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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.
0100
0
T
=2.32.18
2.37.7
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.
621
6
x
=080
037
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
2
6 Let T be the temperature measured.
0100
0
T
=35120
3580
T = 52.9 C
Revision exercise 1 Multiple-choice (p. 19) 1 D
2 B
3 B
4 C
2580
25
R
=0100
040
R = 47 units
5 B
6 C
0100
0
T
=1090
1040
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 0100
0
x
=7.36.24
7.30.12
(1M)
x = 39.7 C (1A)
3 (a) Let T be the temperature when the length
of the alcohol column is 15.6 cm.
0100
0
T
=2.44.18
2.46.15
(1M)
T = 80.3 C (1A)
(b) Let x be the length of the alcohol column
at 30 C.
2.44.18
2.4
x
=0100
030
(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 32FT
100
33 (1M)
= 32FT 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
3
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.
0100
0
T
=8.34.16
8.38.8
(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 =area sectional cross
increased volume
=01.001.0
0013.0
= 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
=10002
1000420
= 210 s =60
210min = 3.5 min
6 B
P =t
Q=
605
2000 30
= 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=
6015
1000900
= 1000 W
10 Power of the heater
=t
Q=
6010
1000600
= 1000 W
Time needed =P
Q=
1000
0001100 1 = 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
New Senior Secondary Physics at Work Oxford University Press 2009
2
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 = mcT,
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
m1c1T1 = m2c2T2
3 c (100 31.7) = 5 4200 (31.7 27)
c = 482 J kg1 C1
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
m1c1T1 = m2c2T2
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
=change etemperaturmass
nsferredenergy tra
= 15252
60400
= 1200 J kg1 C1
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)
=
605.1
4200109045.0
(1M)
= 1680 W (1A)
2 c =Tm
Q
(1M)
=5 3
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
New Senior Secondary Physics at Work Oxford University Press 2009
3
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
=605
000 269
(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 kg1 C1(1A)
The specific heat capacity of the noodles
is 1440 J kg1 C1.
8 (a) By Q = mcT, (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 = mcT (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
New Senior Secondary Physics at Work Oxford University Press 2009
4
(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 = mcT, 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 = mcT, (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 = mcT,
T =mc
Q (1M)
42005.0
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
= mcT
= 0.5 4200 (100 20)
= 168 000 J
By E = ml,
Maximum amount of water boiled away
=l
E=
610262
000 168000 600
.= 0.191 kg
3 65 C
4 Incorrect
5 Energy required
= mlf + mcT
= 108 3.34 105 + 108 4200 (4 – 0)
= 3.51 1013 J
6 Energy needed
= mcT 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
= mcT = 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
= mcT = 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 + mcT
= 1 3.34 105 + 1 4200 (15 – 0)
= 397 000 J
Minimum amount of ice needed
=000397
21561= 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
= mcT = 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 + mcT
= 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 kg1 C1 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
(42003.0
1068.6 4
=) 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
New Senior Secondary Physics at Work Oxford University Press 2009
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
=6060
1013.1 6
= 314 W
8 (a) Energy required
= mlv = 0.2 2.26 106 = 4.52 105 J
(b) By E = mcT,
decrease in temperature
=mc
E=
350050
000 452
= 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
= mcT + 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
= mcwaterT + mlf + mciceT (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)
=6060
0004101
= 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
= mcT + mlv
= 0.1 2000 (110 – 100) + 0.1 2.26 106
= 228 000 J (1M)
Power supplied by the cooker
=t
E (1M)
=60
000 228
= 3800 W (1A)
10 (a) Energy needed
= mcT + 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 = mcT, 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
New Senior Secondary Physics at Work Oxford University Press 2009
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
10853.4 6
= 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 ExampleGood
conductor Good
insulatorSolid (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
New Senior Secondary Physics at Work Oxford University Press 2009
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
New Senior Secondary Physics at Work Oxford University Press 2009
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 kg1 C1
Specific heat capacity of water
= 4200 J kg1 C1)
(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
= mcT (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
5
(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
6
(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 = 21
1 TT
V
= 11727327273
1
= 1.3 m3
Its volume is 1.3 m3 at 117 C.
4 By general gas law,
1
11
T
Vp=
2
22
T
Vp
V2 =2
2
1
11
p
T
T
Vp
=5
35
100.1
273
20273
100.5109.0
= 4.19 10–3 m3
The volume of the balloon is 4.19 103 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 = 21
1 TT
p
= 8027340273
10120 3
= 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
11
V
RTn=
1
15.1V
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
22
V
RTn=
1
1
2
4.25.1
V
TR=
1
195.1V
RT
Percentage change =1
12
p
pp 100%
=5.1
5.195.1 100% = 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 bombard-
ments 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
=AmN
RT3
=
2326 1002.6106.5
2732531.83
= 469 m s1
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
3nRT,
increase in total KE
= 258031.81002.6
1028.7
2
323
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
=change for the interval time
momentumin change
=
v
lmv2
2=
l
mv 2
Pressure exerted on surface W due to this
molecule
=area
force=
2
2
ll
mv
=3
2
l
mv
The pressure p due to N molecules
= 222
213
... Nvvvl
m = 2
3vN
l
m
where 2v 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 =AmN
RT3,
the ratio of crms at 80 C to that at 20 C
=
A
A
mN
R
mN
R
)20273(3
)80273(3
= 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
vp 005 =
nR
vp 005
TY =nR
vp 00 33 =
nR
vp 009
TZ =nR
vp 00 5=
nR
vp 005
ZXY TTT
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)
= 2732731.8
510100 3
= 201 mol (1A)
(b) By general gas law,
1
11
T
Vp=
2
22
T
Vp (1M)
27327
510100 3
=2737
1080 23
V
V2 = 5.83 m3 (1A)
3 Root-mean-square speed
=AmN
RT3 (1M)
=
2327 1002.61035.3
2733131.83
= 1940 m s1 (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
21n
n 100% (1M)
=
1
11
2
22
1
RT
VpRT
Vp
100%
=
1
21p
p 100% (1M)
=
3
3
10220
101001 100%
= 54.5% (1A)
6 (a) (i) Decreases (1A)
(ii) Remains unchanged (1A)
(iii) Remains unchanged (1A)
1 Heat and Gases Chapter 5 Gases
New Senior Secondary Physics at Work Oxford University Press 2009
4
(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
1cNm
Root-mean-square speed 2c
=Nm
nRT3 (1M)
=m
n
NRT
3
=AmN
RT3 (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
1cmN (1M)
From (a), nRT = 2
3
1cNm
Rearrange the terms,
2
2
1cmN 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
= nRT2
3 (1M)
= 2737031.85.32
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 = 21
1 TT
V
= 27310027320
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=
201.0π
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
11
p
Vp
=180 103
7.6310100 3
= 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 = 2222
211 ...
1xNNxx vmvmvm
V
= 2xmv
V
N
(instead of 2xvm
V
N) (1A)
Similarly, the total pressure exerted on the
walls in y- and z-directions will become
2ymv
V
N and 2
zmvV
N respectively.
(1A)
As a result we will derive the equation
pV = 2
3
1mcN instead of pV = 2
3
1cNm .
(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.