Section 7: Statistical Thermodynamics

Question 1

consider balloon of volume V held within thermally-isolated vacuum chamber of volume 2V, although the gas expands in volume,

Answer 1

no work is done on the surroundings

Question 2

entropy in a joule-free expansion: balloon bursts and

Answer 2

the gas is free to expand

Question 3

entropy in a joule free expansion: for adiabatic walls

Answer 3

δQ is also zero so Uf=Ui

Question 4

entropy in a joule free expansion - for an ideal as, U=

Answer 4

U(T) so there is no temperature change

Question 5

entropy in a joule free expansion - in practice most gases

Answer 5

cool slightly and we can measure the joule coefficient

uj=(dT/dV)u

Question 6

joule free expansion - initial and final states are

Answer 6

equilibria so delta s is well defined

Question 7

joule free expansion - for an ideal gas with s(T,V), delta s=

Answer 7

cvlnT2/T1 + Rln V2/V1 = Rln2

so can have a change of entropy without heat input or thermal changes

Question 8

in the context of the kinetic theory of gases, entropy is often described as

Answer 8

the disorder of the particles

hard to quantify

Question 9

consider two dice, each with six degenerate states or microstates.

the macrostate in this scenario is

Answer 9

the sum of the dice

eg for macrostates =3 can have microstates (1,2) or (2,1)

Question 10

in a fluctuating system which macrostate is the most probable

Answer 10

the one with the largest number of microstates

Question 11

the joule free expansion suggests that entropy depends on

Answer 11

how a gas is distributed

boltzmann then formulated the link as S=S(omega)

where omega = number of microstates

Question 12

as entropy is extensive, the entropy of the total system should be

Answer 12

the sum of the two entropies SAB=SA+SB

Question 13

if the two systems are independent then the number of microstates should be

Answer 13

the product of the individual numbers

omegaAB=omegaAomegaB

Question 14

for the composite system, Boltzmann’s hypothesis should obey

Answer 14

S(ΩAB) = S(ΩAΩB) = S(ΩA)+S(ΩB)

Question 15

S=

Answer 15

KB ln Ω

Question 16

microstates for monatomic gas

Answer 16

for gas particles in a box, each particle has position r and momentum p

Question 17

the likeliness of a microstate i is proportional to its

Answer 17

boltzmann factor e^-Bei

B is the thermodynamic beta used to simplify notation

Question 18

lower energy states are more likely since

Answer 18

the exponent is negative

Question 19

the actual probability of each state is also correlated to

Answer 19

how many states the system has

this means the probability is given by Pi=e^-Bei /Z

Z norm const

Question 20

to quantify the normalsiation constant, use

Answer 20

the property that all probabilities have to sum to unity

Question 21

normalisation constant Z is called

Answer 21

the partition function

Question 22

the expectation value for a property A can be found by

Answer 22

averaging over the allowed microstates

Question 23

if you interpret different microstates as fluctuations, then you could consider this average as a

Answer 23

time-average over the fluctuations

Question 24

link between thermodynamic potential and the partition function

Answer 24

F=-KBT ln Z

from F we can derive most other thermodynamic properties

link between statistical and classical approaches

Question 25

calculation strategy

Answer 25

identify energy levels determine partition function determine helmholtz function use standard relations to obtain U,S,P etc

Question 26

a change in internal energy can be achieved by

Answer 26

changing the probabilities or the energy levels

Question 27

energy levels of a system are goverend by

Answer 27

external parameters such as volume and applied fields

Question 28

when energy levels are fixed (dei=0 for all i) the only way to change the internal energy is

Answer 28

change the distribution

Question 29

adding heat or adding thermal energy,

Answer 29

shifts the probabilities and hence the population of the energy levels

Question 30

a classical particle in a parabolic potential executes

Answer 30

simple harmonic motion

Question 31

spacing of energy levels in QHO

Answer 31

evenly spaced

Question 32

energy of the ground state of QHO

Answer 32

not=0 e0=1/2 h bar w even at T=0K, QHO is not motionless - compatible with the uncertainty principle

Question 33

the harmonic oscillator is the basis for

Answer 33

models of molecular vibrations

Question 34

anharmonic terms can be added to the potential to improve

Answer 34

the approximation

Question 35

if the separation of energy levels is much larger than KBT

Answer 35

populating upper levels is unlikely would expect to have heat capacity CV tending to zero

Question 36

einstein model of heat capacity

Answer 36

modelling the bonds between atoms in a crystal lattice as springs so that each atom can undergo harmonic oscillations in 3D

Question 37

heating the crystal raises

Answer 37

vibrational energy which characerises frequency wE

Question 38

in the limit T approaching infinity, CV=

Answer 38

3R agrees with the Dulong-Petit law

Question 39

two critical limitations of the einstein model

Answer 39

there is a single frequency of oscillation the vibrational modes are independent

Question 40

debye model fixes limitations by considering

Answer 40

a spectrum of frequencies that describe coupled vibrations of atoms known as phonons

Question 41

einstein model made the assumption that

Answer 41

only single frequency of oscillations inside the crystal but higher harmonic frequencies would also be supported by the lattice structure

Question 42

beware! for particle in a box

Answer 42

only for a single particle gibbs paradox problem with particle indistinguishability

Question 43

Gibbs paradox

Answer 43

whilst the pressure of N particles is just a multiple, neither U nor S for a gas are N times the above resutls

Question 44

counting many particles states is impractical - its simpler to

Answer 44

consider energy levels

Question 45

in a small system, the number of states of a givn energy is

Answer 45

the degeneracy

Question 46

in a large systemm it is better to move to

Answer 46

integrating over a density of states

Question 47

blackbody radiation has the following characteristics

Answer 47

absorbs all incient photon radiation without reflection at eqbm radiates as much energy as it absorbs re-radiates at energies characterised by the body's temp

Question 48

main problem with classical approach

Answer 48

every mode and hence every energy is assumed to have the same probability

Question 49

when N=1 it is easy to

Answer 49

assign independent quantum states and if atoms are locked in a crystal lattice, it's still easy to distinguish different states

Question 50

unconstrained atoms or molecules of the same type are

Answer 50

indistinguishable

Question 51

a system of N particles ultimately has

Answer 51

one combined wavefunction

Question 52

for two particles, Ψ(x1,x2) describes

Answer 52

particle 1 at position x1 and particle 2 at position x2 exchanging the particles gives Ψ(x2,x1)

Question 53

we measure |Ψ|^2 and if the particles are indistinguishable the measurement should give

Answer 53

the same result under a swap of the particles

Question 54

for theta=pi, the wavefunction is

Answer 54

antisymmetric under exchange of the particles fermions all with half-integer spin

Question 55

for theta=0, the wavefunction is

Answer 55

symmetric under exchange of particles bosons integer spin

Question 56

the N-particle wavefunction can be expressed as

Answer 56

a combination of one-particle wavefunctions

Question 57

setting i=j yeilds what for fermnions

Answer 57

zero pauli exclusion principle

Question 58

is i=j allowed for bosons

Answer 58

yes

Question 59

for non-interacting particles, the energy eigenvalue is

Answer 59

jsut the summation of the individual one-particle energies

Question 60

changes to the partition function

Answer 60

1. there is only one i,j quantum state so need to avoid double counting 2. fermions can't all be in the same quantum state