OCR Advancing Physics B Rise and Fall of the clockwork universe matter very simple

1. Rise and fall of the clockwork
universe
Matter in extremes r2

2. •Matter in extremes considers how kinetic theory
explains the behaviour of matter in probabilistic and
mechanical terms. These ideas are extended to high
and low temperatures. The beginnings of the basis of
thermodynamic thinking appear in the form of the
Boltzmann factor. These are all fundamental ideas of
importance in understanding matter.

3. Candidates should be able to:plan and perform experiments to confirm or to determine mathematical relationships, eg systems which may or may not produce
proportional, exponential, simple harmonic, inverse or inverse square relationships; these should incorporate the skills and techniques
developed throughout the AS course to ensure that the quality of data obtained is taken into account;
• identify, given appropriate data, the consequences of systematic error and uncertainty in measurements and the need to reduce these, eg in
measurement of astronomical distances related to calculations of the size or age of the universe, and the effect of increased resolution and
greater range of observations on those calculations;
•use computers to create and manipulate simple models of physical systems and to evaluate the strengths and weaknesses of the use of
computer models in analysis of physical systems, eg the approximations and simplifications necessary in computer models, the ability of
powerful computers to model very complex systems;
•plan, conduct, evaluate and further develop the practical study of a problem in an extended, open-ended investigation of a physical situation
chosen by the student (this investigation may be done within the teaching of this unit or within that of G495, at a time convenient for the
centre).
Candidates should be able to identify and describe:
the nature and use of mathematical models, eg models of random variation producing mathematical relationships, iterative and iconic
models to predict behaviour in systems too complex for simple analysis; systematic analysis, using physical principles, to produce
mathematical relationships;
changes in established scientific views with time, eg the explanatory power of Newton’s contributions to mechanics and gravitation, the
implications of special relativity, the cosmological view of the universe (which is being continually refined), the statistical nature of the kinetic
theory of gases and of the Boltzmann factor;
an issue arising from scientific research and development, stating more than one viewpoint that people might have about it, eg the
understanding gained from space exploration and its incidental benefits, such as prestige, international co-operation and development in
related disciplines, and drawbacks, such as cost and diversion of finds from other research.

4. •This module covers the core physics of radioactive decay
and decay of charge on a capacitor, energy and
momentum, the harmonic oscillator and circular orbits.
The field model is developed through consideration of
gravitational fields. The idea of differential equations and
their solution by numerical or graphical methods is built
up gradually using finite difference methods.
•Opportunities arise to discuss the place of mathematics
in physics: is mathematics part of the nature of the world
or an artefact of our way of doing things? There are
opportunities too for candidates to pursue their own
interests when considering applications of these ideas.

5. Recommended prior knowledge
•The work is a continuation of work from earlier
in the A2 course as well as picking up on some
ideas from GCSE. Candidates are expected to:
•• show knowledge and understanding of
conservation of energy and momentum (A2
RF1.2);
•• know about energy transfers as a result of
temperature differences (Sc

6. Momentum is always conserved in a closed system
Momentum has direction
Due to the conservation of momentum an equal and opposite force is always produced
Temperature measures the amount of KE in a system (AVERAGE KE)
-Heat always flows from hot to cold
Energy is assumed to be conserved in most equations.
Usually this therefore only involves the transfer of energy from one form to another

7. •RF 2.1 Matter: very simple
•The behaviour of an ideal gas is explained in terms
of kinetic theory. Its behaviour is understood as the
result of averaging over a very large number of
individual particle interactions.
•There are opportunities to consider the work in its
historical context, especially the resistance to ideas
about the existence of atoms and the nature of a
vacuum

10. Any object in a fluid that
displaces a volume of the fluid
that weighs more than the
object does will float upwards in
the gravitational field of earth.
Gases are a simple form of matter. Most of their behaviour depends on the motion of their
molecules.
pressure is due to the collisions of gas molecules with the walls of the container
the motion of molecules in a gas is random, gases have low density. Differences in density due
to differences in temperature are what drive the weather, through convection in the
atmosphere.

11. The Earth’s atmosphere provides an example of molecules gaining energy to diffusing upwards in the atmosphere against the
gravitational potential gradient.

12. Each of the gas laws applies to a fixed mass of gas
It stays the same
A theoretical gas that obeys the gas law at all
temperatures is called an ideal gas
They are good model for real gases.
pressure = force/area
Pressure α density
Pressure α Number of
molecules
at constant temperature
Pressure is greater when the molecules bump
harder and faster as more force and greater
momentum and work done
As you go to higher altitudes above the
earth the pressure and temperature
usually decrease
Atmospheric pressure decreases exponentially
with height. If we know the temperature at a
certain height we can calculate the air pressure at
that height
Pressure in a gas is caused by particles
bouncing off the walls. Particles are, however,
unpredictable. What appears to be a steady
state at the macroscopic level arises from
many unpredictable events at the microscopic
level. So you need to allow time for
equilibrium to be reached. Injecting a larger
number of atoms disturbs that equilibrium, so
this will be one thing to watch for.

13. Pressure p N/m2 or Pa
Volume V m3
Temperature T Degrees C or K
Mass M kg
Density ρ kg/m3
Although ideal gases obey the laws perfectly, they do not exist
they are a useful approximation for real gases and useful for devising
a mathematical model for the behaviour of gases.
R = 8.31 J K-1mol-1
NA = 6.022 × 1023 particles per mole
so k = R/NA = (8.31 J K-1mol-1)/( 6.022 × 1023 particles per mole) = 1.38 × 10-23 J K-1 per
particle
The number of particles in a mole of a substance is 6.022 × 1023 which isthe Avogadro number.
(1 dm3 = 1 litre).

14. We assume the relationship is
linear, without another
thermometer we cant check if
linear.
The different thermometers based on different quantities agree at fixed points but not necessarily at other
temperatures: this is because they assume a linear relationship.
Therefore its necessary for a universal temperature scale that uses the same temperature as a dependent
quantity.
Absolute zero
•The lowest possible temperature theoretically possible wow
•Zero kelvin
•At this temperature all particles have the minimum possible kinetic energy
•With the kelvin scale a particles energy is proportional to its temperature
P = 0 v =0 for a
fixed amount and
volume
Thermometric property
•All equations in thermal physics use k , +273

15. These are the fixed points used for the Kelvin Scale (or ideal gas scale)
0 Kelvin or Absolute Zero The triple point of water
about -273 degrees Centigrade about 273K or 0 degrees Centigrade
The temperature at which the p against T graph cuts
the x axis. Theoretically the coldest possible
temperature
The only temperature at which water, at standard pressure,
can exist in equilibrium in its 3 states
- The quantity we use for the scale uses the
product of pressure and volume of an ideal gas.
-Therefore using ideal gas pV proportional to T
- These thermometers agree well
To measure the temperature you need
something that varies with temperature-
Thermometric property
Examples are a volume of mercury, the resistance
of a piece of wire, pressure in a gas of fixed
volume
Usually the particles which have higher
energies pass to ones with lower.
(heat flows from hot to cold)This
happens until thermal equilibrium-
average temp- reached.

16. if volume decreases the pressure increases - They are inversely proportional
•-temperature and amount (mass) of gas unchanged, a closed system.
pV= constant
1
V
p1V1= p2V2
p α
This graph doesn’t represent
boyles law because v increases so
p should decrease- the opposite
The higher the temperature of the
gas the further the curve is from the
origin, never touch any axis
To see how hard the air pushes back.
More air into a fixed volume also increases the pressure so obviously
p α density (at constant temperature)
p α N (at constant temperature) which is still the density
V
P = constant x n/v
The gases exert a pressure due to small molecules rapidly moving.
Technically constant, is a constant x
N where N= number of molecules
p α m / V
for mass m or p µ N / V for N
molecules
Boyles law
Through origin
If p  0 v∞
Doesn’t touch
axis ever
PV = constant as
either changes
So you have Changed the density of gas
For a mass of gas m we have p  m / V and for N
molecules we have p  N / V.

17. Boyles law
Gases obey Boyles law very well except
for at high pressures ( as particles are
forced too close together intermolecular
forces become more significant and
affect this)
Boyle’slawandgasdensity
Boyle’slaw:
compressgastohalfvolume:
doublepressureanddensity
halfasmuchgasinhalfvolume:
samepressureanddensity
doublemassofgasinsamevolume:
doublepressureanddensity
Boyle’slawsaysthatgaspressureisproportionaltodensity
temperatureconstantineachcase
pressurep
2
34
1
volumeV
massm
densityd
pressurep
2
34
1
volumeV/2
massm
density2d
pushin
piston
pressure2p
2
34
1
volumeV/2
massm/2
densityd
volumeV
mass2m
density2d
pumpin
moreair
pressure2p
2
34
1
pressure and
density are
connected

18. B o y le ’s la w a n d n u m b e r o f m o le c u le s
Tw o w a y s to d o u b l e g a s p r e s s u r e
m o le c u le s in b o x :
p re ss u re d u e to im p a c ts
o f m o le c u le s w ith w a lls
o f b o x
p is to n s q u a s h e s u p s a m e m o le c u le s in t o
h a lf th e v o lu m e , s o d o u b le s th e n u m b e r p e r
u n it v o lu m e
a d d e x tra m o le cu le s t o d o u b le th e n u m b e r,
s o d o u b le th e n u m b e r p e r u n it v o lu m e
If p re s su re is pr op ortio na l to n u m ber o f im p ac ts
on w a ll p er s e c o n d
an d if nu m b er o f im pac ts o n w a ll p er s e co nd is
pr op o rtio na l to n u m be r of m o le c ul es p er u ni t
v ol um e
T h en p re s s u r e is p ro po rtio na l to n um be r o f
m ol ec u le s p e r un it v ol u m e
p = c o n s ta nt  N /V
p = c o n s ta n t  N /V
B o y le ’s law in tw o fo r m s
B o y le ’s la w s a y s t h a t p r e s s u r e is p r o p o r t io n a l to c r o w d in g o f m o le c u le s
p re s s u re p ro po rti on al to 1 /v ol um e
p  1/V
p re s s u re p ro po rtio n al to n u m b er o f m ol ec u le s
p  N
s q u a s h th e g a s
d e c re a s e V in c re a s e N
c ra m in m o r e m o l e c u le s
N m o le c u le s in v o lu m e V
p V = c o n s ta n t  N
2 N m o le c u le s in v o lu m e V
n um be r of m ol ec u les pe r
u n it v o lu m e
n um be r of im pa c ts o n w all
p e r s ec o nd
p re s s ur e
s am e :
N m o le c u le s
in v o lu m e
V /2
P r e s s u r e a n d v o lu m e o f g a s e s in c re a s in g w ith te m p e ra t u re
P re s s u re a n d v o lu m e e x tra p o la te t o z e ro a t s a m e te m p e ra tu re – 2 7 3 .1 6  C
S o d e f in e t h is t e m p e ra tu re a s z e ro o f K e lv in s c a le o f te m p e ra t u re , s y m b o l T
C o n s ta n t v o lu m e
h e a t g a s :
p re s s u re
in c re a s e s
T / C
p re s s u re p
4 5 . 1
C o n s ta n t p re s s u r e
h e a t g a s :
v o lu m e
in c re a s e s
T / C
4 5. 1
– 2 7 3 0
t e m p e ra t u re / C
– 2 7 3 0
t e m p e ra t u re / C
2 7 30
te m p e ra tu re /K
2 7 30
te m p e ra t u re /K
p re s s u re p ro p o rtio n a l t o K e lv in te m p e ra tu re v o lu m e p ro p o rtio n a l to K e lv in te m p e ra tu re
P r e s s u r e a n d v o l u m e a re p r o p o rtio n a l to a b s o lu te te m p e ra tu re
p   T V   T
2
34
1
v o lu m e V

19. Charles law
V= constant
T
V1 V2
T1 T2
V α T
This graph meets absolute
zero.
If the graph had been
plotted in kelvin it would go
through the origin
Absolute zero – v =0
=
•When the temperature increases the
volume also increases - they are directly
proportional to each otherFor a fixed mass of gas and
pressure in a closed system
Temperature is in kelvin!

20. When cooled a real gas condenses to liquid. The
molecules are nearly touching and the liquid
occupies a noticeable volume within the
container. The ideal gas equation assumes that
the molecules occupy no volume.
Therefore real gases behave more like ideal gases
at high temperatures. Because the volume
occupied by the gas molecules becomes
negligible at high temperatures when the gas has
expanded.
Gases like helium, hydrogen nitrogen and
oxygen have low boiling points so these
behave like ideal gases at room
temperature. Camping gas (butane) boils at -
0.5 0C and is easily liquefied at room
temperature by pressurising it. We would
not expect this to behave like an ideal gas at
room temperature.

21. •When the pressure increases the absolute temperature increases-
 they are directly proportional to each other
p= constant
T
p1 p2
T1 T2
p α T
This graph meets
absolute zero.
If the graph had been
plotted in kelvin it
would go through the
origin
=
At constant volume and mass / amount the
pressure increases as temperature increases
At absolute zero pressure = 0
Increasing the temperature increases the speed of
particles and kinetic energy, therefore temp.
The pressure law

22. Combining the laws
Combining p α 1/v and p α N
p α 1/v α N
pv α N or p α N/V
Combining p α T and V α T
pV α T
Combining pV α T and pV α N
pV α NT
We can combine these by thinking of the meaning
- p is proportional to T
- As p increases T increases - by the same factor
-V is proportional to T so as V increases so does T by the same factor
-Therefore if T increases both V and T increase
-The product of V and p increases therefore as T increases
-So product pV is proportional to T
Using if two proportionalities include a common title then the new
combination will be the common title proportional to the product of the
other two titles on the other side of the proportionality sign.
Introducing constant
pV=NkT
pV=NkT
*We don’t know the value for k constant
* We don’t know the value for N as number of molecules wasn’t known
*but by experimenting the value for Nk was found and called the gas
constant
* We replace this with R and include number of moles as n
Giving  Pv=nRT
We assume the gas laws and fixed mass of gas.
For 1 mole of ideal gas the constant of
proportionality is R
So nR for n number of moles.
8.3 J/mol K experimental value universal
molar gas constant
Allowed to rewrite ideal gas formula
Gas constant per mole
NA avogadros number – number of particles in a mole

23. N = nNA Nk =nR
(number of moles n)
Avogadro constant NA = 6.022 x 10 23
R =NA k
Remember often in questions it may be simpler than using the pV=nRt formula, usually it just
uses proportionality, if it’s the same gas, same amount look out for this
Mr x Moles = Mass

24. pV=constant and pv/T = constant (for fixed mass of gas)
For same gas:
In 1827 a Scottish scientist named Robert Brown observed the
random motion of smoke particles under a microscope.
Originally it was explained as the effect of convection but later
it was realised that the smoke particles were being struck by
much smaller particles. So small that they couldn't be seen
with the most powerful microscope, and moving very quickly
in random motion.
Using the gas laws for a fixed amount of same gas
O n e la w fo r a ll g a s e s
2
34
1
v o lu m e V
B o y l e ’s l a w
p re s s u re p
c o m p re s s g a s :
p re s s u re p in c re a s e s
c o n s ta n t te m p e ra t u re T
2
34
1
n u m b e r N
A m o u n t l a w
p re s s u re p
a d d m o re m o le c u le s :
p re s s u re p in c re a s e s
c o n s t a n t te m p e ra tu re T
2
34
1
P r e s s u r e l a w
p re s s u re p
h e a t g a s :
p re ss u re p in c re a se s
c o n s ta n t v o lu m e V
T /K
45 . 1
v o lu m e V
C h a r le s ’ la w
h e a t g a s :
v o lu m e V in c re a s e s
c o n s ta n t p re s s u re p
T /K
45 . 1
p   1 /V
p  N
p  T
V  T
C o m b in e th e
re la tio n s h i p s in to o n e
p  N /V
p V   N
o r
p V   N T
in tro d u c e
c o n s ta n t k :
p V  N k T
c o m b in e :
c o m b in e :
C o m b in e u n k n o w n N
a n d k i n to m e a s u ra b le
q u a n ti ty R
N u m b e r o f m o le c u le s
N n o t k n o w n
c o n s ta n t k n o t k n o w n
N k c a n b e m e a s u re d :
N k = p V /T
F o r o n e m o le , d e fin e
R = N A
k
F o r n m o le s:
p V = n R T
p V  T
k = B o lt zm a n n co n sta n t
N A = A v o g a d ro n u m b e r
(n u m b e r o f m o le cu le s p e r m o le )
R = m o la r g a s co n s ta n t
= 8 .3 1 J K – 1
m o l– 1
m e a su re d from p V /T fo r o n e m o le
W h e n N A
c o u ld b e m e a s u re d :
A v o g a d r o nu m b e r N A
= 6. 0 2  1 0 2 3
p a r tic le s m o l – 1
R = m o la r g a s c o n s ta n t = N A
k = 8 .3 1 J K – 1
m o l – 1
B o ltz m a n n c o n s ta n t k = 1 .3 8  1 0 – 2 3
J K – 1
c o m b in e :
4
5
0
1
2
3
 1 0
5
N m – 2
F o r N m o le c u le s p V = N k T. F o r n m o le s, N = n N A a n d p V = n N A k T = n R T.
The Pressure Law Charles' Law Boyle's Law
If we increase the
temperature the balls whiz
around faster. They hit the
walls with more force, and
more often, so the pressure
increases.
Volume increases
due to force
exerted due to
KE increase from
temperature
increase.
If w make the volume
of the container
smaller then the
particles hit the walls
more often so the
pressure increases.

25. Kinetic theory
We model the particles as balls inside a box. The force they produce
when they collide with the walls results in the pressure of the gas.
The best models are simple so we make assumptions
about our tiny particles
They are perfectly elastic spheres, Collisions are
elastic
They don’t lose kinetic energy when they collide,
otherwise they would slow down.
They do not interact with each other, only with the
walls of the container
The forces between the particles we assume can be
ignored
The sample of gas is large, containing many
molecules – apply statistics
So molecules move randomly in all directions,
different speeds no resultant force eg gravity
So there are equal numbers moving in all directions
producing uniform pressure
The volume occupied by the particles is negligible
compared to the volume of the container. i.e. there
are big spaces
The length of time involved in a collision is negligible
compared to the time between collisions (i.e. we can
ignore the moments when the potential energy
component of the internal energy is not zero).
Forces that act in collisions are instantaneous
Motion of particles follows newtons laws
The Pressure Law Charles' Law Boyle's Law
Model predicts p is
proportional to the mean
KE of the molecules (when
V constant).
Since mean KE of
molecules is proportional
to T we expect p to be
proportional to T.
Model predicts V is
proportional to the
mean KE of the
molecules (when p
constant).
Since mean KE of the
molecules is
proportional to T we
expect V to be
proportional to T.
Model predicts pV will be a constant
(provided the number and speed of
the molecules does not change).
But temperature is linked to the
speed of molecules since
1/2 mv2 = 3/2 kT. To keep speed of
the molecules constant, the
temperature must remain constant –
this is true for Boyle’s law to work.
Evidence for kinetic theory
Diffusion, Brownian motion, vacuum
The collisions are assumed to be elastic, this will be
true on average if the walls are at the same
temperature as the gas,
The model will fit a real gas better if:
Low density – molecules are further apart and the space the
molecules occupy is too small, negligible
Energy of motion of the molecules is large high
temperatures so that any attractions between them can be
ignored (model assumes none)

26. With N molecules of gas…
each hith velocity u1 , u2 …. Un
A single molecule: repeatedly colliding with face (touch, back
touch again)
∆t= 2x / u1 u1 = d/t = (2x) / ∆t as there and back 2x
Impacts per sec: (1/t) = u1 / (2x)
∆mv = 2mu1 (as changes from -mu1 to +mu1 )
Force = ∆mv / ∆t = 2mu1 / (2x/u1) = mu1
2 / x
Pressure = force / area = (mu1
2 / x) / x2 = mu1
2 / x3
For a single molecule though, for all molecules we will need to consider possible directions, on
average a third will go in x direction. All will have separate velocities
All molecules:
Pressure= (1/3) (m/ x3) (u1 + u2 + u3 ….. +u n )
We use the mean square velocity c2 and N , number of molecules:
Pressure = (1/3) (m/ x3) (Nc2 )
As x3 is volume: or as density is m/v:
. .
pV = (1/3) mNC2 p =(1/3) p C2
The volume of the box is inversely
proportional to pressure
Particles will have less distance to travel
so therefore increased impacts per sec.
The smaller area and greater force mean
greater pressure

27. K in e tic m o d e l o f a g a s
U s e c h a n g e o f m o m e n tu mTo s ta rt: o n e m o le c u le in a b o x
z
x
v y
ro u n d trip -tim e b e tw e e n c o llis io n s  t = 2 x / v
c o llis io n s p e r s e c o n d = v / 2 x
e n d w a ll o f b o x
m o m e n tu m + m v b e fo re
m o m e n tu m – m v a fte r
im p u ls e o n w a ll
w a ll h a s c h a n g e in
m o m e n tu m + 2 m v
b a ll h a s c h a n g e in
m o m e n tu m – 2 m v
m o m e n tu m 2 m v g iv e n to w a ll a t e a c h c o llis io n
F o rc e = ra te o f c h a n g e o f m o m e n tu m
fo rc e o n w a ll =
m o m e n tu m p e r c o llis io n  c o llis io n s p e r s e c o n d
fo rc e o n w a ll = m v 2
/ x
2 m v v / 2 x
im p u ls e e a c h tim e m o le c u le re tu rn s
fo rc e
tim e
 t
z
x
y
C a lc u la te p re s s u re = fo rc e o n w a ll / a re a o f w a ll
fo rc e o n w a ll = m v 2
/x
p re s s u re = m v 2
/x y z
p re s s u re p = m v 2
/ V
(a re a = y z )
(V = x y z )
fo rc e
tim e
N tim e s a s m a n y c o llis io n s p e r s e c o n d
N m o le c u le s
p re s s u re p = N m v 2
/V
a re a o f w a ll = y z
x y z = v o lu m e V
p re s s u re p
a d d m a n y m o le c u le s a ll d o in g th e s a m e im p ro v e m o d e l
fo rc e
tim e
1 /3 a s m a n y c o llis io n s p e r s e c o n d1 /3 o f m o le c u le s
in e a c h d ire c tio n ,
o n a v e ra g e
fo rc e
tim e
p re s s u re p = N m v 2
/V
a v e ra g e im p u ls e s ta y s th e s a m e
ta k e a v e ra g e
o v e r v 2
im p ro v e m o d e l
p re s s u re p = N m v 2
/V
a llo w m o le c u le s to m o v e in ra n d o m d ire c tio n s im p r o v e m o d e l
a llo w m o le c u le s to m o v e a t ra n d o m s p e e d s
1
3
1
3
T h e k in e tic th e o ry o f g a s e s p re d ic ts th a t p V = N m v 21
A stripped down view of the kinetic theory
general ideas ideas about a particle in a box
F = –
p
t
force equal to
rate of change
of momentum
p = – 2mv
t =
2c
v
F = mv2
c
P = mv2
abc
P = mv2
V
P = Nmv2
V
P = Nmv2
V
1
3
P = Nmv2
V
1
3
particle mass, m ,
speed, v, travelling
2c between hits
A = a×b
box dimensions:
a×b×c
V = a×b×c
P =
F
A
definition of
pressure
N particles
1
3
in each direction
random
movement
gas molecules moving
at many speeds,
special average, v2

28. This derived equation from the mathematical model works well for real gases.
Very small value for the average value of all the velocities squared
For a very large number of molecules moving randomly the
average velocity is zero.
Because of the vector quantity, they cancel out.
The average speed wont be.
The square root of the average velocity squared is a good
approximation for average speed.
The root mean square velocity
found by taking the average of all the
''squared speeds'' of the molecules.
Squaring the speeds and taking the average
If you take the square root of
this you get the root of
the average squared
speeds or root mean
square speed.
This is not the same as the
mean speed or average speed
of the molecules
Higher temperature
shows higher average
particle speed,
distribution curve
becomes spread out
Constant collisions occur
Energy transferred, some gain or lose
speed, collisions dont affect total
energy of system,
Average speed stays the same
provided temperature does

29. Speedofanitrogenmolecule
AssumewarmroomtemperatureT=300K
massof1moleofN2=2810–3
kgmol–1
AvogadroconstantNA
=61023particlesmol–1
Boltzmannconstantk=1.3810–23
JK–1
kineticenergyofamolecule
fromdynamics fromkineticmodel
mv21
2
kT3
2v2=
3kT
m
massmofN2
molecule calculatespeed
v=500ms–1approximately
Airmolecules(mostlynitrogen)atroomtemperaturegoasfastasbullets
m=
massof1moleofN2
AvogadroconstantNA
m=
m=4.710–26kg
v2
=
31.410–23JK–1300K
4.710–26
kg
v2
=2.7105
Jkg–1
[Jkg–1(ms–1
)2
]
2810–3
kgmol–1
61023
mol–1
The answer given above is the root mean square speed of the gas molecule.

30. Randomly moving molecules are going nowhere in particular.
That is, they go anywhere. A simple general rule is that randomly
moving molecules tend on average to go from where there are a
lot to where there are not.
a single molecule in a gas moves in a
random walk as a result of the many
collisions it makes with other molecules
T o ta l d is ta n c e X tra v e lle d in N s te p s
S im p lifie d o n e - d im e n s io n a l ra n d o m w a lk
R u le : a p a r tic le m o v e s o n e s te p a t a tim e , w ith e q u a l p ro b a b ility to th e rig h t o r le ft
N o ta t io n : le t th e s te p s b e x 1 ,x 2 ,x 3 e tc . w ith e a c h x e q u a l to + 1 o r – 1
X = (x 1 + x 2 + x 3 + ... + x i + ...x N )
E x p e c te d v a lu e E (X ) o f X
Va ria tio n o f X a ro u n d t h e e x p e c te d v a lu e
r = X – E (X )
D e p a rtu re r fro m e x p e c te d v a lu e is :
E (r
2
) = E (X
2
)
E (X ) = (0 )
S in c e E ( X ) = 0
E x p e c te d v a lu e o f r2
E (r
2
) = E (X
2
)
= E (x 1 + x2 + x3 + ... + xi + ...xN )2
T h e e x p e c te d v a lu e o f r is z e r o . B u t
th e e x p e c te d v a lu e o f r2 is n o t z e ro .
S in c e E (X ) = 0 th e e x p e c te d v a lu e
o f r 2 is th e s a m e a s th e e x p e c te d
v a lu e o f x 2
s q u a re d te rm s :
x 1
2 + x 2
2 + x 3
2 + ... + x i
2 + ...x N
2
th is c o n ta in s N te rm s
e a c h e q u a l to + 1
e x p e c te d v a lu e = N
( + 1 )2 = + 1
( – 1 )2 = + 1
m ix e d te rm s :
x 1 x 2 + x 1 x 3 + ... + x i x j + ...
th is c o n ta in s
N (N – 1 ) te r m s
e a c h e q u a lly
lik e ly to b e
+ 1 o r – 1
(+ 1 )  (+ 1 ) = + 1
(+ 1 )  (–1 ) = – 1
(–1 )  (+ 1 ) = – 1
(–1 )  (– 1 ) = + 1
T h e re s u lt is s im p le :
m e a n s q u a re v a ria tio n
E (r
2
) = N
ro o t m e a n s q u a re v a ria tio n
 = N
E a c h s te p is e q u a lly lik e ly to b e + 1
o r – 1 . T h us , o n a v era g e , o v e r m an y
ra n d o m w alk s , th e to ta l dis ta n c e w ill
a d d u p to z e ro .
W h e n m u litip lie d o u t th is
g iv e s tw o ty p e s o f te rm :
e x p e c te d v a lu e = 0
c o m pa re ( a + b )2 = a 2 + a b + b a + b 2
R o o t m e a n s q u a r e d is ta n ce tra v e lled in r an d o m w alk = N
Takes longer than predicted as does not go
in straight line. Diffusion is slow even
though they move fast.
Random walks
the average distance R from the origin at
which the walker ends up after N steps of
length 1, is equal to

31. Brownian motionThe movement of gas particles, seen in smoke, vibrating slightly with small movement.
Due to air particles knocking into them at high speed causing slight movements.
Seen in pollen grains and water particles crashing into.
Shows evidence for air particles and their high speed movement , collisions (kinetic
theory). They have internal energy KE
Collisions produce a force!
Evidence for the
kinetic theory of
gases-->
Diffusion,
Brownian motion
Expansion into a
vacuum

32. Energy in a gas
In our simple model the internal energy U of the
gas consists entirely of kinetic energy. We
ignore potential energy, energy of vibration and
energy of rotation.
We consider the internal energy of a substance made up of
two components:
KE and PE
EK is the vibration, rotation or translation of the atoms
But in ideal gases we only consider translational
Ep is due to the interaction of one atom and its neighbours
An ideal gas has no Ep ; all its internal
energy is from Ek
Because in ideal gases each atom acts
independently, other atoms don’t exist
except in respect to collisions, so we
assume no atoms and therefore no force
exerted on neighbours.
For an ideal gas and most equations we ignore PE and
vibrational and rotational KE (NOT TRANSLATIONAL KE)
-doesn’t take into account vibrational energies
- Or rotational energy
- Collisions in all directions equally assumed, randomly, assumed straight
- Completely elastic collisions assumed
It can be shown that the total internal
energy U of an ideal gas is given by
__
U = 1 Nmv2 = 3 NkT = 3 pV
2 2 2

33. REST ENERGY SHOWN BY E = mc2
-applied to rest mass/energy to show
their proportionality E0 = m0c2
- As all objects have internal energy,
depending on their mass
-Therefore another way of changing internal
energy without doing work or heating is by
changing mass
Masses are kept constant when changing internal
energy by heat or work for this reason
The transfer /loss/increase of KE changes the temperature
energy transfer producing a change in temperature
The second law of thermodynamics
(the number of ways W that the molecules and energy they
share) never decreases, generally increases
Heat can never pass from a colder to a warmer
body without some other change
Reactions involve changes in the special arrangements
aswell as energy Entropy, S = klogW
Thermal equilibrium is where two sources do not
change in temperature as they have reached an
equilibrium
This can occur with more than two sources

34. So during times of increase of temperature (e.g. a, c and d from the graph)
the heat energy supplied is going into the Ek component of the internal
energy.
And during changes of state (e.g. b and d) it's going to Ep. Because Ek isn't
changing during these times, the temperature remains the same.
the energy of an atom is made up of two components - Ek and Ep
The Ek component relates to the temperature of the substance
The Ep component relates to the state of the substance.
_
1 mc2 = 3 nRT
2 2 N
_
1 mc2 = average kinetic energy of an individual particle
2
So as average kinetic energy is directly
proportional to absolute temperature.
Rise in temp affects internal energy
Derived from previous
equation when looking
at the energy of an
individual particle

35. The average translational energy KE of a particle is kT
Where k is the boltzmann constant and T is the tem in k
Its considered an assumption as very very rough estimate
For N particles NkT is
average energy
In gases we use 3/2 as they move around
therefore we consider volume  3/2 NkT
The total internal energy U of an ideal
gas is given by U = (3/2) NkT if energy
of vibration and rotation can be
ignored
If considered as an ideal gas
This means it only has translational KE
Not PE or vibration, rotation
Temperature relates to the probability that particles
occupy different energy states
Higher temp means higher average energy
Particles have average energy roughly equal to kT
The kinetic theory of gases shows us that the mean
kinetic energy per particle in an ideal gas is (3/2)kT.
We can extend this idea to all matter and generalise by stating
that the mean energy per particle is approximately kT.
note that the average kinetic energy used here is only the translational kinetic energy
of the molecules. That is, the energy of the molecules moving in a linear path is taken into
account and no account is made of internal degrees of freedom such as molecular rotation and
vibration. collisional transfer.
The kinetic temperature is the
variable needed for subjects like
heat transfer, because it is the
translational kinetic energy which
leads to energy transfer from a hot
area (larger kinetic temperature,
higher molecular speeds) to a cold
area (lower molecular speeds) in
direct
Average energy is proportional to absolute temp

36. By compressing a gas, pushing molecules into a restricted space, their average kinetic energy
increases. You have to push the gas into shape, knocking into the molecules, increasing their
speed.
◦ Allowing a gas to expand, done by permitting gas molecules to transfer some of their
momentum to a piston, results in a reduction of the average kinetic energy of the molecules –
the gas cools
The first law of thermodynamics describes how energy can be added or removed from
matter in two different ways, doing work and heating.
At constant volume, W = 0, and all the
energy goes into changing the internal
energy of the matter. The relationship
can then be written as U = mcθ.

37. As everything has internal energy, the internal energy of a gas can be changed without
removing any atoms.
-By heating or cooling transferring heat ΔQ
By heating up a gas you pass KE to the atoms, heat flowing into the gas Is positive
-By compressing or expanding work done ΔW
If compressed or expands pressure/volume changes. If gas
expands it exerts a force around it. It has to do work (use energy) by pushing
it surroundings, positive .
There are only two ways of
changing the internal energy U,
of a gas (without adding or
removing any atoms)
Changing internal energy
ΔU = ΔQ - ΔW
Work done + thermal transfer
Force x distance + mcΔθ
The internal energy U of a
material can be changed by
doing work W or by thermal
transfer of energy Q
(expresses the law of conservation of
energy- the first law of thermodynamics)
F = P x A as P=F/A
Or ΔQ = ΔU + ΔW
Note: the negative is due to the fact that work
done by gas is positive. The gas loses energy
by doing work. If something else does work
on the gas then positive.
The question will tell you what the work is done by look out for
this.
For constant pressure processes use:
The cylinder has a frictionless piston, of area A is at
pressure P, If it expands and pushes back the piston by
a very small distance x (so small the pressure doesn’t
change) W =f s f = P A
W= P A S W= PΔV for constant pressure
For constant pressure processes use:
ΔU = ΔQ - pΔV

38. Example
200kg of water is heated from 10 degrees to 40 degrees, C= 4180 Jkg-1K-1
Calculate the change of internal energy of the water
ΔE= 200 x 4180 x 30
ΔE= 25080000 J
ΔE= 25.1 MJ
Example
A 1.5 kg stone at 200 degrees is dropped into water at 10 degrees which
is 300kg, calculate change of internal energy of stone.
ΔE= 1.5 x 4180 x 190
ΔE= 171000J
ΔE= 171 kJ
Asks for change in specific
object, Meaning m for that
object, although we use the
specific heat capacity for
that which we are
transferring energy to.
-The energy transfer show closed system
assumption, all energy goes into heating
-mass in kg, Δθ in kelvin
Or ΔQ = ΔU + ΔW ∆Q = ∆ KE = mc∆θ
Note: the value of 'c' for ice isn't the same as that for water or
for steam. However, the value of 'c' is the same if you are cooling
rather than heating the substance.
In other words you get as much heat back out of the substance
when you cool it as you put in when you heated it.
Thermal heat
capacity, C = Jkg-1
k-1
At constant volume no work done so change in energy U is due to change in Q only
Also if other values are kept constant and
temperature is changed if specific heat
capacity isn't given E = NkΔt can be
used
Say temperature is changed mass and
specific heat unknown but increase in KE
due to THERMAL CHANGE found from
change in temperature and number of
particles

39. Amount of energy needed to raise the temperature
of one kilogram of the material by 1 degree.
Important when working with materials that changes temperature or
is designed to retain thermal energy.
The definition of Specific Heat Capacity is:
Water has a very high thermal
heat capacity with important
environmental and practical
consequences

40. Average energy per particle kT
an average energy of thermal activity of the order kT. Values of kT
may be expressed in joule per particle, electron volt, or kJ mol-1.
.
KE = 0.5 mv2 pV = NKT pV = (1/3) N m c2
NKT = (1/3) N m c2
KT = (1/3) m c2
(3/2)KT = (1/2) m c2
EK= (3/2) NKT
(average kinetic energy for a gas)
Average KE of a molecule = (3/2)kT
K in e t ic e n e r g y o f g a s m o le c u le s a n d th e B o lt z m a n n c o n s ta n t
c o m p a re th e s e
K in etic m o d e l
p V = N m v 2
G a s law s
p V = N k T
O n e m o le c u le M a n y m o le c u les
m v2
= kT1
2
3
2
k in e tic e n erg y pe r
m o lec u le = k T3
2
N m v 2
= N k T
1
2
tota l k in etic e n erg y o f
m o le c u le s = N k T3
2
3
2
m v 2
= 3 k T
to ta l k in e tic e n e r g y o f
o n e m o le o f
m o le c u le s U = R T
3
2
In te rn a l k ine tic en e rg y
of m o le c u le s o f o n e
m ole a t T = 3 0 0 K
B o ltz m a nn c on s tan t k
ran d om th e rm al
e ne rgy o f o n e
m o le c ule is of o rd e r
k T
1
3
U = R T
3
2
U = 3.7 k J m o l– 1
fo r o ne m o le
N = N A
R = N A k
R = 8.3 1 J K – 1
m o l– 1
A v e ra g e kin e tic e n e r g y o f a m o le c u le =
3
2 k T
For one mole:
N = NA
R = Nak
R = 8.31 JK-1mol-1
u = (3/2) NkT = (3/2) RT
For one mole only
u =(3/2) nRT for n moles
Average KINETIC ENERGY of the particles
Confirms that the temperature of a gas
depends on only the Ek of the gas molecules
Rest energy = mc2

41. -As temperature is related to average
translational kinetic energy (KE) associated
with the disordered microscopic motion of
atoms and molecules.
-The flow of heat is from a high temperature
region toward a lower temperature region.
The temperature defined from kinetic theory is called the kinetic
temperature. Temperature is not directly proportional to internal energy
since temperature measures only the kinetic energy part of the internal
energy, so two objects with the same temperature do not in general have
the same internal energy
When the sample of water and copper
are both heated by 1°C, the addition to
the average kinetic energy (translational
KE) is the same, since that is what
temperature measures. But to achieve
this increase for water, a much larger
proportional energy must be added to
the potential energy portion of the
internal energy. So the total energy
required to increase the temperature of
the water is much larger, i.e., its specific
heat is much larger.

42. •RF 2.2 Matter: hot or cold
•In this section, temperature is related to the probability
that particles occupy quantum states of different
energies and the Boltzmann factor is introduced as the
link between energy and temperature. The important
idea that differences drive change is developed here.
•Opportunities are provided for considering a range of
contemporary topics such as the state and behaviour of
matter in the early universe, superconductivity at low
temperatures, the behaviour of 'soft matter' (such as
polymers) and rates of reaction

44. Activation processes
A c t i v a t i o n p r o c e s s e s
A n e n e r g y h i ll h a s t o b e c lim b e d
b e f o r e t h e p ro c e s s c a n h a p p e n
T h e e n e r g y n e e d e d h a s t o b e
a c q u ir e d b y c h a n c e fr o m ra n d o m
th e r m a l a g it a t io n o f th e s u r ro u n d in g s
e n e r g y n e e d e d fo r
p ro c e s s t o b e p o s s ib le
E x a m p l e s o f a c t i v a t i o n p r o c e s s e s
V i s c o s i t y o f l i q u i d s
p a r ti c le s n e e d
e n o u g h e n e rg y t o
b r e a k o u t o f ‘c a g e ’
f o r m e d b y t h e ir
n e i g h b o u rs , s o liq u id
c a n f lo w
a t o m s n e e d
e n o u g h e n e r g y t o
b e c o m e io n is e d ,
f r e e in g e le c t ro n s
w h ic h c a n c o n d u c t
I o n i s a t i o n o f s e m i c o n d u c t o r
E v a p o r a t io n o f l i q u i d s I o n i s a t i o n i n a f l a m e
M a n y c h e m i c a l r e a c t i o n s N u c l e a r f u s i o n i n th e S u n
p a rt ic le s n e e d e n o u g h
e n e rg y t o b re a k a w a y
f r o m a t t ra c t io n s o f
p a r t ic le s a t s u rf a c e ,
a n d l e a v e t h e li q u id
m o le c u le s n e e d
e n o u g h e n e rg y t o
d is s o c ia t e in t o io n s ,
w h ic h c a n c o n d u c t a
c u r r e n t
p a r t ic le s m u s t c o lli d e
w it h e n o u g h e n e r g y
t o g e t c lo s e e n o u g h
t o b o n d t o m a k e a
n e w m o le c u le
p r o t o n s m u s t h a v e
e n o u g h e n e rg y t o g e t
c lo s e e n o u g h to f u s e
in t o d e u t e r iu m
a c t iv a ti o n
e n e r g y 
+
_
_
+
+
+
+
_
_
_
+ +
T h e e n e r g y n e e d e d fo r a p r o c e s s to h a p p e n h a s to b e a c q u ir e d b y c h a n c e
f r o m t h e r a n d o m th e r m a l a g i ta t i o n o f th e s u r r o u n d in g s
A c t i v a t i o n p r o c e s s e s
A n e n e r g y h i l l h a s t o b e c l i m b e d
b e f o r e t h e p r o c e s s c a n h a p p e n
T h e e n e r g y n e e d e d h a s t o b e
a c q u i r e d b y c h a n c e fr o m r a n d o m
th e r m a l a g i t a t i o n o f t h e s u r r o u n d i n g s
e n e r g y n e e d e d fo r
p r o c e s s t o b e p o s s i b l e
E x a m p l e s o f a c t i v a t i o n p r o c e s s e s
V i s c o s i t y o f l i q u i d s
p a r ti c l e s n e e d
e n o u g h e n e r g y t o
b r e a k o u t o f ‘ c a g e ’
f o r m e d b y t h e i r
n e i g h b o u r s , s o l i q u i d
c a n f l o w
a t o m s n e e d
e n o u g h e n e r g y t o
b e c o m e i o n i s e d ,
f r e e i n g e l e c t r o n s
w h i c h c a n c o n d u c t
I o n i s a t i o n o f s e m i c o n d u c t o r
E v a p o r a t i o n o f l i q u i d s I o n i s a t i o n i n a f l a m e
M a n y c h e m i c a l r e a c t i o n s N u c l e a r f u s i o n i n t h e S u n
p a r t ic l e s n e e d e n o u g h
e n e r g y t o b r e a k a w a y
f r o m a t t r a c t i o n s o f
p a r t i c l e s a t s u r f a c e ,
a n d l e a v e t h e l i q u i d
m o l e c u l e s n e e d
e n o u g h e n e r g y t o
d i s s o c i a t e i n t o i o n s ,
w h i c h c a n c o n d u c t a
c u r r e n t
p a r t i c l e s m u s t c o l l i d e
w i t h e n o u g h e n e r g y
t o g e t c l o s e e n o u g h
t o b o n d t o m a k e a
n e w m o l e c u l e
p r o t o n s m u s t h a v e
e n o u g h e n e r g y t o g e t
c l o s e e n o u g h to f u s e
i n t o d e u t e r i u m
a c t i v a ti o n
e n e r g y 
+
_
_
+
+
+
+
_
_
_
+ +
T h e e n e r g y n e e d e d f o r a p r o c e s s t o h a p p e n h a s t o b e a c q u ir e d b y c h a n c e
f r o m t h e r a n d o m t h e r m a l a g i t a t i o n o f t h e s u r r o u n d i n g s
Getting a lot of energy just by chance
the idea that only a few particles gain large
amounts of energy. Particles need a
consecutive run of luck to acquire large
amounts of energy (many times the average).
This leads to exponential behaviour,
described by the Boltzmann factor.
‘breakout’ Or
leave the
potential well
by getting lucky –
borrowing
energy from its
immediate
neighbours by
collision.
The higher kT the
more energetic, on
average, each
‘breakout attempt’
is. The temperature
may also increase
the number of
breakout attempts
per second

45. The number of particles acquiring
the extra activation energy is
governed, to a large extent, by the
Boltzmann factor.
Temperature dependence of the
Boltzmann factor explains why many
processes can double rate for a mere
10 -20 K rise
Atoms or molecules in potential wells,
collisions with others cause energy to be
gained at random. So randomly they gain
enough energy to leave the potential well.
The hotter the substance, the more often a
particle will by chance get the extra
energy it needs. Process succeed more
often at higher temperatures.
-when some particles acquire enough energy
by chance. changing
Activation Energy is the energy
required for a process. Particles must
overcome this "activation energy“.
Activation Energy limits whether a reaction can take
place- overcome an energy barrier before reaction or
change can occur, acquired by chance.
Many processes start happening at an appreciable rate when ε / kT is in the range
10 or 15 to 30.
Liquids (viscous liquids don’t flow well) flow is due to potential wells. If they are loosely
bonded (low E) they are more likely to have enough kT per particle to allow flowing. Water
flows freely because particles can have enough energy to overcome the bonds to the next to
flow. By increasing the temp therefore they are more likely to flow
Also by gaining more energy there will be an increase in the
frequency of collisions- more chances to gain yet

46. The magic ratio
ε
Where ε is the activation energy.
From this we see the chance of process occurring
and therefore the rate at which it occurs
kT
If a fraction "f" of molecules has kT + E energy, then a fraction
"f^2" will have kT + 2E energy, and so on.
The ratio  / kT is the deciding factor as to whether a process occurs or not:
 many processes start happening at an appreciable rate when  / kT is in the range 15 to 30.
IF THE TEMPERATURE IS TOO LOW RATIO WILL BE LARGE AND ABOVE RANGE
At high temperatures, bonds break and matter comes apart. Atoms come apart into ions and electrons. The ratio ε /
kT is small even for large ε.
At low temperatures, thermal activity is faint, ε / kT is large except for processes with very small ε.
Matter condenses to solid or liquid, and complex structures form.
Even though the fraction is very small, there are often very large numbers of particles colliding a huge number of
times a second. This means at an e/kT value of 15, the reaction occurs at a fair rate. Values less than 15
mean that the process occurs to quickly and values over 30 mean the process takes too long.
If e/kT gave us a value of 15, then the fraction of
particles with the activation energy will be very small,a
particle must get very lucky to have an energy of e or
higher.

47. Small amounts of energy are
measured in eV
1 eV = 1.6 x 10-19
J
Avogadro's number, particles in
a mole
1 mole = 6.02 x 1023
particles
1000
100
10
1
1/10
1/100
1/1000
1/10 000
0.1 1 10 100 1000 10 000 100 000 1 million 10 million
tem perature/K
1/10
1/1000
1
10
100
1000
10000
100 000
inner electrons in atoms liberated
1/40
kT at room temperature
surface of
Earth
liquid
nitrogen
liquid
helium
surface of Sun;
hot furnace
interior
of Sun
microwave
photons
super-
conductivity
air
liquid
plasmas
form
x-ray
photons
emitted
ultraviolet
photons
em itted
ceramics
evaporate
metals melt
soft matter
exists
water
evaporates
visible
photons
emitted
liquid nitrogen
comes apart
hydrogen bonds broken
van der Waals bonds broken
outer electrons in atoms liberated
ionic bonds broken
covalent bonds broken
metallic bonds broken
Processes and energy = kT per particle
Many processes requiring energy = 15–30 kT happen at an appreciable rate
2.5
If it falls within the 15-30 range it will occur
(evaporate etc) at an appreciative rate
Larger than 30 means none
Smaller than 15 means kT is higher (must be a gas---
> already evaporated)
gas particles collide into each other
exchanging energy. Each time they collide
they may gain or lose energy.
We imagine that this energy comes in
packets, or quanta.
Most of the particles in the gas have
energy k T. Some have more energy.
Knowing that the relationship is
exponential means we can calculate the
fraction of particles which have a certain
amount of additional energy.
e=hf I atom in a mole is
photon wavelength for E = hf = kT

48. The energy required to remove an electron
from platinum is 5.4 eV.
Removing electrons at 2000k an appreciable
rate is thought. Calculate kt then the ratio and
therefore state if right.
5.4 x 1.6 x 10-19 =8.64 x 10-19 J
8.64 x 10-19 / ( 2.76 x 10-20) = 31
As above 30 removing e-
not occuring as the
ratio is too high.
Energy to remove an electron 2.6 eV at what
temperature range is there appreciable emission?
2.6 eV → 4.16 x 10-19
J
4.16 x 10-19
J / kT =15
4.16 x 10-19
J / kT =30
Range 1010K to 2010K
40 kJmol-1 are required to evaporate water.
How many eV per particle is needed?
40000 J per 1 mole and there are 6.02 x 1023
particles
per mole...
40000 / 6.02 x 1023
= 6.64 x 10-20
J particle-1
6.64 x 10-20
/1.6 x 10-19
= 41.5 eV
Water evaporates at 300k, boils at 273k , true?
Calculate kt at 300k
kT= 300 x 1.38 x 10-23 = 4.14 x 10-21
What is the ratio of e to kt worth?
To evaporate water is 40 kJmol-1 and so 6.6 x 10-20 J
particle-1
e/kT = (6.6 x 10 -20 ) / (4.14 x 10-21) = 1.6 x 101 =16
Therefore shows at 300k, water does evaporate an
appreciable rate
A gas is at a temperature of 300K.
•What is the average energy of the particles?
•What fraction of the particles in the gas have 10
times this much energy?

49. The boltzmann factor
Gives the ratio of the numbers of
particles in energy states E joules apart
The rate of reaction with EA =E is
proportional to the Boltzmann factor
The Boltzmann factor is the
ratio of numbers of particles in
two quantum states differing
by energy 
The origin of the Boltzmann factor is the small
probability of repeatedly gaining extra energy at
random from a large collection of other particles.
To a first approximation the rate of a reaction with
activation energy  is proportional to e (– / kT), and
can increase rapidly with temperature.
0 ≤ fB ≥ 1

50. -a small increase in temp means a big
difference to rate
-Use fB to describe rate of reaction, as
both affected by temp
-Used to find probabilities of a particle
having certain energy
Low temp = low fB means slow reaction
-few particles have sufficient energy
(few have EA

51. graphs showing the variation of the Boltzmann factor with energy and
temperature;
The Boltzmann factor increases rapidly
with temperature. It roughly doubles
for about every 20K rise.
Temperature dependence of the
Boltzmann factor explains why
many processes can double rate for
a mere 10 -20 K rise
Change in Boltzmann factor with tem perature
+
+
+
+
+
+
+
+
+
++++++++++++
Boltzmann factor exp(–/kT) / 10–7
When  > kT the Boltzmann factor increases very rapidly with temperature
1.0
0
320300 340 360 380 400
T/K (linear scale)
/kT approximately 20 at 300k
Boltzmann factor
doubles between each
pair of temperatures
Boltzmann factor exp(–/kT)/10–7
Many important changes
start to happen at an
appreciable rate when /kT
is of the order of 15–30
2.0
3.0
low temperature
linear scale
0.0
1.0
high temperature
 = kT

52. B oltzm a nn fa c to r: th e m ino rity w ith e n e rg y e x p (– /k T )
1 0 – 4
1 0 – 8
1 0 – 12
1 0 – 16
e – 1
e – 10
e – 20
e – 3 0
e – 4 0
 / kT (lin e a r s ca le )
1B oltzm a nn
facto r as a
p ow e r of ten
e xp(– /kT )
log arithm ic
scale
C o ld a t this e nd of the sca le
T he te m pe ra tu re T is lo w
T he r atio  /kT is la rg e
T he fra ction exp (– /kT ) is sm all a nd ap pr oa ch es 0
H o t a t this e nd o f th e scale
T he te m p era ture T is h ig h
T he ratio /kT is sm all
T he fra ction exp( – /kT ) is la rge an d ap pr oa che s 1
35 4 03 02 520151051
w h ere th e a ction is:
 /kT fro m 15 to 3 0
+
+
+
+
+
+
+
+
+
B o ltz m a n n fa c to r p l o tte d ag ai n s t te m p e ra tu r e
line ar sca le
 e qu al to kT
B oltzm a nn fa ctor
e xp (– /kT )
1 .0
0 .8
0.6
0.4
0 .2
0 .0
10 0 1 00 0 1 0 0 00
T /K (lo ga rith m ic scale )
w e ake r bo n ds
h ave all co m e
a pa rt
bo nds com e
tog eth er
B o nd e n erg y
van d e r W aa ls
h ydro ge n bo nd
cova len t b o nd
C o ld at this e nd o f th e sca le
T h e tem p er atu re T is lo w
T h e ra tio /kT is lar ge
T h e fractio n e xp( –/kT ) is sm a ll a nd ap pro ach e s 0
H o t at this e nd o f th e sca le
The te m p era ture T is h ig h
The r atio /kT is sm all
The fra ctio n exp (– /kT) is lar ge an d a pp roa che s 1
T he B o ltzm a nn fa c tor e xp (– /kT ) is th e ra tio of nu m b e rs o f p artic le s in
s ta te s d iffe rin g b y e ne rgy . W h en >>k T the B o ltzm an n fa c tor in c re a s e s
v e ry ra p idly w ith tem pe ra tu re .
The Boltzmann factor changes very rapidly
with temperature in the region where
 / kT is of the order 15-30. This is often why
the rates at which various processes proceed
increase very rapidly with temperature.
The higher activation energy  (or higher
 / kT ) the further to the right the graph.
This means that it takes a higher
temperature to reach certain values for fB
Boltzmann factor against temp (k)
Highest EA

53. At high temperatures, the
Boltzmann factor
approaches 1, so nearly
all the particles will have
enough energy to react
and the reaction will be
really fast
This shape shows
that at low
temperatues, the
boltzmann factor is
also very low, so
very few (if any)
particles will have
sufficient energy to
react and the
reaction will be
really slow
In between the boltzmann factor increases rapidly
with temperature. So a small increase in temperature
can make a big difference to the rate

54. T h e B o lt z m a n n f a ct o r a n d t h e a tm o s p h e re
C olu m n of a ir in th e a tm o s ph e re
pre s s u re m us t b e la rg er
low er d o w n b ec a us e o f
ex tra w e igh t of
atm o s p he re a b ov e
n u m b e r p e r u nit v o lu m e n
p re s s u re p
m a s s o f e x tr a la y e r = n m A d h
w e ig ht o f e x t ra la y e r = n m g A d h
n u m b e r p e r u n it v o lu m e n + d n
p re s s u re p + d p
a s s um e: tem pe ra tu re s am e a t a ll h eigh ts
m = m as s o f m ole c u le
n = n u m b er o f m ol ec u le s pe r un it v o lum e
k = B oltz m an n c o ns ta nt
T = tem pe ra tur e
a re a A
low e r
p re s s u re h e re
ex tra w eig ht
hig h er
pr es s u re
he re
h e ig h t h + d h
h e ig h t h
d h
G a s la w s
p V = N k T
p = (N /V ) k T
p = nk T
(p re s s ur e in c re as e s w ith d en s ity )
d iffe re n ce in p re ss u re
d p = k T d n
Ex tra p re s su re d u e to w e ig h t o f e xtra lay er
pr es s u re d iffe re n c e b etw e en lay e rs
dp = w e ig ht of e x tr a la y er /a re a A
dp = – nm g A d h /A
(p res s u re de c re as e s w ith h ei gh t)
d iffe ren c e in p res s u re
dp = – nm g d h
k T dn = – nm g dh
dn /dh = – (m g /kT )n
ra te of c h an ge of nu m b er w ith h eig h t p ro po rtio na l to nu m b er
n /n 0
= ex p (– m g h /kT )
n /n 0
= ex p (– /kT )
 = m g h = di ffer en c e in po ten tial e n erg y
R atio of nu m be rs of par tic les in s tate s d iffer in g b y e ne rg y  is e qu al to th e
B oltz m a n n fa c to r e x p (– /k T )
 Reactions can also involve changes in the
number of spatial arrangements of particles.
At any height the fraction f of
molecules that have by
chance the extra e=mgh f=fb
Due to concentration gradient
in diffusion in atmosphere.
e –  / k T.