Rise and fall of the clockwork universe

Matter in extremes r2

•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.

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.

•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.

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

Momentum is always conserved in a closed systemMomentum 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

•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

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.

The Earth’s atmosphere provides an example of molecules gaining energy to diffusing upwards in the atmosphere against the

gravitational potential gradient.

Each of the gas laws applies to a fixed mass of gasIt stays the same

A theoretical gas that obeys the gas law at all temperatures is called an ideal gasThey are good model for real gases.

pressure = force/areaPressure α densityPressure α Number of moleculesat 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.

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 is the Avogadro number.

(1 dm3 = 1 litre).

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

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 propertyExamples 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.

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 αThis 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 densityV

P = constant x n/vThe 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 / Vfor mass m or p µ N / V for N molecules

Boyles law

Through originIf 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 Nmolecules we have p N / V.

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’s law and gas density

Boyle’s law:

compress gas to half volume:double pressure and density

half as much gas in half volume:same pressure and density

double mass of gas in same volume:double pressure and density

Boyle’s law says that gas pressure is proportional to density

temperature constant in each case

pressure p

2

34

1

volume V

mass mdensity d

pressure p

2

34

1

volume V/2mass mdensity 2d

push inpiston

pressure 2p2

34

1

volume V/2mass m/2

density d

volume Vmass 2m

density 2d

pump in

more air

pressure 2p2

34

1

pressure and

density are

connected

B o y le ’s la w a n d n u m b e r o f m o l e c u l e s

T w o w a y s t o d o u b l e g a s p r e s s u r e

m o l e c u le s i n b o x :

p r e s s u r e d u e to i m p a c ts

o f m o l e c u l e s w i th w a l ls

o f b o x

p i s to n s q u a s h e s u p s a m e m o l e c u l e s i n t o

h a lf th e v o l u 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 t r a m o l e c u l e s t o d o u b le t h e n u m b e r,

s o d o u b l e th e n u m b e r p e r u n it v o lu m e

I f p re s su re is p r o p o r t io n a l t o n u m b e r o f im p a c t so n w a ll p e r s e c o n d

a n d i f n u m b e r o f im p a c t s o n w a ll p e r s e co n d isp r o p o r t io n a l t o n u m b e r o f m o le c u l e s p e r u n i tv o l u m e

T h e n p re s s u r e is p ro p o r t io n a l t o n u m b e r o fm o l e c u le s p e r u n it v o l u m e

p = c o n s ta n t N /V

p = c o n s t a n t N / V

B o y l e ’ s l a w i n tw o fo r m s

B o y l e ’ s l a w s a y s t h a t p r e s s u r e i s p r o p o r t i o n a l to c r o w d i n g o f m o l e c u l e s

p re s s u re p ro p o r t i o n a l t o 1 / v o l u m ep 1 / V

p re s s u re p ro p o r t io n a l t o n u m b e r o f m o l e c u le sp N

s q u a s h t h e g a s

d e c r e a s e V i n c r e a s e N

c r a m in m o r e m o l e c u le s

N m o l e c u l e s i n v o lu m e V

p V = c o n s t a n t N

2 N m o l e c u le s i n v o lu m e Vn u m b e r o f m o l e c u le s p e ru n it v o lu m en u m b e r o f im p a c ts o n w a llp e r s e c o n dp re s s u r e

s a m e :N m o le c u le s

i n 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 i n c r e a s in g w i t h t e m p e r a t u r e

P r e s s u r e a n d v o lu m e e x t r a p o l a te t o z e r o a t s a m e te m p e r a tu r e – 2 7 3 .1 6 C

S o d e f in e t h i s t e m p e r a tu r e a s z e r o o f K e lv i n s c a le o f te m p e r a t u r e , 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 r e s s u r e

i n c r e a s e s

T / C

p r e s s u r e p

4 5 . 1

C o n s t a n t p r e s s u r e

h e a t g a s :

v o l u m e

in c r e a s e s

T / C

4 5. 1

– 2 7 3 0t e m p e r a t u r e / C

– 2 7 3 0t e m p e r a t u r e / C

2 7 30te m p e r a tu r e /K

2 7 30te m p e r a t u r e /K

p r e s s u r e p r o p o r t i o n a l t o K e l v i n t e m p e r a tu r e v o l u m e p r o p o r t i o n a l to K e l v i n te m p e r a tu r e

P r e s s u r e a n d v o l u m e a r e p r o p o r t io n a l t o a b s o lu t e te m p e r a t u r e

p T V T

2

34

1

v o lu m e V

Charles law

V= constantT

V1 V2T1 T2

V α T

This graph meets absolute zero.

If the graph had been plotted in kelvin it would go through the originAbsolute 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!

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.

•When the pressure increases the absolute temperature increases- they are directly proportional to each other

p= constantT

p1 p2T1 T2

p α TThis 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

Combining the lawsCombining 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 constantpV=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 nGiving Pv=nRT

We assume the gas laws and fixed mass of gas.

For 1 mole of ideal gas the constant of proportionality is RSo 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 moleNA avogadros number – number of particles in a mole

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

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 l a w f o r a ll g a s e s

2

34

1

v o l u m e V

B o y l e ’s l a w

p r e s s u r e p

c o m p r e s s g a s :

p r e s s u r e p in c r e a s e s

c o n s ta n t te m p e r a t u r e T

2

34

1

n u m b e r N

A m o u n t l a w

p r e s s u r e p

a d d m o r e m o l e c u l e s :

p r e s s u r e p i n c r e a s e s

c o n s t a n t te m p e r a tu r e T

2

34

1

P r e s s u r e l a w

p r e s s u r e p

h e a t g a s :

p re s s u re p in c r e a s e s

c o n s ta n t v o lu m e V

T /K

45 . 1

v o l u m e V

C h a r le s ’ la w

h e a t g a s :

v o l u m e V i n c r e a s e s

c o n s ta n t p r e s s u r e p

T /K

45 . 1

p 1 / V

p N

p T

V T

C o m b in e t h e

re la t io n s h i p s in t o o n e

p N /V

p V N

o r

p V N Ti n t r o 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 t i t y R

N u m b e r o f m o l e 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 r e d :

N k = p V /T

F o r o n e m o le , d e f i n e

R = NA

k

F o r n m o le s :p V = n R T

p V T

k = B o lt z m a n n c o n s ta n t

N A = A v o g a d r o n u m b e r

( n u m b e r o f m o l e c u le s p e r m o l e )

R = m o l a r g a s c o n s ta n t

= 8 .3 1 J K – 1 m o l– 1

m e a s u r e d f r om p V /T fo r o n e m o le

W h e n NA

c o u l d b e m e a s u r e d :

A v o g a d r o n u m b e r NA

= 6 . 0 2 1 02 3

p a r t ic le s m o l – 1

R = m o la r g a s c o n s t a n t = NA

k = 8 . 3 1 J K – 1 m o l – 1

B o lt z m a n n c o n s t a 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 05

N m– 2

F o r N m o l e c u l e 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.

Kinetic theoryWe 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 particlesThey are perfectly elastic spheres, Collisions are elasticThey 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 containerThe 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 gravitySo 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 instantaneousMotion 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 theoryDiffusion, 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)

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) = mu12 / x

Pressure = force / area = (mu12 / 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 velocitiesAll 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

K i n e t i c 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 t u mT o s ta r t : o n e m o le c u le in a b o x

zx

v y

r o u n d t r ip - t im e b e t w e e n c o l l i s i o n s t = 2 x / v

c o l l i s io n s p e r s e c o n d = v / 2 x

e n d w a l l o f b o x

m o m e n t u m + m v b e f o r e

m o m e n t u m – m v a f t e r

im p u ls e o n w a l l

w a l l h a s c h a n g e inm o m e n t u m + 2 m v

b a l l h a s c h a n g e inm o m e n t u m – 2 m v

m o m e n t u m 2 m v g iv e n t o w a l l a t e a c h c o l l i s io n

F o r c e = r a t e o f c h a n g e o f m o m e n t u m

f o r c e o n w a l l =m o m e n t u m p e r c o l l i s i o n c o l l i s io n s p e r s e c o n d

f o r c e o n w a l l = m v 2 / x

2 m v v / 2 x

im p u l s e e a c h t im e m o le c u le r e t u r n s

f o r c e

t im e

t

zx

y

C a l c u l a t e p r e s s u r e = f o r c e o n w a l l / a r e a o f w a l l

f o r c e o n w a l l = m v 2 / x

p r e s s u r e = m v 2 / x y z

p r e s s u r e p = m v 2 / V

( a r e a = y z )

( V = x y z )

f o r c e

t im eN t im e s a s m a n y c o l l is i o n s p e r s e c o n d

N m o le c u le sp r e s s u r e p = N m v 2 /V

a r e a o f w a l l = y z

x y z = v o lu m e V

p r e s s u r e p

a d d m a n y m o le c u le s a l l d o in g t h e s a m e i m p r o v e m o d e l

f o r c e

t im e1 / 3 a s m a n y c o l l is i o 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 ir e c t io n ,o n a v e r a g e

f o r c e

t im e

p r e s s u r e p = N m v 2 / V

a v e r a g e im p u l s e s ta y s t h e s a m e

ta k e a v e r a g eo v e r v 2

i m p r o v e m o d e l

p r e s s u r e p = N m v 2 / V

a l lo w m o le c u le s t o m o v e in r a n d o m d i r e c t i o n s i m 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 r a n d o m s p e e d s

1

3

1

3

T h e k i n e t i c t h e o r y o f g a s e s p r e d i c t s t h a t p V = N m v21

3

A stripped down view of the kinetic theory

general ideas ideas about a particle in a box

F = – p t

force equal torate of changeof m om entum

p = – 2m v

t = 2cv

F = m v2

c

P = m v2

abc

P = m v2

V

P = Nm v2

V

P = Nm v2

V13

P = Nm v2

V13

particle m ass, m ,speed, v, travelling2c between hits

A = a×b

box dim ensions:a×b×c

V = a×b×c

P = FA

definition ofpressure

N particles

13

in each direction

randomm ovem ent

gas m olecules m ovingat m any speeds,special average, v2

This derived equation from the mathematical model works well for real gases.

Very small value for the average value of all the velocities squaredFor 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 curvebecomes spread out

Constant collisions occurEnergy transferred, some gain or lose speed, collisions dont affect total energy of system,

Average speed stays the same provided temperature does

Speed of a nitrogen molecule

Assume warm room temperature T = 300 K

mass of 1 mole of N2 = 28 10

–3 kg mol–1

Avogadro constant NA = 6 10

23 particles mol–1

Boltzmann constant k = 1.38 10–23 J K

–1

kinetic energy of a molecule

from dynamics from kinetic model

mv212

kT32v2 =

3 kTm

mass m of N2 molecule calculate speed

v = 500 m s–1 approximately

Air molecules (mostly nitrogen) at room temperature go as fast as bullets

m = mass of 1 mole of N2

Avogadro constant NA

m =

m = 4.7 10–26 kg

v2 = 3 1.4 10–23 J K–1 300 K

4.7 10–26 kg

v2 = 2.7 105 J kg–1 [Jkg–1 (m s

–1)2]28 10

–3 kg mol–1

6 1023 mol–1

The answer given above is the root mean square speed of the gas molecule.

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 t a l d i s t a n c e X t r a v e l le d in N s t e p s

S i m p l i f i e d o n e - d im e n s io n a l r a n d o m w a lk

R u l e : a p a r t i c l e m o v e s o n e s t e p a t a t i m e , w i th e q u a l p r o b a b i l it y to th e r i g h t o r l e ft

N o t a t io n : l e t t h e s te p s b e x1

, x2

,x3

e tc . w it h 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 l u e E ( X ) o f X

V a r i a t i o n o f X a r o u n d t h e e x p e c t e d v a l u e

r = X – E (X )

D e p a r t u r e r f r o m e x p e c te d v a l u e i s :

E ( r2

) = E ( X2

)

E ( X ) = (0 )

S i n c e E ( X ) = 0

E x p e c t e d v a l u e o f r 2

E ( r2

) = E (X2

)

= E (x 1 + x 2 + x 3 + . .. + x i + . . .x N )2

T h e e x p e c t e d v a lu e o f r is z e r o . B u tt h e e x p e c te d v a l u e o f r 2 i s n o t z e r o .S i n c e E ( X ) = 0 th e e x p e c te d v a l u eo f r 2 i s t h e s a m e a s th e e x p e c te dv a lu e o f x 2

s q u a r e d te r m s :

x1

2 + x2

2 + x3

2 + . . . + xi2 + . . . x

N2

th i s c o n t a in s N t e r m se a c h e q u a l t o + 1e x p e c t e d v a l u e = N

( + 1 ) 2 = + 1

( – 1 ) 2 = + 1

m i x e d te r m s :

x1

x2

+ x1

x3

+ . . . + xix

j + . . .

t h i s c o n ta i n sN ( N – 1 ) te r m se a c h e q u a ll yl i k e l y t o b e+ 1 o r – 1

( + 1 ) ( + 1 ) = + 1( + 1 ) ( – 1 ) = – 1( – 1 ) ( + 1 ) = – 1( – 1 ) ( – 1 ) = + 1

T h e r e s u lt i s s i m p l e : m e a n s q u a r e v a r i a t io n

E ( r2

) = N

r o o t m e a n s q u a r e v a r i a ti o n

= N

E a c h s t e p i s e q u a l l y l i k e l y t o b e + 1o r – 1 . T h u s , o n a v e r a g e , o v e r m a n yr a n d o m w a l k s , th e t o ta l d i s ta n c e w il la d d u p to z e r o .

W h e n m u li t ip l ie d o u t t h isg i v e s t w o t y p e s o f t e r m :

e x p e c t e d v a l u e = 0c o m p a r e ( 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 i s ta n c e t r a v e l l e d i n r a n d o m w a l k = N

Takes longer than predicted as does not go

in straight line. Diffusion is slow even

though they move fast.

Random walksthe average distance R from the origin at which the walker ends up after N steps of length 1, is equal to

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

Energy in a gasIn 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 atomsBut 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

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 temperatureenergy 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 changeReactions 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 equilibriumThis can occur with more than two sources

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

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 ignoredIf considered as an ideal gas

This means it only has translational KENot 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

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 = mcθ.

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 transferForce 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

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-1k-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

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

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 i n e t ic e n e r g y o f g a s m o le c u le s a n d t h e B o l t z m a n n c o n s ta n t

c o m p a re t h e s e

K i n e t i c m o d e l

p V = N m v2

G a s l a w s

p V = N k T

O n e m o l e c u l e M a n y m o le c u l e s

m v2

= k T12

3

2

k in e t ic e n e r g y p e r

m o le c u le = k T3

2

N m v2

= N k T1

2

t o t a l k in e t ic e n e r g y o f

m o le c u le s = N k T32

32

m v2

= 3 k T

to ta l k i n e ti c e n e r g y o f

o n e m o l e o f

m o l e c u l e s U = R T3

2

In te rn a l k ine t ic en e rg y

o f m o le c u le s o f o n e

m o l e a t T = 3 0 0 K

B o l t z m a n n c o n s t a n t k

r a n d o m th e rm a le n e r g y o f o n em o le c u l e is o f o rd e rk T

1

3

U = R T32

U = 3 .7 k J m o l– 1

f o r o n e m o leN = N A

R = N A k

R = 8 . 3 1 J K – 1 m o l–

1

A v e r a g e k i n e t i c e n e r g y o f a m o l e c u l e =32 k T

For one mole: N = NA

R = NakR = 8.31 JK-1mol-1

u = (3/2) NkT = (3/2) RTFor 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

-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.

•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

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 l l h a s t o b e c l im b e db 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 ea c q u i r e d b y c h a n c e fr o m ra n d o mth e r m a l a g i t 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 rp 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 de n o u g h e n e rg y t ob 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 rn e i g h b o u rs , s o l iq u idc a n f lo w

a t o m s n e e de n o u g h e n e r g y t ob e c o m e io n is e d ,f r e e in g e le c t ro n sw 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 he n e rg y t o b re a k a w a yf r o m a t t ra c t io n s o fp 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 l i q u id

m o le c u le s n e e de n o u g h e n e rg y t od 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 ac u r r e n t

p a r t ic le s m u s t c o l l i d ew i t h e n o u g h e n e r g yt o g e t c lo s e e n o u g ht o b o n d t o m a k e an e w m o le c u le

p r o t o n s m u s t h a v ee n o u g h e n e rg y t o g e tc lo s e e n o u g h to f u s ein t o d e u t e r iu m

a c t iv a ti o ne 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 ef 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 im b e db 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 ea c q u i r e d b y c h a n c e fr o m ra n d o mth e r m a l a g i t 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 rp 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 de n o u g h e n e rg y t ob 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 rn e i g h b o u rs , s o l iq u idc a n f lo w

a t o m s n e e de n o u g h e n e r g y t ob e c o m e io n is e d ,f r e e in g e le c t ro n sw 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 he n e rg y t o b re a k a w a yf r o m a t t ra c t io n s o fp 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 l i q u id

m o le c u le s n e e de n o u g h e n e rg y t od 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 ac u r r e n t

p a r t ic le s m u s t c o l l i d ew i t h e n o u g h e n e r g yt o g e t c lo s e e n o u g ht o b o n d t o m a k e an e w m o le c u le

p r o t o n s m u s t h a v ee n o u g h e n e rg y t o g e tc lo s e e n o u g h to f u s ein t o d e u t e r iu m

a c t iv a ti o ne 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 ef 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

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

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

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.

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

1/40kT at room temperature

surface of

Earth

liquidnitrogen

liquid

helium

surface of Sun;

hot furnaceinteriorof Sun

microw ave

photons

super-conduc tivity

air

liquid

plasmasform

x-ray

photonsemit ted

ultravioletphotons

em itted

ceramicsevaporate

metals melt

soft matter

ex ists

waterevaporates

v isible

photonsemitted

liquid nitrogen

comes apart

hydrogen bonds broken

van der Waals bonds broken

outer elec trons in atoms liberated

ionic bonds brokencovalent 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

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 10-19J

4.16 x 10-19J / kT =15

4.16 x 10-19J / 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 300kkT= 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?

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

-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

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 Boltzm ann factor with tem perature

+

+

+

+

++

++

+++++++++++++

Boltzmann factor exp(–/kT) / 10–7

When > kT the Boltzmann factor increases very rapidly with temperature

1.0

0320300 340 360 380 400

T/K (linear scale)

/kT approximately 20 at 300k

B oltzmann factordoubles between eachpair of temperatures

Boltzmann factor exp(– /kT)/10–7

Many important changesstart to happen at anappreciable rate when /kTis of the order of 15–30

2.0

3.0

low temperature

linear scale

0.0

1.0

high temperature

= kT

B o l tz m a n n fa c to r : th e m i n o r i ty w i th e n e rg y e x p (– / k T )

1 0– 4

1 0– 8

1 0– 1 2

1 0– 1 6

e– 1

e– 10

e– 2 0

e– 3 0

e– 4 0

/ kT ( l in e a r s ca le )

1B oltz m a nnfac to r as ap ow e r o f ten

e x p(– /k T )

log arithm i cs c a le

C o ld a t th i s e nd o f the s c a leT he te m pe ra tu re T i s lo wT he r a t io /k T is la rg eT he f ra c t ion ex p (– /kT ) is s m all a nd ap pr oa ch es 0

H o t a t th is e nd o f th e s c a l eT he te m p era ture T is h ig hT he ra t io /k T i s s m allT he f ra c t ion ex p( – /k T ) is la rge an d ap pr oa c he s 1

35 4 03 02 520151051

w h er e th e a c ti on i s :

/k T f ro m 15 to 3 0

+

+

+

+

+

+

+

+

+

B o l t z m a n n fa c to r p l o t te d a g a i n s t t e m p e ra tu r e

li ne ar s c a le

e qu a l to k T

B oltz m a nn fa c tore xp (– /k T )

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 s c a le )

w e ak e r bo n dsh av e a ll c o m ea pa rt

bo nds c om etog eth er

B o n d e n er g y v an d e r W aa ls h y dro ge n bo nd c ov a len t b o nd

C o ld a t th is e nd o f th e s c a leT h e tem p er atu re T is lo wT h e ra ti o /k T is la r geT h e f rac t io n e x p( – /k T ) is s m a ll a nd ap pro ac h e s 0

H o t a t th is e nd o f th e s c a le

T he te m p era ture T is h ig hT he r a t io /k T i s s m allT he f ra c t io n ex p (– /k T ) is la r ge an d a pp roa c he s 1

T h e B o ltz m a n n fa c t o r e xp ( – /kT ) is t h e r a tio o f n u m b e rs o f p ar tic le s ins ta t e s d if fe rin g b y e n e r g y . W h en > > k T t h e B o lt z m an n fa c t o r in c re a s e sv e ry ra p id ly w ith t em p e r a 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

At high temperatures, the Boltzmann factor approaches 1, so nearly all the particles will have enough energy to reactand 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 rapidlywith temperature. So a small increase in temperaturecan make a big difference to the rate

T h e B o lt z m a n n f a c t o r a n d t h e a tm o s p h e r e

C o lu m n o f a i r in t h e a t m o s p h e re

p re s s u re m u s t b e la rg e rlo w e r d o w n b e c a u s e o fe x t ra w e ig h t o fa t m o s p h e re a b o v e

n u m b e r p e r u n i t v o lu m e n

p r e s s u r e p

m a s s o f e x tr a la y e r = n m A d h

w e i g h t o f e x t r a la y e r = n m g A d h

n u m b e r p e r u n i t v o l u m e n + d n

p r e s s u r e p + d p

a s s u m e : t e m p e ra tu re s a m e a t a ll h e i g h ts

m = m a s s o f m o le c u len = n u m b e r o f m o l e c u le s p e r u n it v o lu m ek = B o lt z m a n n c o n s t a n tT = t e m p e ra t u r e

a r e a A

l o w e rp re s s u re h e re

e x t ra w e ig h t

h ig h e rp r e s s u reh e re

h e i g h t h + d h

h e ig h t h

d h

G a s l a w s

p V = N k Tp = (N / V ) k Tp = n k T(p re s s u r e in c re a s e s w it h d e n s it y )

d i ffe r e n c e in p r e s s u r ed p = k T d n

E x tr a p re s s u r e d u e to w e ig h t o f e x tr a l a y e r

p r e s s u re d if fe re n c e b e t w e e n la y e rsd p = w e ig h t o f e x tr a la y e r /a re a Ad p = – n m g A d h /A(p re s s u re d e c re a s e s w it h h e i g h t)

d i ffe r e n c e i n p r e s s u r ed p = – n m g d h

k T d n = – n m g d h

d n / d h = – (m g / kT )n

ra t e o f c h a n g e o f n u m b e r w it h h e ig h t p ro p o r t io n a l t o n u m b e r

n / n0

= e x p (– m g h / k T )

n / n0

= e x p (– / k T ) = m g h = d i ff e r e n c e in p o t e n t ia l e n e rg y

R a t io o f n u m b e rs o f p a r tic le s in s ta t e s d if f e r in g b y e n e rg y i s e q u a l to th eB o l t z m a n n f a 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.