Brainscape's Knowledge GenomeTM


Top Flashcards (Certified)
All Flashcards

Chemical physics > Statistical Thermodynamics > Flashcards

Statistical Thermodynamics Flashcards

(90 cards)

Start Studying
1
Q

The First Law

A

The conservation of energy: change in internal energy = change in heat transferred to system + work done.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Internal energy of a system is….

A

The mean total kinetic energy of the atoms and the potential energy of interaction between them.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is a state function?

A

Depends on condition of system and not on the path taken to reach conditions. Can write exact differential of state functions.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

For path dependent quantities we use

A

delta rather than exact differential to denote small quanitities

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

A reversible process is one whose direction

A

can be changed by an infinitesimal change in a variable. They take place infinitely slowly, and the system remains at equilibrium at all points.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What is the greatest in the reversible path out of all the possible paths between 2 thermodynamic states?

A

the work done by the system (delta w’ = - delta w)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

The second law

A

the entropy of the universe increases in a spontaneous process. dS (univ) >= 0)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Can splitup changes in entropy of universe into

A

changed in entropy of system + changes in entropy of surroundings

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

If the system is isolated from the surroundings…

A

any change in the system does not affect entropy of surroundings

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

A small change in entropy is defined by

A

reversible heat required for the change / temperature

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

If the change in entropy occurs irreversibly, dS is defined by the

A

heat that would be involved if the same change were carried our reversibly. So the entropy change is a state function. (independent of path taken)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

The Helmholtz energy

A

A = U - TS.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Helmholtz has the property of decreasing in any spontaneous proces that takes place at…..

A

constant volume and temperature. Therefore it’s minimised at equilibrium as A is the thermodynamic potential at constant V AND T.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

For a change at constant temperature, dA(sys) =

A

dU(sys) - TdS(sys)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

At constant volume no work can be done so dU(sys) =

A

delta q(sys) = - delta q (surr) because heat is supplied by the surroundings.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

dS(surr) =

A

delta q (surr) / T = - delta q (sys) / T

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

-dA(sys) / T =

A

dS(surr) + dS(sys) (entropy change of universe). So dA < 0 for a spontaneous process in a system at constant V and T. (because of second law). dA = 0 at equilibirum

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Considering a reversible process, reversible work is carried out against an external pressure equal to the

A

pressure of the system itself. deltaw(rev) = -pdV. So dU = Tds -pdV.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

dU = Tds -pdV is also valid for irreversible paths since

A

it involves only state functions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

dA =

A

-pdV-Sdt. d = partial derivative A wrt to v at constant temp. s = - partial derivative of A wrt T at constant volume

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

overall description of the system is a

A

macrostate (small number of overall variables)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

detailed specification of the state of the system at the molecular level is a

A

microstate

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

a large number number of microstates are compatible with any given

A

macrostate

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

number of microstates increase with

A

Total energy of system, number of particless, decreasing spacing of energy levels

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
number of microstates always increases with E because
all microstate at a lower energy can be converted to a higher energy by promoting one or more particles
26
if no level is multiply occupied then the number of choices for the label of the lowest occupied level is N leaving ……….. The number of permutations is therefore W =
N -1 choices for the next occupied level and so on. …..N!
27
if several particles occupy the same level permutation of the labels within that level does not
produce distinguishable states. So overall number of permutations is reduced by the number of ways of permuting labels within each level, giving the multinomial coefficient. Equation on page 10
28
Thenumber of microstates increases for a given N and E if
spacing of the energy levels is decreased.
29
For an isolate system (constnat energy) the number of particles and energy are
fixed
30
Microstate subject are all
equally probably and have the same energy
31
the probability that an isolated system wil be found in any given microstate is
1/number of available microstates
32
energy levels of gas molecules are inversely proportional to a square of the box length so
spacing of levels dcreases as box length increases.
33
Number of microstates increases rapidly as spacing decreases therefore
final number of microstates is much bigger than inutial number for gas expansion
34
entropy of an isolated system is S=
k(B)*ln(number of microstates)
35
spontaneous processes in isolated systems involve
an increase in the number of microstates
36
A properrty is extenisve if its value is
additive when two bodies under same conditions joined.
37
Entropy is
extensive quantity
38
If separate bodies joined then total number of microstates
is product of individual number of microstates.
39
S(tot) in isolated system is
Addition of entropies
40
microstates of supersystem is equally likely if supersystem is
isolated. So probability of system being in a particular state with energy Ei is proportional to number of microstates compatible with system having energy Ei
41
probability of a given state being observed decreases with
its energy
42
constant of proportionaility is determined by the fact that
probabilities must sum to unity
43
canonical means the
number of particles, volume of system and temperature are fixed.
44
probability of a state being occupied at a given temperature decreases
exponentially with the energy of the state
45
Boltzmann distribution only applies to
non-interacting molecules since it is impossible to define independent molecular states if molecules affect each others energy
46
bridge relation : A =
-k(B)*T*ln(Q(N))
47
if molecules do not interact strongly it may be reasonalbe to nelgect interaction so energy of the system is
sum of energiesof constiuent molecules
48
molecules parition function is number of
thermally accessible states- states below K(B)T
49
Q(N) is a sum over
energy levels of entrie system
50
sum of q is over
the levels of a single molecule
51
states above K(B)T have low
Boltzmann factor and less thermally accessible, low probabilty of being occupied
52
q is just number of states below
k(B)T
53
q counts number of significantly populated energy levels at
a given temperature.
54
indistinguishable particles has
same set of quantum numbers but a different order,
55
for indistinguishable particless Q(N) =
q^N / N!
56
If multpile occupancy of levels than overcorrected Q(N) because
number of label permutations is smaller than N!> Howevewr number of levels available to molecules much greater than avogadro so chance of multiple occupancy is low.
57
indistinguishable particless does not affect
internal energy
58
distinguishable particles have a factor of
N! more states and therefore have greater entropy
59
Helmhotz energy depends on
indistinguisability or not
60
small alpha when spacing of energy levels is
small in comparison with k(B)T so a large number of translational levels contribute to q.
61
number of thermally accessible translational levels is much greater than
number of molecules and chances of multiple occupancy of any one level are low.
62
rotoational constant is inversely proportional to
moment of inertia so heavy molecule/big radius has lower rotational constant
63
each state in a degenrate level must be
counted but each one makes same contribution to the sum
64
vibrations of polyatomics can be broken down into
normal modes, each has its own frequency and acts independently from the reast
65
Total vibratrional energies of a polytomic is the
sum of the energies in its normal modes and corresponding partition function is the product of q^vib
66
Linear molecules of n atoms has how many vibrational normal modes?
3n-5, additional mode is a degenerate bend
67
Non-linear molecules of n atoms has how many vibrational normal modes?
3n-6
68
q^elec is degeneracy of ground state because
excited electronic states have energy much larger that k(B)T so hardly contribute to partition
69
equipartition principle states
each quadratic term that appears in the classical expression for the total energy of a system contributes 1/2*k(B)T to the internal energy U of the system
70
Translation contributes
3/2*RT per mole to the internal energy
71
Rotation constributes to the internal energy
3/2*RT per mole for non-linear and RT for linear molecule
72
translational quantum levels are so closely space in comparision with RT so they are
a continuum
73
at low temperature the rotational partition function tends to be 1 because the
ground state always contributes e^0 to the sum. it only beings to rise when excited rotatioanl states become thermally populated
74
the rotational internal energy cannot rise above 0 until the
thermal energy is large enough to promote molecules to at least J = 1 rotational level
75
since the heat capactiy is the derivative of the internal energy it starts at
0. It eventually attains the equipartition value of R but overshhorts at first. This bump is due to the interplay between the increasing spacing of the rotational states( which makes higher J levels increasingly difficult to populate) and the increasing degeneracy of the levels
76
at low temepratures only translational levels contribute to C (v) because
rotational levels are not yet thermally accessible
77
vibrational internal energy tends to 0 because
not enough thermal energy to promote vibrations from the ground state thermal energy is much smaller than the spacing of the energy level at lower temperatures
78
the vibrational heat capacity function
gradully rises from 0 to quipartition result. no overshoot because the vibrational levels are equally spaced and non-degenerate.
79
if predicting internal energy or C(v) from equipartition we should include the tranlation and rotation contributions but not vibrational because
the thermal energy lies in the range between B and hv for most diatomics and small molecules
80
a system is only capable of absorbing heat if there are
thermally accessible energy level
81
a low heat capacity indicates
that energy level spacing is large compared to K(B)T so hardly any new levels become thermally accessible for a small increase in temperature so energy does not rise
82
a constant non-zero heat capacity indicates that
new energy levels are becoming available at a rate determined by a fixed power of T as the temperature is increased
83
A jump in the heat capacity indicates
that a new set of energy levels becomes thermally relevant at that temperature
84
third law of thermodynamics
entropy of all perfect crystalline substance is zero at T = 0
85
to obtain absolute entropy at non-zero temperature we would measure
heat capacity down to absolute zero and then integrate dS from zero to the temperature of interest.
86
for polyatomic molecules each normal mode vibraiton acts as an
independent harmonic oscillator so vibrational entropy is sum for each of normal modes
87
entropy is a measure of the number of
available states
88
if molecules have a bigger dissociation energy than
k(p) is maller, deeper well amd less dissociation so equilibirum shifted left
89
canonical distribution
the probability distribution of states at constant volume, temperature and number of particless
90
states of higher energy become
less occupied in a system held at a constant temperature because all microstates of the supersystem are equally probable, so high energy states of the system are less probably

Chemical physics (3 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2025 Bold Learning Solutions. Terms and Conditions