Printing - Advancing Physics A2 Teacher Edition Advancing Physics 1. What is the probability that a particular particle is to be found in the left-hand compartment a long time after

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Molecular energyQuestion 10W: Warm-up Exercise

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Air contains roughly 80% nitrogen and 20% oxygen. The molar mass of oxygen is about 32 g and themolar mass of nitrogen is about 28 g. The temperature in a warm room on a particular day is 27 °C.

1. What, roughly, is the average kinetic energy of an oxygen molecule in this room?

2. What, roughly, is the average kinetic energy of a nitrogen molecule in this room?

3. What is the ratio of the rms speed of an oxygen molecule to the r.m.s. speed of a nitrogenmolecule in this room?

4. A small amount of bromine (molar mass 70 g) leaks from a container in one corner of the room.Explain why the bromine spreads out very slowly.

5. Later a small amount of hydrogen (molar mass 2.0 g) leaks from apparatus in the same corner ofthe room. This mixes with the air much more rapidly. Why?

6. Suggest and justify a relationship between rate of diffusion and molar mass.

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Evaporation of waterQuestion 20W: Warm-up Exercise


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A small puddle contains about 1.0 litre of water at 17 °C and it requires about 40 kJ mol–1 to changewater from liquid to gas at the surface of the puddle.

1. Calculate the energy required for one molecule of water to leave the surface of the puddle.

2. What is the average thermal energy per molecule at 17 °C?

3. What is the ratio of the energy required to free a molecule from the surface to the averagethermal energy per molecule?

4. Explain why the puddle does evaporate at an appreciable rate despite the fact that the averageenergy per molecule is much less than the energy required for evaporation.

5. Explain why a puddle of the same volume would evaporate more rapidly if its temperature was 25°C.

6. How much energy is required to evaporate all of the water in the puddle?

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7. Where does the energy needed to cause the evaporation of the puddle come from?

Thermal radiationQuestion 30W: Warm-up Exercise


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This question is about rough estimates of the wavelength of radiation emitted by objects at varioustemperatures.

h = 6.6 10–34 J s and c = 3.0 108 m s–1

1. Calculate the energy, in electron volts, of a photon of each of the following electromagneticwaves:

microwave at a wavelength of 5 10–3 m

infrared at a wavelength of 5 10–5 m

visible light at a wavelength of 5 10–7 m

ultraviolet at a wavelength of 5 10–8 m

x-ray at a wavelength of 5 10–10 m

All bodies radiate electromagnetic waves because of the thermal motion of their particles. Assumethat the photons they emit have energies related to this thermal motion and that an appreciableamount of radiation will be emitted up to photon energies of about 20kT.

2. Why do human beings radiate in the infrared but not in the visible part of the electromagneticspectrum? Support your answer with a calculation.

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3. The photosphere (surface) of the Sun is at about 6000 K and the peak of the radiation it emitsconsists of photons with energies of about 3 kT. What is the wavelength of these photons and towhich part of the electromagnetic spectrum do they belong?

4. Would you expect the Sun to emit much ultraviolet radiation?

5. Would you expect the Sun to emit much x-radiation?

6. The red giant star Betelgeuse in the constellation Orion appears red. What can you say about itssurface temperature? Explain.

Calculating thermal energiesQuestion 40W: Warm-up Exercise


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This question will give you practice at manipulating different energy units.

1. Hydrogen bonds have a bond energy of about 20 kJ mol–1. Calculate the bond energy of a singlehydrogen bond in joules and in electron volts.

2. The specific thermal capacity of water is 4200 J K–1 mol–1. What is the increase in average

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thermal energy per particle in joules and in electron volts when the temperature of a body ofwater increases by 1 K?

3. The Large Hadron Collider at CERN will accelerate particles to an energy of over 10 TeV (1 TeV= 1012 eV). It is claimed that this will re-create conditions that existed in the very early universesoon after the Big Bang. At what temperature would the average thermal energy per particleequal 10 TeV?

4. Helium liquefies at about 4 K. What is the mean speed of a helium atom at this temperature?(Take the molar mass of helium to be 4.0 g.)

5. The microwave background radiation provides evidence to support the Big Bang theory. Thistheory predicts that the temperature of the universe is about 3 K. Show that this is consistent withthe background radiation peaking in the microwave region of the electromagnetic spectrum.

Probability and equilibriumQuestion 50W: Warm-up Exercise


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The diagram below shows a container with two compartments. Initially the compartment on the left isfull of gas particles (only a few have been shown for clarity) and the compartment on the right isempty. The particles are moving randomly. The barrier separating the two compartments is suddenlyremoved.

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1. What is the probability that a particular particle is to be found in the left-hand compartment a longtime after the barrier has been removed?

2. What is the probability that a particular particle is to be found in the right-hand compartment along time after the barrier has been removed?

3. If there are N particles in the gas (and N is a large number), roughly how many would you expectto find in each compartment a long time after the barrier has been removed?

4. Sketch a graph to show how you would expect the number of particles in the left-hand side tochange with time from the moment the barrier is removed.

5. Your answer to question 4 should suggest that the system approaches an equilibrium state afterthe barrier is removed. What is special about this equilibrium state?

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Nuclear fusion in the SunQuestion 60W: Warm-up Exercise


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This question is about why nuclear fusion in the Sun is a very slow process. It is a good thing that it isslow, because this means that the Sun will shine for billions of years.

The core of the Sun is at a temperature of about 6.0 106 K. The activation energy for nuclear fusion

between two protons is of the order of 106 eV.

1. Why does nuclear fusion have a high activation energy?

2. What does this imply about the conditions required for nuclear fusion reactions to take place?

3. What is the average thermal energy of a proton in the core of the Sun in eV?

4. What is the ratio of the activation energy for nuclear fusion to the thermal energy of protons in thecore?

5. If you use your calculator and the answer to question 4 to calculate the Boltzmann factor forfusion in the core of the Sun, you will either get the result zero, or a message saying that thequantity is too small to calculate. What does this tell you about the fraction of proton–protoncollisions that result in fusion in the core of the Sun?

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These calculations show that even in the core of the Sun fusion is extremely rare. However, theSun is huge and although the power density is low (about 0.25 W m–3 averaged over the entire

Sun) there is a huge mass and so the power released is significant.

6. Fusion reactors on Earth are designed to work at much higher temperatures than the core of theSun. Suggest a reason for this.

Molecules and changeQuestion 20S: Short Answer


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These questions expect you to think about the random jiggling about of molecules, moving withoutpurpose, acquiring energy from and losing energy to each other, many, many times a second.

Try theseDescribe what is happening in terms of movements of molecules in the following cases:

1. Washing drying in a breeze.

2. Ice cubes melting in a glass of water.

3. Milk being stirred into tea.

4. A radiator warming a room.

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Energy per particleQuestion 25S: Short Answer


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Calculating energy per particle in different waysAnswer the questions below to see how to use and compare different units when calculating theapproximate energy per particle in various changes. You will also find out about some very usefulapproximations.

Energy in joules per particle and in electron voltsAn electron has charge e of magnitude 1.6 10–19 C.

1. What energy is delivered when a charge of 1 mC flows across a potential difference of 10 V?

2. What energy is delivered when a charge of 1.6 10–19 C flows across a potential difference of1000 V?

3. What charge, flowing across a potential difference of 1 V, delivers energy 1.6 10–19 J?

4. What energy is delivered when one electron flows across a potential difference of 1 V?

The answer to question 4, the energy delivered when one electron flows across a potentialdifference of 1 V, is a unit used often by physicists. It is called one electron volt.

5. What energy in joules per particle corresponds to an energy of 1 MeV?

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6. The energy needed to ionise an atom of hydrogen is 13.6 eV. What is this in joules per atom?

Energy in kJ per moleOne mole of particles contains 6.02 1023 particles. Chemists often measure energy changes in

kilojoules per mole (kJ mol–1).

7. What energy in kJ mol–1 is delivered when 6.02 1023 electrons (one mole) flow across apotential difference of 1 V?

8. What percentage error is made in taking 100 kJ mol–1 as an approximation to the answer toquestion 7?

9. The energy needed to ionise an atom of hydrogen is 13.6 eV. What is this in kJ mol–1,approximately?

10. The energy needed to remove an electron from sodium Na, giving an Na+ ion, is 5.1 eV. What is

this in kJ mol–1, approximately?

11. The energy needed to break the H-H bond in H2, giving hydrogen atoms, is 436 kJ per mole ofH2. What is this in electron volts, approximately?


12. The energy needed to evaporate one mole of liquid nitrogen is 2.79 kJ mol–1. What is this inelectron volts, approximately?

Photon energiesThe energy of a photon is given by E = hf, where h = 6.6 10–34 J Hz–1 is the Planck constant and f is

the frequency. The speed of light c = 3.00 108 m s–1.

13. Show that the wavelength of photons gaining energy from electrons accelerated to 1 kV is, within20%, about 1 nm.

14. Use the answer to question 13 to estimate the wavelength of photons of energy 1 eV.

15. Use the fact (question 9) that it takes 13.6 eV to ionise a hydrogen atom to estimate the order ofmagnitude of the wavelength of photons which could be produced when an electron combineswith a hydrogen ion to form a hydrogen atom. In what part of the electromagnetic spectrum doesthis fall?

16. The photons of the cosmic microwave background have wavelength of the order 1 mm. Estimate

their energy in kJ per mole of photons. Compare with the energy 0.08 kJ mol–1 needed toevaporate liquid helium.


Useful approximations17. Write down an approximation, accurate to a few percent, for the energy 1 eV expressed in kJ

mol–1.

18. Write down an approximate value, good to about 20%, for the wavelength of photons of energy 1eV. If the energy increases by a factor 10, how does the wavelength change?

Likely events in timeQuestion 30S: Short Answer

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This pair of questions help you think about the average behaviour of different numbers of randomlyshuffled particles.

Try these1. Here is a set of ‘snapshots’ of marbles being shaken in a container. The marbles are identical

except for their colour. Arrange the snapshots in chronological order, explaining the order youhave chosen.

A B C D

2. Why can you not be sure of the order of the snapshots shown here?


A B C D

Values of the energy = kTQuestion 35S: Short Answer


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Temperature and energy per particleThe energy per particle in matter at temperature T is typically of the order of magnitude kT, where k is

the Boltzmann constant 1.38 10–23 J K–1. These questions will show you how to make roughestimates of the energy kT, in several different units.

Energy in electron volts and in kJ mol–1

One electron volt is equal to the energy 1.6 10–19 J. The Avogadro number, the number of particles

per mole, is 6.02 1023 mol–1.

1. Calculate the energy of one mole of particles, each with energy 1 eV.

2. Show that the error in taking the answer to question 1 as approximately 100 kJ mol–1 is onlyabout 4%.

3. Fill in the table below, approximately translating energies in eV to kJ mol–1.


Energy / eV Energy kT / kJ mol–1

0.0001

0.001

0.01

0.1

1 100

10

100

1000

4. Use the table to estimate the energy in kJ mol–1 corresponding to the energy of an electron froma 1.5 V torch battery.

5. Use the table to estimate the energy in electron volts corresponding to the energy 40 kJ mol–1

needed to evaporate water.

Temperature and energy kT

The Boltzmann constant k = 1.38 10–23 J K–1. One mole of particles contains 6.02 1023 particles.

6. Calculate the energy per particle = kT for a temperature of 1000 K. Express the result in joulesper particle.

7. Calculate the energy kT with T = 1000 K, for one mole of particles. Express the result in kJ mol–1.


8. Show that the error in taking the energy kT at T = 1000 K to be 10 kJ mol–1 is close to 20%.

9. Using the approximation that the energy kT for T = 1000 K is about 10 kJ mol–1, estimate theenergy kT for T = 300 K.

10. Estimate the temperature T for which kT = 1000 kJ mol–1.

11. Fill in the table below, accurate to about 20%, translating absolute temperatures T to energy kT in

kJ mol–1

Temperature T / K Energy kT / kJ mol–1

0.01

0.1

1

1000 10

100

1000

10 000

100 000

Photon energies and wavelengthsThe energy of a photon is given by E = hf, where h = 6.6 10–34 J Hz–1 is the Planck constant and f is

the frequency. The speed of light c = 3.00 108 m s–1.

12. Show that the wavelength of photons gaining energy from electrons accelerated to 1 eV is, within20%, about 1 m.


13. If the photon energy increases by a factor 10, by what factor does the frequency change? Bywhat factor does the wavelength change?

14. Fill in the table below with order of magnitude values for photon wavelengths, for photons withenergy kT.

Temperature T / K Energy kT / eV Energy kT /

kJ mol–1photonwavelength

1 0.0001 0.01

10 0.001 0.1

100 0.01 1

1000 0.1 10

10 000 1 100 1 m

100 000 10 1000

1 million 100 10 000

10 million 1000 100 000

15. Estimate the temperature and energy per particle in eV at which photons of energy kT havewavelength the size of a fairly large molecule, say 1 nm.

More estimates16. Estimate the energy kT in eV and in kJ mol–1 at the melting point of platinum, 2045 K.

17. Estimate the energy kT in eV and in kJ mol–1 at the boiling point of liquid nitrogen, 77 K.


18. Estimate a typical wavelength of the cosmic microwave background radiation, if its temperature is3 K.

19. What temperature is needed for the energy kT to be equal to the energy 1 MeV (needed forexample to create electron–positron pairs).

20. Are there x-rays in the radiation at the temperature, 10 million K, at the centre of the Sun?

Particles spreading outQuestion 40S: Short Answer

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Finding the chance of getting back to an ordered stateThis question calculates the likelihood of 100 particles equally spread across two halves of a box andmoving randomly all being found in one half of the box sometime in the future.

1. Firstly, use a calculator to find 2100.

2. Given that log 2100 = 100 log 2 and log 2 0.3, check your answer to part 1 by expressing 2100 asa power of ten.

3. The huge number you have calculated is the total number of ways of arranging the 100 particlesbetween the two halves of the box. There is only one way of having all 100 in the left half and thatis to have all 100 in the left half! So, the chances of finding all the particles in one half of the box

only are 1 in 2100.


Suppose you look at the box once every microsecond. For how many seconds would you need tolook, on average, before seeing all the particles simultaneously in one half?

4. The Universe is of the order of 1010 years old and one year is about 3 107 seconds. How manyUniverse lifetimes is your answer to 3?

Matter starting to ‘come apart’Question 50S: Short Answer


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Particles with more than the average energy kTThe energy per particle in matter at temperature T is typically of the order of magnitude kT, where k is

the Boltzmann constant 1.38 10–23 J K–1. A few particles will have energies quite a lot greater thanaverage. In these questions you will see examples of processes which happen at an appreciable rate,when the energy required is in the range roughly 10 to 30 times kT.

Energy for a process compared to the energy kTIn these questions you will need to convert energies per particle given in electron volts to energies in

joules per particle. One electron volt is equal to the energy 1.6 10–19 J. You will also need to convertenergies given in kilojoules per mole to energies in joules per particle. The Avogadro number, the

number of particles per mole, is 6.02 1023 mol–1.

Water evaporating1. Energy 40 kJ mol–1 is required to evaporate water molecules from the liquid. Convert this to

joules per particle.

2. Water evaporates at a noticeable rate at 300 K, even though it only boils (at one atmospherepressure) at 373 K. Calculate the value of kT at 300 K. What is the ratio /kT, where is theenergy needed to evaporate one molecule?


Electrons ‘evaporating’ from hot metalsElectrons ‘evaporate’ from heated metals. This is called thermionic emission, used to obtain theelectron beam in an old fashioned television set, or in a dental X-ray tube.

3. The energy required to remove an electron from platinum is 5.4 eV. Express this energy in joulesper electron.

4. At room temperature, the emission of electrons from platinum is negligible. However, at 2000 K,an appreciable thermionic current can be detected. Calculate the value of kT at 2000 K. What isthe ratio /kT, where is the energy needed to ‘evaporate’ one electron from platinum?

5. What would you predict about thermionic emission from thorium-coated tungsten, for which theenergy needed to remove an electron is 2.6 eV?

Sunburn from ultraviolet radiation6. The surface of the Sun is at a temperature of about 6000 K. ‘Typical’ photons from a hot object at

temperature T have energy of approximately 3kT.

Calculate energy 3kT and the wavelength of a ‘typical’ photon from the Sun.

In what part of the electromagnetic spectrum does it fall?

h = 6.6 10–34 J Hz–1 and c = 3.0 108 m s–1.

7. Sunlight does contain a small proportion of ultraviolet radiation. As evidence, you get sunburn indirect sunlight. The wavelength of the ultraviolet radiation may go down to as short as 100 nm.

How many times larger than the ‘typical’ energy 3kT of a photon is the energy of these ultravioletphotons?


Ionisation of hydrogen in the Sun8. At the 6000 K temperature of the surface of the Sun, what is the energy kT? Express this energy

in electron volts.

9. Some of the hydrogen in the surface of the Sun is ionised. The energy needed to ionise ahydrogen atom is 13.6 eV.

How many times larger is this than the average energy per particle, of the order kT?

Firing potteryPotters monitor the temperature of their kilns using a series of pottery cones, which partly melt andsag or bend at given temperatures. Cone 9 for example, falls at about 1500 K, and is typical of thetemperatures used to glaze decorative stoneware.

10. Calculate the energy per particle kT for a temperature of 1500 K. Express the result in joules perparticle.

11. Pottery is a ceramic material, with bond energies of the order several hundred kilojoules per

mole, say 400 kJ mol–1.

Estimate the ratio /kT for a ceramic at the temperature of cone 9.

Washing upGrease can just about be washed off dirty dinner plates using hot water. Greases often stick together

by hydrogen bonds, with energy of the order 20 kJ mol–1.

12. Estimate the ratio /kT for doing the washing up, pulling apart a few hydrogen bonds to move agrease molecule.


‘Dry ice’Solid carbon dioxide, often called dry ice, slowly evaporates (sublimes) at the low temperature of–78ºC, or 195 K.

13. Calculate the energy kT, in units of kJ mol–1, at 195 K.

14. The energy to evaporate solid carbon dioxide is 27 kJ mol–1. What is the ratio /kT at 195 K?

Distributions of particlesQuestion 70S: Short Answer

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1. Particles sharing energy do not share it equally. Imagine a set of particles which can occupy aladder of equally spaced energy levels. Suppose that the number of particles on any one rung isa constant fraction of the number on the rung just beneath. Sketch a histogram showing therelative numbers of particles on several rungs of the ladder.

2. Draw a second sketch to show how the pattern would change, if the same number of particlesshared a larger total energy. Assume that the number of particles on any one rung is a differentconstant fraction of the number on the rung just beneath.


The Boltzmann factor:fB = exp(–/kT)Question 90S: Short Answer

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Questions1. Use a calculator to complete the table:

kT f B

1 0.37

2

5

10

20

50

100

2. What value of / k T corresponds to a Boltzmann factor of 10–6?

3. Find the Boltzmann factor for a system where T = 300 K and the energy spacing is 10–21 J.

4. For one particular system the Boltzmann factor fB = 0.5 (so ln fB = – 0.693). The energy spacing is

10–21 J. What is the temperature of such a system?

5. For energy = 1.6 10–19 J (1 eV), plot a graph of the Boltzmann factor against temperature overthe range 300 K to 350 K. Describe the form of the graph.


Electrons from hot metalsQuestion 150S: Short Answer


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Data from an experimentYou will look at data on the electric current obtained by emission of electrons from a metal wire over arange of temperatures.

Thermionic emissionWhen metals are heated they emit electrons. This is called thermionic emission. The maximumcurrent that can be collected measures the rate at which electrons are being emitted.

Here are some experimental data collected for a heated metal wire.

Temperature and thermionic emission for a platinum wire

1150 1200 1250

temperature / °C

1050 1100 1300

0

5

10

15

20

25

1. Describe in words how the current varies with temperature.


2. Look at the range of temperature used. Show that the absolute (kelvin) temperature increased byabout 15% over the range from 1100 C to 1300 C.

3. Show that the thermionic current increased by a factor of 25 or so over this temperature range.

4. By inspection of the graph, show that the current doubles for each increase of between 30 and 40C. Give a different example of a physical relationship which has this kind of behaviour.

5. Read approximate values of the current at each temperature from the graph, and plot a graph ofthe logarithm of the current against the temperature. What can you infer from this graph?


Going further – a bit harderThe remaining questions use a bit more mathematics. They start from the Boltzmann factor

)exp(kT

. The ratio

BTk

is the temperature TB at which the energy kTB is equal to . So the Boltzmann factor can be written inthe form

)exp( B

T

T

. If the temperature increases by a small fraction

T

T,

then the Boltzmann factor is multiplied by the factor

T

T

T

T Bexp,

where T is now the average temperature in the experiment.

6. You showed in question 2 that the temperature T increased by

15.0T

T

approximately. You showed in question 3 that the factor by which the current increased was 5.The average temperature is about 1500 K (check this). Thus you can write the equation

25K 1500

15.0exp B

T

and find TB for the metal. Take logarithms of both sides of the equation, and find TB.

7. Calculate the energy k TB and express the answer in electron volts. This is the estimate fromthese data of the ‘work function’ of the metal.


8. If you can, prove that the Boltzmann factor

)exp(kT

increases by the factor

T

T

T

T Bexp

if the temperature T increases by a small amount T.

Density where jet planes flyQuestion 100C: Comprehension


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Read the introduction, which explains how the Earth’s atmosphere thins out at greater heights.Answering the questions will enable you to estimate the density or pressure of the atmosphere at aheight of 10 km.

Flying highJet planes fly high because the air is thinner higher up, and makes less drag. But not too high – theplanes need air to burn fuel.

Seen from space, the Earth’s atmosphere is just a thin blanket around it. How does the air thin out asyou go up? What keeps it hugging the Earth? And what keeps it up?

The answer is that gravity pulls the air in towards the Earth, but the tendency of molecules to diffuserandomly from regions of higher to lower concentration stops the air from all falling to the ground.


mountainaircraft

15 km

50 km

Ionosphere

meteors

Stratosphere

ozone layer (30 km)

Gravity hill going down towards Earth

gravity hilldownhilltowards Earth

You are now going to calculate the strength of the Earth’s gravitational field in an unusual new way.The slope of the downward gravity hill is the change in potential energy of one kilogram per metre

change in height. This is simply mgh, where g is 9.81 J kg–1 m–1.

1. Show that 9.81 J kg–1 m–1 is the same thing as 9.81 N kg–1, which is the usual unit forgravitational field g.


2. 9.81 J kg–1 m–1 is the slope of the gravitational hill for a mass of 1 kg. Nitrogen N2 has mass 28 gper mole. Write this mass in kg.

3. Use the answer to question 2 to express the slope of the gravitational hill as joules per mole of

nitrogen per metre (J mol–1 m–1).

4. Show that the answer to question 3 can be rewritten as 0.275 kJ mol–1 km–1, and that therefore

the energy difference per mole for a 10 km difference in height is 2.75 kJ mol–1.

Density hill away from EarthThese first four questions show that the slope of the gravity hill towards the Earth for nitrogen

molecules is 0.275 kJ mol–1 km–1. Nitrogen molecules therefore tend to fall towards the Earth, giving

up this amount of potential energy as they do so. Over 10 km the energy change is 2.75 kJ mol–1.

There is a different hill, sloping down away from the Earth, which balances the gravity hill slopingdown towards the Earth. It is the density gradient in the atmosphere, which decreases with distancefrom the surface. Nitrogen molecules diffuse upwards away from the Earth, travelling down thisdensity or concentration gradient. As the molecules diffuse upwards, they must gain gravitationalpotential energy. They have to do this by random collisions with other molecules in the air, whichhave an average energy of the order kT at temperature T.

gravity hilldownhilltowards Earth

density hilldownhill awayfrom Earth

The ratio of probabilities for particles to be in states differing by energy per particle is given bythe Boltzmann factor


.exp

kT

You have just calculated the value of the energy difference per mole for a 10 km difference in

height. It is 2.75 kJ mol–1

5. Assume that the average temperature in the first 10 km from the Earth’s surface is 273 K. Given

N = 6.02 1023 mol–1 and k = 1.38 10–23 J K–1, show that NkT for a temperature of 273 K is 2.27

kJ mol–1.

6. Calculate the ratio N/NkT.

7. Calculate the Boltzmann factor for this ratio. State the ratio of densities at heights different by 10km in the conditions assumed.

Contaminated surfacesQuestion 140C: Comprehension


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Thinking about examples of molecules sticking to othersRead the passage below, about different examples of adsorption – that is, molecules sticking tosurfaces. Answer the questions to check your understanding.

Clean surfaces?At home, you probably have to clean a lot of surfaces. It might be while washing up, cleaning a bikeor sweeping a floor. You also wash your hands and face regularly. Are these ‘clean’ surfaces reallyclean? That is, is there absolutely nothing on them? The answer is a resounding ‘no’. There arealways other molecules sticking to the surface. This is called ‘adsorption’ (the different process,‘absorption’ is the molecules getting inside the material instead of just covering the surface).

A really clean surface


Adsorption of gas atoms or molecules by a surface

gas atoms or molecules

freshly prepared cleansurface

a few atoms have beenadsorbed

the surface is nearly covered

The only way to get a really clean surface is to scrape a new surface with the material in a highvacuum. Then new material is exposed, and there are few gas atoms to strike the new surface andstick to it. If you cut or scrape a metal in air, the new shiny surface often soon gets dull. This is oftenbecause oxygen molecules stick to it, forming metal oxide molecules on the surface. Silver goes dulland then black, especially if there is a trace of hydrogen sulphide in the air. Iron rusts. Gold andplatinum stay shiny, but collect an invisible layer of gas molecules stuck to their surfaces. Thetitanium surface of the Guggenheim museum in Bilbao, Spain, became discoloured when the naturaloxide surface which usually protects it was attacked by other chemicals.

How hard do they stick?Most molecules attract one another to some extent when close to one another. This is why allmaterials condense to liquid or solid when sufficiently cool. Many such forces are quite weak. Thestrength can be measured by the energy needed to pull the particles apart. The van der Waals bondsthat for example hold wax together have energies in a range close to the value of kT at room

temperature, say 0.1 eV or less (a few kJ mol–1). But oxygen atoms bonding chemically to a metalsurface need much more energy to have these chemical bonds broken – of the order 1 to 10 eV or

100 to 1000 kJ mol–1.

A red-blooded exampleThe red colour of your blood comes from haemoglobin. This protein has four chemical ‘pockets’containing an iron atom each, to which oxygen molecules can bond. Red blood is indeedrust-coloured! In your lungs, where the blood is rich in dissolved oxygen, more oxygen molecules getbound to iron atoms in haemoglobin than happen to hop off the iron back into the blood. But in tissuewhich lacks oxygen, the oxygen molecules which hop off go on to do other important work to keepyou alive, such as ‘burning’ sugars. So you see that haemoglobin needs to hang on to adsorbedoxygen molecules quite tightly, but not too tightly. If the bonding were too weak it wouldn’t pick upenough oxygen in the lungs. If the bonding were too strong, it wouldn’t release enough oxygen inother tissues.

Incidentally, carbon monoxide poisons you because carbon monoxide binds more strongly tohaemoglobin than oxygen does. That’s why carbon monoxide poisoning makes you look blue incolour.

Charcoal cleaners


Surface of charcoal with adsorbed gas atoms

Freshly made ‘activated’ charcoal has a clean carbon surface which is very rough on the atomicscale. So it has a huge surface. It is able to adsorb many gases very well. This is why it was used ingas masks in wartime. And it is why it is used to clean gases and liquids. Of course, in the end thecharcoal surface is covered with contaminant molecules and it stops adsorbing any more. It is said tobe ‘poisoned’. But the good thing about charcoal is that it can be re-activated, made good again. Youjust have to heat it. At the higher temperature, the adsorbed molecules easily get enough energy toleave the surface. So heating the charcoal cleans them off.

It’s all down to the Boltzmann factor

All these adsorption processes depend on the Boltzmann factor kTe , where is the depth of theenergy well in which a particle stuck to the surface sits, and T is the absolute temperature. If is not

too much larger than kT the Boltzmann factor kTe , though small, is not so small that there is nochance of an atom or molecule acquiring enough energy to leave the surface.

This makes the value of the ratio /k an interesting way to describe the depth of the energy well. Theunits of the ratio /k are simply those of temperature, degrees kelvin K. The ratio /k is thetemperature at which is equal to kT.

Questions1. Give an example of a really clean surface. Suggest a likely source of contamination.

2. What would be the effect on the gas pressure of putting a newly made clean surface in a gas in aclosed rigid container?


3. Atmospheric pressure is of the order 100 kPa (100 kN m–2). The random speeds of gas

molecules are of the order 100 m s–1. The mass of a molecule in air is of the order 5 10–26 kg.Estimate the change in momentum of such a molecule rebounding from the surface. Use thevalue of the pressure to estimate the number of collisions per second with one square metre ofsurface. Taking the size of a surface atom as of the order 0.1 nm, show that a surface atom is

struck about 108 times per second.

4. From question 3, a surface atom is struck something like 108 times per second by a gas atom.What would have to be the chance that a gas atom hitting a surface atom sticks to it, if a cleansurface is fully contaminated in a time of 10 seconds?

5. Check the following result given in the passage:The van der Waals bonds that, for example, hold wax together have energies in a range close to

the value of kT at room temperature, say 0.1 eV or less (a few kJ mol–1).

6. Suppose a gas molecule is adsorbed onto a surface at 300 K, and that energy 0.1 eV = 10 kJ

mol–1 is needed to remove it. Suppose it has 108 opportunities per second to leave the surface,through acquiring energy from the thermal motion of atoms around it. Estimate the average time ittakes before it leaves the surface. How will this change if the surface is cooled to 100 K?


7. Write a few sentences explaining, in terms of the Boltzmann factor, why heating ‘poisoned’charcoal can re-activate it.

8. Show that, as stated in the passage, 'the units of the ratio /k are simply those of temperature,degrees kelvin K'. The value of /k for the attraction between two nitrogen molecules is about 90K. Explain why you would expect nitrogen to become liquid at a temperature below 90 K.

Resistance and conductance of thermistorsQuestion 110D: Data Handling


Quick Help

Handling and plotting dataUse commercial thermistor data on the spreadsheet to make a variety of plots to see how theresistance and conductance of thermistors vary with temperature. It is this property that allowsthermistors to be used as temperature sensors.

Thermistor resistance dataFirst, start up the spreadsheet of thermistor resistance data. It shows how the resistance, in k, offive different thermistors varies with temperature over the range –30 °C to 250 °C.

Open the Excel Worksheet

Data from RS Components Data Sheet 232-4538Type Type Type Type Type


five different thermistors varies with temperature over the range –30 °C to 250 °C.


Data from RS Components Data Sheet 232-4538Type Type Type Type Type

198-927 198-933 198-949 198-955 198-961

Temperature Resistance Resistance Resistance Resistance Resistance

T /oC R / k R / k R / k R / k R / k

-30 176.0 352.0 528.0 880.0 1760.0

-20 96.3 192.6 288.9 481.5 962.9

-10 54.9 109.7 164.3 274.3 548.5

0 32.4 64.8 97.2 162.1 324.1

10 19.8 39.6 59.4 99.0 198.0

20 12.5 24.9 37.4 62.4 124.7

25 10.0 20.0 30.0 50.0 100.0

30 8.07 16.13 24.20 40.33 80.66

40 5.34 10.68 16.03 26.71 53.42

50 3.62 7.24 10.85 18.09 36.18

60 2.50 5.00 7.51 12.51 25.02

70 1.76 3.53 5.29 8.82 17.63

80 1.27 2.53 3.80 6.33 12.65

90 0.923 1.845 2.768 4.613 9.226

100 0.683 1.367 2.050 3.417 6.834

110 0.516 1.032 1.547 2.579 5.158

120 0.394 0.788 1.183 1.971 3.942

130 0.305 0.610 0.914 1.524 3.048

140 0.238 0.476 0.715 1.191 2.382

150 0.188 0.376 0.564 0.941 1.881

160 0.150 0.299 0.449 0.748 1.495

170 0.120 0.241 0.361 0.602 1.204

180 0.0982 0.1964 0.2945 0.4909 0.9818

190 0.0809 0.1619 0.2428 0.4046 0.8093

200 0.0674 0.1348 0.2022 0.3370 0.6739


Type Type Type Type Type

198-927 198-933 198-949 198-955 198-961

Temperature Resistance Resistance Resistance Resistance Resistance

T /oC R / k R / k R / k R / k R / k

210 0.0567 0.1133 0.1700 0.2833 0.5665

220 0.0481 0.0961 0.1441 0.2403 0.4805

230 0.0411 0.0822 0.1233 0.2054 0.4109

240 0.0354 0.0708 0.1062 0.1770 0.3540

250 0.0307 0.0614 0.0922 0.1536 0.3072

How resistance and conductance vary with temperature1. Scan the values in the table. What is the largest value of resistance in the table? What is the

smallest value of resistance in the table? What is that in ohm? What is the ratio of the largestresistance in the first column (type 198-927) to the smallest resistance in that column? Does thissuggest anything about how best to plot these data?

2. Use the spreadsheet graphing function to make an x-y plot of the resistance of the first thermistorlisted (type 198-927) against temperature in °C. What does this graph show? Why is it almostuseless?

3. Now alter the resistance scale so that it is logarithmic. You can usually do this by clicking on they-scale on the graph, and selecting ‘logarithmic’ in the options for the y-scale. What do equalintervals on this new scale represent? How has the graph changed in appearance? What doesthe graph show?

4. Use the spreadsheet to calculate the ratio of the resistances of the thermistor (type 198-927) forsuccessive 10 degree intervals. (Be careful where there is an extra entry at 25 °C.) What do you


see? Does this agree with the logarithmic graph?

5. Now make a combined plot, on a logarithmic resistance scale, showing how all five thermistorsbehave. What is remarkable about this plot? What does it mean?

6. What is the conductance in millisiemens of the thermistor type 198–927 at 250 °C when itsresistance is 30.7 ? Predict the shape of the graph of conductance against temperature. Usethe spreadsheet to calculate the conductances of the thermistors over the given temperaturerange, and make a logarithmic plot to check your prediction.

7 The conductance of a thermistor depends mainly on the number of charge carriers which are freeto conduct. Charge carriers can become free if they acquire energy from the thermal motion of

atoms in the material, so their number may be proportional to the Boltzmann factor e–/kT. Showthat if this is so, the conductance G will be given by

constantln

kTG

. Choose one type of thermistor and use the spreadsheet to plot a graph of ln G against 1/T, withabsolute temperature T equal to the celsius temperature t + 273.

8 Find the slope of the graph in question 6, equal to

k

.

Find a value for the activation energy (Boltzmann constant k = 1.3810–23 J K–1). Express theenergy in electron volts. Compare the value of with the value of kT at 300 K.


Vapour pressure of waterQuestion 120D: Data Handling


Quick Help

Analyse these dataHere are data for the vapour pressure of water, at temperatures from 0 °C to 200 °C. You can show

that the variation with temperature reflects the Boltzmann factor e–/kT. The data are provided in thespreadsheet below.

Open the Excel worksheet

Vapour pressure of water: variation with temperature

temperature t / °C vapour pressure p / kPa

0 0.611

10 1.230

20 2.340

30 4.240

40 7.380

50 12.34

60 19.93

70 31.18

80 47.37

90 70.12

100 101.3

110 133.9

120 198.5

130 270.0

140 361.2

150 475.7

160 617.7

170 791.5


Vapour pressure of water: variation with temperature

temperature t / °C vapour pressure p / kPa

180 1002

190 1254

200 1554

The variation of vapour pressure with temperatureThe vapour pressure of a vapour above a liquid is proportional to the density of molecules in thevapour. This reflects the extent to which molecules can acquire enough energy to leave the liquid andjoin the vapour. Thus, at least approximately, you can expect that the vapour pressure p will be

proportional to the Boltzmann factor e–/kT, where is the energy needed to evaporate one molecule,

T is the absolute temperature and k is the Boltzmann constant 1.38 10–23 J K–1.

1. Show that if p = Ae–/kT then a graph of ln p against 1 / T will be a straight line of slope –/k.

2. Calculate /k for water, given that the energy to evaporate one mole (6.02 1023) of molecules of

water is about 41 kJ mol–1. Show that this predicts a slope of about –5000 K for the graph of ln pagainst 1/T, for water vapour.

3. Use the spreadsheet to calculate values for 1/T where T is the Celsius temperature t + 273.

Calculate values of the natural logarithm of pressure, ln p. Plot a graph of ln p against 1/T. Is it astraight line?

4. How near is the slope of the graph to the predicted value, about –5000 K?


5. The vapour pressure is equal to atmospheric pressure at the boiling point of water, 100 °C.

Estimate from the data given the temperature in a pressure cooker when the vapour pressure inthe cooker is twice atmospheric pressure.

Runny liquidsQuestion 130D: Data Handling


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The fluidity of water, alcohol and molten glassThe fluidity of a liquid measure how easily it flows. The quantity usually given in data tables is theviscosity, how badly the fluid flows. The fluidity = 1/viscosity.

Liquid flow can arise from molecules getting enough energy to push past their neighbours. The morethat can do this, the larger the fluidity. The number of molecules with energy of at least at

temperature T is often given at least approximately by the Boltzmann factor e–/kT. Thus the fluidity

can be expected to be proportional to e–/kT. k is the Boltzmann constant 1.38 10–23 J K–1.

You can use data provided in a spreadsheet to test this relationship for water, ethanol and moltenglass.


Fluidity of water

Viscosity and fluidity of water

temperaturet / °C

viscosity /mPa s

fluidity =1/viscosity

0 1.792 0.558

5 1.519 0.658

10 1.307 0.765


Viscosity and fluidity of water

temperaturet / °C

viscosity /mPa s


15 1.138 0.879

20 1.002 0.998

25 0.890 1.124

30 0.797 1.255

40 0.653 1.531

50 0.547 1.828

60 0.467 2.141

70 0.405 2.469

80 0.355 2.817

90 0.315 3.175

100 0.282 3.546

125 0.222 4.505

150 0.182 5.495

This table shows how the fluidity of water increases with temperature from 0 °C to 150 °C. Noticethat the water would have to be under pressure to exist as a liquid above its boiling point.

1. Show that if kTA e then a graph of ln against 1/T will be a straight line of slope -/k.

2. Use the spreadsheet to calculate values for 1/T where T is the Celsius temperature t + 273.Calculate values of the natural logarithm of fluidity, ln . Plot a graph of ln against 1/T. Make arough estimate of the slope of the graph.


3. Use the fact that the slope is equal to –/k to estimate the energy needed for a water moleculeto move past its neighbours. Express the energy in electron volts.

4. The graph is not quite straight. Is it steepest at high or at low temperatures? What might thissuggest about the energy , which is proportional to the slope?

Fluidity of ethanol

Viscosity and fluidity of ethanol

temperature

t / °C

viscosity /mPa s


–100 98.96 0.010

–50 8.318 0.120

0 1.873 0.534

25 1.084 0.923

30 0.983 1.017

50 0.684 1.462

75 0.459 2.179

100 0.323 3.096

The table shows how the fluidity of ethanol (ethyl alcohol) increases with temperature over a wide


range of temperatures.

5 Use the spreadsheet to calculate values for 1/T where T is the Celsius temperature t + 273.Calculate values of the natural logarithm of fluidity, ln . Plot a graph of ln against 1/T. Make anestimate of the slope of the graph.

6. Use the fact that the slope is equal to –/k to estimate the energy needed for an ethanolmolecule to move past its neighbours. Express the energy in electron volts.

Fluidity of molten glass

Viscosity and fluidity of silica glass

temperaturet / °C

viscosity /Pa s


1100 14.6 0.068

1200 12.7 0.079

1300 11.8 0.085

1400 9.7 0.103

1600 8.2 0.122

1800 4.7 0.213

2000 3.4 0.294

The table shows how the fluidity of molten glass increases with temperature, when the glass ismolten. The original data (Kaye and Laby 1986 Tables of Physical and Chemical Constants 15th


edition, Longman) are given to two significant figures only and may not be very reliable.

7 Use the spreadsheet to plot a graph of ln against 1/T. Discuss the shape and slope of thegraph, and compare it with those for water and ethanol. The units of fluidity in the case of glassare 1000 times smaller than the units used for water and ethanol.

Matter ‘comes apart’Question 45X: Explanation–Exposition


Quick Help

Describing what happens to the particlesIn matter at temperature T the particles have energy of the order kT where k is the Boltzmannconstant. They are also bound together in various ways, lying in energy wells which vary in depth.These questions ask you to describe, without any calculations, what happens to the particles inmatter as the temperature changes. Write a brief paragraph in each case.

Changes in matter with temperature1. Plumbers melt solder in joints in copper pipes, by heating the pipe with a blow-torch. As the pipe

cools the solder solidifies again. Describe these events from the point of view of the atoms in thesolder.

2. Why is butter kept in the refrigerator hard to spread?


3. Why are atoms in the Sun mostly ionised?

4. Why do puddles of water left after rain evaporate quite quickly?

5. Why do metals melt in a furnace but the brick lining of the furnace doesn’t?

6. Glaciers made of solid ice nevertheless creep slowly downhill? Why?

7. Liquid helium boils at 4 K. Liquid nitrogen boils at 77 K. What can you suggest about the energyneeded to evaporate an atom or molecule in the two cases?

8. White hot sparks can erode metal from the electrodes across which the spark occurs, leaving themetal pitted. Suggest why.


Thinking about the Boltzmann factorQuestion 60X: Explanation–Exposition


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The Boltzmann factor e–/kT

The Boltzmann factor kT

e gives the ratio of numbers of particles in states differing by energy attemperature T. It is often a good guide to the probability that a particle will have energy or larger,which can have a big effect on the rate at which a process occurs. These questions ask you to thinkabout how the Boltzmann factor changes with temperature, and about some consequences of how itchanges.

Changes in e–/kT with temperature1. Fill in the blanks in the sentence below, with the word ‘increases’ or the word ‘decreases’, as

appropriate:

When the temperature T increases, the ratio / kT __________________, so the expression e/kT

______________, and the Boltzmann factor e– /kT, which is equal to 1 / e /kT therefore_______________.

2. If the Boltzmann factor fB = e– /kT then ln fB = –/kT. Fill in the blanks in the sentences below, with

the word ‘increases’ or the word ‘decreases’, as appropriate:

When the temperature increases, the ratio / kT ________________, so its negative –/ kT_______________. If –/ kT increases, then ln fB _____________. If ln fB increases then fB______________. Thus as the temperature increases the Boltzmann factor ______________.

Rate doubles for a 10 K riseChemists sometimes use the rule of thumb that, for many reactions which require an activationenergy, the rate of reaction doubles for a 10 K rise in temperature.

3. Suppose that in a given process, the ratio /kT = 20, at temperature 300 K. Calculate the

Boltzmann factor fB = e /kT .

4. Now suppose the temperature increases by 10 degrees to 310 K. Calculate the new ratio / kT,assuming that the energy stays the same.

5. Calculate the new value of the Boltzmann factor fB = e /kT . Show that it is roughly double theprevious value.


6. What can you infer from the fact that both values of the Boltzmann factor are very small?

7. What can you infer about a reaction in which the rate of reaction doubles for a ten degree rise intemperature around 300 K?

Cooking8. Food cooks much more quickly in a pressure cooker than in boiling water, even though the

temperature in the pressure cooker is only a few degrees higher. Explain why.

9. If you come home late, you may find the message, 'Your dinner is waiting in the oven'. Why isfood cooked quickly in a hot oven, but can be kept in a warm oven for quite a long time withoutcooking much more?

10. Why can food be kept for a long time in a freezer?

11. Fill one glass with cold water, and another glass with hot water. Drop one sugar cube into each.See how long it takes for the water at the top to taste perceptibly sweet in each case. Would youexpect the two times to be in the ratio of the two temperatures?


Explosions and fires12. Explain how a reaction which releases energy, such as the burning of gunpowder, may go faster

and faster once it is started off, causing an explosion.

13. Suggest why throwing water on a fire can put it out.

Exponential distributionsQuestion 80X: Explanation–Exposition


Quick Help

These questions are about the distribution of energy between particles.

Questions1. If there are half as many particles on each level as on the next lowest level (f = 0.5) and there are

256 particles on the lowest level, calculate the numbers expected on each level up to the eighth.Above what level will it be common to find no particles?

2. How do the answers change if f = 0.25 and f = 0.75?

3. Why does the fact that the number of particles on a given level is a constant fraction of thenumber on the level below mean that the distribution is correctly called exponential? Whatfunction of the number of particles on each level would you plot on a graph to obtain a straightline?


Einstein’s solidReading 10T: Text to Read

Teaching Notes | Key Terms

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The kinetic theory of gases is not the only theory of how energy is distributed amongst the atoms ormolecules in matter. An early model works for solids like copper or aluminium made of single atoms.It was created by Einstein. The model is often called ‘The Einstein solid’.

Oscillating atoms

Einstein solid

solid consists of three-dimensional lattice of atoms

energy of oscillation =kinetic energy +potential energy

each atom can vibrate inthree independent directions

atom behaves likemass and springs

Einstein imagined a solid as a three-dimensional array of atoms or ions. Each atom oscillates about afixed point. The atoms act like masses and the bonds between atoms act like springs. Like allharmonic oscillators, this gives the atoms kinetic and potential energy. The atoms exchange energyrandomly with one another: if one oscillates more, another oscillates less. Each atom can oscillateindependently in all three directions. So the solid behaves like a collection of harmonic oscillators,with three ‘oscillators’ to every atom.

Energy shared around equallyEinstein used his model to calculate the way the energy could be shared amongst the oscillators. Heproved that at high enough temperatures (far from absolute zero) the average energy per oscillator isjust k T. That is ½ k T for kinetic energy and ½ k T for potential energy. Compare an ideal gas, withenergy ½ k T for each of three directions of motion.


Internal energy of a monatomic crystal (‘Einstein solid’)

solid consists of three-dimensional lattice of atoms

each atom can vibratein three independentdirections

energy of oscillation= kinetic energy +potential energy

atom behaves likemass and springs

kT × 2 × 3

kT × 6 = 3kT per atom

Internal energy of one mole of atoms, U = 3NAkTMolar thermal capacity, 3NAk = 25 J K–1 mol–1

Many monatomic solids have molarthermal capacity about25 J K–1 mol–1

This is known as the Law of Dulongand Petit

energy kT for eachindependentcomponent of energy

12

12 1

2

The three directions of motion are still there in the solid. Each atom is like three oscillators vibrating atright angles to one another. So that is 3 2 ½ k T of energy per atom on average, or 3 k T in all.

Thermal capacityIf the energy per atom is 3 k T on average, the internal energy U of N atoms is

.3NkTU

So if the temperature rises by T, the internal energy increases by

.3 TNkU

Thus the thermal capacity of N atoms in the solid (at constant volume), which is the change in internalenergy per degree rise in temperature, is 3 N k.

Result: a prediction:

thermal capacity of an Einstein solid = 3NA k

–1–1–1A

–1–123

–123A

molK J 253

K J 1038.1

mol 1002.6

kN

k

N


Energy in quanta E = h f Actually, Einstein’s model does much more than we have told you. It starts from quantum behaviourof oscillating atoms, and – like photons – gives each an energy E = h f or a multiple of h f. This letEinstein model the sharing of energy between atoms as the exchange of energy in quanta h f. Hecould then calculate the energy an atom was likely to have at any temperature, even close to absolutezero. We have just told you the result for temperatures well above absolute zero.

The energy of the oscillators comes in quanta

spring:bondbetweenatoms

mass:mass ofatom

Einstein’s idea:

Use the quantum relation E = hffor the energy of oscillation

h is the Planck constantf is the frequency of oscillation

Let the oscillators have energyhf, 2hf, 3hf, etcIn general energy E = nhf

Ladder of energy

hf

hf

hf

hf

Equipartition of energyReading 20T: Text to Read


Quick Help

Or ‘equal shares for all’If you assume that at temperature T the internal energy of a substance will be of the order k T perparticle, you may generally expect to be within a factor of two or three of the right answer. Thisreading explains why this is.

Many ways to moveThink about a fairly complicated molecule. Buffeted by others, it may move in a variety of ways.


A complex molecule may:

move as a whole in eachof three directions

spin or rotate around axesin each of three directions

have particles that vibratefrom side to side

have particles that vibratealong bond directions

If you had to write an equation for the total energy, it would be a sum of terms. Each term would

belong to a given way of moving. For example, ½ m v2 for movement in a given direction, or ½ k x2 forpotential energy of a stretched spring-like bond. The total number of such terms which you wouldhave to write down is called the number of degrees of freedom. Roughly speaking, it is the numberof independent ways in which a particle can have energy. The energy available gets shared outamongst them.

Energy per degree of freedomCompare the complicated molecule with a single atom, which only has energy by travelling in threedimensions.

each atom can move ineach of three directions

three degrees of freedom


The total energy of an atom like this is just the sum of three terms:

.2212

212

21

zyx mvmvmv

You should know that from the kinetic theory of gases, the energy per atom comes out as (3/2) kT.There is no reason to expect the average energy of motion in the x-, y- and z-directions to bedifferent. So it looks as if, on average, each of the three degrees of freedom gets energy (1/2) kT.This turns out to be rather generally correct (with an important proviso, mentioned below).

Energy in a solidThink about a monatomic solid, such as copper or gold. Its atoms are arranged in regular array. Theydo not travel about at random at all. But the random thermal agitation is still there. The atoms oscillateabout fixed positions, and have energy in that way instead.

The equation for the energy is:

energy of particle= kinetic energy plus potential energy in x-direction+ kinetic energy plus potential energy in y-direction+ kinetic energy plus potential energy in z-direction

The equation has six terms.There are six degrees of freedom

each atom can oscillatein each of threedirections

If the atoms are rather loosely bound together, so that the vibrations of one do not affect the way its

neighbours can vibrate, the energy equation has six terms. There is (1/2) m v2 kinetic energy and

(1/2) kx2 potential energy, for each of three directions. There are now six degrees of freedom. If eachdegree of freedom gets on average energy (1/2) kT, that gives 3kT in all, per particle. The measuredvalues of the thermal capacities per mole for many monatomic crystalline solids confirm that theirchange in internal energy with temperature change T is indeed approximately 3kT per particle, or3NkT per mole.

This fact was first noticed by Dulong and Petit. 3Nk comes to around 25 J K–1 mol–1, the same for alarge number of different monatomic solids.

The idea does not work if……the smallest possible energy for a degree of freedom is bigger than kT. For example, an H2

molecule can vibrate. But the H atoms are not very massive, and the bond between them is strongand stiff. So the natural frequency of vibration f is very high. The smallest possible energy of vibrationis of the order E = hf, where h is the Planck constant. That is, the same quantum equation as forphotons works here (see chapter 7). This energy E = hf is, for H2 molecules, less than kT up totemperatures of about 1000 K. So the buffeting from the molecules hitting each other in their randomthermal motion is not enough to set the oscillations going. The energy per particle is less than youexpect from the ‘equal sharing out’ principle.

As the temperature falls, the same kind of difficulty applies to other kinds of motion. The spinning


motion of H2 molecules only sets in between 50 K and 100 K, for example. The rotations have aminimum energy too, but it is much smaller than the minimum for vibration.

But it’s still a good principleIn spite of the above proviso, it’s still true that the energy per particle is within a factor of two or threeof the quantity kT, for many substances. The idea goes wrong at very low temperatures, butotherwise is a good guide as to what to expect. It’s rough and ready, but it’s simple and it’s veryuseful.

Flow in liquidsReading 30T: Text to Read


Quick Help

Liquids are runny; solids are stiff. Here you can read more about the differences between liquids andsolids, and how the ease with which some liquids flow has to do with the Boltzmann factor:

.exp

kT

Water and iceEven though you know it is true, it can be hard to believe that the rigid, brittle material which comesout of the freezer as ice cubes is the same stuff as the water which runs from the tap. Their densitiesare very similar; actually ice has the lower density. So it can’t simply be that the water molecules arefurther apart in water, with spaces between them to let them move about. But it must be true that themolecules in ice can’t move about, while the molecules in water can.

If you were a water molecule you would find it very difficult to tell from your immediate surroundingswhether you were part of an ice cube or of a glass of water. Other water molecules would surroundyou at much the same distances in quite similar patterns. Only if you as a molecule had ‘long-rangevision’ could you tell the difference. In ice you would ‘see’ molecules arranged in an orderly spatialstructure. In water, there would not be the same long-range orderliness. The patterning of moleculesat any instant would be similar to that in ice, but if you kept watching the pattern would be changing allthe time.

An aside:H2O is specialAs an aside, we need to say that water is special. It is special because of the structure of the H2Omolecule.


+

+

Water molecule

positive protonsleft almost ‘bare’

negative electronsmostly concentratedaround the oxygenatom

–

Hydrogen bonds between water molecules

hydrogen bonds:‘bare’ protons attract electronsaround oxygen atom of adjacentmolecule

–

–

–

–

repulsion:‘bare’ protons repel‘bare’ protons ofanother molecule

covalent bondsharing twoelectrons

+

+

+

+

+

+ +

+

Oxygen

Hydrogen

Because hydrogen atoms have only one electron, when they bond to oxygen and share this electronwith the oxygen atom, the hydrogen nucleus (proton) is left rather exposed. It can attract or repelcharges on other molecules. Because the two hydrogen atoms in H2O bond on the same side of theiroxygen atom, the H2O molecule is left electrically lopsided. It is a ‘polar’ molecule.

In liquid water, the molecules continually form, break and re-form bonds to other water molecules.These bonds of attraction between a proton on the side of the molecule and the electrons round theoxygen atom of another molecules are called hydrogen bonds. They give water many of its specialproperties, especially remaining liquid at rather high temperatures (compare H2S).


The cage modelIt has proved very difficult to get a successful model of liquids at the molecular level. One reasonablyuseful model is the ‘cage model’. This says that each molecule behaves, just as in the solid, as if‘caged in’ by its neighbours. But, in a liquid, the molecules can break out of one cage and get into anext-door one. This makes the material fluid, whilst remaining approximately as densely packed asthe solid.

Cage model of flow in liquids

molecule caged in by neighbours molecule breaks out into next-door cage

Molecules bumping into one anotherYou can estimate how often molecules in a liquid bump into the neighbours that cage them in. Fromthe kinetic theory of gases, you know that molecular speeds at around 300 K are of the orderhundreds of metres per second. The molecules in a liquid are packed close together at distances of

the order of the diameter of the molecule, say a few tenths of a nanometre, or around 10–10 m. For amolecule to bump into a neighbour, it would need to travel say a tenth of the distance between the

centres of molecules, if there is not much empty space. So it needs to go say 10–11 m at a speed of

the order of 102 m s–1. This suggests that molecules in liquids bump into the walls of their cages

roughly 1013 times a second. Another way to the same answer is to calculate the frequency ofoscillation of a molecule, thinking of the cage as like a restraining spring.

Frequency of breakoutsExperimental evidence suggests that at room temperature a water molecule oscillates inside its

temporary cage for about 4 ps, that is 4 10–12 s, before succeeding in breaking out and entering

another cage. If the molecule goes back and forth in the cage 1013 times a second (see above), thenit stays in the cage for about 40 back and forth trips. These molecular cages are clearly verytransitory. But it still makes sense to think of caged in molecules, if they have to make about 40breakout attempts on average in order to get out.


Flow in liquids: a simplified model

Each molecule is ‘caged in’ by its neighbours

speed of molecule 100 m s–1

distance to break out 0.01 nm

time taken 0.01 nm

= 10–13 s

frequency of ‘break-out attempts’ 1013 s–1

Liquid flows if molecules can break out of one cage into the next.To break out, a molecule must first climb over a potential energy hill.

Flow in liquids is an activation process: temporary energy of activationmust be borrowed from energy of thermal agitation

moleculecaged in

energy tobreak out

moleculecaged in

molecule pushes neighbours aside

about 1/10 atom diameter0.01 nm

displacement relative to neighbours

100 m s–1

Getting luckyTo get lucky, one time in 40 or so, and break out, a water molecule must probably find favourablecircumstances, such as a temporary gap in the wall of the cage just where it happens to go. But it willalso need energy to shoulder other molecules aside (and energy will be needed momentarily to openup a gap). You can estimate the energy involved, using the Boltzmann factor exp(– / kT ).

Try it with a calculator. What value of x = / kT gives exp(x) equal to 40, and so exp(– / kT ) equal to1 / 40?

You’ll find that it is between 3 and 4, about 3.7. This is around 1 / 10 eV, or about 10 kJ mol–1. It isless, but not much less, than the energy to break a hydrogen bond, which is typically twice as much.However, the calculations are very rough, and it would be wrong to pay much attention to a merefactor of two. You can see that if a water molecule happens by chance to get energy equal to just afew times the average energy of the order kT, it can break out of its cage.

If you don’t have a calculator handy, here’s a table to help:

Ratio k T exp ( / k T)

1 2.7

1.5 4.5


Ratio k T exp ( / k T)

2 7.4

2.5 12.2

3 20.1

3.5 33.1

4 54.6

4.5 90.0

5 148.4

Why liquids flowMany liquids, such as water, flow because the patterns of arrangements of molecules don’t holdtogether for very long. The patterns survive, in the case of water at room temperature, for 40 or so‘attempts’ to break them up. On our human time scale of seconds or longer, structures like cagesaround molecules last for only picoseconds. But from the molecular point of view they last quite awhile. Few prisoners would count 40 escape attempts before getting out as an easy option. But ofcourse the molecules aren’t penalised for trying!

Getting hotter, flowing betterAt the macroscopic, human scale, molecules changing places are reflected as the ease with which aliquid flows. Syrup flows very slowly, water flows rather easily. The difficulty of flow is measured by aquantity called the viscosity. The discussion here should suggest to you that the viscosity will get lessas the temperature rises. You would be right if you guessed that, for many liquids, the viscosity is ingreat part determined by the Boltzmann factor exp(– / kT ). This is of great importance to thedesigners of motor oils. They need the oil to flow well enough for a cold start, but not to get so runnyin the hot engine that the lubrication fails. The label such as ‘15W–40’ on a can of motor oil indicatesthe variation it can tolerate.

Things to do, things to think about1. Write out for yourself the calculation which suggests that molecules in a liquid make about 1013

attempts per second to break out of the cage formed by their neighbours. Think about theassumptions being made.

2. If / k T is about 3.7 at T = 300 K, what is it at T = 310 K? Show that this suggests that theviscosity of water might decrease by about 10% for such a temperature rise of only 3%.

3. Suggest why the arguments here would not work for polymer liquids made of long chainmolecules slithering amongst one another.

Why you can’t get to absolute zeroReading 40T: Text to Read


Teaching Notes | Key Terms Quick Help

You can find out here why it is impossible to get to absolute zero. The reason is that it is infinitely faraway! In finding out you will learn more about the Boltzmann factor exp(–/kT).

The meaning of hot and coldIn something very very cold, nearly all the particles have the least possible energy. Only a very fewhave more than the least possible energy. The Boltzmann factor, which is the ratio of the number ofparticles in one state to the number in a state with energy less, is essentially zero.

(We have said ‘least possible energy’, not ‘zero energy’ because in quantum physics the lowestpossible energy is not zero. But since it is the lowest possible energy, it behaves just like zero energy– there’s nothing less. So from now on we shall say ‘zero energy’ for the lowest possible state of aparticle, without embarrassment.)

As matter gets hotter, more and more particles occupy higher and higher energy states. The averageenergy per particle gets bigger. Since particles are continually exchanging energy, some going up alevel and others going down, there are always still some particles in the lower energy states, howeverhot the material gets. In fact, the hottest it can get and still be in equilibrium is with equal numbers ofparticles in each state with a given energy.

The Boltzmann factor, again the ratio of numbers of particles in one state to the number in a statewith energy less, is now nearly equal to 1.

Here is a graph of the values of the Boltzmann factor, at different ratios kT/, as the temperature Tvaries.

Boltzmann factor plotted against T

Absolute zero Infinitely hot

1.0

0.8

0.6

0.4

0.2

0.0

0 1 2 3 4

kT /

Most particleshave zeroenergy. Fewhave higherenergies

Particles existat all energiesin equalnumbers

It is a graph of the Boltzmann factor against temperature T, but with T scaled so as to show the ratioof the thermal agitation energy k T to the difference in energy between pairs of states. The positionsof the ‘hot’ and ‘cold’ ends of the scale are marked. The pattern of spreading of particles amongstenergy levels at the extreme ends is pictured.

An S-shaped graphThe graph goes from 0 when T is zero to 1 as T becomes bigger. Only as the temperature tends to


infinity does the graph tend to 1. But it is already a good way there, reaching 0.8 in the graph shownfor kT only four times .

At the low end, where temperature T goes towards zero, the graph creeps towards zero. As thetemperature rises, the graph gets steeper and steeper at first, before starting to level off.

This is quite a complicated graph, and it is not easy to see how the S-shape arises from the equationexp (–/ kT ). Luckily there’s a simpler way of looking at it.

A simpler graphThe graph of exp (–x) is just a simple decay curve. But in the Boltzmann factor / kT is just equivalentto x in this expression. So a plot of the Boltzmann factor exp (–/ kT ) against / kT should be asimple decay curve. And here it is:

Boltzmann factor plotted against 1/T

Infinitely coldInfinitely hot

1.0

0.8

0.6

0.4

0.2

0.0

0 2 4

/ kT



6 8 10

The graph is a nice recognisable shape. It starts at 1 and falls, but ever more slowly, towards zero. Itnever quite reaches zero, though it gets as close as you like.

But there’s something peculiar. Because it is in effect a graph against 1 / T , not against T , thetemperature scale runs from ‘hot’ at the low end to ‘cold’ at the high end. Near the origin, T is verylarge and 1 / T is very small. If the temperature were infinite, 1 / T would be zero.

Similarly, the ‘cold’ end of the scale is at the large values on the axis. As temperature T gets less, so1 / T gets bigger. If the temperature could reach zero, then 1 / T would become infinite.

In this simpler graph absolute zero is infinitely far away. But it seems to be in the wrong place, at thetop end of the scale!

Yet another graphIf you would prefer to have absolute zero at the low end of the scale, there’s a simple way to do it.Just plot the Boltzmann factor against –1 / T instead of against 1 / T . Here it is:


Boltzmann factor plotted against –1/T

Infinitely cold Infinitely hot

1.0

0.8

0.6

0.4

0.2

0.0

–10 –8 –6

– / kT



–4 –2 0

The graph is exactly the same as the last one, but just flipped over right to left. Now absolute zero isinfinitely far away at the lower end of the scale. Hotness increases from left to right, reaching infinitehotness when the scale stops being negative and becomes zero.

So it really does seem to be true that you can’t get to absolute zero because it is infinitely far away.

Try it yourselfYou can use a spreadsheet to plot the graphs above.


Not just mathematical gamesThis playing around with graphs probably strikes you as a mathematical game that proves very little.In a way you would be right. For example, you wouldn’t want to say that superconductivity isunattainable just because the conductance becomes infinite if the resistance becomes zero.

But there is a reason why we have not just been playing mathematical games. It is to do with how tochoose a good scale of coldness to hotness. On such a good scale, absolute zero belongs naturallyat minus infinity. And ‘as hot as can be’ belongs naturally at zero.

What does ‘extremely hot’ mean?You learned as a child that hot things burn you. That’s because hot things give up energy very easily.It is easy for chance to find a particle that gives up its energy, because there are a lot of particles withplenty of energy, for chance to choose from. Roughly speaking, it makes very little difference to a hotobject to give up a bit of energy. Nor for that matter to acquire a bit more. It’s like a rich person: givingaway or getting a small amount of money makes no real difference. So for hot things, gaining orlosing energy is a case of ‘almost no change’.

What does ‘extremely cold’ mean?Contrast something very cold. It is rare for it to give up energy by chance, because chance has a hardtime finding a particle with any energy to give up. And giving up energy is a big deal, with so little toshare around amongst the particles. Similarly, a very cold object gaining energy is also a big deal.Now it’s like a very poor person losing or gaining money. Losing a small amount may be a disaster;


gaining a small amount may be a huge help. What would make no difference to the rich person is ofgreat importance to the poor one. For cold objects gaining or losing energy is a case of a really bigchange.

Very cold and very hot compared

Very cold Very hot

Most particles have zero energy.Few have higher energies

Particles exist at all energies inalmost equal numbers

energy easilygoes in thisdirection byrandomexchanges

What does temperature difference mean?Here’s the bottom line in the meaning of a temperature difference: energy goes from hotter to colder.That is, energy goes spontaneously, just by random exchange of energy between particles, from hotto cold: from the haves to the have-nots. And it happens because it’s easier for chance to find aparticle to give up energy in the hotter object, and harder for chance to find one to do it in the colderobject.

A good scale of temperature will show that exchanging energy at random makes almost zerodifference to a very hot object, but makes a huge difference to a cold one. So on a ‘good’ scale oftemperature, ‘hot’ should correspond to near zero, and ‘cold’ should correspond to a large magnitude.That’s just what the scale in which hotness and coldness is measured by –1 / (conventionaltemperature) achieves. So there’s some sense in it after all.

Each step to absolute zero is harder than the lastThink about a substance near zero temperature. The majority of particles have no energy. Just a fewhave just a little. To make it colder, somehow one of the tiny minority must give up energy (and noneof the majority must gain any). That’s hard to fix. And if it succeeds, it gets even harder next timebecause there are even fewer particles with energy to give up.

This is the sense in which absolute zero is infinitely far away. You can always get closer, and indeedpeople have got well below millionths of a kelvin. But the next step is always harder than the last, soyou can’t ever quite get there.

A last word:EntropyRead on only if you want to…

Removing energy decreases entropy; adding energy increases it.

But adding or removing energy to or from hot things hardly changes their entropy at all.

Adding or removing energy to or from cold things changes their entropy a lot.

As a consequence, energy going from something hot to something cold reduces the entropy of


the hot thing only a little and increases the entropy of the cold thing a lot.

Result: the total entropy increases. And that’s what entropy does: it never ever decreases andessentially always increases, in every possible change.

The story of rubberReading 50T: Text to Read


Quick Help

Gases exert a pressure because of the thermal agitation of their molecules. Rubber is stretchy andpulls back as you stretch it, also because of thermal agitation. Here’s why.

Wriggling chainsRubber is twitchy stuff. Its molecules are long chains, but the links in the chains can twist and turn.Where a gas molecule is free to move with energy of the order kT , a link in a rubber chain can twistand turn with similar energy.

A rubber molecule: a linked chain

isoprene molecule

every link cantwist and turn

rubber is the polymerpolyisoprene

Raw rubber, latex from the rubber tree, is a milky liquid. The twisting and turning of links in the chainscoils up each molecule into a ball. Only when a further ingredient is added do you get the familiarstretchy stuff of rubber bands.

VulcanisationThe eccentric American inventor Charles Goodyear drove his family to distraction, indeed to penury.In 1839 on his wife’s kitchen stove, he made a discovery which created a whole new industry.Cooking raw rubber with sulphur, he found that the material stopped being wet and sticky, andbecame dry, solid and stretchy. The more sulphur there was in the brew, the stiffer the rubberbecame. A make of car tyre is named after him. More important, whole forests in Asia and SouthAmerica were planted with rubber trees. Colonies were founded on this substance.

Goodyear called his process vulcanisation, after Vulcan the Roman god of fire. What happens is this.The polyisoprene chains get cross linked by sulphur bridges (something similar happens to hair if yougive it a ‘permanent wave’). The cross-links pin the chains together every so often. Now stretching the


rubber becomes pulling the cross-linked places further apart, straightening the chains between them.

Two cross-linked chains

sulphur bridges

Feel the pullPull a rubber band in your hands. Feel it pull back on you. You are feeling the equivalent, in rubber, ofthe pressure of air when you blow your cheeks out. But it’s a pull, not a push.

The rubber pulls back because each link in the chain is twisting and turning, with energy of the orderk T. Most random twists or turns will shorten the chain between cross-links, because only a few twistsor turns set a link pointing in the special direction need to lengthen it in the direction you happen to bepulling. Most are in some other direction. So chance twisting and turning sees to it that the rubberpulls back.

Stretching vulcanised rubber

unstretched stretched

greater distancebetween cross-links

Measure the number of isoprene links between sulphur bridgesTake a rubber band and measure its natural unstretched length. Now pull it as hard as you can, untilit almost feels like steel and won’t stretch any further. Get someone to measure the fully stretchedlength (quickly, before it breaks).

When fully stretched, the chains between sulphur bridges must almost run straight. If there are Nisoprene links between pairs of sulphur bridges, and each link has length L, the distance between


sulphur bridges has been pulled out until it is equal to NL.

When the rubber is relaxed, the isoprene links meander randomly between sulphur bridges. Nothingtells them to go one way or the other. This is a well known problem, the ‘drunkard’s walk’. You mayhave met it in discussing diffusion of a gas. In a random walk of N steps, the average (strictly the rootmean square) distance between one end of the walk and the other is the equivalent of N steps. Sothe sulphur bridge cross links will on average be LN apart.

Thus you have:

stretched length of rubber proportional to LN

unstretched length of rubber proportional to LN

ratio of stretched length to unstretched length = LN / LN = N.

For example, if you can stretch a rubber band by five times, there are about 25 isoprene linksbetween sulphur bridges in it.

Motor tyres have much more sulphur than rubber bands, and stretch much less. N is much smaller.

Hotter rubber pulls harderIf a rubber band is pulled gently and made hotter, the tension increases. This is because the links inthe chain twist and turn more vigorously and pull back harder for a given extension. Theory showsthat rubber behaves rather like an ideal gas:

gas: pressure is proportional to absolute temperature at constant volume

rubber: tension is proportional to absolute temperature at constant extension.

You might like to try to test this experimentally. It’s quite a challenge. If the temperature is changed by10 K from say 300 K to 310 K, the tension should increase by only 3% (equal to the percentagechange of temperature). It isn’t easy to measure so small a change. And it isn’t easy to do it whilekeeping the rubber at a constant known temperature above room temperature. Maybe something likethis would be a starting point:

Tension and temperature

laser beam detects deflection of support

screw toadjusttension

thermocoupleto measuretemperature

warm air tocontroltemperature

rubber band


Liquid crystalsReading 60T: Text to Read


Quick Help

Your digital watch, a lap-top computer screen, a beetle’s back, a cell in your body, a thermometerwhich changes colour. What do they all have in common? Answer: all contain liquid crystals. Thisreading tells you what liquid crystals are and describes some of their uses.

Rod-like moleculesIt’s fair enough to think of many small molecules as more or less spherical. But some organicmolecules are much more cigar-shaped, perhaps ten times as long as they are wide. In the liquidform, the rod-like molecules have a tendency to line up, despite the general thermal agitation.

Nematic liquid crystal

ordered inorientation

not ordered in three-dimensional spatialposition

Imagine a shoal of fish. They all swim in roughly the same direction. But the distances between fishvary more or less randomly, and change all the time. But the direction of swimming of the shoal andeach fish in it, stays the same. It’s a pattern in orientation, not in relative positions.

A substance with molecules arranged in this way is called a liquid crystal.

It has the disorder in spatial position of a liquid, and the order – even though only in the way themolecules point – of a crystal. This type of liquid crystal is called ‘nematic’ (from the Greek wordnematos, a thread). The ordering is sensitive to temperature. For example the organic substancep-azoxyanisole is crystalline below 118.2 C, becomes a nematic liquid crystal at that temperature,and then becomes a liquid at 135.3 C. Liquid crystal behaviour typically exists over just a narrowranges of temperatures. Their fragile ordering is very easily disrupted by too much thermal agitation.

Twist gives colourThere is a variation on the lined-up order of nematic liquid crystals, which can make them coloured.The colour can be extremely temperature sensitive, changing over a range of only a degree or so.This variation is for the direction of lining up to twist gradually over distance.


Cholesteric liquid crystal

pitch

If the pitch is of the order of the wavelength of visible light, interferencecan occur between similarly oriented layers, giving reflected colours

Instead of all being lined up in one direction, the direction of lining up gradually changes. Thedistance over which the lining up takes one complete turn is called the pitch. This kind of liquid crystalis called cholesteric, after the substance cholesterol benzoate in which it was first discovered.

The pitch may be many times the length of a molecule. If it is of the order of the wavelength of visiblelight, the material behaves as if it contains layers spaced at that distance. This can give interferenceeffects. The light reflected by the liquid crystal can be coloured.

The pitch can be very sensitive to temperature. One type of cholesteric liquid crystal changes fromred to blue when the temperature rises by only 3 C (note: red is cold, blue is hot!). It is used to makevisual thermometers. You may be given such a material to detect, for example, the slight cooling ofwater on evaporation.

Wristwatch and computer displaysIn wristwatch and computer displays, a liquid crystal is sandwiched between transparent surfaces.The alignment of the molecules is controlled by switching a potential difference between oppositepatches of the surface on or off. With polarising filters, the patch can be made to look transparent ordark. In a computer screen, each patch is one pixel.

Advantages of such displays are the small potential difference needed and the negligible currentdrawn. Your watch battery can last for years. A disadvantage is the rather slow reaction time of theliquid crystal, which makes animating the display difficult.

Liquid crystals inside youThe walls of the cells of your body are made of liquid crystal. The walls consist of a fence of parallelrod-like molecules. Such a wall is flexible and partially permeable, but protects the contents of the cellfrom the outside.


Phospholipid molecule

water-attractingoxygen-rich end

water-repellinghydrocarbon tail

watery surroundings of cell

watery interior of cell

Cell wall

Cell wall isa layer ofmolecules,with water-attractingends facingoutwards

Revision ChecklistI can show my understanding of effects, ideas and relationships bydescribing and explaining cases involving:the ratios of the numbers of particles in a body that exist in states at different energy, and thechanges in these ratios when the temperature changes

A–Z references: energy kT, Boltzmann factor

Summary diagrams: Climbing a ladder by chance, Temperature and the average energy perparticle, Different processes related to the average energy per particle, The Boltzmann factor,How the Boltzmann factor changes with temperature

the idea of activation energy, and link this idea to what happens in various processes when thetemperature changes (e.g. changes of state, thermionic emission, ionisation, conduction insemiconductors, viscous flow)

A–Z references: thermal activation processes

Summary diagrams: The Boltzmann factor, How the Boltzmann factor changes withtemperature, Examples of activation processes

I can use the following words and phrases accurately whendescribing effects and observations:Boltzmann factor

A–Z references: Boltzmann factor

Summary diagrams: The Boltzmann factor


activation energy

A–Z references: energy kT, thermal activation processes

Summary diagrams: Examples of activation processes

I can sketch, plot and interpret:graphs showing how the Boltzmann factor varies with energy and temperature

A–Z references: thermal activation processes

Summary diagrams: Different processes related to the average energy per particle, How theBoltzmann factor changes with temperature

I can make calculations and estimates involving:ratios of characteristic energies (energies of a particle at which changes might occur) to theapproximate mean energy per particle kT

A–Z references: energy kT

Summary diagrams: Temperature and the average energy per particle, Different processesrelated to the average energy per particle

the Boltzmann factor exp(– / kT)

A–Z references: Boltzmann factor, thermal activation processes

Summary diagrams: How the Boltzmann factor changes with temperature, Examples ofactivation processes

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Printing - Advancing Physics A2 Teacher Edition Advancing Physics 1. What is the probability that a particular particle is to be found in the left-hand compartment a long time after