1 Advancing Physics

Absolute temperatureQuestion 10W: Warm-up Exercise

1. Mercury at air temperature is 2 cm above the level for ice and water at 0 °C. Mercury for boilingwater at 100 °C is 10 cm above level for 0 °C. Thus room air temperature is (2 cm/10 cm) 100 °C= 20 °C

2. You had to assume that the change in length of the mercury column is proportional to the changein temperature. In other words, the expansion coefficient of the mercury is a constant.

3.

0temperature/ºC

20 40 60 80 100 120 140

0

20

40

60

80

100

120

140

4. The pressure increases by 141 kPa – 93 kPa = 48 kPa for a rise in temperature of 140 °C. Thusfor a rise of 27 °C the pressure rises by 48 kPa (27 °C / 140 °C) = 9.2 kPa. The pressure at 27°C is thus 93 kPa + 9.2 kPa = 102 kPa to three significant figures.

5.

–300temperature/ºC

–250 –200 –150 –100 –50 0 –50 –100 –150

0

20

40

60

80

100

120

140

6. Pressure 0 kPa; temperature – 273 °C

7. Temperature – 273 °C

8. + 273 K

9. + 273 K + 100 K = 373 K

10. + 273 K + 27 K = 300 K

2 Advancing Physics

Boyle's lawQuestion 20W: Warm-up Exercise

1.

V

2.

1 / V

3.

3 Advancing Physics

V

4. p is inversely proportional to V

5.

kPa 2045.0

1 kPa 102

2

112

V

Vpp

6. The result is lower than expected. Perhaps some gas escaped.

7. The result is higher than expected. Work done on the gas as it was compressed caused itstemperature to increase.

Molecular motionQuestion 30W: Warm-up Exercise

1. The mean kinetic energy of the molecules increases.

2. U = NE

3. U = NkT = 6.02 1023 mol–1 1.38 10-23 J K–1 293 K = 2400 J mol–1

4.

35113

m J10mol J 2400mol m 024.0

1

5. 5.0 m 8.0 m 3.0 m 105 J m–3 = 1.2 107 J

6. kTmv 221

leading to

1s m 4002 m

kTv

7. You can’t walk straight because you keep colliding with dancers and this repeatedly deflects you

4 Advancing Physics

from your path.

8. For example, 4 m.

9. For a distance of 4 m between collisions this gives a time of 2 s and a frequency of 0.5 s–1

10. Time between collisions = 10–7 m / 400 m s–1 = 2.5 10–10 s

11. f = 1 / t = 1 / 2.5 10–10 s = 4 109 collisions s–1

12. Molecules undergo repeated collisions and so follow a ‘random walk’. This means that theaverage speed with which they mix or spread out (diffuse) is much lower than their averagestraight line speed.

The ideal gas equationQuestion 40W: Warm-up Exercise

1. p is pressure; V is volume; n is number of moles; R is the molar gas constant; T is thetemperature.

2. T must be in Kelvin. Only on the Kelvin scale will the two gas laws involving temperature show arelationship directly proportional to temperature.

3. Real gases approximate to this behaviour but show significant deviations from ideal behaviour athigh pressures and close to their liquefaction temperature.

4. If n and T are constant then nRT is a constant and the equation reduces to pV = constant, whichis Boyle’s law.

5.

p

nR

T

V

so if n and p are constant then V / T = constant (Charles’s law).

6.

V

nR

T

p

so if n and V are constant then p / T = constant (pressure law).

7.

3365

365

2 cm 1.4m 10056.4Pa 10

K 289

K 285

m 10Pa 100.4

V

8.

Pa 103.2K 293

K 311Pa 102.2 55

2

V

Vp

9.

K 437m 101Pa 101.1

K 293m 1025.0Pa 106.6335

335

2

T

5 Advancing Physics

= 437 – 273 °C = 164 °C

Floating or sinking?Question 10S: Short Answer

1. 7 kN – 1 kN = 6 kN

2.

N 101.1

kg N 8.9m 10m kg 09.02.1

)ρ(ρ

ρρ

ρ displaced air of weight upthrust

6

–1353-

airshipgasairload

loadairshipgasairshipair

airshipair

gVgm

gmgVgV

gV

3.

N 100.1

kg N 8.9m 10m kg 18.02.1

)ρ(ρ

ρρ

ρ displaced air of weight upthrust

6

–1353–

airshipgasairload

loadairshigasairshipair

airshipair

gVgm

gmgVgV

gV

p

4.

N.87.0

kgN8.9)m104.7(mkg2.1

Upthrust1223

VgF

5.

kN.725.0

kgN8.9)m104.7(mkg1000

Upthrust1223

VgF

6. The table-tennis ball, because it will have a small mass. If the mass of air displaced by the objectis a significant fraction of the mass being measured, it should be taken into account.

7.

N 100

kg N 8.9kg15kg N 8.9m 25m kg 18.02.1

)ρ(ρ

ρρ

rope in tension balloonempty of weightgas of weight upthrust

–1–133–

balloonballoongasair

balloonballoongasballoonair

gmgVT

TgmgVgV

8. Pine floats because a volume smaller than its own displaces enough water to equal the weight ofthe pine. Even fully submerged, rosewood will displace a weight of water less than its own weight,

6 Advancing Physics

and will therefore sink.

9.

kN.35kgN8.9m3000mkg2.1

Upthrust123

VgF

The hot air is less dense so weighs less than the air it displaces.

10. No. Because the boat displaces a volume of water equal to its own weight, the load on theaqueduct remains the same, provided the displaced water has time to flow out of the aqueduct.

11. The upthrust must remain the same (equal to the weight of the ship). Let A be the area of thehorizontal section of the ship in water:

m.6.20m0.20kgN1000

kgN1030

kgN8.9kgN1000kgN8.9m0.20kgN1030

1

1

1111

x

xAA

12. The surface level of the drink remains exactly the same. Because the ice floats, the weight ofwater it displaces is equal to the weight of ice in the drink. When the ice melts, its volume isidentical to the water previously displaced.

Using the ideal gas relationshipsQuestion 40S: Short Answer

Solutions1.

.m103.0Pa000150

K373KmolJ31.85

/

311

pnRTV

2.

.m113.0Pa000100

K273KmolJ31.85

/

311

pnRTV

3.

.litres14.7K320

Pa00080

K280

litres5Pa000100

const/

VV

TpV

4. If the lab is roughly 3 m high 10 m 10 m, volume = 300 m3, then about 12 kmol (with a massof 360 kg) because:

mol 12000K 300K mol J 8.31

m 300Pa 100000–1–1

3

RT

pVn

7 Advancing Physics

5. Use 1 mole in the calculation for convenience; it always has a mass of 0.045 kg. Choose a

volume of 1 m3. So:

mol 8.52K 273K mol J 8.31

m 1Pa 120000–1–1

3

RT

pVn

The mass of 52.8 mol is 52.8 0.045 kg = 2.38 kg. The density of the gas is 2.38 kg m–3.

6. At sea level 1 m3 has a mass of 1.25 kg. This mass will have expanded to a new volume on top ofthe mountain. This new volume is:

.m46.2

)Pa00033/K223()K273/Pa000100(3

V

So the density is now 1.25 kg / 2.46 m3 = 0.51 kg m–3.

Momentum and collisions with a wallQuestion 50S: Short Answer

1. mv = 2 kg × 12 m s–1 = 24 kg m s–1

2. 0 kg m s–1

3. 24 kg m s–1

4. 24 kg m s–1

5. 24 000 kg m s–1

6. 24 000 kg m s–1

7.

kN 4.2s10

s m kg 1024 –13

F

mvFt

8. 48 kg m s–1

9. 48 000 kg m s–1

10. 4.8 kN

Kinetic theory by numerical exampleQuestion 60S: Short Answer

Solutions1.

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Pa 800m 6

N 48002

A

Fp

2. d = 12 m s–1× 10 s = 120 m

3. 8 m

4. 15

5.

Pa 12m 6

N 72

N 72s 10

s m kg 720

s m kg 720s m 12kg 2215

2

–1

–1–1

A

Fp

t

mvF

mv

6. mv = 5 × 10–26 kg × 500 m s–1 = 2.5 10–23 kg m s–1

7. – 2.5 10–23 kg m s–1

8. – 5.0 10–23 kg m s–1

9. 5000 m

10. 8 m

11. 625

12. mv = 625 × 5.0 10–23 kg m s–1 = 3.1 10–20 kg m s–1

13. mv = 2 1026 3.1 10–20 kg m s–1 = 6.2 106 kg m s–1

14.

N 102.6s 10

s m kg10.26 5–16

t

mvF

15.

Pa 100.1m 6

N 10.26 52

5

A

Fp

Kinetic theory algebraicallyQuestion 70S: Short Answer

1. mv

2. mv

3. mv2

4. vt

9 Advancing Physics

5. a2

6. avt 2/

7. atmv /2

8. 3/N

9. atNmv 3/2

10. aNmv 3/2

11. abcNmv 3/2

12. abc

13. 3/2NmvpV

More about the kinetic theory of gasesQuestion 80S: Short Answer

Solutions1. There are n molecules per unit volume, and a volume A x, so the number is n A x.

2. If the gas is not drifting towards or away from the wall, half the molecules must be going towardsit and half away from it. Alternatively, if the velocities are directed at random, half must bedirected towards the wall and half away from it.

3. In time t a molecule with velocity component vx travels x = vx t in the x-direction. So all

molecules within this distance will reach the wall inside time t.

4. The number of molecules hitting the wall in time t is half the number inside the distance x = vx

t. The number inside distance x is n A x, so the number hitting area A in time t is.

21 tnAv x

5. The x-component of momentum is m vx.

6. The momentum delivered in time t is the momentum m vx delivered per particle multiplied by thenumber of particles

tnAv x 21

arriving in time t, giving:

.221 tnAmv x

7. Divide

tnAmv x 221

by t to get the rate of arrival of momentum:

.221

xnAmv

10 Advancing Physics

8. The recoil momentum is + m vx, opposite in direction to the momentum – m vx of the molecule.

9. Number leaving in time t is the same as the number arriving, that istnAv x 2

1

because the wall is not on average absorbing or releasing molecules. Dividing by t gives thenumber leaving per second:

.21

xnAv

10. The momentum delivered per particle is mvx and the number leaving per second is

xnAv21

so momentum delivered per second is

.221

xnAmv

11. The momentum per second is2

21

xnAmv

for arriving particles and2

21

xnAmv

for departing particles, giving2

xnAmv

in total.

12. Because force is equal to rate of change of momentum. Here the rate of change is a rate ofdelivery, a flow of momentum.

13. The force on area A is2

xnAmv

so dividing by A gives the pressure

.2xnmvp

14. Momentum per particle is proportional to vx. So is the number of particles per second. Thus

momentum per second (= force) is proportional to vx2.

15. Since2

312 vv x

then2xvnmp

becomes2

31 vnmp

16. Writing

V

Nn

in2

31 vnmp

gives

11 Advancing Physics

2

3

1vm

V

Np

which gives2

31 vNmpV

17.

231

231

231

ρ

and

ρ

gives

into for ngsubstituti

ρdensity

is gas the of gas the of mass total the is Since

vp

vVpV

vNmpV

NmV

Nm

Nm

The speeds of gas molecules:Some questionsQuestion 90S: Short Answer

1.

30

25

20

15

10

5

0

100 150 200 250 300 350 400 450 500 550 600

A histogram showing number of moleculesmoving at that speed against molecule speed

speed / m s–1

2. 250 m s–1 – the most common one.

3. [(100 m s–1 8) + (150 m s–1 17) + (200 m s–1 22) +…+ (600 m s–1 1)] / (8 + 17 + 22 +…+ 4+ 1 + 1) = 285 m s–1

12 Advancing Physics

4. {[(1002 m2 s–2 8) + (1502 m2 s–2 17) + (2002 m2 s–2 22) +…+ (6002 m2 s–2 1)] / 150} = 304 m s–1

5. Some molecules reach the far corner almost immediately. If scent only travelled by diffusion itwould take a long time for it to cross the room. There must be an air current that carries the scentmolecules, indeed the initial 'burst' from the aerosol may provide that current. . The nose is verysensitive and can detect small concentrations of air freshener. The concentration builds upgradually (and unnoticeably) over a few minutes.

6. A typical speed of about 300 m s–1 for 120 seconds gives a total distance of about 36 km. Therehave been many collisions that deflected it off course. In fact this time is much too short fordiffusion alone. Air currents must have helped.

7. Earth mass = 6 1024 kg; r E = 6 Mm; r M = 1.8 Mm; Moon mass = 7 1022 kg.

8. Substitute into (2 G m / r ) to yield [(2 6.67 10–11 N2 kg2 6 1024 kg) / (6 106 m)] = 11500 m s–1.

9. 2

31 vp

10.

V

nRT

vp

231

mM

RTm

nRT

V

nRTv

3

3

32

11.

mM

kt

v

3

speedrms 2

.sm109.1kg002.0

K300KJ31.83:H 13

1

2

.sm104.1kg004.0

K300KJ31.83:He 13

1

.sm102.5kg028.0

K300KJ31.83:N 12

1

2

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.sm108.4kg032.0

K300KJ31.83:O 12

1

2

12. All less, but H2 and He only less by a factor of 5–10 times.

13. These are ‘average’ molecular speeds. Some molecules move considerably faster than this; thetwo light gases have a considerable number of molecules at speeds about the escape speed.These are lost from the atmosphere. The atmosphere then redistributes its energy so thatmolecules of lighter gases again have these fast speed ranges in order to keep the overalldistribution correct. Again these faster molecules are lost, and so on. The heavier gases have asmaller number at the escape speed. However, this begs the question of why they do noteventually go too….

14. The escape speed from the Moon is considerably less than that from the Earth – a good studentmight work out the figure (about 2000 m s–1). At this escape speed all the molecules have beenlost.

Speed of sound and speed of moleculesQuestion 100S: Short Answer

1. Helium, at 972.5 m s–1 at 273 K and 1019 m s–1 at 300 K.

2. Helium at 4 g mol–1.

3. Carbon dioxide, at 257.4 m s–1 at 273 K and 269.8 m s–1 at 300 K.

4. Carbon dioxide, at 44 g mol–1.

5. The speed of sound is larger in all cases at the higher temperature 300 K.

6. From the kinetic theory, the kinetic energy and so the speed of the molecules will be higher at thehigher temperature, 300 K.

7. 2

31 vNmNkt

Divide both sides by N giving:

231 vmkt

Multiply both sides by 3 and rearrange, obtaining:

m

kTv

32

8.

kg1064.6mol106.02

mol kg104 27123

13

m

9.

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22627

1232 sm1087.1

kg 1064.6

K300)K J1038.1(3

v

whence taking the square root the rms speed is 1370 m s–1.

10. Since

m

kTv

32

then the ratio of the mean square speeds is 4 / 28.

11. The rms speed for nitrogen will be (4 / 28) = 0.378 of the rms speed for helium, giving a speed

of 0.378 1370 m s–1 = 517 m s–1.

12. The factor is now (4 / 44) = 0.301 giving a speed of 0.301 1370 m s–1 = 413 m s–1.

13. The table is now:

Gas Molar mass /

g

Speed of sound at

273 K / m s–1Speed of sound at

300 K / m s–1rms speed ofmolecules at 300

K / m s–1

Helium 4 972.5 1019 1370

Nitrogen 28 337.0 355.5 517

Carbon dioxide 44 257.4 269.8 413

14. The squares of the speeds are proportional to the temperature, so if the temperature falls from300 K to 273 K the speeds fall in the ratio (273 / 300) = 0.954.

15. Yes. The ratios 972.5 / 1019, 337.0 / 355.5 and 257.4 / 269.8 are all in the ratio 0.954approximately.

Specific thermal capacity:Some questionsQuestion 110S: Short Answer

1. There are a number of factors which determine the time it takes the toffee to cool sufficiently toeat:

The boiling point of the sugar solution is higher than that of water so the toffee is cooling from ahigher temperature.

The sugar solution has a higher specific thermal capacity than pure water. So for an equivalenttemperature drop, more energy has to be lost.

Perhaps most significant is the amount of energy that has to be lost for the liquid toffee mixture tosolidify, cooling and then changing from a liquid to a solid and cooling some more takes a longtime.

2. The specific thermal capacity of water is large; water is cheap and liquid, it has a reasonabletemperature range before boiling. It is pressurised to elevate the boiling point – but, as important,

15 Advancing Physics

also to retain the material.

3. Coolants used include: water, heavy water (D2O), liquid sodium, pressurised carbon dioxide.They need a high specific thermal capacity and, ideally, should not absorb neutrons, and not beexplosive or flammable.

4. Area of mirror is about 3 m2. So 3 kW delivered. One litre has a mass of 1 kg, assume an 80degree temperature rise. Minimum time taken is (4200 J C–1 80 C) / 3000 J s–1 = 112 s butthis does not allow for losses to the surroundings from the container which the Sun has to makeup. This is an estimate of the energy needed to raise the temperature to 100 C, it does not boilthe water. Assuming no energy losses, it takes a further 800 s (minimum) to vaporise the liquid.An equally valid estimate – this time experimental – is to time your household kettle having readits electrical energy input from the base plate and then scale up or down its time to boilingaccordingly.

5. Assume the laboratory is 3 m 10 m 10 m, this leads to a volume of 300 m2. Assume furtherthat the heating is just a question of warming up the air. The density of air is approximately 1 kgm–3 so energy = 300 kg 1000 J kg–1 C–1 10 C, i.e. 3 MJ. A reasonable heater might deliverthis in 1000 s (about 20 minutes). Most people would guess that the heating time would be muchlonger. This estimate ignores heating the contents of the room, other laboratories, etc. Ourperception of temperature is affected both by the humidity of the air and by the cooling effect ofany draughts. It would take much longer in reality.

Thermal transfers in the homeQuestion 130S: Short Answer

1.

minutes.2

s140

W105.2

C15100KkgJ4200kg1

Time

21

3

11

P

mct

2. The temperature of a larger mass of water will fall less for the same transfer of thermal energy tothe pasta, so the water temperature drops less when the pasta is added.

3. Measure the mass of pasta used with a balance. Calculate the mass of water from its volume.Ignore the saucepan and thermal transfers to the surroundings. Measure the room temperature(assumed to be the same as the starting temperature of the pasta), r, water temperature atboiling and the temperature immediately after adding the pasta and stirring. Then the specificthermal capacity can be found by calculating

rfinalpasta

finalboiling11

waterpasta

KkgJ4200

m

mc

.

4. energy lost by the hot water = energy gained by the cold water

16 Advancing Physics

kg14

C35

C8kg100

C)1045(KkgJ4200C4550KkgJ4200kg100 1111

m

m

5. Water has the highest specific thermal capacity of commonly occurring liquids, and so can carrymost thermal energy away from the engine at a temperature which does not pose seriousdangers for a person working under the bonnet.

6.

MJ.9.2

C1080CkgJ4200kg10

Energy11

mcE

Brownian motionQuestion 140S: Short Answer

1. Using average kinetic energy = (3/2) kT, T = 27 °C = 300 K, so (3/2) kT = 6.3 10–21 J.½ 10–12 kg v 2 = 6.3 10–21 J solving for v ~ 0.1 mm s–1 which is too slow.

2. ½ m (10–3 m s–1)2 = 6.3 10–21 J; m ~ 1.3 10–14 kg

3. mass = density volume so 1.3 10–14 kg = 103 kg m–3 d 3 and d ~ 2.4 10–6 m.

Gases and massQuestion 20E: Estimate

1. Mass of air = volume density so:

.kg9.2mkg29.1m25.2mass

m2.25

m5.2m2.1m75.0volume

33

3

3. mass of helium = volume density so:mass of helium = 5000 m3 0.179 kg m–3 = 895 kgmass of air = volume density so:mass of air = 5000 m3 1.29 kg m–3 = 6450 kgthe upward force (due to displaced air) is greater than the weight of the balloon;net force = (6450 kg × 9.8 N kg–1) – (6000 kg × 9.8 N kg–1) = 4400 N.

4. mass = density volume

17 Advancing Physics

tonnes 2000kg 102.0m 300m 126m 126 m kg 29.1 air of mass

so

heightarea base pyramid of volume

6313–

31

No. The mass of air as calculated is about 2000 tonnes, which is less than a third of the mass ofiron in the tower.

The wonderful oddity of waterQuestion 120C: Comprehension

1. If water melted at 170 K no ice would exist under current climate conditions. There would havebeen no ice ages and no glaciers. The landscape and flora and fauna would be different.

2. With a boiling point of 200 K there would not be liquid water on Earth: no seas to act asthermostatic controls on extreme temperature changes; no marine life and probably no life as weknow it.

3. Accept any reasonable attempt to fit triangular shapes onto a tetrahedron; look kindly on attemptsto fit larger numbers of molecules together.The ecological significance is that solid water is less dense than liquid. So ice floats above waterand ponds, lakes and seas freeze from the top and rarely freeze through, so life can survive thecold winter.

4. The extra energy is needed because of the hydrogen bond linking separate molecules; not quiteas strong as the full covalent but stronger than ordinary van der Waals forces.

5. Volume = ocean area × mean depthvolume = 3.6 ×1014 m2 × 3900 m = 1.40 1018 m3

6. Mass = volume ×densitymass = 1.4 ×1018 m3 × 1030 kg m–3 = 1.45 1021 kg

7. Energy = mass × specific thermal capacity × temperature riseenergy = 1.45 × 1021 kg × 4200 J kg–1 K–1 × 1 Kenergy = 6.1 1024 J

8. One per cent of half of 1.4 × 103 W m–2 = 0.01 × 0.5 × 1.4 × 103 W m–2 = 7 W m–2. Assume thathalf the area of the oceans is in sunlight at any one time. Then the additional rate of arrival ofenergy is 0.5 × 3.6 × 1014 m2 × 7 W m–2 = 1.3 × 1015 W.

9. Energy = power × time6.1 1024 J = 1.3 × 1015 J s–1 × tt = 4.7 × 109 s = 150 years (1 year = 3.1 × 107 s approximately). Note however that as the surface warms, the Earth radiates faster. A 1 K rise from 288 K to 289 Kincreases the rate of radiation by about 1% (Stefan's T4 law) so any rise in temperature will beslower than this. Note also that a rise in temperature may lead to more evaporation from theoceans and so to more cloud cover, increasing the reflection of sunlight directly back into spaceand so moderating the greenhouse effect which caused it.

10. The top layers would become warmer more quickly, because warm water floats on denser cold

18 Advancing Physics

water, and the warm and cold layers only mix slowly through global ocean currents. So therecould be earlier substantial environmental effects due to this surface warm water, before thewhole of the oceans became warmer.

11. Cross section of Florida Straights = 200 × 103 m × 75 mflow rate = 200 × 103 m × 75 m × 6.5 × 103 m h–1

flow rate = 2.7 × 107 m3 s–1

12. Mass flow rate = 2.7 × 107 m3 s–1 × 1030 kg m–3 = 2.8 1010 kg s–1

13. Hurricanes and tornadoes in southwestern USA; cyclonic features (depressions) further north,providing typical British weather; keeping Narvik ice free; fogs near Labrador; melting glaciersand sea ice from polar ice cap (sinking the Titanic), etc.

Thermal changesQuestion 140D: Data Handling

Solutions

1. kJ.210K10KkgJ4200kg0.5 11 mcE

2. J.13K10KkgJ385kg0035.0 11 mcE

3. J.65K10KkgJ1300kg005.0 11 mcE

4. kJ.8.6K10KkgJ450kg5.1 11 mcE

5. The fruit, largely sugar and water, has a high thermal capacity and therefore stays hot longer.

6. Both air and cake have low specific thermal capacities and are poor conductors, so little energy istransferred; metals have higher specific thermal capacities but are good conductors, so theenergy can be rapidly conducted to your hand.

7. Low specific thermal capacity and density mean little energy is stored in the pumice; because it isa poor conductor, there is little energy transfer from below the surface.

8.

K 23.0

K kg J4200

m 100kg N 8.9

θ

θ

–1–1

–1

mc

mgh

mcmgh

9.

19 Advancing Physics

.KkgJ960

K)2029(kg0.1

s603W48

11

m

Ptc

10. Overestimate. Because of thermal transfer to the surroundings, less energy goes into thealuminium than this calculation assumes.

11. Assume the freezer volume is half air (the other half is the contents of the freezer) and that all theair escapes when the door is opened. If the room temperature is 20 C, the temperaturedifference is 40 K:

kJ 39

K 40K kg J 10002

m 1.5m kg 3.1

θ

–1–13

3–

E

mcE

If the door were opened frequently, the power demand would be significant. But perhaps this is acontribution to the air-conditioning of the supermarket.

12.

K.17.0

KkgJ4200sm1100mkg1000

W10800

)/(

11131

6

ctV

P

Pumping up my tyresQuestion 30X: Explanation–Exposition

1. As you pack more and more molecules into the same volume (the tyre) so the number ofcollisions between the tyre wall and the molecules increases.

2. The extra number of molecules injected into the tyre is proportional to the number of pumps,provided each pump injects an equal volume of air into the tyre. The pressure in a fixed volume,at a constant temperature, is proportional to the number of moles present.

3. The predicted relationship is linear, so:

20 Advancing Physics

number of pumps

4. Probably not. A good answer would back this up by referring to the increased leakage at higherpressures and to the incomplete emptying of the cylinder at higher pressures.

5. Find the volume, then the mass of gas

kg.102.4

)m105.3(mkg2.1

m105.3

)m10500(2

m1030

4

343

34

3

23

VM

V

Air is mostly nitrogen, so assume 0.028 kg mol–1:

21

–1232

2

–1

–4

100.9

mol 106.02mol 105.1 molecules of number

mol 105.1

mol kg 0.028

kg 104.2 moles of number

6. Treating the tyre as a cylinder:

.m102.8

)m10650(2

m1040

34

3

23

V

7. To fill the tube to 1 atmosphere takes

.m105.3

m102.834

34

To fill to 4 atmospheres takes four times this amount, so:

21 Advancing Physics

.4.94m105.3

m102.834

34

Say 10 pumps, so about 1 mol of air.