10x that between the vibrational ground and first excited state

Question 68

in many cases, q^E = 1

Answer 68

unless a molecule/atom has a particularly low lying electronic state and T>>298K, we only need to consider the occupancy of the ground electronic state such that: q^E = g0 the ground electronic state of most simple diatomic molecules is non-degenerate, so q^E = 1 - q^E does not contribute to the overall partition function of the molecule

Question 69

determining degeneracy from the molecular term symbol of the electronic state

Answer 69

a common ground state term symbol for a diatomic molecule is: 1Σ degeneracy is determined from the superscript (2S+1) value (multiplicity) - if molecular orbital is full (no unpaired electrons) S = 0 - hence, g0 = 1 notable exceptions: O2 (g0 = 3) - molecular ground state term symbol is 3Σ NO (g0 =2) - molecular ground state term symbol is 2Π

Question 70

q^E for atomic species

Answer 70

atoms frequently have degenerate electronic ground states - degeneracy is determined by the total angular momentum quantum number J Li ground state term symbol is: 2S 1/2 subscript 1/2 denotes J (J=1/2) degeneracy is given by 2J +1 2(1/2)+1 = 2 g0=2 hence, if no excited electronic states are occupied q^E=2 in contrast to diatomic molecules, it is relatively common for atoms to have low-lying electronic states that are accessible at elevated temperatures

Question 71

examples of two-level (2L) chemical systems

Answer 71

two non-degenerate energy levels separated by an energy difference: ε1 - ε2 = ε - two conformations of a molecule with different energy (proteins with different tertiary structure conformations) - an electorn spin in a magnetic field (Zeeman effect) - protons in a magnetic field (NMR) - atoms or molecules with a low-lying electronic excited state

Question 72

when kT << ε, the temperature dependence of q2L and population (ni/N) is

Answer 72

Question 73

when kT >> ε, the temperature dependence of q2L and population (ni/N) is

Answer 73

Question 74

the general temperature dependence of q2L is:

Answer 74

partition function for a two-level system

Question 75

energy for a sample size, N

Answer 75

at T = 0K N< ε > = 0 as T → ∞ N< ε > → Nε/2

Question 76

heat capacity for two-level systems

Answer 76

plot of dimensionless Cv against dimensionless T - illustrates how heat capacity changes with temperature for a 2L system heat capacity is the change in internal energy as temperature is changed - for a 2L system, it tells us the rate of population of the upper level as a function of temperature at very low temperatures (kT << ε) - Cv is small as promotion to the higher level is not possible when T increases - there is sufficient thermal enegry to populate the upper level - Cv increases rapidly as the internal energy increases

Question 77

a maximum is observed for graph of heat capacity for a 2L system

Answer 77

there is a tendency for upper and lower levels to become equally populated rate of change of internal energy slows down as you are reaching equal population of the two states rate of internal energy (dU/dT) slows down, hence Cv goes through a maximum

Question 78

internal energy tends towards a constant value of Nε/2

Answer 78

hence, Cv drops to zero - there can be no further change in internal energy as T increases Schottky anomaly

Question 79

worked problem 6.1

Answer 79

Question 80

worked problem 6.2

Answer 80

Question 81

instantaneous configuration

Answer 81

consider N molecules that can occupy a range of states, s0, s1, s2...etc the number of molecules in s0 is n0 the number of molecules in s1 is n1 the population of these states at a given time (instantaneous configuration) is expressed in the form: {n0, n1, n2...} instantaneous configuration fluctuates continuously with time as the relative populations of the states change for a large number of N molecules, the populations of the different energy states are changing but the total energy remains unchanged - i.e. continuous gas collisions cause atoms to change energy state

Question 82

assume that all states available to the molecule have exactly the same energy

Answer 82

- allows us to assume that there is no restriction on the number of molecules that occupy each state - overall energy will be the same for every possible configuration

Question 83

N = 3 molecules - each molecule can occupy one of three degenerate states, s0, s1, s2

Answer 83

instantaneous configuration: {n0, n1, n2} there are three possible configurations of this system when N = 3 {3, 0, 0} - only 1 way this configuration can be achieved {2, 1, 0} - 3 ways this configuration can be achieved {1, 1, 1} - 6 ways this configuration can be achieved

Question 84

{2, 1, 0}

Answer 84

label 3 molecules A, B and C

Question 85

{1, 1, 1}

Answer 85

label 3 molecules A, B and C the system is statistically more likely to adopt confirmation 3 over 2 and 1 in the ratio 6:3:1 for larger systems of N molecules, (with many available degenerate states) system more skewed towards one dominating configuration

Question 86

weight (of the configuration)

Answer 86

the number of ways in which a configuration can be achieved for a large system of N molecules, the dominating configuration is found by determining for which values of ni the value of W is a maximum - achieved by varying ni and determining those values for which dW=0 (maxima of a function)

Question 87

constraints on real systems

Answer 87

the total numer of molecules and their total energy remain constant for a real system, the available states can have different energies

Question 88

value of ni for the dominating (most probable) configuration

Answer 88

Question 89

total energy for an independent (non-interacting) molecule

Answer 89

each of the energy terms is determined by summing over all occupied states i εtot is independent of how many molecules are present

Question 90

total energy of a collection of molecules

Answer 90

determined by multiplication of the mean total molecule energy < εtot > by the number of molecules N

Question 91

total energy for interacting molecules

Answer 91

additional term for molecular interacting energy, which is a function of N Ei(N) is an energy state (of all forms of energy) that can be occupied by a molecule in the canonical ensemble we have defined canonical ensemble therefore consists of a range of energy states, Ei that molecules can occupy - the energy of these states can change as a function of how many molecules are present N

Question 92

fixing T in the canonical ensemble

Answer 92

in the canonical ensemble, we define a constant N and V - so the energies of the states are fixed if T is also fixed, then the total energy remains constant - as the number of energy states that are accessible to the molecules only depends on T

Question 93

if states are equally thermally accessible

Answer 93

there will be a dominating configuration which determines the thermodynamic properties of the ensemble

Question 94

the number of states at a particular energy level rise sharply as a function of energy

Answer 94

at the same time, the occupany of energy levels will decrease sharply as a function of energy (Boltzmann distribution) to determine the distribution of ensemble members across the available states: multiply occupancy of energy level by the number of states at that energy level this product is a sharply peaked function at the mean energy - shows that most members of the ensemble have an energy that deviates little from the mean energy

Question 95

worked problem 7.1

Answer 95

Question 96

worked problem 7.2

Answer 96

Question 97

the Boltzmann formula

Answer 97

S - statistical entropy

Question 98

Answer 98

situation A - increasing T at constant V - energy levels are unchanged but populations change - W will also increase - therefore, S will increase as T → 0, every atom must occupy the lowest energy level - hence, W=1 and S=0 as T → 0, S → 0 (Third Law of thermodynamics: the entropies of all perfect crystals approach zero as T → 0) situation B - decreasing V at constant T (compression) - as volume decreases, spacing between energy levels increases and so fewer energy levels are occupied - W will decrease - therefore, S will decrease

Question 99

deriving an expression that relates q (the partition function) to S (entropy)

Answer 99

Question 100

for independent, indistinguishable molecules W is reduced by a factor of N! because there are N! permutations among the energy states that would result in the same system

Answer 100

lnW and therefore S are reduced by klnN! from the distinguishable value

Question 101

for interacting molecules

Answer 101

insert canonical partition function in place of the molecular partition function

Question 102

residual entropy

Answer 102

where the entropy at T=0K is greater than zero (the value predicted by the third law of thermodynamics)

Question 103

origin of residual entropy

Answer 103

consider a crystal composed of A-B molecules A and B have similar sizes and electronegativity, hence molecule has a small dipole moment there is little energy difference between A-B A-B A-B A-B A-B B-A B-A A-B and other possible arrangements during crystallisation, molecules adopt random A-B and B-A arrangements in the solid - assuming the two orientations are equally probable and the system consists of N molecules the total number of ways of achieving the same energy is: W = 2^N hence, molar residual entropy for molecule A-B that can randomly adopt 2 orientations at T=0 K is 5.8 J K-1 mol-1

Question 104

is molecule A-B has a large dipole moment, there is a strong enthalpic drive for the molecules to order A-B A-B A-B A-B in the solid

Answer 104

this results in a single orientation for all the molecules W = 1 AND S = 0

Question 105

worked problem 8.1.1

Answer 105

Question 106

worked problem 8.1.2

Answer 106

Question 107

worked problem 8.1.3

Answer 107

Question 108

worked problem 8.2

Answer 108

Question 109

worked problem 8.3

Answer 109