Phy Bk Ans 1

1 Heat and Gases Chapter 1 Temperature and Thermometers
New Senior Secondary Physics at Work  Oxford University Press 2009
1
1 Temperature and Thermometers
Practice 1.1 (p. 10)
1 B
2 D
3 A
4 Temperature is a measure of the degree of hotness of an object.
5 (a) On the Celsius temperature scale, the lower fixed point is the ice point (0 C) and the upper fixed point is the steam point (100 C).
(b) We can reproduce the lower and upper fixed points by using pure melting ice and pure boiling water at normal atmospheric pressure respectively.
6 (II), (IV), (V), (I), (III)
7 Let T be the temperature when the thread is
7.7 cm long.
100 0
0

T  =
18.2 3.2
7.7 3.2


T = 30 C
8
The length of the mercury column at 100 C is
25 cm.
9 Let x be the column length for 37 C.
21 6
6
 x  =
80 0
37 0

 x = 12.9 cm
10 According to the kinetic theory, all matter is made up of particles. For solids, the particles are held in position by strong forces and so they have fixed shapes.
For liquids and gases, the particles are held by weaker forces and can move from one place to another. Therefore, they do not have fixed shapes. Practice 1.2 (p. 16)
1 D
2 C
3 A
4 (a) Thermistor thermometer/ liquid-in-glass thermometer (b) Resistance thermometer/ alcohol-in-glass thermometer (c) Resistance thermometer
(d) Liquid-in-glass thermometer/ infra-red thermometer/ thermistor thermometer/ liquid crystal thermometer
5 (a) The curvature of the bimetallic strip.
(b) It consists of a bimetallic strip which is made up of two strips of different metals.
The metals expand at different rates as they are heated. The different expansions of strips make the bimetallic strip bend one way. As a result, a particular curvature of the bimetallic strip represents a particular temperature.
1 Heat and Gases Chapter 1 Temperature and Thermometers
New Senior Secondary Physics at Work  Oxford University Press 2009
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6 Let T be the temperature measured.
100 0
0

T  =
120 35
80 35


T = 52.9 C
Revision exercise 1
Multiple-choice (p. 19)
1 D
2 B
3 B
4 C
80 25
25

R  =
100 0
40 0


R = 47 units
5 B
6 C
100 0
0

T  =
90 10
40 10


T = 37.5 C
7 A
8 A
Conventional (p. 20)
1 Choose the ice point and the steam point as the lower fixed point and the upper fixed point respectively. (1A)
Then divide the range between these fixed points into 100 equal divisions. (1A)
Each division is 1 C. (1A)
The lower fixed point is taken as 0 C and the upper fixed point is taken as 100 C. (1A)
2
100 0
0
 x  =
24.6 3.7
12.0 3.7


(1M)
x = 39.7 C (1A)
3 (a) Let T be the temperature when the length of the alcohol column is 15.6 cm.
100 0
0

T  =
18.4 4.2
15.6 4.2


(1M)
T = 80.3 C (1A)
(b) Let x be the length of the alcohol column at 30 C.
18.4 4.2
4.2
 x  =
100 0
30 0


(1M)
x = 8.46 cm (1A)
4 (a) TN = TC 
100
33 (1M)
= 250 
100
33
= 82.5 N (1A)
(b) TN = TC 
100
33
=9
5  TF  32
100
33 (1M)
=
TF  32
60
11 (1A)
5 (a)
(Correct labelled axis) (1A)
(Correct points) (1A)
(A smooth curve passing through all data points) (1A)
(b) 32 C (1A)
1 Heat and Gases Chapter 1 Temperature and Thermometers
New Senior Secondary Physics at Work  Oxford University Press 2009
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6 (a) A rotary thermometer measures temperature by measuring the curvature of the bimetallic strip. (1A)
The bimetallic strip consists of two metal strips which expand at different rates when heated to cause a change in curvature of the strip. (1A)
If a strip with only one kind of metal is used, it would only expand but not bend when heated. (1A)
(b) Zinc (1A)
Figure a shows that zinc expands more when heated. Therefore, zinc corresponds to metal A which expands more as shown in Figure b. (1A)
7 (a) Water freezes at temperatures below the ice point and vaporizes at temperatures above the steam point. (1A)
The working range of a water-in-glass thermometer is much narrower than that of a mercury-in-glass thermometer. (1A)
(Or other reasonable answers)
(b) Let T be the temperature measured when the length of the column is 8.8 cm.
100 0
0

T  =
16.4 3.8
8.8 3.8


(1M)
T = 39.7 C (1A)
(c) Mercury is toxic. (1A)
8 (a) In a certain range of temperature, the volume of mercury is proportional to the temperature. (1A)
(Or other reasonable answers)
(b) Volume increased
= 0.0748  0.0735 = 0.0013 cm3 (1M)
Length increased = cross sectional area volume increased
=
0.01 0.01
0.0013

= 13 cm (1M)
Length of the mercury column
= 13 + 3.6 = 16.6 cm (1A)
9 (a) To measure temperature
(b) Any two from: (2A)
• Clinical thermometer has a smaller range.
• Clinical thermometer is more accurate/reads to more (significant) figures/decimal places/more sensitive/has a narrower column.
• Clinical thermometer can maintain reading/temperature. • Clinical thermometer has a kink/constriction/button to reset
Physics in articles (p. 22)
(a) Energy of infra-red radiation emitted (1A)
(b) It takes less time to obtain the results. (1A)
(c) Doing experiment in laboratory (1A)
(Or other reasonable answers)
(d) It is less accurate. (1A)
If it is used in a hospital, doctors may not be able to determine the patient’s condition correctly and may miss noticing a dangerous situation instantly. (1A)
1 Heat and Gases Chapter 2 Heat and Internal Energy
New Senior Secondary Physics at Work  Oxford University Press 2009
1
2 Heat and Internal Energy
Practice 2.1 (p. 26)
1 D
2 A
3 D
4 C
5 The internal energy of a body is the sum of the kinetic energy and potential of all its particles.
It is the total energy stored in the body.
Temperature is a measure of the degree of hotness of an object. When it increases, the kinetic energy, and hence the internal energy, of a body increases.
6 This statement is incorrect. Temperature accounts for the average kinetic energy of the particles in an object, while internal energy is the sum of the kinetic and potential energy of all the particles in an object. A drop of hot water has a higher temperature than water in an ocean, but the latter has more internal energy than the former since it contains much more particles.
Practice 2.2 (p.31)
1 C
2 B
3 A
4 D
5 A
Time needed
=
2 1000
420 1000


= 210 s =
60
210 min = 3.5 min
6 B
P = t Q =
5 60
30 000 2


= 200 W
7 Heat is the energy transferred from one body to another as a result of a temperature difference, while internal energy is the energy stored in a body.
8 Energy transferred
= Pt
= 5  1000  30  60
= 9 000 000 J (= 9 MJ)
9 Power = t Q =
15 60
900 1000


= 1000 W
10 Power of the heater
=
t
Q =
10 60
600 1000


= 1000 W
Time needed =
P
Q =
1000
1 1001000
= 1100 s
11 The statement is incorrect. Heat always flows from a body with higher temperature to a body with lower temperature. However, an object having more internal energy does not mean that it has a higher temperature.
12 (a) B.
It has a higher power and transfers more energy to the water in a fixed time.
(b) By E = Pt, for electric kettle A,
E = 1500  5  60
= 450 000 J
= 450 kJ for electric kettle B,
E = 2000  5  60
= 600 000 J
= 600 kJ
(c) Boiling the same amount of water requires the same amount of energy. Therefore, the kettles cost the same.
1 Heat and Gases Chapter 2 Heat and Internal Energy
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Practice 2.3 (p.47)
1 C
2 B
3 D
4 C = mc = 5 × 480 = 2400 J °C–1
5 Copper has a higher temperature rise than water. 6 Let T be the temperature of the soup after
5 minutes.
By E = Pt = mcT,
200  5  60 = 0.5  3500  (T  20)
T = 54.3 C
7 Let c be the specific heat capacity of the metal block. Energy lost by the metal block
= energy gained by the water bath m1c1T1 = m2c2T2
3  c  (100  31.7) = 5  4200  (31.7  27) c = 482 J kg1 C1
The heat capacity of the metal block is
482  3 = 1450 J C–1.
8 Let T be the final temperature of the mixture.
Energy lost by the 80 C water
= energy lost by the 30 C water m1c1T1 = m2c2T2
2  4200  (80  T) = 5  4200  (T  30)
T = 44.3 C
9 Since water has a very high specific heat capacity, it can absorb a lot of energy with only a small temperature rise. Hence water is suitable to be used as a coolant in motor cars and air-conditioners.
Revision exercise 2
Multiple-choice (p. 50)
1 B
2 A
3 B
Let m be the mass of the water. m  4200  (35  20) = 2  480  (100  35) m = 0.990 kg
4 D
5 B
6 C
7 B
8 C
Specific heat capacity of the liquid
=
mass temperature change energy transferred

=2 25 15
400 60
 

= 1200 J kg1 C1
9 D
10 (HKCEE 2002 Paper II Q20)
11 (HKCEE 2002 Paper II Q21)
12 (HKCEE 2007 Paper II Q10)
13 (HKCEE 2005 Paper II Q27)
Conventional (p. 52)
1 P = t Q
(1M)
=  
1.5 60
0.45 90 10 4200

  
(1M)
= 1680 W (1A)
2 c = m T
Q

(1M)
=
3 5
6750

= 450 J kg–1 °C–1 (1A)
C = mc (1M)
= 3 450
= 1350 J °C–1 (1A)
1 Heat and Gases Chapter 2 Heat and Internal Energy
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3 Let T be the initial temperature of the iron sphere. 1.2  480  (T – 15) = 3  4200  (15 – 12)
(1M)
T = 80.6 °C (1A)
4 (a) Let T be the final temperature of the water.
2  450  (90 – T) = 3  4200  (T – 10)
(1M)
T = 15.3 °C (1A)
(b) Water has a high specific heat capacity. It can absorb a large amount of heat from engines without rising to a high temperature. (1A)
5 Energy released
= 0.3  (80 – 65)  4200 (1M)
= 18 900 J (1A)
6 (a) The specific heat capacity of pottery is greater than that of metal. (1A)
As the bowl and the mug have the same mass, the heat capacity of the mug is greater than that of the bowl. (1A)
For the same temperature rise, the energy absorbed by the mug is greater than that by the bowl. (1A)
To reach thermal equilibrium (same temperature), more energy is transferred to the mug than to the bowl. (1A)
(b) The soup in the metal bowl has a higher final temperature. (1A)
7 (a) (i) Let T be the temperature of the
‘mixture’ before it is heated.
0.8  4200  (20 – T)
= 2  0.08  2400  (T – 2)
(1M)
T = 18.2 C (1A)
The temperature of the ‘mixture’ before heated is 18.2 C.
(ii) Energy needed
= (0.8  4200 + 2  0.08  2400) 
(90  18.2)
= 269 000 J (1M)
Power of the stove
=
5 60
269 000

(1M)
= 897 W (1A)
(b) Energy provided by the stove
= 897  60
= 53 820 J (1M)
Let c be the specific heat capacity of the noodles. 0.5  (90  15)  c = 53 820 (1M) c = 1440 J kg1 C1(1A)
The specific heat capacity of the noodles is 1440 J kg1 C1.
8 (a) By Q = mcT, (1M)
Energy gained by the water
= 0.45  4200  (35  15)
= 3.8  104 J (1A)
(b) (i) 3.8  104 J (1A)
(ii) 0.12  390  T = 3.8  104 (1M)
T = 812 C (1A)
(iii) 812 + 35 = 847 C (1A)
9 (a) (i) 0.75 hour (45 minutes) (1A)
(ii) The biggest temperature difference is
12 C (at 3:30 pm). (1A)
(b) E = mcT (1M)
= 10  4000  20
= 800 kJ (1A)
The water in the radiator takes 800 kJ to raise its temperature by 20 C.
(c) The radiator loses energy to the surroundings and so it needs energy greater than that in (b). (1A)
1 Heat and Gases Chapter 2 Heat and Internal Energy
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(d) The student is incorrect. (1A)
From 7:00 pm to 8:00 pm, the temperature drop of water is 22 C and that of oil is
33 C. Since the specific heat capacity of water is twice that of oil, (1A) by E = mcT, water gives out more energy than oil. (1A)
10 (a) By E = Pt and P = VI, (1M)
E = VIt
= 12  4.2  5  60
= 15 120 J (1A)
(b) (i) By E = mcT, (1M)
15 120 = 0.8  c  19 c = 995 J kg–1 C–1 (1A)
The specific heat capacity of aluminium is 995 J kg–1 C–1.
(ii) (1) Some energy is lost to the surroundings. (1A)
(2) Wrap the aluminium block with cotton wool. (1A)
Physics in articles (p. 54)
1 (a) When the water is stirred, the average kinetic energy of the water particles increases. (1A)
(b) By Q = mcT,
T = mc Q
(1M)
0.5 4200
1050


= 0.5 C
The temperature of the water increases by
0.5 C. (1A)
(c) The statement is correct. (1A)
A liquid with a lower specific heat capacity will have a larger temperature change for the same energy transferred to it. (1A)
As a result, the temperature change can be measured more accurately. (1A)
2 (a) The body temperature is higher than the temperature of cold water, (1A) so heat flows from his/her body to the water. (1A)
(b) Water has a very high heat capacity. The temperature of water rises very little after it has absorbed a large amount of energy from a human body. (1A)
Therefore, people keep on losing energy at a high rate in cold water. (1A)
This may cause hypothermia and damage important organs, causing unconsciousness or even death. (1A)
(c) A high percentage of the mass of our body is made up of water. (1A)
Water has a high specific heat capacity, so the temperature of the body only changes slowly when the temperature of the surroundings changes. (1A)
1 Heat and Gases Chapter 3 Change of State
New Senior Secondary Physics at Work  Oxford University Press 2009
1
3 Change of State
Practice 3.1 (p. 70)
1 B
2 A
Energy provided by the heater
= Pt
= 1000  (10  60)
= 600 000 J
Energy for heating up the water to 100 C
= mcT
= 0.5  4200  (100  20)
= 168 000 J
By E = ml,
Maximum amount of water boiled away
=
l
E = 2 26 106
600 000 168 000


.
= 0.191 kg
3 65 C
4 Incorrect
5 Energy required
= mlf + mcT
= 108  3.34  105 + 108  4200  (4 – 0)
= 3.51  1013 J
6 Energy needed
= mcT  mlv
= 0.2  4200  (100  10)  0.2  2.26  106
= 5.28  105 J
7 Energy needed to change water at 0 C to water at 100 C
= mcT = 1  4200  (100 – 0) = 420 000 J
Let m be the amount of 100 C steam needed.
Energy lost by steam
= energy taken up by water m  2.26  106 = 420 000 m = 0.186 kg
8 A control is needed because ice absorbs energy from the surroundings and melts at room temperature.
If the energy absorbed from the surroundings is ignored, by E = ml, the specific latent heat of fusion of ice found would be smaller than it should be.
9 Energy lost by the coke
= energy needed to melt the ice
+ energy needed to raise the temperature of water from 0 C to T
0.3  5300  (25  T)
= 0.1  3.34  105 + 0.1  4200  T
T = 3.16 C
The final temperature is 3.16 C.
T is higher in reality.
10 Energy released by the juice
= mcT = 0.3  3850  (68 – 15) = 61 215 J
Energy needed to change 1 kg of ice at
0 C to water at 15 C
= mlf + mcT
= 1  3.34  105 + 1  4200  (15 – 0)
= 397 000 J
Minimum amount of ice needed
=
397 000
61 215 = 0.154 kg
11 Let c be the specific heat capacity of the coconut milk.
Energy gained by the ice
= energy lost by the milk
0.17  3.34  105 = 0.2  c  (70 – 0) c = 4060 J kg–1 C–1
12 Energy released by cooling water at
20 C to 14 C
= mcT = 0.3  4200  (20 – 14) = 7560 J
1 Heat and Gases Chapter 3 Change of State
New Senior Secondary Physics at Work  Oxford University Press 2009
2
Let l be the specific latent heat of fusion of ice. Energy absorbed by changing melting ice to water at 14 C
= ml + mcT
= 0.02  l + 0.02  4200  (14 – 0)
= 0.02 l + 1176
Energy lost by water = energy gained by ice
7560 = 0.02 l + 1176 l = 3.19  105 J kg–1
13 Assume the mass, the temperature and the specific heat capacity of hot drink are 0.3 kg,
50 C and 4200 J kg1 C1 respectively.
To cool the cup of hot drink, the water/ice absorbs energy from the hot drink. If all the
0.2-kg ice at 0 C just melts to become water at 0 C, it will absorb 3.34  105  0.2
= 6.68  104 J of energy. This can cool down
0.3-kg of hot drink by a temperature of
(
0.3 4200
6.68 104


=) 53.0 C. However, since the initial temperature of the hot drink is 50 C only, the drink will become 0 C.
On the other hand, the hot drink cooled by the
0-C water bath must have a final temperature higher than 0 C. Therefore, ice can cool the hot drink to a lower temperature in this case.
(One may get a different conclusion if different assumptions on the mass, the temperature and the specific heat capacity of the cup of hot drink are made.)
14 (a) None
(b) c
(c) c
(d) lf
(e) lf
(f) lv, lf, c
(g) lv, c
(h) None
(i) lf
(j) lf
Practice 3.2 (p. 80)
1 A
2 C
3 C
4 When we get out of a swimming pool, water on our skin absorbs energy from our bodies and evaporates. Therefore, we lose energy and feel cold.
If it is windy, the evaporation rate, and therefore the rate at which our bodies lose energy, will be higher. We would feel much cooler. 5 On a humid day, the rate of evaporation is lower. As a result, less energy is taken away by evaporation. Therefore, the soup cools down slower.
6 When vapour meets a cool surface, it will condense on the surface and release energy.
(a) The glasses are cooler than the surroundings when the person has just got out of the car. Therefore, water vapour in air condenses on them.
(b) The reason is similar to (a). The water vapour in the steamy bathroom is at a higher temperature than the glasses. And the large amount of vapour enhances the condensation. 7 (a) Assume that the latent heat of vaporization is solely obtained from his body.
E = mlv
= 0.5  2.26  106
= 1.13  106 J (= 1.13 MJ)
1 Heat and Gases Chapter 3 Change of State
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3
The maximum amount of energy removed is 1.13  106 J.
(b) 1.13  106 J is removed from his body in 1 hour. Therefore, rate of cooling by sweating
=
60 60
1.13 106


= 314 W
8 (a) Energy required
= mlv = 0.2  2.26  106 = 4.52  105 J
(b) By E = mcT, decrease in temperature
=
mc
E =
50 3500
452 000

= 2.58 C
9 Vaporization.
The refrigerating liquid absorbs energy in the process of vaporization, so that the food is cooled. 10 As glass A is half covered by a plastic sheet, some water vapour that evaporates from the hot water is trapped in the glass and the air becomes humid. Therefore, the rate of evaporation is slower in glass A and so the water level in glass A drops more slowly than that of glass B.
Revision exercise 3
Multiple-choice (p. 84)
1 B
2 A
3 D
4 B
5 B
6 B
7 B
8 C
Let m be the mass of steam and km be the mass of ice.
Energy lost by steam = energy gained by ice m  2.26  106 + m  4200  (100  50)
= km  3.34  105 + km  4200  (50  0) k = 4.5
The ratio of the mass of ice to the mass of steam is 4.5 : 1.
9 A
10 A
11 (HKCEE 2005 Paper II Q9)
12 (HKCEE 2006 Paper II Q11)
13 (HKCEE 2007 Paper II Q7)
14 (HKCEE 2007 Paper II Q34)
15 (HKCEE 2004 Paper II Q43)
Conventional (p. 86)
1 Energy has to be removed from the water
= mcT + mlf (1M)
= 0.2  4200  30 + 0.2  3.34  105
= 92 000 J (= 92 kJ) (1A)
2 When snow melts, it absorbs latent heat of fusion from its surroundings. (1A)
Therefore, the surrounding air loses energy
(1A)
and the air temperature drops. (1A)
3 (a) Melting point (1A)
(b) Room temperature (1A)
(c) (i) The average KE of the ice molecules increases in this period. (1A)
(ii) The average PE of the ice molecules increases in this period. (1A)
4 The water on a wet finger absorbs energy from the finger and evaporates. The finger would feel cold and the cooling effect increases in the wind. (1A)
1 Heat and Gases Chapter 3 Change of State
New Senior Secondary Physics at Work  Oxford University Press 2009
4
Therefore, from the side of the finger we feel cooler, the wind direction can be told. (1A)
5 Let T be the final temperature of the ‘mixture’.
Energy lost by water = energy gained by ice
(1M)
0.5  4200  (30  T)
= 0.1  3.34  105 + 0.1  4200  T (1M)
T = 11.7 C (1A)
6 (a) Total energy released by the water
= mcwaterT + mlf + mciceT (2M)
= 3  4200  (30 – 0) + 3  3.34  105
+ 3  2060  [0 – (–5)] (1M)
= 1 410 900 J
= 1.41 MJ (1A)
(b) Effective power of the refrigerator
=t
E (1M)
=
60 60
1 410 000

= 392 W (1A)
7 Let T be the final temperature of the ‘mixture’.
Energy lost by water
= energy gained by ice (1M)
1  4200  (20  T)
= 0.5  3.34  105 + 0.5  4200  T (1M)
T = 13.2 C (1A)
Since the final temperature should be between
0 C and 20 C, the result 13.2 C shows that not all the ice melts. Therefore, the final temperature of the ‘mixture’ is 0 C. (1A)
8 (a) Temperature is higher on a sunny day.
(1A)
Therefore, particles move faster on average and they can break free from liquid more easily. (1A)
(b) On a windy day, particles in the vapour are blown away. (1A)
This makes fewer particles in the vapour ready to return to the liquid. As a result, the rate of evaporation increases. (1A)
(c) Water vapour condenses to droplets on the cold surfaces. (1A)
Since the temperature is lower than the freezing point, the droplets solidify to form frost. (1A)
9 When 0.1 kg of steam at 110 C condenses to water at 100 C, energy released
= mcT + mlv
= 0.1  2000  (110 – 100) + 0.1  2.26  106
= 228 000 J (1M)
Power supplied by the cooker
=
t
E (1M)
=
60
228 000
= 3800 W (1A)
10 (a) Energy needed
= mcT + mlv (1M)
= 0.5  4200  (100 – 25)
+ 0.5  2.26  106 (1M)
= 1 287 500 J
= 1.29 MJ (1A)
(b) Steam reaching the cover condenses to water droplets, which may drip back to the wok. (1A)
Thus the energy required is larger than the result in (a). (1A)
1 Heat and Gases Chapter 3 Change of State
New Senior Secondary Physics at Work  Oxford University Press 2009
5
11
(Slopes larger than those in the old curve)
(1A)
(Shorter horizontal line) (1A)
(Horizontal lines at the same height) (1A)
12 Energy absorbed by the liquid
= Energy supplied by the heater
= Pt
= 10  3  60
= 1800 J (1M)
Let l be the specific latent heat of vaporization of the liquid.
E = mlv (1M)
1800 = 0.01  l l = 180 000
= 1.8  105 J kg–1 (1A)
13 (a) Freezing water tends to warm the surroundings. (1A)
(b) Since water has high specific heat capacity and latent heat of fusion, (1A) it can release a large amount of energy before it freezes. Therefore, spraying water on fruit trees can protect fruits from freezing. (1A)
14 (a) In heating the water, the balance reading remains unchanged at first. This shows that the water temperature is below the boiling point and keeps increasing. (1A)
The balance reading then drops. It shows that the boiling point is reached. Water temperature keeps steady. (1A)
(Temperature increases at first.) (1A)
(Curve remains steady afterwards.) (1A)
(b) Since the beaker has more water, the initial balance reading is larger. (1A)
And by Pt = mcT, it will take a longer time for the heater to boil the water. (1A)
Hence, the horizontal line of the new curve is longer and is above the old curve.
After the water boils, the rate of vaporization of water depends on the power of the heater, which is unchanged.
(1A)
Hence, the slope of the curve remains the same. (Longer horizontal line above the old curve, same slope) (1A)
0
balance reading / kg time / s
1 Heat and Gases Chapter 3 Change of State
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6
15 (a) (i) When the water particles near the surface absorb enough energy from the surroundings, they can escape into the space above the sea and form water vapour. (1A)
(ii) The vapour is heated up by the sun and it rises. (1A)
The temperature is lower in higher altitude. Therefore, the vapour condenses back to water droplets.
(1A)
Droplets accumulate and form larger droplets. The larger droplets then fall as rain. (1A)
(iii) Condensation of vapour releases heat.
(1A)
Surrounding air is warmed. (1A)
(b) If the temperature is low enough, water droplets freeze to ice and fall as snow.(1A)
16 (a) (i) The air current brought by the fan removes water molecules in the air around the wick. (1A)
Hence, it is easier for the water in the wick to evaporate. (1A)
(ii) Yes, it is true. (1A)
As relative humidity increases, the rate of evaporation decreases. (1A)
It is harder for the water to evaporate from the wick. (1A)
(b) (i) I would feel cooler. (1A)
This is because the water in the wick absorbs energy from the surroundings to vaporize. (1A)
(ii) I will feel warmer. (1A)
It would become more humid after using the humidifier for a long time.
It is harder for sweat on our skin to evaporate. (1A)
17 (HKCEE 2005 Paper I Q3)
18 (HKCEE 2006 Paper I Q10)
19 (HKCEE 2007 Paper I Q4)
Physics in articles (p. 90)
1 (a) By E = mlv (1M) lv of water = m E
=
2
4.853106
= 2 426 500
= 2.43  106 J kg–1 (1A)
(b) The result is larger than the standard value.
This shows that more energy is needed to vaporize the water. (1A)
Vapour of sweat can condense on the clothes of an athlete and drip back to the skin. (1A)
(c) The more humid the air, the lower the rate of evaporation. (1A)
As a result, the efficiency of cooling by sweating is lower. Therefore, marathon runners feel hotter if the air is humid.(1A)
2 (a) They would evaporates gradually. (1A)
(b) Any two of the following: temperature / wind speed / humidity. (2A)
(c) Vapour condenses on mirrors when we are having showers.
Water condenses in a dehumidifier. (2A)
(Or other reasonable answers)
1 Heat and Gases Chapter 4 Transfer Process
New Senior Secondary Physics at Work  Oxford University Press 2009
1
4 Transfer Process
Practice 4.1 (p. 100)
1 C
2 C
A: The fur is not a source of heat. It cannot raise the temperature of the thermometer.
C: The fur slows down the rate of heat transfer
(from the bulb to the surroundings).
3 C
4
Material Example Good conductor Good insulator Solid
(metal) copper 
Solid
(non-metal) plastic 
Liquid water 
Gas air 
5 Metal is a good conductor. When we touch a metal railing, it conducts energy away from our hands quickly. However, wood is a good insulator. When we touch a wooden railing, only little energy is conducted away from our hands. Therefore, a metal railing feels colder than a wooden railing even if they have the same temperature.
6 Goose-down in the jacket traps air. Since air is a poor conductor of heat, energy cannot easily escape by conduction.
7 In conduction, particles vibrate and transfer energy when they collide with each other.
Since there is no particle in a vacuum, conduction does not occur.
8 Since fat is a good insulator, it reduces heat transfer from the body to the surroundings by conduction. Hence, fat people feel warmer than thin people in winter.
9 Since copper is a better conductor of heat than stainless steel, it conducts heat faster and avoids the accumulation of heat at a point.
This ensures even heating.
10 (a) Water is a poor conductor of heat. Heat is conducted slowly from the top of the boiling tube down to the ice.
(b) We cannot get the same result if we replace the boiling tube by a metal tube.
Metal is a good conductor of heat. Heat is conducted along the tube to the ice quickly. 11 (a) Heat flows from inside to outside.
(b) Polyfoam is a poor conductor of heat. Heat flows from inside through the polyfoam container to outside slowly, so the food is kept warm.
(c) In summer, the temperature outside the container is higher than the temperature of the cold food. Therefore, heat flows from outside to inside. The polyfoam container keeps the food cool by slowing the flow of heat. Practice 4.2 (p. 108)
1 A
2 A
3 B
4 The cover of the container reduces energy loss by convection.
5 In a convection current, hot water rises and cold water sinks. If the heating element of an electric kettle is fixed at the top, water at the bottom cannot be heated through convection.
1 Heat and Gases Chapter 4 Transfer Process
New Senior Secondary Physics at Work  Oxford University Press 2009
2
6 In convection, particles of a fluid carry energy from the hot region to the cold region and the particles in the cold region move along the convection current to the hot region to absorb energy. Since no particles exist in a vacuum, no convection current can be formed and convection does not occur.
7 The warm air around the light bulb rises. The opening lets the warm air flow out to prevent overheating. 8 The air trapped in cotton and feathers can not move around and so it reduces heat loss through conduction and convection. On the other hand, the air surrounding a hot pan can move freely and it can carry the energy away from the pan through convection.
Practice 4.3 (p. 119)
1 D
2 D
3 B
4 C
5 Heat travels from the sun to the earth by radiation. 6 Since black paper is a better absorber of radiation than white paper, snow under the black paper melts faster than that under the white paper.
7 It is very cold in the space. The spacesuit is white in colour to reduce energy loss through radiation. If it were dull black, the astronaut would radiate energy quickly and feel very cold. On the other hand, the temperature of the side facing the sun would increase rapidly.
8 Light-coloured surfaces are poor radiators of heat. The hot air inside the hot air balloon cools down more slowly. This saves fuel.
9 Thermometer A will have the highest reading.
Since dull black surfaces are good absorbers of radiation, the blackened foil will absorb the largest amount of radiation from the sun.
10 For keeping warm, John should wear the transparent plastic coat on the outside. The transparent plastic coat lets the radiation from the sun pass through. The radiation is then absorbed by the black coat, which is a good absorber of radiation. The plastic coat also reduces energy loss from the black coat to outside by conduction and convection.
Revision exercise 4
Multiple-choice (p. 125)
1 C
2 B
3 D
4 B
5 C
6 A
7 C
8 D
9 (HKCEE 2005 Paper II Q7)
10 (HKCEE 2005 Paper II Q8)
11 (HKCEE 2006 Paper II Q9)
12 (HKCEE 2007 Paper II Q9)
Conventional (p. 126)
1 Black surfaces are good absorbers of radiation and become hot under sunlight. (1A)
There is a risk of explosion of the fuel in the tanks under high temperature. (1A)
1 Heat and Gases Chapter 4 Transfer Process
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3
2 Shiny surfaces are poor absorbers and poor radiators of heat. (1A)
The shiny foil blanket reduces energy loss by radiation and avoids the runner cooling down too quickly. (1A)
3 Air is a poor conductor of heat. (1A)
The layer of air reduces the amount of energy entering the food compartment. (1A)
4 (a) Aluminium (1A)
(b) Since aluminium conducts heat better than stainless steel does, it conducts heat away from the hot part faster. (1A)
This can reduce the accumulation of heat and ensure even heating. (1A)
5 (a) The transparent plate allows sunlight to pass through. (1A)
Furthermore, it reduces energy loss by the panel through convection by trapping the warm air between the plates. (1A)
(b) Since a black surface is a good radiation absorber, (1A) a lower plate with a black surface can be heated up by radiation more quickly. (1A)
The surface should be dull in colour because a dull surface is a better radiation absorber than a shiny surface. (1A)
6 (a) Radiation from the sun can pass through the car windows. (1A)
This warms up the air and other materials inside the car. The warm air and materials emit infra-red radiation which cannot escape through the windows easily. (1A)
Moreover, since all the windows of the car are closed, the car has no energy loss through convection. (1A)
Hence, the temperature inside the car rises drastically. (b) The shield covering the car windows can reflect sunlight away. This avoids the air and materials inside the car being warmed up. (1A)
7 (a) There is a layer of trapped air in a double-glazed window. Since air is a better insulator than glass, (1A) it reduces heat loss by conduction. (1A)
(b) Since conduction cannot occur in a vacuum, (1A) a double-glazed window with a vacuum between the panes can stop heat loss by conduction. (1A)
Therefore, it performs better than that with air. (1A)
8 (Design for preventing heat loss by conduction,
e.g. walls made of a good insulator or double walls.) (1A)
(Design for preventing heat loss by convection,
e.g. airtight lid.) (1A)
(Design for preventing heat loss by radiation,
e.g. outside walls with light or silver colour.)
(1A)
(Appropriate drawing of the container) (1A)
(Appropriate labels of the components) (1A)
1 Heat and Gases Chapter 4 Transfer Process
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4
9 (a) Radiation (1A)
(b) The specific heat capacity of sea water is much higher than that of sand. (1A)
For similar amount of energy lost to the surroundings, the temperature drop of sea water is mild, while that of sand is huge.
(1A)
(For teacher’s reference: specific heat capacity of sand
= 835 J kg1 C1
Specific heat capacity of water
= 4200 J kg1 C1)
(c)
(Correct cold and warm regions) (2A)
(Correct movement of air) (1A)
10 (a) Radiation (1A)
(b) The fan speeds up the movement of air.
(1A)
This reduces the temperature difference at different points. (1A)
Therefore, the food can be cooked more evenly. (1A)
(c) As a good insulator, the vitreous enamel reduces energy loss to the outside of the oven. (1A)
(Or it prevents the outside of the oven from being too hot to touch.)
(d) Since a shiny surface is a poor absorber of radiation, (1A) it reduces energy loss by radiation. (1A)
11 (a) (i) Radiation (1A)
(ii) Conduction (1A)
(b) This creates an environment similar to a greenhouse for the heater. (1A)
Hence, it raises the temperature of the air around the heater and allows the water inside the heater to absorb more energy.
(1A)
(c) This is because black objects are good absorbers of radiation. (1A)
(d) Copper is more suitable for making pipes.
(1A)
Energy is transferred from the pipes to water by conduction and copper is a better conductor than plastic. (1A)
12 (a) The glass plate traps the air inside and reduces energy loss by convection. (1A)
Also, it absorbs most of the infra-red radiation emitted by the warm objects inside and reduces the energy loss by radiation. (1A)
(b) Amount of energy absorbed by the water
= mcT (1M)
= 0.5  4200  (60  20)
= 84 000 J (1A)
(c) Add reflectors to the inside walls of the cooker (1A) to reflect additional sunlight into the cooker. (1A)
13 (a) A dull surface is better absorber of radiation than a shiny surface. (1A)
Therefore, the dull surface should face outwards during baking because it can absorb more energy and reduce the baking time. (1A)
1 Heat and Gases Chapter 4 Transfer Process
New Senior Secondary Physics at Work  Oxford University Press 2009
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(b) No, the shinny surface should face outwards in this case. (1A)
Since a shiny surface is a poor radiator of heat, this arrangement can reduce energy loss from the food through radiation. (1A)
(c) Evaporation takes away a large amount of energy from the food. (1A)
The aluminium foil traps the steam evaporated from the food. (1A)
This reduces the rate of evaporation and thus the rate of heat transfer. (1A)
14 The steam of herbal tea rises. Glass plates can reduce the energy loss due to evaporation(1A) and convection. (1A)
Cold air from freezers sinks and the food in freezers can be kept frozen even if they do not have covers. (1A)
15 In winter, because of the large specific heat capacity of water, the land loses energy more quickly than the sea. (1A)
The warm air above the sea rises (1A) and the cool air blows in from the land to replace it. This results in winter monsoons.
(1A)
16 (a) Curve B (1A)
(b) Dull black objects are good radiators.(1A)
The beaker wrapped in dull black paper loses energy through radiation more quickly. (1A)
Therefore, curve B, which shows a more rapid drop in temperature, corresponds to this beaker. (1A)
(c) This is because the temperature difference between the water and the room temperature decreases. (1A)
17 (a) Black ‘fuel effect’ lumps burn and release heat. Air around the lumps is heated. It expands and rises. (1A)
Hot air leaves the fire at A. (1A)
Cold air flows in from C to replace the hot air. (1A)
The air forms a convection current flowing from C to A. (1A)
(b) (i) Radiation (1A)
(ii) The lumps should be dull black in colour. (1A)
18 (a) (i) Since black bodies absorb radiation better, (1A) pipes painted in black heat water at a higher rate. (1A)
(ii) Copper is a good conductor of heat, but plastic is not. (1A)
(b) Advantage: It saves energy. (1A)
Disadvantage: It can only be used in sunny days. (1A)
(c) Water at the top of the tank is warm (1A) while water at the bottom is cold. (1A)
19 (a) (i)
(Air above the heater rises.) (1A)
(Rest of circulation) (1A)
(ii) Convection (1A)
1 Heat and Gases Chapter 4 Transfer Process
New Senior Secondary Physics at Work  Oxford University Press 2009
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(iii) The heater heats up the air around it.
The warm air expands (1A) and becomes less dense. (1A)
Therefore, it rises. (1A)
The surrounding cold air moves in to replace the warm air. (1A)
(b) Since metal foil is shiny, (1A) it is a poor absorber of radiation which reduces the energy loss from the heater to the surroundings. (1A)
(c) Any two of the following or other reasonable answers: (2A)
Use double-glazed window.
Paint the walls of the room with light colours. Add insulating materials to the walls.
Add insulating materials to the ceiling.
Put the heater away from the walls to the outside. 20 (a) Their house acts as a control set-up. (1A)
(b) (i) The silver foil is poor radiator of heat. (1A)
This reduces energy loss by radiation. (1A)
(ii) The cotton wool is a poor conductor.
(1A)
This reduces energy loss by conduction. (1A)
(iii) After the polythene is taken away, the air can flow freely into and out of the house. (1A)
This increases the energy loss by convection. (1A)
Physics in articles (p. 132)
1 (a) Convection (1A) and radiation (1A)
(b) We will feel hot first and then get burnt after a long time. (1A)
This is because heat is transferred to our hand by convection and radiation. (1A)
(c) Bamboo is a poor conductor of heat while metal is a good conductor of heat. (1A)
Bamboo conducts less heat to our hand. and so it is safer. (1A)
2 (HKCEE 2007 Paper I Q3)
1 Heat and Gases Chapter 5 Gases
New Senior Secondary Physics at Work  Oxford University Press 2009
1
5 Gases
Practice 5.1 (p. 150)
1 B
2 The reading on the syringe gives the volume of air inside the syringe but does not include that in the rubber tubing. The shorter the rubber tubing, the smaller the error in measuring the volume.
3 By Charles’ law,
1
1
T
V
=
2
2
T
V
V2 = 2
1
1 T
T
V 
= 273 117
273 27
1  

= 1.3 m3
Its volume is 1.3 m3 at 117 C.
4 By general gas law,
1
1 1T p V =
2
2 2T p V
V2 =
2
2
1
1 1 p T
T
p V 
= 5
5 3
1.0 10
273
273 20
0.9 10 5.0 10



   
= 4.19  10–3 m3
The volume of the balloon is 4.19  103 m3.
5 (a) Any two of the following:
Immerse the whole flask, including the neck, in water.
Use a very short rubber tubing to connect the Bourdon gauge and the flask.
Wait until the pressure and the temperature have become steady before taking the readings.
Do not allow the flask and the thermometer to touch the bottom of the beaker. (b) By pressure law,
1
1
T
p =
2
2
T
p p2 = 2
1
1 T
T
p 
= 273 80
273 40
120 103
 


= 135 000 Pa = 135 kPa
The pressure of the gas is 135 kPa.
6 Let p1 and p2 be the gas pressure in X before and after the tap is opened respectively.
By the general gas law, p1 =
1
1 1
V
n RT =
1
1.5 1
V
RT 
The temperature of the gas does not change after the tap is opened, and the pressure in X and that in Y are the same after the tap is opened. Therefore, p2 =
2
2 2
V
n RT =  
1
1
2
1.5 2.4
V
  RT
=
1
1.95 1
V
RT 
Percentage change =
1
2 1 p p  p
 100%
=
1.5
1.95 1.5
 100% = 30%
Practice 5.2 (p. 164)
1 C
2 C
3 D
4 B
5 (a) The smoke particles move about constantly along zigzag paths.
(b) They are bombarded by a large amount of air molecules around them. The bombardments come from all sides but not in equal numbers. This results in a random motion
1 Heat and Gases Chapter 5 Gases
New Senior Secondary Physics at Work  Oxford University Press 2009
2
(c) Slower motion of smoke particles
6 (a) Increases
(b) Increases
7 The number of air particles inside the tyre increases, so there are more frequent bombardments on the tyre wall. As a result, the pressure of the tyre increases.
8 Root-mean-square speed
=
mNA
3RT
=  
5.6 10 26 6.02 1023
3 8.31 25 273
  
  

= 469 m s1
9 Particles of the perfume vapour move at high speeds and travel in all directions. This makes the smell of the perfume spread.
10 Since total KE =
2
3 nRT, increase in total KE
= 8.31 80 25
6.02 10
7.28 10
2
3
23
24
  



= 8290 J
11 (a) The root-mean-square speed becomes 3 times its original value.
(b) The root-mean-square speed triples.
(c) The root-mean-square speed remains unchanged. 12 Average force exerted on surface W due to a molecule with speed v
=
time interval for the change change in momentum
=
v l mv
2
2 = l mv2
Pressure exerted on surface W due to this molecule = area force = 2
2
l l mv
= 3
2
l mv The pressure p due to N molecules
=  2 2 
2
2
3 1 v v ... vN l m    = 2
3 Nv l m where v 2 is the mean value of v2 of all the molecules. Revision exercise 5
Multiple-choice (p. 168)
1 A
2 D
3 D
By crms = mNA 3RT , the ratio of crms at 80 C to that at 20 C
=
A
A
mN
R
mN
R
3 (273 20)
3 (273 80)


= 1.10
4 B
Let v0 the volume of gas at X.
Let p0 the pressure of gas at Z.
By general gas law, T = nR pV .
Temperature at X, Y, Z can be expresses as:
TX = nR 5p0  v0
=
nR
5 p0v0
TY = nR 3p0 3v0
=
nR
9 p0v0
TZ = nR p0 5v0
=
nR
5 p0v0
TY  TX  TZ
5 A
By general gas law, pV = nRT
V = nR 
1




T p 1 Heat and Gases Chapter 5 Gases
New Senior Secondary Physics at Work  Oxford University Press 2009
3
The volume of gas is inversely proportional to




T
p , which is the slope of the line connecting the point to the origin as shown below.
The slope for stage X is greater than that for stage Y, i.e.
X
X
T
p
>
Y
Y
T p , so VY > VX.
6 C
7 (HKCEE 2002 Paper II Q22)
8 (HKCEE 2002 Paper II Q23)
9 (HKCEE 2003 Paper II Q24)
10 (HKALE 2005 Paper II Q21)
Conventional (p. 169)
1 When the light bulb is switched on, the temperature inside the bulb increases. Since the volume of gas inside a light bulb is fixed, the pressure is then increased. (1A)
The bulb will burst if the pressure is too high.
Filling it with a gas at low pressure can prevent the bulb from bursting. (1A)
2 (a) By general gas law, pV = nRT
Number of moles of gas
=
RT pV (1M)
=8.31 27 273
100 103 5
 
 
= 201 mol (1A)
(b) By general gas law,
1
1 1T p V
=
2
2 2T p V (1M)
27 273
100 103 5

  =
7 273
80 10 2
3

 V
V2 = 5.83 m3 (1A)
3 Root-mean-square speed
=
mNA
3RT (1M)
=  
3.35 10 27 6.02 1023
3 8.31 31 273
  
  

= 1940 m s1 (1A)
4 While rising to the water surface, n and T of the air inside his lungs remains constant.
By Boyle’s law, p1V1 = p2V2 (1M)
2p2  V1 = p2V2
V2 = 2V1
Therefore, the volume of his lungs would double. (1A)
5 By general gas law, n =
RT
pV
Percentage of air escape
=  


 


1
1 2 n n  100% (1M)
=
   


   



1
1 1
2
2 2
1
RT p V
RT
p V
 100%
=  


 


1
1 2 p p  100% (1M)
= 



 




 3
3
220 10
1 100 10  100%
= 54.5% (1A)
6 (a) (i) Decreases (1A)
(ii) Remains unchanged (1A)
(iii) Remains unchanged (1A)
1 Heat and Gases Chapter 5 Gases
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(b) As the number of air molecules inside the carton decreases, the number of bombardments on the walls of the carton decreases. (1A)
As a result, the pressure inside the carton decreases (1A) and the carton collapses due to the larger pressure outside. (1A)
(c) Heat the sealed carton in a water bath.(1A)
The average kinetic energy of the air molecules inside the carton increases; thus the number of bombardments on the wall of the carton increases. (1A)
As a result, the pressure inside the carton increases and the carton gradually regain its original shape due to the smaller pressure outside. (1A)
7 (a) (i) (1) The piston floats lower. (1A)
(2) The piston floats higher. (1A)
(ii) The ball-bearings in the tube represent the gas molecules; the number of cardboard discs, the voltage applied to the motor and the height of the piston represent the pressure, temperature and the volume of the gas respectively. (1A)
When the voltage is gradually increased, the piston floats higher.
(1A)
This shows that the gas volume increases as the temperature increases.
This simulates Charles’ law. (1A)
(b) (i) Wait until the volume and temperature have become steady before taking the readings. (1A)
(ii) The open reservoir of oil keeps the air inside the tube at atmospheric pressure throughout the experiment.
(1A)
(iii) (1) The length of the air column remains unchanged no matter what temperature of the water bath is. (1A)
(2) According to pressure law, at a constant volume, the gas pressure increases as the temperature increases. (1A)
8 (a) By general gas law, pV = nRT. (1M)
Therefore,
nRT = 2
3
1 Nmc
Root-mean-square speed c 2
=
Nm
3nRT (1M)
=
m n N
RT

3
= mN A
3RT (1A) where m is the mass of a gas molecule and
T is the absolute temperature of the gas.
(b) Total kinetic energy of the molecules
= 2
2
N  1 mc (1M)
From (a), nRT = 2
3
1 Nmc
Rearrange the terms,
2
2
N  1 mc nRT
2
 3 (1A) where n is the number of moles and T is the absolute temperature of the gas.
1 Heat and Gases Chapter 5 Gases
New Senior Secondary Physics at Work  Oxford University Press 2009
5
(c) By the equation in (b),
Total kinetic energy of the molecules
= nRT
2
3 (1M)
= 3.5 8.31 70 273
2
3    
= 15 000 J (1A)
9 (a) Gas particles gain more kinetic energy and
(1A)
bombard on the piston more frequently and more violently. (1A)
Therefore, the gas pressure inside the syringe becomes larger than that outside
(1A)
and so the piston moves outwards.
(b) (i) As the piston moves outwards, the frequency of bombardment decreases,
(1A)
and so does the pressure inside. (1A)
When the pressure inside and outside the syringe become the same, the piston stops. (1A)
(ii) By Charles’ law,
1
1
T
V =
2
2
T
V
(1M)
V2 = 2
1
1 T
T
V 
= 100 273
20 273
50  

= 63.7 cm3 (1A)
(c) (i) Before the syringe is turned vertical, the pressure inside balances the pressure outside. After turning vertical, the weight of the piston is not balanced. This becomes a net force and so the piston drops. (1A)
When the piston drops, the volume of the gas decreases, and the frequency of bombardment increases, and so does the pressure inside. (1A)
When the pressure inside balances both the pressure outside and the pressure provided by the weight of the piston, the piston stops. (1A)
(ii) Pressure provided by the weight of the piston
=A
F = π 0.012
1

= 3180 Pa (1M)
New pressure of carbon dioxide
= 100  103 + 3180
= 103 180 Pa (1A)
By Boyle’s law, p1V1 = p2V2 (1M)
V2
=
2
1 1p p V
=
103 180
100 103  63.7
= 61.7 cm3 (1A)
(iii) By Boyles law, the pressures of carbon dioxide are 102.1 kPa and
101.1 kPa when its volume are
62.4 cm3 and 63 cm3 respectively.
(Correct labelled axes) (1A)
1 Heat and Gases Chapter 5 Gases
New Senior Secondary Physics at Work  Oxford University Press 2009
6
(A smooth curve with correct starting point and ending point) (1A)
10 (a) If the collisions are not perfectly elastic, energy will be lost during the collisions and the speeds of the gas molecules will become slower and slower. (1A)
We cannot get a definite value for the change in momentum when a molecule collides with a wall. (1A)
Therefore, we cannot find the pressure exerted on the wall by the molecule and can no longer make the derivation. (1A)
(b) If the gas molecules are not in random motion, the mean values of the velocities of the molecules in x, y and z directions may not be the same. (1A)
Therefore, the average force, hence the pressure, exerting on each wall may not be the same. (1A)
In this case, a single value of pressure p is not well defined. (1A)
(c) If the gas molecules have different mass, the total pressure exerted on the wall in the x-direction by all the molecules will become: p =  2 2
2 2
2
1 1 1 ... m vx m vx mN vxN
V
  
= 2 mvx V
N
(instead of 2 mvx V
N ) (1A)
Similarly, the total pressure exerted on the walls in y- and z-directions will become
2
mvy
V
N and 2 mvz V
N respectively.
(1A)
As a result we will derive the equation pV = 2
3
1 Nmc instead of pV = 2
3
1 Nmc .
(1A)
11 (HKALE 2001 Paper I Q10)
12 (HKALE 2004 Paper II Q5)
13 (HKALE 2006 Paper I Q5)
Physics in articles (p. 173)
(a) The gas pressure inside a ‘vacuum balloon’ is zero. (1A)
The atmospheric pressure acts on the
‘balloon’ and so the ‘balloon’ collapses.(1A)
(b) The temperature of the air inside a hot air balloon is higher than that outside. (1A)
Therefore, the kinetic energy of the air molecules is higher. (1A)
These air molecules move faster and collide with the wall of the balloon more frequently and more violently. (1A)
As a result, the air inside the balloon can provide the same pressure as the surroundings with less air particles.
(c) By general gas law, pV = nRT (1M)
90  103  10 = n  8.31  (70 + 273) n = 316 mol (1A)
There are 316 mol of air particles inside the
balloon.