1
(a)
State two assumptions made about the motion of the molecules in a gas in the derivation
of the kinetic theory of gases equation.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
Use the kinetic theory of gases to explain why the pressure inside a football increases
when the temperature of the air inside it rises. Assume that the volume of the ball remains
constant.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(c)
The ‘laws of football’ require the ball to have a circumference between 680 mm and
700 mm. The pressure of the air in the ball is required to be between 0.60 × 105 Pa and
1.10 × 105 Pa above atmospheric pressure.
A ball is inflated when the atmospheric pressure is 1.00 × 105 Pa and the temperature is 17
°C. When inflated the mass of air inside the ball is 11.4 g and the circumference of the ball
is 690 mm.
Assume that air behaves as an ideal gas and that the thickness of the material used for the
ball is negligible.
Deduce if the inflated ball satisfies the law of football about the pressure.
molar mass of air = 29 g mol–1
(6)
(Total 11 marks)
Page 1 of 78
2
The diagram below shows a number of smoke particles suspended in air. The arrows indicate
the directions in which the particles are moving at a particular time.
(a)
(i)
Explain why the smoke particles are observed to move.
...............................................................................................................
...............................................................................................................
(1)
(ii)
Smoke particles are observed to move in a random way. State two conclusions about
air molecules and their motion resulting from this observation.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(b)
A sample of air has a density of 1.24 kg m–3 at a pressure of 1.01 × 105 Pa and a
temperature of 300 K.
the Boltzmann constant = 1.38 × 10–23 J K–1
(i)
Calculate the mean kinetic energy of an air molecule under these conditions.
(2)
(ii)
Calculate the mean square speed for the air molecules.
(3)
Page 2 of 78
(iii)
Explain why, when the temperature of the air is increased to 320 K, some of the
molecules will have speeds much less than that suggested by the value you
calculated in part (b)(ii).
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 10 marks)
3
Which one of the graphs below shows the relationship between the internal energy of an ideal
gas (y-axis) and the absolute temperature of the gas (x-axis)?
(Total 1 mark)
Page 3 of 78
4
The diagram shows the p-V diagram of an ideal hot-air engine. WX and YZ are isothermal
changes.
Which line of the table below correctly indicates the nature of the work done on or by the air in
each part of the cycle?
WX
XY
YZ
ZW
A
zero
by
zero
on
B
by
zero
on
zero
C
zero
on
zero
by
D
on
zero
by
zero
(Total 1 mark)
5
In the diagram the dashed line X shows the variation of pressure, p, with absolute temperature,
T, for 1 mol of an ideal gas in a container of fixed volume.
Which line, A, B, C or D shows the variation for 2 mol of the gas in the same container?
(Total 1 mark)
Page 4 of 78
6
At a certain temperature, the root-mean-square speed of the molecules of a fixed volume of an
ideal gas is c. The temperature of the gas is changed so that the pressure is halved. The
root-mean-square speed of the molecules becomes
A
B
C
D
2c
(Total 1 mark)
7
The diagram below shows an apparatus used to demonstrate that electron beams possess
energy and momentum.
Electrons are accelerated by the electron gun and travel through the space in the tube which has
been evacuated to a low pressure. They collide with the mica wheel and cause it to rotate.
(a)
(i)
Explain how the electrons would lose energy if the tube were not evacuated.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
Page 5 of 78
(ii)
The pressure of the gas in the tube is 0.20 Pa when the room temperature is 300 K.
The volume of the tube is 3.5 × 10–2 m3. Determine the number of moles of gas in the
tube.
universal gas constant, R = 8.3 J mol–1 K–1
(2)
(iii)
The Avogadro constant is 6.0 × 1023 mol–1.
How many gas molecules are there in the tube?
...............................................................................................................
(1)
(b)
In the electron gun, the electrons of mass 9.1 × 10–31 kg and charge 1.6 × 10–19 C are
accelerated to a speed of 2.7 × 107 m s–1. The beam current is 5.0 mA.
(i)
Calculate, in J, the energy of each electron in the beam.
(1)
(ii)
Use the definition of the volt to determine the potential difference through which the
electrons are accelerated.
(2)
(iii)
How many electrons are accelerated each second?
(2)
(iv)
What happens to the energy of the electrons when they strike the mica target?
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(v)
Determine the force exerted on the paddle wheel by the electrons. Assume that the
electrons are brought to rest when they collide.
(3)
(vi)
The moment of this force about the axle of the paddle causes rotational acceleration
of the paddle. What is the magnitude of this moment?
(2)
(Total 18 marks)
8
Nitrogen at 20°C and a pressure of 1.1 × 105 Pa is held in a glass gas syringe as shown in
Figure 1. The gas, of original volume 8.5 × 10–5 m3, is compressed to a volume of 5.8 × 10–5 m3
by placing a mass on to the plunger of the syringe. The change in pressure of the gas is
adiabatic. The new pressure of the gas is 1.9 × 105 Pa.
Page 6 of 78
Figure 1
(a)
(i)
Calculate the new temperature of the nitrogen. Give your answer in °C.
(3)
(ii)
Calculate the number of moles of nitrogen present in the syringe.
molar gas constant, R = 8.3 J mol–1 K–1
(3)
(iii)
The mass of the nitrogen in the syringe is 1.1 × 10–4 kg. Calculate the mean square
speed of the molecules when the gas has been compressed.
(2)
(iv)
Explain why the change in pressure of the nitrogen is adiabatic.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 7 of 78
(v)
Explain, in terms of the behaviour of the nitrogen molecules, how the gas exerts a
greater pressure than it did before it was compressed.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(b)
After the adiabatic compression, the nitrogen is allowed to cool at constant volume.
Figure 2 shows the variation of pressure with volume for the adiabatic compression and
the subsequent cooling. The dotted line represents the isothermal compression that would
have achieved the same final state.
Figure 2
(i)
Draw arrows on the graph to show the directions of the changes. Label your arrows
adiabatic compression and cooling as appropriate.
(1)
(ii)
State the significance of the shaded area of the graph.
...............................................................................................................
...............................................................................................................
(1)
(Total 15 marks)
Page 8 of 78
9
The graph shows the relation between the product pressure × volume, pV, and temperature, θ, in
degrees celsius for 1 mol of an ideal gas for which the molar gas constant is R.
Which one of the following expressions gives the gradient of this graph?
A
B
C
D
R
(Total 1 mark)
Page 9 of 78
10
X and Y are two gas bottles that are connected by a tube that has negligible volume compared
with the volume of each bottle.
Initially the valve W is closed.
X has a volume 2V and contains hydrogen at a pressure of p.
Y has a volume V and contains hydrogen at a pressure of 2p.
X and Y are both initially at the same temperature.
W is now opened. Assuming that there is no change in temperature, what is the new gas
pressure?
A
B
C
D
(Total 1 mark)
11
A fixed mass of an ideal gas initially has a volume V and an absolute temperature T. Its initial
pressure could be doubled by changing its volume and temperature to
A
V/2 and 4T
B
V/4 and T/2
C
2V and T/4
D
4V and 2T
(Total 1 mark)
Page 10 of 78
12
The diagram shows a p-V graph for a fixed mass of gas. The volume increases from V1 to V2
while the pressure falls from p1 to p2.
Which one of the paths A, B, C or D will result in the greatest amount of work being done by the
gas?
(Total 1 mark)
13
The temperature of a room increases from 283K to 293K. The r.m.s. speed of the air molecules
in the room increases by a factor of
A
1.02
B
1.04
C
1.41
D
2.00
(Total 1 mark)
14
(a)
Define the Avogadro constant.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(b)
(i)
Calculate the mean kinetic energy of krypton atoms in a sample of gas at a
temperature of 22 °C.
mean kinetic energy ........................................... J
(1)
Page 11 of 78
(ii)
Calculate the mean-square speed, (crms)2, of krypton atoms in a sample of gas at a
temperature of 22 °C.
State an appropriate unit for your answer.
mass of 1 mole of krypton = 0.084 kg
mean-square speed..................................... unit ..........................
(3)
(c)
A sample of gas consists of a mixture of krypton and argon atoms.
The mass of a krypton atom is greater than that of an argon atom.
State and explain how the mean-square speed of krypton atoms in the gas compares with
that of the argon atoms at the same temperature.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 7 marks)
15
(a)
Outline what is meant by an ideal gas.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 12 of 78
(b)
An ideal gas at a temperature of 22 °C is trapped in a metal cylinder of volume 0.20 m3 at a
pressure of 1.6 × 106 Pa.
(i)
Calculate the number of moles of gas contained in the cylinder.
number of moles ....................................... mol
(2)
(ii)
The gas has a molar mass of 4.3 × 10–2 kg mol–1.
Calculate the density of the gas in the cylinder.
State an appropriate unit for your answer.
density ......................... unit ..............
(3)
(iii)
The cylinder is taken to high altitude where the temperature is −50 °C and the
pressure is 3.6 × 104 Pa. A valve on the cylinder is opened to allow gas to escape.
Calculate the mass of gas remaining in the cylinder when it reaches equilibrium with
its surroundings.
Give your answer to an appropriate number of significant figures.
mass .......................................... kg
(3)
(Total 10 marks)
Page 13 of 78
16
The graph in the figure below shows the best fit line for the results of an experiment in which the
volume of a fixed mass of gas was measured over a temperature range from 20°C to 100°C. The
pressure of the gas remained constant throughout the experiment.
(a)
Use the graph in the figure to calculate a value for the absolute zero of temperature in °C.
Show clearly your method of working.
(4)
Page 14 of 78
(b)
Use data from the graph in the figure to calculate the mass of gas used in the experiment.
You may assume that the gas behaved like an ideal gas throughout the experiment.
gas pressure throughout the experiment
=
1.0 × 105 Pa
molar gas constant
=
8.3 J mol–1 K–1
molar mass of the gas used
=
0.044 kg mol–1
(5)
(Total 9 marks)
Page 15 of 78
17
When a deuterium nucleus and a tritium nucleus overcome their Coulomb barrier and fuse
together they may be considered as charged spheres in contact. The constant ro which relates
the nuclear radius, R, to the cube root of the mass number A may be assumed to be 1.3 fm.
(a)
(i)
Calculate the radius of the deuterium nucleus and, R, the radius of the tritium nucleus
before fusion.
RD ......................................................................................................
............................................................................................................
RT .......................................................................................................
............................................................................................................
(ii)
Calculate the minimum energy, in MeV, which must be supplied to the deuterium
nucleus and the tritium nucleus when they fuse together.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
(b)
Estimate the temperature at which deuterium and tritium nuclei would have enough kinetic
energy to undergo fusion.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 8 marks)
Page 16 of 78
18
Figure 1 shows a p–V graph that you are to use to illustrate the process of a gas undergoing two
changes.
In its initial state, the gas has a pressure of 50 kPa and a volume of 1.5 m3; this is plotted on the
graph.
First, the gas undergoes an isothermal change from an initial volume of 1.5 m3 to 0.85 m3
followed by a compression at constant pressure to a volume of 0.35 m3.
Figure 1
(a)
Show that the final pressure of the gas is about 90 kPa.
(2)
(b)
Complete the graph in Figure 1 to show both changes.
(2)
Page 17 of 78
(c)
(i)
Use your graph to estimate the work done during the whole process.
(3)
(ii)
State and explain whether the work in part (c)(i) is done on or by the gas.
...............................................................................................................
...............................................................................................................
(1)
(Total 8 marks)
19
The pressure exerted by an ideal gas in a container of volume 1.2 × 10–5 m3 is 1.5 × 105 Pa at a
temperature of 50°C.
(a)
Calculate the number of molecules of gas in the container.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(b)
The pressure of the gas is measured at different temperatures whilst the volume of the
container and the mass of the gas remain constant. Draw a graph on the grid to show how
the pressure varies with the temperature.
(3)
Page 18 of 78
(c)
The container described in part (a) has a release valve that allows gas to escape when the
pressure exceeds 2.0 × 105 Pa. Calculate the number of gas molecules that escape when
the temperature of the gas is raised to 300°C.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 9 marks)
20
Most cars manufactured in recent years have been fitted with air bags as a safety feature. In a
collision the bag inflates automatically to protect the driver as air is released from a compressed
air cylinder.
(a)
Explain why the driver would be less seriously injured in a collision if the air bag inflates
than he would be if unrestrained.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(5)
(b)
Why is the driver of a car fitted only with seat belts more likely to be injured than if an air
bag was fitted? Ignore the different deceleration times and the difference in the materials
used.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 19 of 78
(c)
With reference to pressure, volume and temperature, discuss what happens to the air as
the bag inflates.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 10 marks)
21
(a)
(i)
State what is meant by thermal equilibrium.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Explain thermal equilibrium by reference to the behaviour of the molecules when a
sample of hot gas is mixed with a sample of cooler gas and thermal equilibrium is
reached.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(b)
A sealed container holds a mixture of nitrogen molecules and helium molecules at a
temperature of 290 K. The total pressure exerted by the gas on the container is 120 kPa.
(i)
molar mass of helium
=
4.00 × 10–3 kg mol–1
molar gas constant R
the Avogadro constant NA
=
=
8.31 J K–1 mol–1
6.02 × 1023 mol–1
Calculate the root mean square speed of the helium molecules.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 20 of 78
(ii)
Calculate the average kinetic energy of a nitrogen molecule.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(iii)
If there are twice as many helium molecules as nitrogen molecules in the container,
calculate the pressure exerted on the container by the helium molecules.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(Total 9 marks)
22
(a)
Use the kinetic theory of gases to explain why
(i)
the pressure exerted by an ideal gas increases when it is heated at constant volume.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
the volume occupied by an ideal gas increases when it is heated at constant
pressure.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(b)
A quantity of 0.25 mol of air enters a diesel engine at a pressure of 1.05 × 105 Pa and a
temperature of 27°C. Assume the gas to be ideal.
(i)
Calculate the volume occupied by the gas.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 21 of 78
(ii)
When the gas is compressed to one twentieth of its original volume the pressure rises
to 7.0 × 106 Pa. Calculate the temperature of the gas immediately after the
compression.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(Total 8 marks)
23
(a)
The diagram shows curves (not to scale) relating pressure p, and volume, V, for a fixed
mass of an ideal monatomic gas at 300K and 500K. The gas is in a container which is
closed by a piston which can move with negligible friction.
molar gas constant, R, = 8.31 J mol–1 K–1
(i)
Show that the number of moles of gas in the container is 6.4 × 10–2.
...............................................................................................................
...............................................................................................................
(ii)
Calculate the volume of the gas at point C on the graph.
...............................................................................................................
...............................................................................................................
(3)
Page 22 of 78
(b)
(i)
Give an expression for the total kinetic energy of the molecules in one mole of an
ideal gas at kelvin temperature T.
...............................................................................................................
(ii)
Calculate the total kinetic energy of the molecules of the gas in the container at point
A on the graph.
Explain why this equals the total internal energy for an ideal gas.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(c)
Defining the terms used, explain how the first law of thermodynamics, ΔQ = ΔU + ΔW,
applies to the changes on the graph
(i)
at constant volume from A to B,
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
at constant pressure from A to C.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
(d)
Calculate the heat energy absorbed by the gas in the change
(i)
from A to B,
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 23 of 78
(ii)
from A to C
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(Total 15 marks)
24
Figure 1
Figure 1 shows a cylinder, fitted with a gas-tight piston, containing an ideal gas at a constant
temperature of 290 K. When the pressure, p, in the cylinder is 20 × 104 Pa the volume, V, is
0.5 × 10–3 m3.
Page 24 of 78
Figure 2 shows this data plotted.
Figure 2
(a)
By plotting two or three additional points draw a graph, on the axes given in Figure 2, to
show the relationship between pressure and volume as the piston is slowly pulled out.
The temperature of the gas remains constant.
(3)
(b)
(i)
Calculate the number of gas molecules in the cylinder.
answer = ...................................... molecules
(2)
Page 25 of 78
(ii)
Calculate the total kinetic energy of the gas molecules.
answer = .................................................... J
(3)
(c)
State four assumptions made in the molecular kinetic theory model of an ideal gas.
(i)
.............................................................................................................
.............................................................................................................
(ii)
.............................................................................................................
.............................................................................................................
(iii)
.............................................................................................................
.............................................................................................................
(iv)
.............................................................................................................
.............................................................................................................
(4)
(Total 12 marks)
25
(a)
State two quantities which increase when the temperature of a given mass of gas is
increased at constant volume.
(i)
................................................................................................................
(ii)
...............................................................................................................
(2)
(b)
A car tyre of volume 1.0 × 10–2 m3 contains air at a pressure of 300 kPa and a temperature
of 290K. The mass of one mole of air is 2.9 × 10–2 kg. Assuming that the air behaves as an
ideal gas, calculate
(i)
n, the amount, in mol, of air,
...............................................................................................................
...............................................................................................................
Page 26 of 78
(ii)
the mass of the air,
...............................................................................................................
...............................................................................................................
(iii)
the density of the air.
...............................................................................................................
...............................................................................................................
(5)
(c)
Air contains oxygen and nitrogen molecules. State, with a reason, whether the following are
the same for oxygen and nitrogen molecules in air at a given temperature.
(i)
The average kinetic energy per molecule
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
The r.m.s. speed
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(Total 11 marks)
Page 27 of 78
26
(a)
Write down four assumptions about the properties and behaviour of gas molecules which
are used in the kinetic theory to derive an expression for the pressure of an ideal gas.
Assumption 1 .................................................................................................
........................................................................................................................
Assumption 2 .................................................................................................
........................................................................................................................
Assumption 3 .................................................................................................
........................................................................................................................
Assumption 4 .................................................................................................
........................................................................................................................
(4)
(b)
(i)
A cylinder, fitted with a pressure gauge, contains an ideal gas and is stored in a cold
room. When the cylinder is moved to a warmer room the pressure of the gas is seen
to increase. Explain in terms of the kinetic theory why this increase in pressure is
expected.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
After a time, the pressure of the gas stops rising and remains steady at its new value.
The air temperature in the warmer room is 27°C. Calculate the mean kinetic energy
of a gas molecule in the cylinder.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(Total 10 marks)
27
(a)
(i)
Write down the equation of state for n moles of an ideal gas.
.............................................................................................................
Page 28 of 78
(ii)
The molecular kinetic theory leads to the derivation of the equation
pV =
,
where the symbols have their usual meaning.
State three assumptions that are made in this derivation.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(b)
Calculate the average kinetic energy of a gas molecule of an ideal gas at a temperature
of 20 °C.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(c)
Two different gases at the same temperature have molecules with different mean square
speeds.
Explain why this is possible.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 9 marks)
Page 29 of 78
28
(a)
A cylinder of fixed volume contains 15 mol of an ideal gas at a pressure of 500 kPa and a
temperature of 290 K.
(i)
Show that the volume of the cylinder is 7.2 × 10–2 m3.
.............................................................................................................
.............................................................................................................
(ii)
Calculate the average kinetic energy of a gas molecule in the cylinder.
.............................................................................................................
.............................................................................................................
(4)
(b)
A quantity of gas is removed from the cylinder and the pressure of the remaining gas falls
to 420 kPa. If the temperature of the gas is unchanged, calculate the amount, in mol, of gas
remaining in the cylinder.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(c)
Explain in terms of the kinetic theory why the pressure of the gas in the cylinder falls when
gas is removed from the cylinder.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(4)
(Total 10 marks)
29
(a)
The air in a room of volume 27.0 m3 is at a temperature of 22 °C and a pressure of
105 kPa.
Calculate
(i)
the temperature, in K, of the air,
.............................................................................................................
Page 30 of 78
(ii)
the number of moles of air in the room,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
the number of gas molecules in the room.
.............................................................................................................
.............................................................................................................
(5)
(b)
The temperature of an ideal gas in a sealed container falls. State, with a reason, what
happens to the
(i)
mean square speed of the gas molecules,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
pressure of the gas.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 9 marks)
Page 31 of 78
30
A life jacket inflates using gas released from a small carbon dioxide cylinder. The arrangement is
shown in the following figure.
(a)
The cylinder initially contains 1.7 × 1023 molecules of carbon dioxide at a temperature of
12 °C and occupying a volume of 3.0 × 10−5 m3.
(i)
Calculate the initial pressure, in Pa, in the carbon dioxide cylinder.
(ii)
When the life jacket inflates, the pressure falls to 1.9 × 105 Pa and the final
temperature is the same as the initial temperature. Calculate the new volume of the
gas.
(iii)
Calculate the mean molecular kinetic energy, in J, of the carbon dioxide in the
cylinder.
(6)
(b)
(i)
Explain, in terms of the kinetic theory model, why the pressure drops when the
carbon dioxide is released into the life jacket.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 32 of 78
(ii)
Explain why the kinetic theory model would apply more accurately to the gas in the
inflated life jacket compared with the gas in the small cylinder.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(6)
(c)
Explain, in terms of the first law of thermodynamics, how the temperature of the gas in the
system can be the same at the beginning and the end of the process.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(4)
(Total 16 marks)
Page 33 of 78
31
A diffraction grating was used to measure the wavelength of a certain line of a line emission
spectrum.
(a)
The grating had 600 lines per millimetre. The angle of diffraction of the second order line
was 35.8°.
(i)
Calculate the wavelength of this line.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the energy, in eV, of a photon of this wavelength.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
Page 34 of 78
(b)
The line emission spectrum observed in part (a) was produced by a hot gas.
(i)
The energy level diagram for the atoms that produced the line spectrum is shown in
the diagram below. Mark on the diagram a vertical arrow to show the electron
transition between the two levels that produced photons of energy 6.8 eV.
(ii)
The temperature of the gas was 5000K. Show that the mean kinetic energy of a gas
atom at this temperature is 0.65 eV.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 35 of 78
(iii)
Describe how the atoms of a gas produce a line emission spectrum and explain why
the gas at a temperature of 5000K can produce a line of the wavelength calculated in
part (a)(i).
You may be awarded marks for the quality of written communication in your answer.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(Total 11 marks)
32
(a)
(i)
One of the assumptions of the kinetic theory of gases is that molecules make elastic
collisions. State what is meant by an elastic collision.
.............................................................................................................
.............................................................................................................
(ii)
State two more assumptions that are made in the kinetic theory of gases.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
Page 36 of 78
(b)
One mole of hydrogen at a temperature of 420 K is mixed with one mole of oxygen at
320 K. After a short period of time the mixture is in thermal equilibrium.
(i)
Explain what happens as the two gases approach and then reach thermal
equilibrium.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Calculate the average kinetic energy of the hydrogen molecules before they are
mixed with the oxygen molecules.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 7 marks)
Page 37 of 78
33
The graph shows how the pressure of an ideal gas varies with its volume when the mass and
temperature of the gas are constant.
(a)
On the same axes, sketch two additional curves A and B, if the following changes are
made.
(i)
The same mass of gas at a lower constant temperature (label this A).
(ii)
A greater mass of gas at the original constant temperature (label this B).
(2)
(b)
A cylinder of volume 0.20 m3 contains an ideal gas at a pressure of 130 kPa and a
temperature of 290 K. Calculate
(i)
the amount of gas, in moles, in the cylinder,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
the average kinetic energy of a molecule of gas in the cylinder,
.............................................................................................................
Page 38 of 78
(iii)
the average kinetic energy of the molecules in the cylinder.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(Total 7 marks)
34
The number of molecules in one cubic metre of air decreases as altitude increases. The table
shows how the pressure and temperature of air compare at sea-level and at an altitude of
10 000 m.
(a)
altitude
pressure/Pa
temperature/K
sea-level
1.0 × 105
300
10 000 m
2.2 × 104
270
Calculate the number of moles of air in a cubic metre of air at
(i)
sea-level,
.............................................................................................................
.............................................................................................................
(ii)
10 000 m.
.............................................................................................................
.............................................................................................................
(3)
(b)
In air, 23% of the molecules are oxygen molecules. Calculate the number of extra oxygen
molecules there are per cubic metre at sea-level compared with a cubic metre of air at an
altitude of 10 000 m.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 5 marks)
Page 39 of 78
35
The pressure inside a bicycle tyre of volume 1.90 × 10–3 m3 is 3.20 × 105 Pa when the
temperature is 285 K.
(i)
Calculate the number of moles of air in the tyre.
answer = ................................... mol
(1)
(ii)
After the bicycle has been ridden the temperature of the air in the tyre is 295 K.
Calculate the new pressure in the tyre assuming the volume is unchanged.
Give your answer to an appropriate number of significant figures.
answer = ...................................... Pa
(3)
(b)
Describe one way in which the motion of the molecules of air inside the bicycle tyre is
similar and one way in which it is different at the two temperatures.
similar ............................................................................................................
........................................................................................................................
different ..........................................................................................................
........................................................................................................................
(2)
(Total 6 marks)
Page 40 of 78
36
In the research into nuclear fusion one of the most promising reactions is between deuterons
and tritium nuclei
in a gaseous plasma. Although deuterons can be relatively easily
extracted from sea water, tritium is difficult to produce. It can, however, be produced by
bombarding lithium-6
with neutrons. The two reactions are summarised as:
+ energy
+ energy
Masses of reactants:
= 1.008665u
= 2.013553u
= 3.016049u
= 4.002603u
= 6.015122u
1u is equivalent to 1.66 × 10–27 kg or 931 MeV
(a)
(i)
Explain why the atomic mass unit, u, may be quoted in kg or MeV.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(2)
Page 41 of 78
(ii)
Calculate the maximum amount of energy, in MeV, released when 1.0 kg of lithium-6
is bombarded by neutrons.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
energy released ...................................... MeV
(5)
(iii)
Suggest why the lithium-6 reaction could be thought to be self-sustaining once the
deuteron-tritium reaction is underway.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(1)
(b)
(i)
In order to fuse, a deuteron and a tritium nucleus must approach one another to
within approximately 1.5 × 10–15 m.
Calculate the minimum total initial kinetic energy that these nuclei must have.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
minimum total kinetic energy of nuclei ............................................. J
(3)
Page 42 of 78
(ii)
Show that a temperature of approximately 4 × 109 K would be sufficient to enable this
fusion to occur in a gaseous plasma.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(iii)
Explain in terms of the forces acting on nuclei why the deuteron-tritium mixture must
be so hot in order to achieve the fusion reaction.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 18 marks)
Page 43 of 78
Mark schemes
1
(a)
The molecules (continually) move about in random motion✓
Collisions of molecules with each other and with the walls are elastic✓
Time in contact is small compared with time between collisions✓
The molecules move in straight lines between collisions✓
ANY TWO
Allow reference to ‘particles interact according to Newtonian
mechanics’
2
(b)
Ideas of pressure = F / A and F = rate of change of momentum✓
Mean KE / rms speed / mean speed of air molecules increases✓
More collisions with the inside surface of the football each second✓
Allow reference to ‘Greater change in momentum for each collision’
3
(c)
Radius = 690 mm / 6.28) = 110 mm or T = 290 K ✓ seen
volume of air = 5.55 × 10-3 m3✓
n × 29(g) = 11.4 (g)✓ n = 0.392 mol
Use of pV = nRT =
✓
p = 1.70 × 105 Pa ✓
Conclusion: Appropriate comparison of their value for p with the requirement of the
rule, ie whether their pressure above 1 × 105 Pa falls within the required band✓
Allow ecf for their n V and T✓
6
[11]
Page 44 of 78
2
(a)
(i)
collisions with/bombardment by air molecules
(condone particles)
B1
1
(ii)
motion of air molecules (“they are“) random
(in all directions)
B1
fast moving
B1
max 2
air molecules small or much smaller than smoke
particles
B1
(b)
(i)
3/2kT or substituted values (independent of powers)
do not allow all equations written
C1
6.21 × 10–21 J
A1
2
(ii)
pV = 1/3 Nm<c2>
C1
relates Nm/V to ρ
C1
2.4 × 105 m2s–2
A1
(allow compensation of ½ m<c2> for 1)
3
Page 45 of 78
(iii)
there will be a range of speeds
B1
there will be molecules with lower speeds
than mean /average
means higher and lower values
B1
2
[10]
3
4
5
6
7
A
[1]
B
[1]
A
[1]
C
[1]
(a)
(i)
they would collide with atoms of gas
M1
losing energy by:
ionising the gas
A1
exciting gas molecules
A1
allow 1 for stating inelastic collisions
(3)
(ii)
pV = nRT
C1
n = 2.8 × 10–6
A1
(2)
(iii)
1.7 × 1018
B1
(their (ii) × 6.0 × 1023)
(1)
(b)
(i)
energy ½mv2 = 3.3 × 10–16 J
B1
(1)
Page 46 of 78
(ii)
1 volt = 1 Joule per coulomb
C1
or p.d. = energy gained by electron / charge on electron
2070 V or 2100 V
A1
(2)
(iii)
number of electrons = current / e
C1
3.1 × 1016
A1
(2)
(iv)
it produces heating of the mica
B1
it causes (rotation) kinetic energy of the mica paddle
B1
(2)
(v)
force = change in momentum per second
C1
change in momentum per collision = 2.46 × 10–23 (N s)
C1
force = their (iii) × their momentum (7.6 × 10–7 N)
A1
(3)
(vi)
moment = force × perpendicular distance
C1
1.9 × 10–8 N m
A1
(2)
[18]
8
(a)
(i)
pV / T = constant in any form
C1
correct substitution including absolute temperatures / 345K
C1
72°C not 345K
condone no unit and condone just °
A1
(3)
Page 47 of 78
(ii)
pV = nRT
C1
correct substitution: n =
or
C1
3.8(5) × 10–3 (mol) or 3.8(4) × 10–3 (mol)
A1
e.c.f. for their (i)
(3)
(iii)
pV = Nm<c2 or p = ρ<c2
C1
3.0 × 105 m2 s–2 condone subsequent calculation of rms speed
A1
(2)
(iv)
no heat transfer / ΔQ = 0 / no energy loss
B1
process too quick (for conduction to take place) /
glass is poor (thermal) conductor / the system is isolated
B1
(2)
(v)
molecules move faster / have more KE
B1
greater number of collisions (per second) (between molecules and wall)
not between molecules
B1
greater (rate of) change in momentum in each collision
B1
(3)
(b)
(i)
anticlockwise arrows correctly labelled - both arrows needed
B1
(3)
(ii)
work done on the gas (during compression)
B1
(1)
[15]
9
10
11
D
[1]
C
[1]
B
[1]
Page 48 of 78
12
13
14
B
[1]
A
[1]
(a)
the number of atoms in 12g of carbon-12
or the number of particles / atoms / molecules in one mole of substance ✓
not – NA quoted as a number
1
(b)
(i)
mean kinetic energy ( = 3 / 2 kT) = 3 / 2 × 1.38 × 10−23 × (273 + 22)
= 6.1 × 10−21 (J) ✓
6 × 10−21 J is not given mark
1
(ii)
mass of krypton atom
= 0.084 / 6.02 × 10+23 ✓
( = 1.4 × 10−25 kg)
( = 2 × mean kinetic energy / mass
= 2 × 6.1 × 10−21 / 1.4 × 10−25)
= 8.7 - 8.8 × 104 ✓
m2 s−2 or J kg−1 ✓
1st mark is for the substitution which will normally be seen within a
larger calculation.
allow CE from (i)
working must be shown for a CE otherwise full marks can be given
for correct answer only
no calculation marks if mass has a physics error i.e. no division by
NA note for CE
answer = (i) × 1.43 × 1025
3
Page 49 of 78
(c)
(at the same temperature) the mean kinetic energy is the same
or
gases have equal
or
mass is inversely proportional to mean square speed / m ∝ 1
✓
or mean square speed of krypton is less ✓
1st mark requires the word mean / average or equivalent in an
algebraic term
2nd mark ‘It’ will be taken to mean krypton. So, ‘It is less’ can gain a
mark
allow ‘heavier’ to mean more massive’
allow vague statements like speed is less for 2nd mark but not in
the first mark
2
[7]
15
(a)
molecules have negligible volume
collisions are elastic
the gas cannot be liquified
there are no interactions between molecules (except during collisions)
the gas obeys the (ideal) gas law / obeys Boyles law etc.
at all temperatures/pressures
any two lines
a gas laws may be given as a formula
2
(b)
(i)
n (= PV / RT) = 1.60 × 106 × 0.200 / (8.31 × (273 + 22))
(130.5 mol)
= 130 or 131 mol
2
(ii)
mass = 130.5 × 0.043 = 5.6 (kg)
(5.61kg)
allow ecf from bi
density (= mass / volume) = 5.61 / 0.200 = 28
(28.1 kg m−3)
kg m−3
a numerical answer without working can gain the first two marks
3
Page 50 of 78
(iii)
(V2 = P1 V1 T2 / P2 T1)
V2 = 1.6 × 106 × .200 × (273 – 50) / 3.6 × 104 × (273 + 22) or 6.7(2) (m3)
allow ecf from bii
[reminder must see bii]
look out for
mass remaining = 5.61 × 0.20 / 6.72 = 0.17 (kg)
or
(0.167 kg)
n = (PV / RT = 3.6 × 104 × 0.200 / (8.31 × (273 − 50)) = 3.88(5) (mol)
mass remaining = 3.885 × 4.3 × 10−2 = 0.17 (kg)
2 sig figs
any 2 sf answer gets the mark
3
[10]
16
(a)
0.001 (cm2 °C−1)
m = gradient = 0.035
B1
c = intercept = 10.5
0.1 (cm2)
B1
absolute zero = -c/m
C1
= -300
10 °C some relevant working must be seen
A1
4
or
a meaningful attempt to use similar triangles
M1
an accurate value for the base found
C1
adjustment to the base (if appropriate)
C1
answer = -300
10 °C
A1
4
Page 51 of 78
or
relevant use of V1/T1 = V2/ T2
M1
valid introduction of unknown temperature
A1
consistent solution of the equation
C1
answer = -300
10 °C
C1
4
(b)
statement of or use of pV = nRT
C1
accurate reading from the line
C1
Celsius converted to Kelvin (+273 or + answer to part (a))
C1
n = 4.52 × 10–4 or 4.11 × 10–4 or 4.2 × 10–4 (∆V/∆T) (mol)
A1
m = n × 0.044 (m = 4.52 × 0.044 = 2.0 × 10–5 kg)
B1
5
[9]
17
(a)
(i)
RD = 1.3 × 21/3 = 1.64 fm (1) RT = 1.3 × 31/3 = 1.64 fm (1)
Page 52 of 78
(ii)
energy at ‘contact’ =
(1)
= 6.56 × 10–14 J (1)
= 4.10 MeV (1)
(max 5)
(b)
energy of nucleus = 3/2 kT (1)
6.56 × 10–14 = 3/2 × 1.38 × 10–23 × T (1)
gives T = 3.2 × 109 K (1) (marks available for alternative
sensible use of energy data)
reference to range of speeds (or energies) of nuclei (or atoms) (1)
(max 3)
[8]
18
(a)
pV = constant seen
C1
p = 88 kPa
A1
2
(b)
completes correct shape curve to (0.85,88 000 or 90000),
B1
then horizontal to 0.35 m3
B1
2
(c)
attempts to measure area [graph evidence or words]
C1
correct use of graph scale
C1
answer in range (80 – 91) Kj
A1
3
Page 53 of 78
(d)
done on gas because it is compressed
B1
1
[8]
19
(a)
(1) = (6.71 × 104 mol)
number of molecules = nNA (1) = 6.71 × 10–4 × 6.02 × 1023
= 4.04 × 1020 (1)
[or equivalent solution using pv = NkT]
(3)
(b)
straight line with positive gradient (1)
through (50, 1.5) (1)
crosses temperature axis between –250 and –300°C (1)
(3)
(c)
number of moles left in container after valve opens
(1) (= 5.04 × 10–4 mol)
∴ number of molecules left in container = 5.04 × 10–4 × 6.02 × 1023
= 3.03 × 1020 (1)
∴ number of molecules that escape
= 4.04 × 1020 – 3.03 × 1020
= 1.01 × 1020 (1)
Page 54 of 78
[alternative (c)
∴ number of moles that escape (= 6.71 × 10–4 – 5.04 × 10–4)
= 1.67 × 10–4
∴ number of molecules that escape
= 1.67 × 10–4 × NA
= 1.01 × 1020 (1)]
(3)
[9]
20
(a)
moving driver has momentum (1)
in sudden impact momentum must be lost in v. short time (1)
F = Δ(mv) / Δt (or F = ma) (1)
air bag increases stopping contact time (1)
hence reduces force (or reduces force by decreasing the deceleration) (1)
(max 5)
(b)
seat belt applies force to smaller area than air bag (1)
causing greater pressure on parts of body (1)
(2)
(c)
air initially at v. high pressure occupies small volume (1)
on expansion volume increases and pressure decreases (1)
temperature of gas decreases as it expands (1)
pressure caused by momentum change in molecular collisions (1)
(max 3)
[10]
21
(a)
(i)
no net flow of (thermal) energy (between two or more bodies) (1)
bodies at same temperature (1)
(ii)
(kinetic) energy is exchanged in molecular collisions (1)
until average kinetic energy of all molecules is the same (1)
max 3
Page 55 of 78
(b)
(i)
(1)
= 1340m s–1 (1)
(ii)
average k.e. of nitrogen molecules = average k.e. of helium molecules (1)
=
×
(1340)2 = 5.97× 10–21 J (1)
alternative schemes for (ii):
average k.e. =
=
kT (1)
× 1.38 × 10–23 ×290
= 6.00 × 10–21 J (1)
or
average k.e. =
=
(1)
×
= 6.00 × 10–21 J (1)
(iii)
use of p =
(1)
pHe =
or equivalent [or, at same temperature, p ∝ no. of molecules]
× 120 = 80 kPa (1)
6
[9]
22
(a)
(i)
more collisions (with wall) per second (1)
and more momentum change per collision (1)
greater force because more momentum change per second (greater pressure) (1)
(ii)
(same pressure) faster molecules so more momentum change per collision (1)
greater volume, fewer collisions per second [or greater volume
to maintain same pressure] (1)
max 4
(b)
(i)
(1) = 5.94 × 10–3 m3 (1)
Page 56 of 78
(ii)
(1)
= 1000 K (727 °C) (1)
4
[8]
23
(a)
(i)
correct p and V from graph (1)
(ii)
V2 = V1
= 3.3 × 10–3 m3 (1)
(3)
(b)
(i)
(ii)
RT or
NAkT (1)
total kinetic energy
= 1.5 × 8.3 × 0.064 × 300 (1) = 239 J (1)
molecules have no potential energy (1)
no attractive forces [or elastic collisions occur] (1)
(max 4)
(c)
ΔQ = heat entering (or leaving) gas
ΔU = change (or increase) in internal energy
ΔW = work done
[(1) (1) for three definitions, deduct one for each incorrect or missing]
(i)
ΔQ = ΔU (1)
temperature rises but no work done (1)
(ii)
ΔQ = ΔU +ΔW (1)
temperature rises and work done in expanding (1)
(max 5)
(d)
(i)
ΔU =
nR(500 – 300) = 159 J (1) (= ΔQ)
(ii)
pΔV = 8.0 × 104 × (3.3 – 2.0) × 10–3 = 104 J (1)
∴ΔQ = ΔU + pΔV = 263 J (1)
(3)
[15]
Page 57 of 78
24
(a)
3
curve with decreasing negative gradient that passes through the given point
which does not touch the x axis (1)
designated points
pressure/104 Pa
volume/10–3 m3
10
1.0
5.0
2.0
4.0
2.5
2.5
4.0
2 of the designated points (1)(1) (one mark each)
(b)
(i)
N = PV/kT = 5 × 104 × 2 × 10–3/1.38 × 10–23 × 290 (1)
[or alternative use of PV = nRT
5 × 104 × 2.0 × 10–3/8.31 × 290 = 0.0415 moles]
= 2.50 × 1022 molecules (1)
2
Page 58 of 78
(ii)
(mean) kinetic energy of a molecule
(1) (= 6.00 × 10–21 J)
=
(total kinetic energy = mean kinetic energy × N)
= 6.00 × 10–21 × 2.50 × 1022 (1)
= 150 (J) (1)
3
(c)
all molecules/atoms are identical
molecules/atoms are in random motion
Newtonian mechanics apply
gas contains a large number of molecules
the volume of gas molecules is negligible (compared to the volume
occupied by the gas) or reference to point masses
no force act between molecules except during collisions or the
speed/velocity is constant between collisions or motion is in a
straight line between collisions
collisions are elastic or kinetic energy is conserved
and of negligible duration
any 4 (1)(1)(1)(1)
max 4
[12]
25
(a)
(i)
pressure (1)
(ii)
(average) kinetic energy
[or rms speed] (1)
(2)
(b)
(i)
pV = nRT (1)
= 1.20 (mol) (1) (1.24 mol)
(ii)
mass of air = 1.24 × 29 × 10–3 = 0.036 kg (1)
(allow e.c.f from(i))
Page 59 of 78
(iii)
(5)
(c)
(i)
same (1)
because the temperature is the same (1)
The Quality of Written Communication marks were awarded primarily for the
quality of answers to this part.
(ii)
different (1)
because the mass of the molecules are different (1)
(4)
[11]
26
(a)
number of molecules in a gas is very large
duration of collision much less than time between collisions
total volume of molecules small compared with gas volume
molecules are in random motion
collisions are (perfectly) elastic
there are no forces between molecules (any four) (4)
(any 4)
(b)
(i)
heat (energy) transferred to gas from warmer air outside
mean kinetic energy of gas molecules increases
or molecules move faster
momentum of molecules increases
more collisions per second
each collision (with container walls) transfers more momentum
force (per unit area) on container wall increases (any four) (4)
The Quality of Written Communication marks were awarded primarily for the
quality of answers to this part.
(ii)
T = 273 + 27 = 300K (1)
mean kinetic energy =
= 1.5 × 1.38 × 10–23 × 300 = 6.2 × 10–21 J (1)
(allow e.c.f. for incorrect T)
(6)
[10]
Page 60 of 78
27
(a)
(i)
pV = nRT (1)
(ii)
all particles identical or have same mass (1)
collisions of gas molecules are elastic (1)
inter molecular forces are negligible (except during collisions) (1)
volume of molecules is negligible (compared to volume of
container) (1)
time of collisions is negligible (1)
motion of molecules is random (1)
large number of molecules present
(therefore statistical analysis applies) (1)
monamatic gas (1)
Newtonian mechanics applies (1)
max 4
(b)
Ek =
or
=
(1)
(1)
= 6.1 × 10–21 J (1)
(6.07 × 10–21 J)
3
(c)
masses are different (1)
hence because Ek is the same,
mean square speeds must be different (1)
2
[9]
28
(a)
(i)
p V = nR T (1)
(1) (gives V = 7.2 × 10–2m3)
V=
(ii)
(use of Ek =
kT gives) Ek =
× 1.38 × 10–23 × 290 (1)
= 6.0 × 10–21 (J) (1)
4
(b)
(use of pV = nRT gives)
[or use p
n]
(1)
n = 13 moles (1) (12.5 moles)
2
Page 61 of 78
(c)
pressure is due to molecular bombardment [or moving molecules] (1)
when gas is removed there are fewer molecules in the cylinder
[or density decreases] (1)
(rate of) bombardment decreases (1)
molecules exert forces on wall (1)
is constant (1)
Nm (c2) (1)
[or pV =
V and m constant (1)
(c2) constant since T constant (1)
p
N (1)]
[or p =
p(c2) (1)
explanation of ρ decreasing (1)
(c2) constant since T constant (1)
p (c2) ρ (1)]
max 4
[10]
29
(a)
(i)
T (=273 + 22) = 295 (K) (1)
(ii)
pV = nRT (1)
105 × 103 × 27 = n × 8.31 × 295 (1)
n = 1160 (moles) (1)
(1156 moles)
(allow C.E. for T (in K) from (i)
(iii)
N = 1156 × 6.02 × 1023 = 7.0 × 1026 (1)
(6.96 × 1026)
5
(b)
(i)
decreases (1)
because temperature depends on mean square speed (or
[or depends on mean Ek] (1)
)
Page 62 of 78
(ii)
decreases (1)
as number of collisions (per second) falls (1)
rate of change of momentum decreases (1)
[or if using pV = nRT
decreases (1)
as V constant (1)
as n constant (1)]
[or if using p = 1/3ρ
decrease (1)
as ρ is constant (1)
as
is constant (1)]
max 4
[9]
30
(a)
(i)
PV = NkT (1)
223 × 105 Pa (1)
2
(ii)
pV = const or repeat calculation from (i) (1)
3.5 × 10-3 m3 (1)
2
(iii)
kinetic energy = 3/2 kT (1)
5.9(0) × 10-21 J (1)
2
(b)
(i)
volume increase (1)
)
time between collisions increases (1)
)
speed constant as temp constant (1)
)
rate of change of momentum decreases (1)
)
max 3
(ii)
volume smaller in cylinder (1)
molecules occupy significantly greater proportion of the volume (1)
molecules closer so intermolecular forces greater (1)
3
Page 63 of 78
(c)
internal energy stays the same (1)
gas does work in expanding so W is negative (1)
gas must be heated to make U positive (1)
U and W equal and opposite (1)
4
[16]
31
(a)
(i)
(use of d sin θ = nλ gives)
2λ = d sin 35.8° (1)
(= 1.67 × 10–6)
= 4.9 × 10–7m (1)
(4.87 × 10–7m)
(ii)
E (= hf = 6.63 × 10–34 × 6.16 × 1014) = 4.1 × 10–19(J) (1) (4.0(8) × 10–19(J))
= 2.6 (eV) (1)
(2.55 (eV)
(for E = 4.1 × 10–19(J) = 2.56 (eV)
5
(b)
(i)
from C to A (1)
(ii)
(use of Ek = 3 / 2kT gives)
Ek = 1.5 × 1.38 × 10–23 × 5000 = 1.0(4) × 10–19J
[or = 0.64(7) eV] (1)
(iii)
some gas atoms have enough kinetic energy to cause excitation by
collision (1)
photons (of certain energies) only released when de-excitation
or electron transfer to a lower level, occurs (1)
gas atoms have a spread of speeds / kinetic energies (1)
mean Ek (of gas atoms) proportional to T (1)
excitation can occur to level C (1)
de-excitation from C to B produces 2.6 eV photon / light
of this wavelength (1)
(max 6)
QWC 1
[11]
Page 64 of 78
32
(a)
(i)
a collision in which kinetic energy is conserved (1)
(ii)
molecules of a gas are identical
[or all molecules have the same mass] (1)
molecules exert no forces on each other except during impact (1)
motion of molecules is random
[or molecules move in random directions] (1)
volume of molecules is negligible (compared to volume of
container)
[or very small compared to volume of container or point
particles] (1)
time of collision is negligible (compared to time between
collisions) (1)
Newton‘s laws apply (1)
large number of particles (1) (any two)
3
(b)
(i)
the hot gas cools and cooler gas heats up
until they are at same temperature
hydrogen molecules transfer energy to oxygen molecules
until average k.e. is the same
(any two (1) (1))
(ii)
(use of Ek =
kT gives) Ek =
× 1.38 × 10–23 × 420 (1)
= 8.7 × 10–21 J (8.69 ×10–21 J)
4
[7]
33
(a)
(i)
curve A below original, curve B above original (1)
(ii)
both curves correct shape (1)
2
Page 65 of 78
(b)
(i)
(use of pV = nRT gives) 130 × 103 × 0.20 = n × 8.31 × 290 (1)
n = 11 (mol) (1) (10.8 mol)
(ii)
(use of Ek =
kT gives) Ek =
× 1.38 × 10–23 × 290 (1)
= 6.0 × 10–21 J (1)
(iii)
(no. of molecules) N = 6.02 × 1023 × 10.8 (= 6.5 × 1024)
total k.e. = 6.5 × 1024 × 6.0 × 10–21 = 3.9 × 104 J (1)
(allow C.E. for value of n and Ek from (i) and (ii))
(use of n = 11 (mol) gives total k.e. = 3.9 (7) × 104 J)
5
[7]
34
(a)
(i)
(use of
gives)
(1)
= 40(.1) moles (1)
(ii)
= 9.8(1) moles (1)
3
(b)
(total) = (40 × 6 × 1023) − (9.8 × 6 × 1023) = 1.8(1) ×1025 (1)
(allow C.E. for incorrect values of n from (a))
(oxygen molecules) = 0.23 × 1.8 ×1025 = 4.2 × 1024 (1)
2
[5]
35
(a)
(i)
n = PV/RT = 3.2 × 105 × 1.9 × 10-3/8.31 × 285
n = 0.26 mol
(0.257 mol)
1
Page 66 of 78
(ii)
3.31 × 105 Pa
3 sig figs
(allow 3.30-3.35 × 105 Pa)
sig fig mark stands alone even with incorrect answer
3
(b)
similar
-( rapid) random motion
- range of speeds
different
- mean kinetic energy
- root mean square speed
- frequency of collisions
2
[6]
36
(a)
(i)
mass and energy have equivalent values
B1
E = mc2 mentioned
B1
MeV is energy unit (and kg that of mass)
B1
max2
Page 67 of 78
(ii)
clear attempt to substitute amu values into equation
C1
5.135 × 10–3 (u) or 4.78 (MeV) seen
C1
mass of 1 lithium nucleus = 9.98 × 10–27 (kg)
C1
total number of nuclei in 1 kg = 1.00 × 1026
C1
total energy given out = 4.78 × 1026 MeV
A1
5
(iii)
neutrons needed (for the lithium reaction) can
come from the other (deuterium-tritium) reaction
B1
1
(b)
(i)
potential energy equation (E =
or used
) quoted
C1
correct substitutions
C1
1.5(3) × 10–13 (J)
A1
3
Page 68 of 78
(ii)
ke = 3/2 kT
C1
0.75/0.765 × 10–13 (J) or half of (b) (i)
or 4 × 109 (K) used
C1
3.7 × 109 (K) or total energy 1.6 × 10–13 (J)
A1
3
(iii)
each nucleus carries a positive charge
B1
(electrostatically) repel each other
B1
strong nuclear force
B1
this has a range of nucleus diameters
B1
high temperature needed for high kinetic energy
B1
max4
[18]
Page 69 of 78
Examiner reports
2
(a)
(b)
(i)
Most candidates recognised that the random motion of the smoke particles was due
to collision with air molecules. A minority simply mentioned Brownian movement
without any explanation.
(ii)
The majority of candidates suggested that air molecules moved randomly but few
were able to link a second property, which could be inferred from the observation of
the smoke particles.
(i)
Most candidates were able to quote the correct equation and substitute appropriate
values to gain the answer 6.21 × 10–21 J. A minority of candidates used
missed the minus sign for the power of 10.
(ii)
Although most candidates recognised the equation pV =
kT or
Nm<c2>, a high
proportion failed to see how the density related to this equation and then either
fudged its substitution, or else made no progress beyond quoting the equation. A
significant number of candidates went on to quote the rms speed and were then
penalised if they had not quoted the mean square value with its unit.
8
(a)
(iii)
Few candidates gave a totally convincing argument that there was a distribution of
molecular speeds and that at a higher temperature there would still be molecules
moving with speeds well below the mean speed.
(i)
Most candidates quoted the correct equation. The majority also carried on to
complete the calculation correctly but a significant number used centigrade
temperature instead of absolute temperature.
(ii)
This was done correctly by most candidates. Some of those who had previously used
centigrade temperatures corrected their mistakes here.
(iii)
Although the correct process was carried by most candidates, the use of incorrect
data was common. Unit penalties and significant figure penalties were common on
this part of the question.
(iv)
Many candidates explained what is meant by the word adiabatic but fewer could go
on to say why the process described in the question would be approximately
adiabatic.
(v)
Candidates tended to mention either the more frequent collisions with the wall of the
syringe or the fact that the molecules moved faster. Relatively few mentioned both
factors. Even fewer went further and explained how these factors would affect the
rate of change of momentum of the molecules and, hence, the pressure exerted on
the walls of the syringe.
Page 70 of 78
(b)
14
(i)
This was generally well done.
(ii)
This was also answered correctly by many candidates but some were insufficiently
clear about the direction of the process.
A majority of candidates obtained the mark for part (a) by stating it was equal to the number of
atoms in one mole of substance. Very few gave the full definition which relates to carbon-12.
Many candidates must have been aware of the definition because they tried to incorporate it into
what they put down. For example, 'The number of atoms in one mole of carbon-12'. Sometimes
the halfway approach went wrong and we saw, 'The number of particles in 1 atom of carbon-12',
or similar. There were also a few candidates who took a kinetic theory equation, which had
Avogadro constant, which they then rearranged to make the constant the subject. This was not
regarded as a definition.
In part (b)(i) was an easy substitution into an easy equation and most candidates scored the
mark.
By contrast in (b)(ii) it was only the very best candidates who completed the whole of the
question. The main problem was that the majority did not appreciate that the mass of an
individual krypton atom was required in the equation for mean kinetic energy. The other
surprising difficulty was the unit of mean square speed. Only about a quarter of candidates got
this correct.
In part (c) most candidates scored the mark for krypton's mean square speed being less. As
expected the most common error in the explanation was to suggest the kinetic energies of both
gases are the same rather than having the same means for their kinetic energies.
15
A majority of candidates referred to obeying a gas law in answer to part (a). A second marking
point was often missed out, wrong or vague. This is illustrated in the two answers that follow: ‘It
has properties of a gas such as Brownian Motion’, and ‘The gas obeys the assumptions of the
kinetic theory’.
Parts (b)(i)+(ii) were done well by most. Only a few did not convert the temperature to Kelvin
before performing a calculation. Again very few did not know the unit for density.
In part (b)(iii) more than half the candidates could perform the calculation but a significant
number of those did not quote the answer to 2 significant figures. Of those missing out on the
calculation many did score the first mark but then went wrong by using the wrong density or by
not finding the proportion of the gas still in the container.
Page 71 of 78
16
This was expected to be a challenging question so it is especially pleasing that so many
candidates gained at least half of the marks and some of them all of the marks.
(a)
The examiners anticipated several ways of successfully finding a value for absolute zero
from the data presented in the graph. The candidates used all of these and some others as
well. Candidates who attempted the V/T = constant route often got bogged down in their
figures and at some stage anticipated the answer by adding 273 to the unknown
temperature.
(b)
This question was usually answered well with the most common errors being a failure to
convert from Celsius to Kelvin and/or an incorrect conversion from cm3 to m3.
18
19
(a)
This straightforward application of Boyle’s law was done well by the vast majority of
candidates.
(b)
Many went on to complete the graph accurately. Some lost one mark by drawing the
isothermal change as a straight line on the pressure–volume graph. The horizontal section
of the graph was usually correct.
(c)
(i)
Whilst the estimate of work done during the process was often done well,
many
correct solutions were marred by a lack of descriptive detail. It was sometimes very
unclear whether a candidate had used the graph, as there was a complete lack of
method or clear statement of the equivalence between the graph area and a
computed energy. Most estimates (derived from a correct assessment of the graph
area) led to values within the range allowed by examiners. However, some scripts
showed (often on the graph itself) that the candidate concerned did not understand
which area to compute.
(ii)
A large majority of candidates recognised that work was done on the gas because it
is being compressed during the process.
Answers to this question showed yet again how important it is for candidates to appreciate the
distinction between number of molecules and number of moles and to understand that absolute
temperatures are required when carrying out calculations on gases. Very many of the attempts at
part (a) succeeded only in determining the number of moles of gas in the container, gaining just
one of the three marks. A substantial number of candidates who made this omission later
recovered their thoughts, giving a correct answer in part (c).
Most of the graphs in part (b) were straight lines with a positive gradient. Few candidates took
the trouble to show their line passing through the data point (50°C, 1.5 ×105 Pa) and several who
tried to do this had difficulty with the scales on the grid. Making the line meet the temperature
axis at an acceptable value for absolute zero also proved to be challenging for many candidates.
Page 72 of 78
A pleasing number of correct solutions to part (c) showed a clear understanding of how to find
the number of molecules left in the fixed volume container after the release valve had opened
and, hence, the number that had escaped. However, this strategy did not come readily to the
weaker candidates.
21
The anticipated emphasis in answers to part (a)(i) was on how thermal equilibrium is manifested,
whilst in part (a)(ii) it was on a molecular explanation. In general, higher marks would have been
awarded had the candidates read the question more carefully. Many answers simply fudged the
two different aspects. Most candidates seem to have gained only the most basic understanding
of molecular behaviour, a problem which pervaded the whole question. In particular, candidates
should realise that the molecules in any system have a range of energies and that temperature is
a measure of their mean kinetic energy. Thermal equilibrium may be satisfactorily explained by
considering two systems in contact. When in equilibrium there is no net flow of energy between
them because both systems are at the same temperature. However, there is a microscopic flow
of energy in each direction between them. It is untrue that “every molecule has the same kinetic
energy” when equilibrium has been established, as was stated in most answers.
Very few candidates were able to calculate the r.m.s. speed of the helium molecules in part (b)(i).
The main problems, also seen in other similar questions in the recent past, were confusion
between molar and molecular quantities and an inability to differentiate between mean square
values and r.m.s. values. Many candidates recovered in part (b)(ii) to gain a couple of marks by
using 3k T / 2 or its molar counterpart. Only a handful of candidates spotted that the average
kinetic energy of a nitrogen molecule would be the same as that of a helium molecule, and that
the result from part (b)(i) could be combined with ½ mʋ2 to arrive at the answer. There were
many correct numerical answers to part (b)(iii) but most were arrived at instinctively; justification
of (2 / 3) of 120KPa was expected for one of the two marks.
22
Part (a) proved to be difficult. In part (a)(i) the effect of the increase in the number of collisions
per second with the container was often appreciated, although in too many answers ‘more’ was
written instead of ‘more frequent’. Mention of momentum was less common and a precise
statement that the change in momentum per collision increased was rare. Increased force per
collision was common but gained no credit.
This lack of understanding made part (a)(ii) even more difficult, as even those candidates who
said that for constant pressure the volume had to increase to reduce the number of collisions per
second could not say why, and many thought that the number of collisions had to remain
constant. Many candidates tried to answer macroscopically, used the gas law in terms of
referred to the first law of thermodynamics.
or
In contrast, most candidates scored full marks in part (b). Some candidates unwisely introduced
other units than m3 , e.g. dm3 and cm3 , with varying success. Only a small proportion of the
candidates forgot to change the temperature to K. A surprising number of candidates misread
one-twentieth, mostly as one-twelfth.
Page 73 of 78
23
It was surprising to find that this was a popular question. The large majority of answers
suggested that it had been a poor choice of question for many candidates. However, part (a) was
almost universally answered correctly. For a significant number of candidates the three marks
scored represented the total mark for the whole question. Very few correct answers were seen to
part (b)(i). The words “one mole” were largely ignored and, for those candidates who did not
ignore them, they were read as “one molecule” and the Avogadro constant appeared in the
denominator of the equation.
Errors abounded in part (b)(ii) because candidates had not read the question sufficiently
carefully. Very few candidates were able to go back to part (a) and identify correctly the
temperature and the number of moles present. In consequence, the large majority of answers
related either to one mole or one molecule. A good number of candidates knew exactly why the
calculated value represented the total internal energy, i.e. because there are no intermolecular
forces (or zero potential energy), but a larger number had absolutely no idea and either left the
part blank or simply restated the question.
Parts (c) and (d) attracted the poorest answers in the entire section. Confusion abounded over
the meanings of the terms in the First Law of Thermodynamics. The only term which could be
defined with any certainty was that for external work done, ΔW. Most answers had many
contradictions. In part (c)(ii) candidates could not come to terms with the fact that heat energy
input had to be shared between changes in both internal energy and work done. Most
candidates thought that, in part (c)(ii), the internal energy remained constant in spite of the fact
that there was an obvious temperature change. There was widespread evidence that candidates
had only a very superficial grasp of thermodynamics and acquired little skill in the interpretation
of p−V graphs, which was at the heart of this question. Very few correct answers were seen to
part (d) and it was very disappointing to see that there are candidates at Advanced level who still
confuse heat and temperature. Thus, a significant number of candidates gave 200 J as the
answer to part (d)(i). It was almost inevitable that those candidates who could not answer part (c)
would also have difficulty with part (d), but some were able to give a partially correct answer in
calculating the work done. For many candidates, this was their complete answer to part (d)(ii).
Only a very small minority could calculate the changes in internal energy.
Page 74 of 78
24
Part (a) proved difficult for less able candidates. Some drew straight lines and others tried to
force the curve to intercept the volume axis. The less able candidates sometimes marked correct
points on the grid but did not draw a line. It seemed that some less able candidates followed the
wrong order in tackling this part. They drew the curve before they marked points on the grid. As a
result the points were just randomly placed on the curve they had drawn.
Part (b) (i) was done well by most. Candidates who used the alternative equation PV = nRT often
stopped when they had found the number of moles of gas. Part (b) (ii) was much more
discriminating with less than 50% of candidates obtaining the correct answer. Many candidates
did not have a clue whereas others could find the mean kinetic energy but then did not follow this
up by finding the total kinetic energy.
Although part (c) looks like a basic question it did discriminate well. It was only the more able
candidates who scored full marks. Many did not know what the question was getting at and
guessed. Sometimes these candidates did score the mark associated with molecules moving in
random motion. In other cases candidates did not complete their statements fully. For example,
stating ‘atoms travel in straight lines’, rather than, ‘atoms travel in straight lines between
collisions’.
25
In part (a) most candidates could only give one correct quantity. This was usually pressure. Very
few gave the average kinetic energy. The calculation in part (b) was generally well done, with the
most common error being the inability to convert kPa to Pa. This however only incurred a
one-mark penalty.
Candidates found part (c) much more difficult and even those who knew the correct answer were
often not able to express themselves clearly. It was apparent from many answers that candidates
could remember the appropriate equations but were not as secure in their conceptual
understanding.
26
A small minority of candidates clearly were not prepared for kinetic theory in any detail and
produced nonsense answers in part (a). Most candidates remembered two of the assumptions
and either left the final two unanswered or, quite commonly, were content to rephrase their first
two assumptions. On the other hand many candidates did get all four assumptions correct.
Very few candidates achieved all four marks for the explanation in part (b)(i). Many made no
reference to molecules or to the kinetic theory but invoked the general gas law. Most candidates
knew that gas molecules move faster on average when heat energy is transferred into the gas
from the warmer surroundings and related this to an increased collision rate with the container
wall. Few candidates made any mention of momentum exchange. Most candidates used the
correct expression
kT for the mean kinetic energy in part (ii), but few used the absolute
temperature in the calculation. A surprising number of candidates, having arrived at a correct
answer, lost the mark by omitting the unit.
Page 75 of 78
27
This question also gave good discrimination. In part (a) the better candidates made very few
errors and scored high marks but the less able candidates found it more difficult and were unable
to come up with three assumptions of the kinetic theory.
The calculation in part (b) caused problems and even candidates who selected the correct
expression had difficulties. The commonest of these was failing to convert the temperature to
Kelvin.
Part (c) was answered correctly by only the best candidates and although many did identify that
molecular mass was important, they could not quite explain why.
28
29
There were significant variations between candidates. Most were able to correctly apply the
equation of state for an ideal gas but were less confident when it came to calculating in part (a)
the average kinetic energy of a gas molecule in the cylinder. It was evident that a significant
number of candidates did not know the appropriate formula. In part (c) explaining, in terms of the
kinetic theory, why the pressure of the gas fell was well answered. The only common error was
assuming that the kinetic energy of molecules fell even though the temperature remained
constant.
Again candidates found this a demanding question and a significant minority were not even able
to convert the temperature from Celsius to Kelvin in part (a). In the same section, the calculation
of number of moles of gas and the number of gas molecules also proved difficult and significant
figure errors were common.
Part (b) realised better answers, although many candidates did not seem to appreciate that
temperature is related to the mean kinetic energy and not to the kinetic energy of an individual
molecule.
31
Many candidates scored full marks in part (a), although some lost a mark as a result of not
converting d from mm into m and also giving 4.87 × 10−4 m as the wavelength. In part (ii), a
significant number of candidates failed to convert correctly from joules into eV, often multiplying
rather than dividing by 1.6 × 10–19.
In part (b) (i), most candidates knew that the required transition occurred between energy levels
A and C, but many indicated that the transition was either upwards or in both directions. Again,
most candidates were able to show that the mean kinetic energy in part (ii) was 0.65 eV. In part
(iii), only a small number of candidates realised that excitation of the gas atoms is due to
collisions between the atoms, and that the spread of speeds of the gas atoms at 5000 K resulted
in collisions which caused electrons in the atoms to move to the higher energy levels, including
C. Some did realise that a 2.6 eV photon was emitted when an electron transition occurred from
C to B. Sadly, the only credit gained in part (iii) by most candidates was for stating that a photon
was emitted when de-excitation of a gas atom occurred. Some candidates thought that excitation
in a heated gas is due to atoms absorbing photons rather than collisions between gas atoms.
Page 76 of 78
32
33
Candidates answered this question confidently although there was still some confusion as to
acceptable assumptions made in the kinetic theory. The term ‘thermal equilibrium’ did cause
some confusion to weaker candidates but full marks for this question were quite common.
Candidates found this question quite demanding and although in part (a) most were able to draw
the two graphs correctly, the answers were spoilt due to a lack of care in the sketches.
The calculations in part (b) were more discriminating and arithmetic errors were common. The
correction to part (b) (iii) did not seem to cause problems, although only the more able
candidates were able to come up with a correct response. A significant proportion of candidates
answered the question correctly but then lost the final mark due to a significant figure penalty.
34
35
Again many candidates scored high marks in this question. There was a tendency to incur a
significant figure penalty for either quoting an answer to too many significant figures or rounding
down to one significant figure. Part (b) caused problems for less able candidates who, although
correctly determining the change in number of moles, were then either unable to calculate 23%
of the number or else did not multiply the change in number of moles by Avogadro’s number.
Part (a)(i) was an easy introductory question, which most students got correct. Part (a)(ii) was
also successfully attempted in a majority of scripts. Use of the ideal gas equation again was
more popular than using pressure is proportional to temperature. A small percentage of papers
gave answers to only 2 significant figures rather than the 3 required. A majority of students only
scored one mark out of two for part (b). They correctly referred to the random motion but failed to
refer to a mean when giving some quantity, such as kinetic energy, that increases with
temperature.
Page 77 of 78
36
Credit was not given for a statement that mass and energy are the same thing in answer to part
(a) (i).
The most common mark for this part was one out of the two available.
Only a limited number of candidates gained all five marks in part (a) (ii), but several gained two
for finding a value of 4.78 MeV by converting the mass defect from values in terms of u to MeV.
Although the mark to part (a) (iii) was frequently gained, there was considerable benefit of doubt
applied with candidates not saying that it was the deuterium-tritium reactions which supplied the
neutrons needed for the lithium reactions.
Part (b) (i) was answered poorly. Most candidates did not use the electric potential energy
equation – force or potential were more common.
A limited number of candidates used the 3/2 kT equation in answer to part (b) (ii).
A few candidates started with 4 × 109 K and worked backwards but rarely saw that this gave
approximately half the ke in (b) (i) as would be expected. Part (b) (iii) was generally answered
well.
Page 78 of 78