Printing - Advancing Physics A2 Teacher Edition14...3 Advancing Physics 5. Assume hc / kT which gives = 5 10–3 m; millimetre wavelengths are in the microwave part of the spectrum

1 Advancing Physics

Molecular energyQuestion 10W: Warm-up Exercise

1. E = kT = 1.38 10–23 300 = 4 10–21 J. (This answer uses the rule of thumb that molecular

energies are of the order E = kT. For a gas, a more exact relationship is E = (3 / 2)kT.)

2. The same as the average kinetic energy of an oxygen molecule, 4 10–21 J (they are in thermal

equilibrium).

3.

2NN

2OO 2

1

2

1cmcm

so the ratio of r.m.s. speeds is equal to the inverse ratio of the square roots of the masses:

94.032

28

r.m.s.N

r.m.s.O c

c

4. Individual bromine molecules are continually colliding with air molecules so their paths are likerandom walks. The diffusion rate will be much less than the molecular speed.

5. Hydrogen molecules have a molar mass of 2 g compared with 70 g for bromine so their rmsspeeds will be 35 = 6 times greater as will their diffusion rate.

6. The diffusion rate is inversely proportional to the square root of the molar mass.

Evaporation of waterQuestion 20W: Warm-up Exercise

1.

J 106.6mol 1002.6

mol J 100.4 20123

13

2. = kT = 1.38 10–23 J K–1 290 K = 4 10–21 J.

3.

5.16J 100.4

J 106.6/energyratio

21

20

kT

4. Activation processes go at an appreciable rate if act is less than about 30 kT. In this case it isabout 16.5 kT, so it should proceed at a reasonable rate. At this temperature the fraction ofmolecules with enough energy to escape at any particular moment is small, but there are somany molecules that the flux from the surface is significant.

5. At a higher temperature act / kT is smaller and a greater fraction of molecules have enoughenergy to escape, so the rate of evaporation increases.

6.

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J 102.2mol J 1040mol g 18

g 1000 6131

7. Energy must come from the surroundings.

Thermal radiationQuestion 30W: Warm-up Exercise

1. E = hf = hc / in J = hc / [(1.6 10–19) ] in eV = 1.24 10–6 / : microwave 2.5 10–4 eV;

infrared 2.5 10–2 eV; visible 2.5 eV; ultraviolet 25 eV; x-ray 2500 eV.

2. Body temperature is about 310 K. At this temperature kT = 2.7 10–2 eV which corresponds tothe energy of infrared photons, so we radiate a spectrum of electromagnetic waves dominated bythe infrared.

3. hf = hc / = 3kT, thus = hc / 3kT = 8.0 10–7 m, which is in the near-infrared.

4. kT = 0.53 eV. If we take the wavelength of the near-ultraviolet to be 400 nm the correspondingphoton energy is hc / = 3.1 eV, which is about 6 kT so the Sun will emit plenty of ultravioletradiation.

5. No. The energy of an x-ray photon is several thousand times kT so only a tiny fraction of theSun’s energy will be radiated as x-radiation.

6. It does not emit very much at shorter visible wavelengths and so appears red. It must have asurface temperature significantly below 6000 K.

Calculating thermal energiesQuestion 40W: Warm-up Exercise

1.

eV0.21orJ103.3mol 1002.6

mol J 1020 20123

13

2. The energy per particle per kelvin is

eV044.0J100.7mol 1002.6

K 1mol K J 4200 21123

11

3. kT = (1013 eV) (1.6 10–19 J eV–1) so T = 1.6 1017 K

4. ½mv 2 = kT from which

12 s m 1302 m

kTv

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5. Assume hc / kT which gives = 5 10–3 m; millimetre wavelengths are in the microwave partof the spectrum.

Probability and equilibriumQuestion 50W: Warm-up Exercise

1. 0.5

2. 0.5

3. N / 2

4. Decreasing with time from N initially, tending towards N / 2 after a long time.

5. It has the same unchanging distribution of particles on either side of the barrier. It is also thearrangement that can occur in the largest number of distinct ways.

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

1. Protons are positively charged and repel one another. To get them close they have to climb up anelectrical ‘potential energy hill’ in order to get close enough to interact via the strong nuclear forceand fuse.

2. High temperatures are required so that some protons have enough thermal kinetic energy tonegotiate this electrostatic potential energy barrier when they collide.

3. kT = [(1.38 10–23 J K–1) (6.0 106 K)] / 1.6 10–19 J = 520 eV

4. 106 eV / 520 eV = 2000, approximately

5. Very very few. (Note there is also a quantum tunnelling effect that contributes to the rate offusion.)

6. Fusion reactors on Earth require a much higher power density because they cannot be built onthe scale of the Sun! This means they must operate at much higher temperatures so that asignificant fraction of the interacting particles have enough energy to achieve fusion.

Molecules and changeQuestion 20S: Short Answer

1. The wetness is due to water molecules bonded to the fabric and to other water molecules.Random collisions can lead to water molecules having enough energy to break these bonds andescape into the air. In a breeze they would then be swept away.

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2. A water molecule bound in an ice lattice will have a set of permanent neighbours. As thetemperature rises bonds will break and the water molecules will move about more chaotically andfreely, though there are still intermolecular forces present. A molecule’s neighbours too are nowmoving about!

3. The particles which make up the milk become dispersed in the tea, mixing with the othermolecules already present (such as tannin). Eventually, when fully stirred in, the milk particles willbe equally dispersed throughout the liquid mixture.

4. Thermal energy is transferred from the radiator by the emission of infrared radiation and byconvection, where energy is given directly to nearby, freely moving air molecules.

Energy per particleQuestion 25S: Short Answer

1. Energy = qV = 1.0 10–3 C 10 V = 1.0 10–2 J = 10 mJ

2. Energy = qV = 1.6 10–19 C 1000 V = 1.6 10–16 J

3. Energy = qV, so q = Energy/V, giving q = 1.6 10–19 J/ 1 V = 1.610–19 C

4. Energy = qV = 1.6 10–19 C 1 V = 1.6 10–19 J

5. 1 eV = 1.6 10–19 J, so 1 MeV = 106 1.6 10–19 J = 1.6 10–13 J

6. 1 eV = 1.6 10–19 J, so 13.6 eV = 13.6 1.6 10–19 J = 21.8 10–19 J

7. Energy = qV = 6.02 1023 1.6 10–19 C 1 V = 96300 J = 96.3 kJ

8. Error as percent = 100 (100 kJ – 96.3 kJ)/96.3 kJ = 3.8%, or 4% approximately

9. Taking 1 eV 100 kJ mol–1, 13.6 eV 1300 kJ mol–1

10. Taking 1 eV 100 kJ mol–1, 5.1 eV 500 kJ mol–1

11. Taking 100 kJ mol–1 1 eV, 436 kJ mol–1 4.4 eV

12. Taking 100 kJ mol–1 1 eV, 2.79 kJ mol–1 0.03 eV

13. Energy E = 1000 eV = 1000 1.610–19 J. Frequency f = E/h, so wavelength = c/f = hc/E = 6.6

10–34 J Hz–1 3.0 108 m s–1 / 1000 1.6 10–19 J = 1.2 10–9 m = 1.2 nm 1 nm to within20% error.

14. Energy reduced by factor 1000, frequency reduced by factor 1000, so wavelength increased byfactor 1000. Thus wavelength is roughly 1000 nm = 1 m.

15. Energy is 13.6 eV, of order 10 eV, so wavelength is of order 1000 nm / 10 = 100 nm. This is in theultraviolet region.

16. If wavelength is 1 mm, energy per photon is 1000 times less than energy for wavelength 1 m,which is 1 eV approximately (see answer to question 14). Energy per photon = 1/1000 eV. Since

1 eV is close to 100 kJ mol–1, the energy per mole of photons is approximately 0.1 kJ mol–1. This

is about 10 times larger than the energy 0.08 kJ mol–1 to evaporate liquid helium.

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17. 1 eV 100 kJ mol–1

18. To within 20%, a photon energy of 1 eV corresponds to a wavelength of 1000 nm = 1 m. If theenergy increases by a factor 10, the frequency increases by the same factor, and the wavelengthdecreases by a factor 10.

Likely events in timeQuestion 30S: Short Answer

1. Chronologically, the sequence is most likely to run B, C, D, A, with decreasing amounts of orderas time goes on.

2. The second sequence with fewer marbles is more difficult to establish because each of thesnapshots shown is quite likely to occur as there are fewer possible arrangements altogether.

Values of the energy = kTQuestion 35S: Short Answer

Solutions1. Energy = 6.02 1023 mol–1 1.6 10–19 J = 96300 J mol–1 = 96.3 kJ mol–1

2. Error as percent = 100 (100 kJ – 96.3 kJ)/96.3 kJ = 3.8%, or 4% approximately

3.

Energy / eV Energy kT / kJ mol–1

0.0001 0.01

0.001 0.1

0.01 1

0.1 10

1 100

10 1000

100 10000

1000 10 0000

4. 1.5 eV 100 kJ mol–1 eV–1 = 150 kJ mol–1

5. 40 kJ mol–1 / 100 kJ mol–1 eV–1 = 0.4 eV

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6. Energy = kT = 1.38 10–23 J K–1 1000 K = 1.38 10–20 J

7. Energy per mole = 1.38 10–23 J K–1 1000 K 6.02 1023 mol–1 = 8.31 103 J mol–1 = 8.3 kJ

mol–1

8. Error in percent = 100 (10 – 8.3)/8.3 = 20%

9. At T = 300 K, the energy kT will be 300/1000 of its value at 1000 K. Thus kT for T = 300 K is

(300/1000) 10 kJ mol–1, or 3 kJ mol–1.

10. If kT = 10 kJ mol–1 for T = 1000 K, then for kT = 1000 kJ mol–1, 100 times greater, T = 100 000 K,also 100 times greater.

11.

Temperature T / K Energy kT / kJ mol–1

1 0.01

10 0.1

100 1

1000 10

10 000 100

100 000 1000

1 000 000 10 000

10 000 000 100 000

12. Energy E = 1.0 eV = 1.610–19 J. Frequency f = E/h, so wavelength = c/f = hc/E = 6.6 10–34 J

Hz–1 3.0 108 m s–1 / 1.6 10–19 J = 1.2 10–6 m = 1.2 m 1 m to within 20% error.

13. Since E= hf if the energy increases by a factor 10, the frequency increases by a factor 10. Since = c/f, if the frequency increases by a factor 10, the wavelength decreases by a factor 10.

14.

Temperature T / K Energy kT / eV Energy kT /

kJ mol–1photonwavelength

1 0.0001 0.01 10 mm

10 0.001 0.1 1 mm

100 0.01 1 0.1 mm

1000 0.1 10 0.01 mm

10 000 1 100 1 m

100 000 10 1000 100 nm

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Temperature T / K Energy kT / eV Energy kT /

kJ mol–1photonwavelength

1 million 100 10 000 10 nm

10 million 1000 100 000 1 nm

15. From the table, T 10 million K, and the energy 1 keV for wavelength 1 nm.

16. At 2045 K, the energy kT is about 0.2 eV and 200 kJ mol–1

17. At 77 K, the energy kT is about 0.008 eV and 0.8 kJ mol–1

18. Wavelength at 10 K is of order 1 mm. Wavelength at 3 K is of order 1 mm 10 K / 3 K = 3 mmapproximately

19. Temperature for kT = 1 MeV is 1000 times greater than 10 million K for kT = 1 keV, sotemperature is 10 billion K

20. The wavelength 10 nm for kT = 1 keV, temperature 10 million K, just approaches the x-ray region.In fact, however, there are photons with energy several times kT so there are certainly x-rays inthe centre of the Sun.

Particles spreading outQuestion 40S: Short Answer

1. 2100 1030

2. log 2100 = 100 log 2 30; so 2100 1030

3. 1024 seconds

4. 3 106

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

1. Energy = 40 103 J mol–1/ 6.02 1023 mol–1 = 6.6 10–20 J per particle.

2. At 300 K, kT = 1.38 10–23 J K–1 300 K = 0.4110–20 J per particle. Ratio /kT = 6.6 10–20 J

per particle / 0.4110–20 J per particle = 16 approximately.

3. Energy per electron = 5.4 eV 1.6 10–19 J eV–1 = 8.6 10–19 J per electron.

4. At 2000 K energy kT = 1.3810–23 J K–1 2000 K = 2.8 10–20 J per electron. Ratio /kT = 8.6 10–19 J per electron / 2.8 10–20 J per electron = 30 approximately.

5. The energy is less by a factor 2, so the material might emit electrons appreciably at about half

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the temperature, say 1000 K.

6. Energy 3kT = 3 1.38 10–23 J K–1 6000 K = 2.5 10–19 J per photon. Wavelength = c/f =

hc/ = 6.6 10–34 J Hz–1 3 108 m s–1/2.5 10–19 J = 800 nm approximately. This is in theinfrared.

7. Energy of a photon of wavelength 100 nm is 8 times larger than energy of photon wavelength 800nm.

8. At 6000 K, energy kT = 1.38 10–23 J K–1 6000 K = 8.3 10–20 J. Expressed in eV, energy kT =

8.3 10–20 J / 1.6 10–19 J eV–1 = 0.52 eV approximately.

9. If energy = 13.6 eV, then /kT = 13.6 eV / 0.52 eV = 26 approximately.

10. At 1500 K, energy kT = 1.38 10–23 J K–1 1500 K = 2.1 10–20 J per particle.

11. = 400 kJ mol–1 = 400 10–3 J mol–1 / 6.02 1023 mol–1 = 6.6 10–19 J per particle. Ratio /kT =

6.6 10–19 J per particle / 2.1 10–20 J per particle = 30 approximately.

12. At 60 ºC 330 K, kT = 1.38 10–23 J K–1 330 K = 0.45 10–20 J per particle. Expressed as kJ

mol–1 kT = 0.45 10–20 J per particle 6.02 10–23 mol–1 / 1000 = 2.7 kJ mol–1 for one hydrogen

bond. Ratio /kT = 20 kJ mol–1 / 2.7 kJ mol–1 = 7.4 for one bond, or perhaps 15–20 for severalbonds.

13. At 195 K, energy kT = 1.38 10–23 J K–1 195 K 6.02 1023 mol–1 / 1000 = 1.6 kJ mol–1.

14. Ratio /kT = 27 kJ mol–1 / 1.6 kJ mol–1 = 17 approximately.

Distributions of particlesQuestion 70S: Short Answer

1. It is an exponential distribution so that the ratios of the lengths (representing number) of adjacentlines are in a fixed ratio.

level

number of particles

2. In this case, where more energy is being shared by the same number of particles, more are onhigher levels, fewer on the lowest level.

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level

number of particles

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

1.

kT f B

1 0.367 879

2 0.135 335

5 0.006 738

10 4.54 10–5

20 2.06 10–9

50 1.93 10–22

100 3.72 10–44

2.

.14

)10ln(

exp10

exp

6

6

B

kT

kT

kT

kTf

3.

.77.0

K300KJ1038.1

J10exp

exp

B

1-23

21

B

B

f

f

kTf

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4.

K.105

KJ1038.1693.0

J10

KJ1038.1

J10693.0

ln

exp

1-23

21

123

21

B

B

T

T

T

kTf

kTf

5. The Boltzmann factor increases very rapidly with temperature.

0300 310 320 330 340 350

T / K

5 × 10-15

4 × 10-15

3 × 10-15

2 × 10-15

1 × 10-15

Electrons from hot metalsQuestion 150S: Short Answer

1. The current increases at an increasing rate. This fact is not by itself enough to call the variation‘exponential’, of course.

2. Increase in temperature = 200 K at an average of about 1500 K, an increase of 13%.

3. Current at 1100 C is about 1.0 10–8 A, and at 1300 C is 25 10–8 A, a factor of 25.

4. For example the current doubles between 1250 C and 1280 C. It also doubles between justover 1200 C and just under 1250 C.

5. The graph is a straight line. The current increases approximately exponentially with temperature.

6. 150015.025ln BT

, therefore TB = 32 000 K.

7. The energy kT in electron volts is 1.38 10–23 J K–1 32 000 / 1.6 10–19 J eV–1 = 2.8 eV.

8. Ratio of Boltzmann factors =


T

T

T

T

TTTT

T

T

TT

TB

BB

B

exp11

exp

exp

exp

Density where jet planes flyQuestion 100C: Comprehension

1. A force of 9.81 N moving through a displacement of 1 m changes energy by 9.81 J. Thus 9.81 J

m–1 is the same thing as 9.81 N. So 9.81 J kg–1 m–1 is the same thing as 9.81 N kg–1.

2. 1 g = 0.001 kg, so 28 g = 0.028 kg.

3. 11111 mmolJ275.0molkg028.0mkgJ81.9

4. Just multiply by 1000. So 0.275 J mol–1 m–1 is equal to 0.275 kJ mol–1 km–1. For 10 km the energydifference is 10 times larger, 2.75 kJ mol–1.

5. 1123123 molkJ27.2K273KJ1038.1mol1002.6

6. 21.1molkJ27.2/molkJ75.2 11

7.

.3.0

36.3/1

)21.1exp(/1)21.1exp(

36.3)21.1exp(

The ratio of densities of molecules is the same as the ratio of the probabilities to be at these twoheights.

Contaminated surfacesQuestion 140C: Comprehension

1. A possible example might be a freshly grown and polished slice of silicon crystal, for makingmicrochips. The silicon surface is deliberately ‘contaminated’ with oxygen, to form a layer of inertsilicon oxide on the surface to protect it. Great care is taken to avoid other contaminating vapourscoming into contact with the fresh surface.

2. The gas pressure would be reduced as molecules are removed from the gas. But the reduction isvery small, because the volume of a layer only a few molecules thick over the surface is a verysmall fraction of the volume of the gas.

3. Change in momentum = 2mv = 2 510–26 kg 100 m s–1 = 10–23 kg m s–1. Force = rate of change of momentum.


Number of collisions per second per unit area = force per unit area / momentum change percollision = 10–5 N m–2 / 10–23 N s = 1028 s–1 m–2. Area of one atom = (10–10 m)2 = 10–20 m2. Collisions per atom = 1028 s–1 m–2 10–20 m–2 = 108 collisions per second.

4. 1 in 109

5. At room temperature T = 300 K, the value of kT = 1.3810–23 J K–1 300 K = 4.14 10–21 J. In electron volts this is 4.14 10–21 J / 1.610–19 J eV–1 = 0.025 eV or about 1/40 eV. In kJ mol–1

the energy is 4.14 10–21 J 6.02 1023 mol–1 = 2500 J mol–1 = 2.5 kJ mol–1. Thus van der Waals energies are a few times kT, as stated.

6. Using the result of question 5, at 300 K, kT = 1/40 eV, so that /kT = 4. Thus e–/kT = e–4 = 0.018or 0.02 approximately. Of 108 opportunities per second on average 0.02 108 = 2 106 persecond will be successful. So it will take a typical adsorbed atom about 0.5 s to acquire the energy to leave the surface. If T = 100 K, then /kT becomes 3 times larger, with /kT = 12. e–12 = 6 10–6, so now only 6 10–6 108 = 600 opportunities per second will be successful. The atom will stay on the surface forroughly 2 ms.

7. Heating the charcoal increases kT and so increases the Boltzmann factor e–/kT. The probabilitythat a gas molecule will acquire enough energy to leave the surface becomes larger, and sohappens more often.

8. The units of /k are J / J K–1 = K. The ratio /k is the temperature at which kT = . When kT = theprobability that a molecule will gain extra energy is quite large, so happens often. Thereforesuch bonds break as soon as they form. If /k = 90 K for N2 molecules sticking together, thetemperature will have to be less than 90 K for the molecules to stay together for a reasonabletime. So nitrogen will liquefy at a temperature below 90 K.

Resistance and conductance of thermistorsQuestion 110D: Data Handling

1. Largest value is 1760 k. Smallest value is 0.0307 k, or 30.7 . Ratio of largest to smallestvalues in first column is 1760 k / 0.0307 k = 57000 approximately. When the ratio of largest tosmallest values in a batch of data much exceeds 10, a simple linear plot will rarely be helpful. Alogarithmic plot is probably called for. It is good practice to look first of all at the largest andsmallest values, when looking at a new set of data.

2. Here is the linear plot of resistance against temperature for the type 198-927.


temperature / C

200.0

0.00 50 100 150 200 250–50

150.0

100.0

50.0

The graph shows that the resistance falls with temperature, rapidly at first and then more slowly.The graph is almost useless because, to fit in the values of about 100 k, the scale has to makevalues less than about 5 k invisible. The variation above 100 °C cannot be seen at all.

3. Here is what the logarithmically scaled graph looks like.

temperature / C

1000.0

10.0

1.0

0.1

0.010 50 100 150 200 250–50

100.0

Equal intervals on the resistance scale now represent equal multiples of resistance (x 10). Thegraph now shows the variation of resistance over the full temperature range. The resistancealways decreases with temperature. The graph is much less curved than before. If it were straightit would show that the resistance decreases by a constant factor for each equal increase intemperature. Because the graph curves upwards, the ratio of resistances between pairs ofequally spaced temperatures must decrease somewhat as the temperature rises.

4. The ratio gradually reduces, from 1.8 between –30 °C and –20 °C to just over 1.1 between 240°C and 250 °C. This fits with the logarithmic graph curving upwards: that is, decreasing moregradually as the temperature rises.

5. Here is what the combined graph looks like. All the five graphs are pretty well parallel. This


means that the resistances of all five thermistors vary in the same fundamental way withtemperature. There is such a thing as ‘thermistor behaviour’, not just the behaviour of this or thatthermistor.

10000

temperature / °C

1000

100

10

1

0.1

0.01–50 0 50 100 150 200 250 300

6. The conductance of the type 198-927 at 250 °C is 32.5 mS. The conductance should rise withtemperature, rather faster for low temperatures than for high temperatures.

7. Let G = Ae–/kT. Taking natural logarithms of both sides gives

kTAG

lnln

If ln A is constant, a graph of ln G against 1/T will be a straight line of slope –/k. The graph of lnG against 1/T for thermistor type 198-927 looks like this:


0.001

–12

–6

–8

–2

–4

0.002 0.003 0.004 0.005

–10

(1/T) / K–1

8 Slope of graph above is approximately –4000 K. Thus = 4000 K 1.38 10–23 J K–1 = 5.5 10–20 J. Expressed in electron volts this is = 5.5 10–20 J / 1.6 10–19 J eV–1 = 0.34 eV. At 300K the value of kT = 1.38 10–23 J K–1 300 K = 4.110 –21 J or 0.026 eV. Thus is about 13 kT.

Vapour pressure of waterQuestion 120D: Data Handling

1. If kTAp e

then taking natural logarithms of both sides

AkT

p lnln

If A is constant then the graph of ln p against 1/T is a straight line of slope –/k.

2. Energy = 41 103 J mol–1/ 6.02 1023 mol–1 = 6.8 10–20 J. Ratio /k = 6.8 10–20 J / 1.38 10–23 J K–1 = 4900 K. The slope of the graph of ln p against 1/T is –/k, so the slope is about–5000 K.

3. The graph of ln p against 1/T is like this:


8

7

6

5

4

3

2

1

0

–1

0.002 0.0025 0.003 0.0035 0.004

(1/T) / K–1

4. On this graph, when ln p decreases by 9, the value of 1/T increases by approximately 0.0018 K–1.Thus the slope is approximately equal to –9 / 0.0018 K–1 = –5000 K.

5. From the table of data, the pressure at 100 °C is 101.3 kPa, equal to atmospheric pressure.Twice this is 202.6 kPa. By 120 °C the pressure has risen to 198.5 kPa, nearly doubleatmospheric pressure. In this region the pressure is rising at about 6 kPa per degree, so thetemperature at which the pressure is twice atmospheric pressure is close to 121 °C.

Runny liquidsQuestion 130D: Data Handling

1. If kTA e then taking natural logarithms of both sides

AkT

lnln

.

If A is constant then the graph of ln against 1/T is a straight line of slope –/k.

2. For water, the graph of ln against 1/T is as shown below. When ln falls by 2, 1/T increases byabout 0.001 K, very roughly. Thus the slope = –2 / 0.001 K–1 = 2000 K approximately


0.0030 0.00350.0020 0.0025 0.0040

–1.0

–0.5

0.0

0.5

2.0

1.0

1.5

(1/T) / K–1

3. On this graph, if the slope is 2000 K, the value of in electron volts is given by 2000 K 1.38 10–23 J K–1 / 1.610–19 J eV–1 = 0.17 eV or about 0.2 eV.

4. The graph is steepest at low temperatures (large 1/T). This suggests that the energy decreasesslightly with temperature. Notice that the temperature range takes water well above its normalboiling point, so that it must have been under pressure. There is no information as to whether thevalues tabulated are all for the same pressure, or not.

5. For ethanol, the graph of ln against 1/T is as shown below. When ln falls by 5, 1/T increasesby about 0.0028 K–1, roughly. Thus the slope = –5 / 0.0028 K–1 = 1800 K approximately, similar tothe value for water.


0.004 0.0050.002 0.003 0.006

–5

–4

–2

–1

2

0

1

–3

(1/T) / K–1

6. If the slope is 1800 K, the value of in electron volts is given by 1800 K 1.38 10–23 J K–1 /1.610–19 J eV–1 = 0.15 eV or in the range 0.1 eV to 0.2 eV.

7. For molten glass, the graph of ln against 1/T looks like this:

0.00040 0.00045

–3.0

–2.0

–2.5

–1.0

–1.5

0.00050 0.00055 0.00060 0.00065 0.00070 0.00075

(1/T) / K–1

The graph is rather irregular, though it still has a negative slope as expected. The value ofviscosity 8.2 Pa s at 1600 °C seems out of line. If it is a misprint for 7.2, the graph shapebecomes simpler. However, such an error cannot be assumed without checking, just to ‘tidy up’data. But if the value is correct there must be a mechanism for making the fluidity of glass less


than expected in the region of 1600 °C. It is far from clear to us what such a mechanism could be.Taking the broad trend the slope of the graph is very roughly 5500 K, more than double the valuefor water or ethanol. Particles in glass have a harder job pushing past their neighbours, which iswhy glass is only fluid at higher temperatures, and why the fluidity increases very rapidly withtemperature. The energy might be of the order 0.2 eV (5500 K / 2000 K) = 0.5 eV approx.

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

1. A good answer will bring out how all the atoms have energy, randomly distributed amongst them,and mention the energy needed for an atom to push others aside, so that flow can occur. Notethat the expansion on melting is very small, and does not account for melting by ‘providing morespace for atoms to move about in’.

2. Butter is an emulsion of fat and water. At low temperatures, the fat is very viscous, making thebutter very hard. For the butter to spread the molecules must be able easily to move past oneanother. If their thermal energy is small, this does not happen often.

3. A good answer will mention the considerable energy needed to remove electrons from atoms, toionise them, and the very high temperatures inside the Sun, giving each atom a large randomthermal energy.

4. A good answer will mention that the molecules in the water all have energy of random thermalmotion, and that a few near the surface will occasionally acquire enough energy to leave theliquid. Moving at random, these molecules will mostly move away from the water, and not oftenre-enter it.

5. A substance becomes liquid when the possibility of particles moving past one another is greatenough to destroy the regular order of the solid phase. This occurs at lower temperatures formany metals than for many ceramics, because the energy needed for a particle to move past itsneighbours is smaller. But even ceramics melt at high enough temperatures: pottery is made bymelting part of the material in a kiln.

6. Even though ice is solid, there is still some small chance that molecules will acquire enoughenergy to hop past one another when the material is under stress. This is how creep occurs inmany materials, including lead on church roofs and glass in ancient windows.

7. The data suggest that the energy to evaporate an atom of helium from the liquid is less than theenergy needed to evaporate a molecule of nitrogen.

8. Locally, in a spark, a small region of metal may get hot enough that a small fraction of its atomsby chance acquire enough energy to evaporate from the metal.

Thinking about the Boltzmann factorQuestion 60X: Explanation–Exposition

1. When the temperature T increases, the ratio /kT decreases, so the expression e /kT decreases,

and the Boltzmann factor e– /kT, which is equal to 1/ e /kT therefore increases.


2. “When the temperature increases, the ratio /kT decreases, so its negative –/kT increases. If–/kT increases, then ln fB increases. If ln fB increases then fB increases. Thus as the temperatureincreases the Boltzmann factor increases.”

3. If /kT = 20 then e– /kT= e–20 = 2.06 10–9.

4. At 310 K the value of /kT = 20 (300/310) = 19.35.

5. The Boltzmann factor is now e–19.35 = 3.93 10–9. This is an increase of 3.93 10–9/2.06 10–9 =1.9 or approximately 2.

6. That the Boltzmann factor is small (10–9) shows that only a very small faction of particles have theenergy or greater.

7. If the rate of a reaction increases by a factor 2 for a 10 K rise in temperature near 300 K, theactivation energy is roughly 20 times the energy kT of random thermal motion.

8. The rate of cooking may depend on a Boltzmann factor, and the Boltzmann factor increasesrapidly with temperature, often doubling for a 10 K rise in temperature. You could check thisagainst advice on cooking times with pressure cookers. Notice that it is not the increase inpressure which speeds up the cooking. The increase in pressure is caused by the small rise intemperature. Typically the pressure is doubled.

9. In a hot oven, say 200 ºC or 470 K the reactions in which proteins are denatured (‘cooking’)proceed rapidly, though the cooking time also depends on how rapidly the temperature of theinterior of the food rises. In a warm oven (say 100 ºC or 370 K) the reactions proceed much moreslowly, because the Boltzmann factor is very much reduced.

10. At low temperatures the metabolism of bacteria proceeds very slowly, so their waste products(which are what spoil food) are not produced at an appreciable rate. A small reduction intemperature (typically from room temperature to about –20 ºC) is enough for a very largereduction in rate of reaction.

11. It is not really possible to say, without more information. If the sugar molecules simply becomedispersed in the water, there is no good reason to expect an important effect of temperature. Butin fact, sugar dissolves much faster in hot water than in cold water, suggesting that there is anactivation energy involved.

12. The explosion starts when part of the gunpowder is made hot enough for the reaction to proceedat an appreciable rate. The energy released raises the temperature further, making the reactiongo faster, which raises the temperature faster still, which…and the gunpowder explodes as thereaction ‘runs away’ exponentially.

13. Water thrown on a fire evaporates, making steam. A large amount of energy is needed toevaporate water, so this energy taken from the fire cools it down, slowing or stopping the burningreaction. The water or steam may also prevent oxygen reaching the combustible material.

Exponential distributionsQuestion 80X: Explanation–Exposition

1.

Level Number (f = 0.5) Number (f = 0.25) Number (f = 0.75)


Level Number (f = 0.5) Number (f = 0.25) Number (f = 0.75)

0 256 256 256

1 128 64 192

2 64 16 144

3 32 4 108

4 16 1 81

5 8 1 ~ 61

6 4 1 ~ 45

7 2 1 ~ 34

8 1 1 ~ 26

2. See table above.

3. The distribution is exponential if, for equal steps in energy level, the function changes by aconstant factor. A graph of the logarithm of the number of particles on each level against levelwould be linear.

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Printing - Advancing Physics A2 Teacher Edition14...3 Advancing Physics 5. Assume hc / kT which gives = 5 10–3 m; millimetre wavelengths are in the microwave part of the spectrum