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MYP 3 · Science

Particle Model of Matter

60 questions across 3 sub-topics

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States of Matter and the Particle Model Changes of State Diffusion and Particle Motion

States of Matter and the Particle Model 20 questions

QUESTION 1 3 marks Criterion A
Easy
X Y Z

The diagrams show the particle arrangement in three samples of the same substance, X, Y and Z.

a. Identify the state of matter shown in each diagram.
[1]
b. For diagram Z, describe (i) the arrangement of the particles and (ii) how the particles move.
[2]
Show complete worked solution
(a)
X is a solid (particles in a regular, tightly packed pattern). Y is a gas (particles far apart, scattered randomly, filling the whole box). Z is a liquid (particles close together but arranged randomly, not in a fixed pattern).
(b)
(i) In Z the particles are close together, touching their neighbours, but arranged randomly rather than in a regular repeating pattern. (ii) The particles can move around and slide past each other, changing position while staying close together, which is why a liquid can flow and take the shape of its container.
QUESTION 2 2 marks Criterion A
Easy

State two properties of a gas that are different from a solid, in terms of shape and volume.

Show complete worked solution

A gas has no fixed shape and no fixed volume — it spreads out to completely fill whatever container it is put in. A solid, in contrast, has both a fixed shape and a fixed volume, which do not change unless a force is applied to it.

QUESTION 3 2 marks Criterion A
Easy

In terms of the particle model, explain why solids cannot be compressed (squashed into a smaller volume) easily.

Show complete worked solution

In a solid, the particles are already packed very closely together, touching their neighbours, with almost no empty space between them. Since there is very little room left for the particles to be pushed any closer together, solids strongly resist compression.

QUESTION 4 4 marks Criterion A
Medium
P Q

The diagrams show the same substance as a liquid (P) and as a gas (Q).

a. Compare the spacing between particles in P and Q.
[1]
b. Compare the strength of the forces of attraction between particles in P and Q.
[1]
c. Use your answers to (a) and (b) to explain why gas Q can be compressed much more than liquid P.
[2]
Show complete worked solution
(a)
The particles in Q (gas) are spaced much farther apart than in P (liquid), with large empty gaps between them, while in P the particles are close together, almost touching.
(b)
The forces of attraction between particles are much weaker (almost negligible) in gas Q than in liquid P, where noticeably stronger forces hold the particles close together.
(c)
Because there are large empty gaps between the particles in a gas, the particles can be pushed much closer together when the gas is squashed. In the liquid, the particles are already close together (almost touching), so there is very little space left to reduce further — which is why liquids (like solids) cannot be compressed much.
QUESTION 5 3 marks Criterion A
Medium

Substance XDensity (g/cm³)
Solid2.70
Liquid2.40
Gas0.0012
This table shows the density of substance X in each of its three states.

a. Which state has particles that are closest together? Use the density data to justify your answer.
[1]
b. Explain why the gas has such a much lower density than the solid or liquid.
[2]
Show complete worked solution
(a)
The solid, since it has the greatest density ($2.70\,\text{g/cm}^3$) — the most mass is packed into each $\text{cm}^3$, which happens when the particles are spaced most closely together.
(b)
In the gas, the particles are spread very far apart with mostly empty space between them, so a given volume of gas contains far fewer particles (and much less mass) than the same volume of solid or liquid — giving the gas a much lower density.
QUESTION 6 3 marks Criterion A
Medium

State whether each description applies to a solid, a liquid, or a gas.

a. Particles vibrate about fixed positions but do not change places with each other.
[1]
b. Particles are close together and constantly move around each other, slipping and sliding.
[1]
c. Particles move quickly and randomly in all directions, spreading out to fill all the available space.
[1]
Show complete worked solution
(a)
Solid.
(b)
Liquid.
(c)
Gas.
QUESTION 7 4 marks Criterion A
Hard

A solid substance is heated until it becomes a gas, without changing its chemical identity.

a. Describe how the arrangement and movement of the particles change as the solid becomes a gas.
[2]
b. Explain, in terms of forces of attraction and particle energy, why this change happens as the substance is heated.
[2]
Show complete worked solution
(a)
The particles change from a fixed, closely packed, regular arrangement (only vibrating on the spot) into particles that are free to move throughout the whole container, spread far apart with no regular order, moving quickly and randomly in all directions.
(b)
Heating gives the particles more kinetic energy, so they vibrate and move faster. Eventually they gain enough energy to overcome the forces of attraction holding them together — first partially (becoming a liquid, able to move around while staying close), and then almost completely (becoming a gas, able to move freely far apart) — which is why the particles end up much further apart and moving much faster once the substance is a gas.
QUESTION 8 7 marks Criterion B
Medium

A student wants to investigate whether gases or liquids can be compressed (squashed into a smaller volume) more easily. They plan to use two identical sealed syringes: one filled completely with air, one filled completely with water, with the outlet of each blocked so nothing can escape.

a. State the independent variable and the dependent variable for this investigation.
[2]
b. State one variable that should be kept the same (controlled) between the two tests, and explain why.
[2]
c. Describe a method for carrying out this investigation, including how you would measure the results.
[3]
Show complete worked solution
(a)
Independent variable: which substance is in the syringe (air or water). Dependent variable: the change in volume (how far the plunger can be pushed in) for a given force.
(b)
Use identical syringes (same starting volume/size) and apply the same force to the plunger in each test. If a bigger force, or a bigger syringe, were used for one test, any difference in how much the volume changed could be caused by that rather than by whether the substance is a liquid or a gas, making the comparison unfair.
(c)
  1. Fill syringe A completely with air and syringe B completely with water, then seal the outlet of each (e.g. with a blocked nozzle) and record the starting volume from the scale on the side of each syringe.
  2. Push the plunger of syringe A in as far as possible, using the same measured force each time (e.g. hanging the same fixed weight from the plunger), and record the new volume reading.
  3. Repeat for syringe B, using the identical force.
  4. Record the decrease in volume for each syringe and compare the two results.
QUESTION 9 7 marks Criterion B
Medium

A student wants to find out whether a balloon filled with helium gas has a different density from an identical balloon filled with air, to help explain why helium balloons float.

a. State the independent and dependent variables for this investigation.
[2]
b. Describe how the student could make sure the comparison between the two balloons is fair.
[2]
c. Explain how the results (mass of each balloon) could be used to determine which gas is less dense, and relate this to why a helium balloon floats.
[3]
Show complete worked solution
(a)
Independent variable: the type of gas used to fill the balloon (helium or air). Dependent variable: the mass of the filled balloon, for the same volume.
(b)
Use two identical balloons, inflated to exactly the same volume (e.g. checked by measuring the diameter of each with a tape measure, or by comparing how much water each displaces), and measure the mass of each on the same balance.
(c)
Since both balloons have the same volume, the gas with the smaller mass is the less dense one (density $=$ mass $\div$ volume, and volume is equal for both). If the helium balloon has a smaller mass than the air balloon of the same volume, helium is less dense than air. A gas that is less dense than the surrounding air experiences a greater upward push than its own weight, so the balloon rises and floats.
QUESTION 10 8 marks Criterion B
Hard

Repeating the sealed-syringe compression test (air vs water) described above, a student got inconsistent results for the air syringe: sometimes the plunger felt easy to push in, sometimes it felt hard, even though they tried to use "the same push" by hand each time.

a. Identify the main source of error in pushing the syringe plunger by hand, and explain how it affects the reliability of the results.
[2]
b. Suggest an improved method that would make the applied force more consistent, and explain how it works.
[3]
c. Explain how using repeated readings would further improve the reliability of the conclusion.
[3]
Show complete worked solution
(a)
The main source of error is that a force applied by hand cannot be made exactly the same every time (natural variation in how hard a person pushes). This means the amount of compression measured varies between repeats for a reason unrelated to the substance being tested, which reduces the reliability of the results.
(b)
Instead of pushing by hand, hang a fixed mass (e.g. a $500\,\text{g}$ weight) from a lever or clamp arm pressing on the plunger, or use a force meter/spring balance to push the plunger to the same reading each time. Because the same mass or force is used for every repeat, the push is identical every time, so any difference in how far the plunger moves is due only to the substance being tested (air or water), not to variation in the applied force.
(c)
Repeat each test (for both air and water) at least three times using the fixed force, and calculate a mean decrease in volume for each substance. Taking a mean reduces the effect of any small random errors (e.g. a slightly misread scale, a tiny air leak) in a single reading, giving a more reliable comparison of how compressible air is compared with water.
QUESTION 11 3 marks Criterion C
Easy

A solid block has a mass of $54\,\text{g}$ and a volume of $20\,\text{cm}^3$.

a. Calculate the density of the block. State the formula you use.
[2]
b. Is this density value more typical of a solid or a gas? Explain briefly using the particle model.
[1]
Show complete worked solution
(a)
$$ \text{density} = \frac{\text{mass}}{\text{volume}} = \frac{54}{20} $$ Answer: $2.7\,\text{g/cm}^3$
(b)
More typical of a solid — solids have particles packed closely together, giving a relatively high density; gases (with particles spread far apart) typically have densities around a thousand times smaller, e.g. around $0.001\,\text{g/cm}^3$.
QUESTION 12 5 marks Criterion C
Medium

SamplePQRS
Density (g/cm³)2.700.00132.650.0680
Typical densities are roughly $1$–$10\,\text{g/cm}^3$ for solids and liquids, and roughly $0.0005$–$0.002\,\text{g/cm}^3$ for gases.

a. Using the typical values given, classify samples P, Q and R as likely solid/liquid or likely gas.
[2]
b. Explain why sample S's density value ($0.0680\,\text{g/cm}^3$) is difficult to classify as clearly solid/liquid or clearly gas.
[2]
c. Suggest what should be done to check whether S's value is a measurement error.
[1]
Show complete worked solution
(a)
P ($2.70$) and R ($2.65$): likely solid or liquid (particles closely packed). Q ($0.0013$): likely a gas (particles far apart).
(b)
$0.0680\,\text{g/cm}^3$ does not match either typical range: it is roughly $50$ times denser than a typical gas, but roughly $40$ times less dense than a typical solid or liquid, so it sits between the two groups and cannot be classified from density alone.
(c)
Repeat the mass and volume measurements for sample S to check that the density value is consistent; if repeats keep giving a similarly unusual value, other evidence (e.g. its state at room temperature, or a particle diagram) would be needed alongside density to classify it.
QUESTION 13 5 marks Criterion C
Medium

Force applied (N)05101520
Volume of trapped air in syringe (cm³)2016131110
A student pushed the plunger of a sealed syringe containing air, recording the volume of trapped air at increasing forces.

a. Describe the pattern shown by the data as force increases.
[1]
b. Calculate the percentage decrease in volume from $0\,\text{N}$ to $20\,\text{N}$.
[2]
c. Explain, using the particle model, why the volume decreases less for each additional $5\,\text{N}$ of force applied.
[2]
Show complete worked solution
(a)
As the force increases, the volume decreases, but by smaller and smaller amounts each time ($20\to16$ is $-4$, $16\to13$ is $-3$, $13\to11$ is $-2$, $11\to10$ is $-1$) — the air becomes progressively harder to compress further.
(b)
$$ \text{decrease} = 20 - 10 = 10\,\text{cm}^3 $$ $$ \%\ \text{decrease} = \frac{10}{20}\times100 = 50\% $$
(c)
As the gas is compressed, its particles are pushed closer together; with less empty space left between them, it becomes increasingly difficult to squeeze the particles even closer, so each equal increase in force produces a smaller further decrease in volume.
QUESTION 14 5 marks Criterion C
Medium

Temperature (°C)20406080
Volume of gas in balloon (cm³)240255270285
A fixed mass of gas is sealed inside a flexible balloon and heated.

a. Describe the pattern shown by the data.
[1]
b. Calculate the increase in volume per $1\,^\circ\text{C}$ rise in temperature.
[2]
c. Explain, using the particle model, why the volume increases as the gas is heated.
[2]
Show complete worked solution
(a)
The volume increases steadily as temperature increases — by $15\,\text{cm}^3$ for every $20\,^\circ\text{C}$ rise, a constant rate.
(b)
$$ \text{rate} = \frac{15\,\text{cm}^3}{20\,^\circ\text{C}} = 0.75\,\text{cm}^3\ \text{per}\ ^\circ\text{C} $$
(c)
Heating increases the particles' kinetic energy, so they move faster and collide with the inside of the balloon more forcefully and more often, pushing the flexible balloon wall outward until the gas occupies a larger volume.
QUESTION 15 5 marks Criterion C
Medium
Before: 40 cm³ After: 20 cm³

The diagram shows the same number of gas particles in a syringe before and after the plunger is pushed in.

a. What has happened to the volume of the gas?
[1]
b. What has happened to the number of gas particles?
[1]
c. Calculate the percentage decrease in volume, and explain in terms of particle spacing why this is possible for a gas but would not be possible for a solid.
[3]
Show complete worked solution
(a)
It has decreased (halved, from $40\,\text{cm}^3$ to $20\,\text{cm}^3$).
(b)
It has stayed the same — the same $9$ particles are shown in both diagrams; none have been added or removed.
(c)
$$ \%\ \text{decrease} = \frac{40-20}{40}\times100 = 50\% $$ Gas particles are normally spread far apart with a lot of empty space between them, so pushing them closer together (reducing that empty space) is possible. A solid's particles are already touching and tightly packed, with almost no empty space left, so there is very little room to compress it further.
QUESTION 16 7 marks Criterion C
Hard

Trial12345
Volume of metal block (cm³)12.412.612.515.012.3
A student measured the volume of a metal block five times by water displacement. The block's mass was measured separately as $33.75\,\text{g}$.

a. Identify the anomalous result and explain how you identified it.
[2]
b. Calculate the mean volume, excluding the anomalous result.
[2]
c. Use the mass and your mean volume to calculate the density of the block, and state whether this is consistent with the block being a solid.
[3]
Show complete worked solution
(a)
Trial $4$ ($15.0\,\text{cm}^3$) is anomalous. The other four readings cluster consistently between $12.3$ and $12.6\,\text{cm}^3$, while $15.0\,\text{cm}^3$ is well outside this range.
(b)
$$ \text{mean} = \frac{12.4+12.6+12.5+12.3}{4} = \frac{49.8}{4} = 12.45\,\text{cm}^3 $$
(c)
$$ \text{density} = \frac{33.75}{12.45} = 2.71\,\text{g/cm}^3\ (\text{3 s.f.}) $$ This is a typical solid density (particles packed closely together), consistent with the block being a solid metal (e.g. similar to aluminium, which has a density of about $2.7\,\text{g/cm}^3$).
QUESTION 17 6 marks Criterion C
Hard

SubstanceStateDensity (g/cm³)
IronSolid7.90
EthanolLiquid0.79
OxygenGas0.0013
HeliumGas0.00018
A student claims: "This data proves that every solid is denser than every liquid, and every liquid is denser than every gas."

a. Using the data, evaluate whether the student's conclusion is fully supported.
[2]
b. Explain, using the particle model, why gases are typically far less dense than solids or liquids of the same substance, regardless of which specific substance is chosen.
[2]
c. Suggest one additional piece of data that would make the student's general conclusion better supported.
[2]
Show complete worked solution
(a)
The data shown does follow the pattern (iron $>$ ethanol $\gg$ oxygen and helium). However, this data set only covers four particular substances — it cannot prove a universal rule about every solid, liquid and gas that exists. Some unusual solids and liquids exist that could break such a strict pattern, so the claim is not fully justified by this small amount of data.
(b)
In a gas, particles of any substance spread far apart from each other, leaving huge empty spaces between them, compared with the same substance as a solid or liquid, where the particles are packed closely together. The same mass of particles spread over a much bigger volume in a gas gives a much lower density.
(c)
Density data for many more, more varied, solids, liquids and gases (not just these four substances), to check whether the pattern truly holds in every case tested, rather than only for this small sample.
QUESTION 18 5 marks Criterion D
Medium

Unlike most substances, water is unusual: ice (solid water) is less dense than liquid water, so ice floats. This happens because water particles arrange into a more open pattern, with more space between them, when they freeze.

Discuss one benefit and one drawback of this unusual property.

Show complete worked solution

Benefit: Because ice floats and forms an insulating layer on the surface of lakes and rivers, the liquid water underneath stays several degrees warmer than the freezing air above, allowing fish and other aquatic organisms to survive through winter. If ice sank instead, bodies of water could freeze from the bottom up, likely killing most aquatic life.

Drawback: Because water expands as it freezes (its particles spread into a less dense arrangement), water trapped inside pipes, engine blocks, or cracks in rock can burst them when it freezes, causing damage to buildings, plumbing and roads (frost damage) — a costly problem in cold climates.

QUESTION 19 5 marks Criterion D
Medium

Because gases can be compressed into a much smaller volume, large amounts of gas (e.g. oxygen for scuba diving, or propane for camping stoves) can be stored under high pressure in relatively small metal cylinders.

Discuss one benefit and one drawback of storing compressed gases in this way.

Show complete worked solution

Benefit: Compressing a gas squeezes its particles much closer together, allowing a far larger mass of gas (measured at normal pressure) to be stored in a small, portable cylinder than could otherwise fit. This makes it possible for a diver to carry hours of breathable air, or a camper to carry enough fuel gas in a small canister.

Drawback: Gas cylinders are under very high pressure, so if they are damaged, punctured, or heated too much (increasing the pressure further as the particles move faster and collide more forcefully with the container walls), they can rupture or explode violently. This poses a serious safety risk, which is why compressed gas cylinders must be stored, transported and used carefully (e.g. away from heat, secured upright).

QUESTION 20 6 marks Criterion D
Hard

Natural gas is normally transported as a gas through pipelines. For countries not connected by pipeline, natural gas is instead cooled to about $-162\,^\circ\text{C}$, turning it into a liquid (LNG) that takes up about $\dfrac{1}{600}$ of the volume of the same mass of gas. The liquid is shipped by sea in insulated tankers, then turned back into a gas before use.

Evaluate the impact of liquefying natural gas in this way, discussing both a benefit and a concern, and referring to the particle model in your answer.

Show complete worked solution

Benefit: Cooling the gas removes enough particle kinetic energy for the forces of attraction between particles to pull them from being widely spread apart (as a gas) into a closely packed liquid arrangement, shrinking its volume by around $600$ times. This makes it possible to transport a genuinely useful amount of energy by ship to places with no pipeline, greatly increasing global access to natural gas as a fuel.

Concern: The cooling process itself uses a large amount of energy, and the specialised insulated tankers and cooling/re-heating plants are expensive to build and run, meaning LNG has a bigger carbon footprint than piped gas before it is even burned. In addition, if a tanker or storage tank were damaged, the very cold liquid would rapidly turn back into a large volume of flammable gas, which is a serious safety hazard that must be carefully managed with strict safety regulations.

Changes of State 20 questions

QUESTION 1 4 marks Criterion A
Easy

State the name of the change of state described in each case.

a. A solid turns into a liquid.
[1]
b. A liquid turns into a gas.
[1]
c. A gas turns into a liquid.
[1]
d. A liquid turns into a solid.
[1]
Show complete worked solution
(a)
Melting.
(b)
Boiling / evaporating (vaporisation).
(c)
Condensation.
(d)
Freezing (solidification).
QUESTION 2 2 marks Criterion A
Easy

State the term for a solid changing directly into a gas, without becoming a liquid first, and give an example substance where this happens at room temperature.

Show complete worked solution

This is called sublimation. Example: solid carbon dioxide ("dry ice"), or iodine crystals, both turn directly into a gas at room temperature and pressure without passing through a liquid stage.

QUESTION 3 2 marks Criterion A
Easy

The melting point of pure ice is $0\,^\circ\text{C}$. State the freezing point of pure water, and explain the relationship between a substance's melting point and its freezing point.

Show complete worked solution

The freezing point of pure water is also $0\,^\circ\text{C}$. For a pure substance, the melting point and freezing point are always the same temperature — melting (solid $\to$ liquid) and freezing (liquid $\to$ solid) are simply the reverse of each other, occurring at the one temperature where solid and liquid can exist together in balance.

QUESTION 4 4 marks Criterion A
Medium

SubstanceMelting point (°C)Boiling point (°C)
Ethanol-11478
Oxygen-218-183
Mercury-39357

a. State the state of matter (solid, liquid or gas) of each substance at room temperature ($20\,^\circ\text{C}$).
[2]
b. Explain how you used the melting and boiling points to decide the state of oxygen at room temperature.
[2]
Show complete worked solution
(a)
Ethanol: liquid ($-114 < 20 < 78$). Oxygen: gas ($20$ is above its boiling point of $-183$). Mercury: liquid ($-39 < 20 < 357$).
(b)
Room temperature ($20\,^\circ\text{C}$) is above oxygen's boiling point ($-183\,^\circ\text{C}$). This means the temperature is high enough that oxygen particles have enough energy to completely overcome the (very weak) forces holding them together as a liquid, so oxygen exists as a gas at that temperature.
QUESTION 5 4 marks Criterion A
Medium
0481216-20-100204060 Time (min) Temperature (°C)

The graph shows a substance being heated at a constant rate, starting as a solid.

a. What is happening to the substance during the flat section between $t=4$ and $t=9$ minutes?
[1]
b. What is the melting point of this substance, according to the graph?
[1]
c. Calculate the rate at which the temperature rises while the solid is being heated, between $t=0$ and $t=4$ minutes.
[2]
Show complete worked solution
(a)
The substance is melting — changing from a solid to a liquid — at its melting point.
(b)
$0\,^\circ\text{C}$ (the temperature of the flat section).
(c)
$$ \text{rate} = \frac{0-(-20)}{4-0} = \frac{20}{4} = 5\,^\circ\text{C/min} $$
QUESTION 6 3 marks Criterion A
Medium

Explain why the temperature of boiling water stays constant at $100\,^\circ\text{C}$ while it is boiling, even though a Bunsen burner keeps supplying heat energy underneath it.

Show complete worked solution

While the water is changing state (boiling), all of the extra heat energy supplied is being used to give the particles enough energy to completely overcome the forces of attraction holding them together as a liquid, rather than to increase their average kinetic energy any further. Since temperature is a measure of average particle kinetic energy, and that is not increasing during the change of state, the temperature stays constant until all of the liquid has turned into gas.

QUESTION 7 5 marks Criterion A
Hard

Both evaporation and boiling turn a liquid into a gas, but they are different processes.

a. State two differences between evaporation and boiling.
[2]
b. Explain, in terms of particle energy, why evaporation can still happen at temperatures well below the boiling point.
[3]
Show complete worked solution
(a)
Evaporation happens only at the surface of the liquid and can occur at any temperature below the boiling point. Boiling happens throughout the whole liquid (bubbles of gas form inside it) and only occurs at one specific temperature, the boiling point.
(b)
Even below the boiling point, particles in a liquid do not all have exactly the same energy — some move faster than others. At the surface, the fastest-moving, most energetic particles occasionally have enough energy on their own to completely escape the pull of neighbouring particles and enter the air as gas, even though the average particle in the liquid does not have enough energy to do so. This is why evaporation is a gradual surface process that can occur at any temperature.
QUESTION 8 7 marks Criterion B
Medium

A student wants to investigate how temperature affects the rate at which water evaporates from a shallow dish.

a. State the independent variable and the dependent variable for this investigation.
[2]
b. State two variables that should be controlled, and explain why one of them matters.
[2]
c. Describe a method for carrying out this investigation.
[3]
Show complete worked solution
(a)
Independent variable: temperature of the water/surroundings. Dependent variable: rate of evaporation (e.g. mass of water lost per minute, or time taken for all the water to evaporate).
(b)
Keep the volume of water, the surface area of the dish, and the air flow around the dish the same in every test. Changing the size of the dish between tests would also change the surface area, which independently affects evaporation rate, making the comparison unfair.
(c)
  1. Measure an identical volume of water (e.g. $20\,\text{cm}^3$) into identical shallow dishes for each temperature to be tested.
  2. Place each dish at a different set temperature (e.g. $20\,^\circ\text{C}$, $40\,^\circ\text{C}$, $60\,^\circ\text{C}$, using a water bath or heat mat), and record the starting mass on an electronic balance.
  3. Record the mass every $5$ minutes for a set time (e.g. $30$ minutes) — the loss in mass shows how much water has evaporated.
  4. Compare the rate of mass loss (g per minute) at each temperature.
QUESTION 9 7 marks Criterion B
Medium

A student wants to investigate whether the surface area of a puddle of water affects how quickly it evaporates, using containers of different shapes but the same volume of water.

a. State the independent variable and the dependent variable for this investigation.
[2]
b. State two variables that must be controlled, and explain why one of them matters.
[2]
c. Describe how the student could carry out this investigation.
[3]
Show complete worked solution
(a)
Independent variable: surface area of the water (using a wide, shallow dish vs a narrow, tall container). Dependent variable: time taken for the water to fully evaporate (or mass lost in a fixed time).
(b)
Control the volume of water used and the room temperature/location for every test. If the room temperature were different between tests, faster evaporation could be caused by higher temperature rather than by the larger surface area, giving a misleading conclusion about the effect of surface area alone.
(c)
Pour an identical, measured volume of water (e.g. $50\,\text{cm}^3$) into containers of different surface areas (e.g. a wide tray, a medium dish, a narrow beaker), keeping the starting volume and location the same for all. Measure the mass of each container and water at the start, then again every hour, calculating the mass lost. Compare the mass lost per hour (evaporation rate) for the different surface areas.
QUESTION 10 8 marks Criterion B
Hard

In an evaporation investigation, water in identical dishes was placed at $20\,^\circ\text{C}$ and $40\,^\circ\text{C}$ to compare evaporation rates. The $20\,^\circ\text{C}$ dish was left on a windowsill in direct sunlight, and the $40\,^\circ\text{C}$ dish was placed inside a warm, dark cupboard.

a. Identify a flaw in this method that would make it an unfair test.
[2]
b. Explain how sunlight and air movement, if not controlled, could each independently affect the results, separately from temperature.
[3]
c. Suggest an improved method that would fix this problem.
[3]
Show complete worked solution
(a)
Two variables are changing at once between the dishes — temperature is not the only difference, since the windowsill dish is also in direct sunlight and possibly a draughtier spot, while the cupboard is dark and enclosed. Any difference in evaporation rate cannot be confidently attributed to temperature alone.
(b)
Direct sunlight can heat the water directly (radiant heat) beyond just the surrounding air temperature, and it may also increase local air movement (convection currents) — both of which are known to speed up evaporation on their own. A draughtier position generally removes water vapour from just above the surface faster, maintaining a bigger concentration gradient and further speeding up evaporation, regardless of temperature.
(c)
Test both dishes in the same enclosed, dark space (e.g. a cupboard or box) with no direct sunlight, using a thermostatically controlled heat mat or water bath to hold each dish precisely at its intended temperature ($20\,^\circ\text{C}$ or $40\,^\circ\text{C}$). If possible, keep the dishes side by side so air movement around them is as similar as possible, changing only temperature between the two tests.
QUESTION 11 3 marks Criterion C
Easy

SubstanceMelting point (°C)Boiling point (°C)
Water0100
Ethanol-11478
Mercury-39357

a. Which substance is liquid over the widest range of temperatures?
[1]
b. Explain why mercury thermometers can be used to measure much higher temperatures than water-based ones.
[2]
Show complete worked solution
(a)
Mercury — its liquid range is $357-(-39)=396\,^\circ\text{C}$, wider than water's $100\,^\circ\text{C}$ range or ethanol's $192\,^\circ\text{C}$ range.
(b)
Mercury stays liquid up to $357\,^\circ\text{C}$ (its boiling point) before turning to gas, whereas water would already have boiled into gas by $100\,^\circ\text{C}$. Since a thermometer's liquid must remain a liquid to expand and give a reading, mercury allows accurate readings at much higher temperatures than water could.
QUESTION 12 5 marks Criterion C
Medium
04812162024-10020406080100120 Time (min) Temperature (°C)

The graph shows a substance being heated at a steady rate, starting as a solid.

a. What is the melting point of this substance?
[1]
b. What is the boiling point of this substance?
[1]
c. Calculate the rate of temperature increase, in $^\circ\text{C}$ per minute, while the solid is being heated, before melting begins ($t=0$ to $t=4$ minutes).
[3]
Show complete worked solution
(a)
$0\,^\circ\text{C}$ (the first flat section).
(b)
$100\,^\circ\text{C}$ (the second flat section).
(c)
$$ \text{rate} = \frac{0-(-10)}{4-0} = \frac{10}{4} = 2.5\,^\circ\text{C/min} $$
QUESTION 13 5 marks Criterion C
Medium
04812162035506580 Time (min) Temperature (°C)

The graph shows molten (liquid) candle wax cooling down at a steady rate as it loses heat to its surroundings.

a. At what temperature does the wax freeze?
[1]
b. How long does it take for all the wax to freeze (the flat section)?
[1]
c. Explain, in terms of particle energy, why the temperature does not fall during this time even though the wax is still losing heat to the surroundings.
[3]
Show complete worked solution
(a)
$55\,^\circ\text{C}$ (the flat section of the graph).
(b)
$6$ minutes (from $t=5$ to $t=11$ minutes).
(c)
During freezing, particles are still losing energy to the surroundings, but that energy loss corresponds to the particles slowing down and settling into a fixed, ordered arrangement (releasing the energy that had been keeping the forces of attraction overcome), rather than a further drop in their average kinetic energy. Since temperature depends on average kinetic energy, it stays constant until all of the particles have settled into the solid arrangement, after which the now-solid wax can cool further.
QUESTION 14 5 marks Criterion C
Medium

Trial12345
Melting point of naphthalene (°C)79.880.180.084.579.9

a. Identify the anomalous reading in this table.
[1]
b. Calculate the mean melting point, excluding the anomalous reading.
[2]
c. Suggest a reason for the anomalous trial 4 result, and state what should be done before finalising a result.
[2]
Show complete worked solution
(a)
Trial $4$ ($84.5\,^\circ\text{C}$) is anomalous — it stands out from the consistent cluster of $79.8$–$80.1\,^\circ\text{C}$.
(b)
$$ \text{mean} = \frac{79.8+80.1+80.0+79.9}{4} = \frac{319.8}{4} = 79.95\,^\circ\text{C} $$
(c)
Possible reasons include misreading the thermometer, or taking the reading before the sample had actually started/finished melting (timing error). The anomalous reading should be discarded, and, if time allows, the measurement repeated to confirm the more consistent mean of $79.95\,^\circ\text{C}$.
QUESTION 15 5 marks Criterion C
Medium

Time (min)0102030
Mass of water remaining at 20°C (g)50484644
Mass of water remaining at 40°C (g)50443832

a. Calculate the rate of mass loss (evaporation rate) at $20\,^\circ\text{C}$ and at $40\,^\circ\text{C}$, in g/min.
[2]
b. Which temperature gives the faster evaporation rate, and by what factor?
[1]
c. Explain this difference using the particle model.
[2]
Show complete worked solution
(a)
$20\,^\circ\text{C}$: $\dfrac{50-44}{30}=0.2\,\text{g/min}$. $\quad 40\,^\circ\text{C}$: $\dfrac{50-32}{30}=0.6\,\text{g/min}$.
(b)
$40\,^\circ\text{C}$ is faster — $0.6 \div 0.2 = 3$ times faster than at $20\,^\circ\text{C}$.
(c)
At the higher temperature, particles at the surface of the water have more kinetic energy on average, so more of them have enough energy to escape the liquid's surface and become gas particles per minute, giving a faster rate of evaporation (greater mass loss per minute).
QUESTION 16 7 marks Criterion C
Hard
048121620-102060100140170 Time (min) Temperature (°C)

The graph shows substance X being heated at a steady rate from a solid at $-10\,^\circ\text{C}$ to a gas at $170\,^\circ\text{C}$.

a. Describe fully what is happening to substance X during each of the five sections of the graph ($0$–$3$, $3$–$7$, $7$–$13$, $13$–$19$, $19$–$22$ minutes).
[3]
b. Calculate the total time spent changing state (melting plus boiling), and the total time spent heating within a single state (not changing state).
[2]
c. Calculate the rate of temperature increase while the liquid alone is being heated ($7$ to $13$ minutes).
[2]
Show complete worked solution
(a)

$0$–$3\,\text{min}$: the solid is heating up — particles vibrate faster, temperature rises from $-10$ to $20\,^\circ\text{C}$.

$3$–$7\,\text{min}$: the substance is melting at $20\,^\circ\text{C}$ — particles gain enough energy to break free of fixed positions and start moving around each other; temperature stays constant.

$7$–$13\,\text{min}$: the liquid is heating up — particles move faster on average, temperature rises from $20$ to $150\,^\circ\text{C}$.

$13$–$19\,\text{min}$: the substance is boiling at $150\,^\circ\text{C}$ — particles gain enough energy to completely overcome the remaining forces of attraction and become a gas; temperature stays constant.

$19$–$22\,\text{min}$: the gas is heating up — particles move even faster, temperature rises from $150$ to $170\,^\circ\text{C}$.

(b)
Changing state: melting ($4$ min) $+$ boiling ($6$ min) $= 10\,\text{min}$. Heating within a state: $(3-0)+(13-7)+(22-19) = 3+6+3 = 12\,\text{min}$.
(c)
$$ \text{rate} = \frac{150-20}{13-7} = \frac{130}{6} = 21.7\,^\circ\text{C/min}\ (\text{3 s.f.}) $$
QUESTION 17 7 marks Criterion C
Hard

Time (min)024681012
Liquid A temperature (°C)80655050503825
Liquid B temperature (°C)80604230302214
Two different liquids, A and B, were cooled from $80\,^\circ\text{C}$ and their temperature recorded every 2 minutes.

a. State the freezing point of Liquid A and of Liquid B, using the flat sections of the data.
[2]
b. A student claims: "Liquid B freezes faster than Liquid A." Evaluate this claim using the data.
[2]
c. Suggest one way to improve this investigation to more confidently compare how quickly each liquid freezes.
[3]
Show complete worked solution
(a)
Liquid A: $50\,^\circ\text{C}$ (constant across $t=4,6,8$). Liquid B: $30\,^\circ\text{C}$ (constant across $t=6,8$).
(b)
If "freezes faster" means reaching its freezing point sooner: Liquid A reaches its freezing point ($50\,^\circ\text{C}$) by $t=4\,\text{min}$, while Liquid B only reaches its freezing point ($30\,^\circ\text{C}$) by $t=6\,\text{min}$ — so A actually reaches its freezing point sooner, not B. Readings are only taken every $2$ minutes, so it is not possible to know exactly when each plateau truly ends without more frequent data — the claim is not clearly supported by this data.
(c)
Take temperature readings much more frequently (e.g. every $30$ seconds instead of every $2$ minutes), especially around the suspected plateau region, so the exact moment freezing starts and finishes can be identified precisely for both liquids, allowing a fair, well-supported comparison of freezing rate.
QUESTION 18 5 marks Criterion D
Medium

Refrigerators and air conditioners work using a special fluid (refrigerant) that is made to evaporate (absorbing heat energy, cooling the inside of the fridge) and then condense again (releasing that heat outside) in a repeating cycle.

Discuss one benefit and one drawback of this evaporation–condensation technology.

Show complete worked solution

Benefit: Because evaporation absorbs a large amount of heat energy from its surroundings as the refrigerant particles use that energy to overcome the forces holding them together as a liquid, this cycle can continuously remove heat from a small enclosed space (a fridge) or a room (air conditioning). This keeps food safe from bacterial growth for much longer, and makes hot climates safe and comfortable to live and work in.

Drawback: Running the compressor needed to keep this cycle going uses a significant amount of electricity, contributing to a building's energy consumption and carbon footprint. In addition, some refrigerant gases (such as older CFCs, and some HFCs still used today) are powerful greenhouse gases, or damage the ozone layer, if they leak into the atmosphere — which is why safer refrigerants and proper disposal or recycling of old fridges are now important.

QUESTION 19 5 marks Criterion D
Medium

Scrap metal (such as aluminium cans or steel) can be recycled by melting it down and recasting it into new products, instead of mining and refining new metal ore.

Discuss one benefit and one drawback of recycling metal this way.

Show complete worked solution

Benefit: Melting down existing metal to reuse it uses far less energy than extracting and refining brand-new metal from ore (for aluminium, recycling can use around $5\%$ of the energy of producing new metal). It also reduces the amount of mining needed, conserving natural resources and reducing the habitat destruction and pollution associated with mining.

Drawback: Melting metal (heating it past its melting point) still requires a large amount of energy and typically involves high-temperature furnaces, which can release significant emissions if the electricity or fuel used comes from burning fossil fuels. Collecting, sorting, and transporting scrap metal to recycling facilities also uses energy and resources.

QUESTION 20 6 marks Criterion D
Hard

In some water-scarce countries, seawater is turned into drinking water using thermal desalination: the seawater is heated until it evaporates (leaving the dissolved salt behind), and the water vapour is then cooled so it condenses back into pure liquid water.

Evaluate the impact of using this evaporation–condensation process to produce drinking water, discussing both a benefit and a concern.

Show complete worked solution

Benefit: This process provides a reliable source of clean, salt-free drinking water in regions with very little natural fresh water, since evaporation naturally separates water particles from the dissolved salt (only the water turns into vapour, not the salt), and condensing that vapour gives pure water. This allows large, dry coastal populations to have a stable water supply.

Concern: Heating a huge volume of seawater until it evaporates requires an enormous amount of energy, making thermal desalination expensive and, if powered by fossil fuels, a significant source of carbon emissions. The leftover concentrated salty water (brine) also has to be disposed of, and pumping it back into the sea in large quantities can harm marine ecosystems near the outflow point by making the local water far saltier and warmer than normal.

Diffusion and Particle Motion 20 questions

QUESTION 1 2 marks Criterion A
Easy

Define the term diffusion.

Show complete worked solution

Diffusion is the net (overall) movement of particles from a region of higher concentration to a region of lower concentration, caused by the particles' own random motion, continuing until the particles are evenly (uniformly) spread out.

QUESTION 2 2 marks Criterion A
Easy

State two states of matter in which diffusion happens easily, and explain why it does not happen (or happens only extremely slowly) in a solid.

Show complete worked solution

Diffusion happens easily in gases and liquids, because their particles are free to move around and change position. In a solid, particles are held in fixed positions by strong forces of attraction and can only vibrate on the spot — they cannot move from place to place to spread out, so diffusion essentially does not happen in solids.

QUESTION 3 2 marks Criterion A
Easy

A bottle of perfume is opened in the corner of a room. After a few minutes, people across the room can smell it, even though the air was completely still (no draught or fan).

Explain, using the particle model, why the smell spreads across the room without any stirring or wind.

Show complete worked solution

Gas particles (perfume vapour) are constantly moving randomly and quickly in all directions. Through this random motion, particles naturally spread out from where they are concentrated (near the bottle) into the surrounding air where there are fewer of them, eventually reaching all parts of the room. This is diffusion, and it happens on its own, without needing to be stirred.

QUESTION 4 4 marks Criterion A
Medium
t = 0 s t = 30 s

The diagrams show the same closed container of gas particles at two different times.

a. Describe how the arrangement of particles has changed between the two diagrams.
[1]
b. Name this process.
[1]
c. Explain, in terms of particle movement, why this happens, even though no one stirred or pushed the particles.
[2]
Show complete worked solution
(a)
The particles have spread out from being clustered in one area of the container (at $t=0\,\text{s}$) to being spread randomly throughout the whole container (at $t=30\,\text{s}$).
(b)
Diffusion.
(c)
Gas particles are always moving randomly and quickly in all directions, colliding with each other and the container walls. Over time, this random motion causes a net movement of particles away from the crowded area toward the emptier areas, simply because there is more empty space to randomly move into there, until the particles become evenly spread throughout the container.
QUESTION 5 3 marks Criterion A
Medium

Explain why diffusion happens much faster in a gas than in a liquid, even though both involve random particle movement.

Show complete worked solution

In a gas, particles are much further apart, move much faster, and collide with each other far less often, so they can travel large distances quickly and freely. In a liquid, particles are packed close together and constantly collide with their close neighbours, which slows down how quickly they can spread through the liquid — so diffusion in a gas is much faster than in a liquid.

QUESTION 6 3 marks Criterion A
Medium

Explain what Brownian motion is, and what it provides evidence for.

Show complete worked solution

Brownian motion is the continuous, random, jittery movement of small particles (such as pollen grains or smoke particles) that can be observed under a microscope, caused by these particles being constantly and unevenly bombarded by fast-moving, invisibly small particles (air or water molecules) around them. It provides strong evidence that gas and liquid particles are in constant, random motion, even though the individual gas or liquid particles themselves are far too small to see directly.

QUESTION 7 4 marks Criterion A
Hard

Diffusion happens faster under some conditions than others.

a. State two factors that increase the rate of diffusion, and explain how one of them has this effect.
[2]
b. Predict and explain whether smaller, lighter particles or larger, heavier particles diffuse faster, if all other conditions are the same.
[2]
Show complete worked solution
(a)
Factors: a higher temperature, and a bigger concentration difference (concentration gradient) between the two regions (also: smaller/lighter particles). Explanation for temperature: raising the temperature gives particles more kinetic energy, so they move faster on average, spreading out (diffusing) more quickly.
(b)
Smaller, lighter particles generally diffuse faster. For the same amount of energy (at a given temperature), a lighter particle can be accelerated to a higher speed than a heavier particle, so lighter particles move — and therefore spread out (diffuse) — more quickly on average.
QUESTION 8 7 marks Criterion B
Medium

A student wants to investigate how water temperature affects the rate at which a coloured dye diffuses through still water, by placing a small crystal of potassium permanganate at the bottom of a beaker of water and timing how long it takes for the purple colour to spread a fixed distance up the beaker.

a. State the independent variable and the dependent variable for this investigation.
[2]
b. State two variables that should be controlled, and explain why not stirring matters.
[2]
c. Describe a method for carrying out this investigation.
[3]
Show complete worked solution
(a)
Independent variable: temperature of the water. Dependent variable: time taken for the purple colour to spread a fixed distance (e.g. $5\,\text{cm}$) up the beaker.
(b)
Use the same size of potassium permanganate crystal, the same volume/depth of water, and the same size/shape of beaker for every temperature tested. Do not stir the water in any test — stirring would spread the colour by mixing (convection currents), not by diffusion alone, giving misleading results.
(c)
  1. Fill identical beakers with the same volume of water at different set temperatures (e.g. $10\,^\circ\text{C}$, $25\,^\circ\text{C}$, $40\,^\circ\text{C}$, $55\,^\circ\text{C}$, checked with a thermometer).
  2. Using forceps, carefully drop one identical-sized crystal of potassium permanganate into the bottom-centre of each beaker at the same moment, without stirring.
  3. Start a stopwatch immediately, and record the time taken for the purple colour to reach a mark $5\,\text{cm}$ above the crystal in each beaker.
  4. Repeat each temperature at least twice and take a mean time.
QUESTION 9 7 marks Criterion B
Medium
Ammonia-soaked cotton wool HCl-soaked cotton wool sealed glass tube 0 cm 50 cm white ring (NH?Cl) forms where the two gases meet by diffusion

A long glass tube is set up with a piece of cotton wool soaked in concentrated ammonia solution pushed into one end, and a piece of cotton wool soaked in concentrated hydrochloric acid pushed into the other end, at the same moment. Both liquids release gas that diffuses along the tube; where the two gases meet, they react to form a white solid ring (ammonium chloride).

a. State the independent variable and the dependent variable for this investigation.
[2]
b. State one variable that must be controlled for this to be a fair test, and explain why.
[2]
c. Describe how the position of the white ring could be used to determine which gas diffuses faster.
[3]
Show complete worked solution
(a)
Independent variable: which gas is being tracked (ammonia or hydrogen chloride). Dependent variable: the distance travelled by each gas from its end of the tube to where the white ring forms.
(b)
Both pieces of cotton wool must be pushed in at exactly the same time, into an identical tube (same length/diameter) with both ends sealed at the same moment. If one gas were given a head start, it would appear to have diffused further even if it wasn't actually faster, giving a misleading comparison.
(c)
Measure the distance from each end of the tube to where the white ring forms. The gas that has travelled the greater distance in the same amount of time has diffused faster. Since ammonia gas has lighter, smaller particles than hydrogen chloride gas, the ring is expected to form closer to the hydrochloric acid end, showing that ammonia diffused faster.
QUESTION 10 8 marks Criterion B
Hard

In the potassium permanganate diffusion investigation, one student's crystal was noticeably larger than the crystals used by other students in the class, but everyone shared the same overall results for comparing temperatures.

a. Explain why using a larger crystal in one test would make the comparison between temperatures unreliable.
[2]
b. Suggest a way to make sure the crystal size is properly controlled across all tests.
[3]
c. The ammonia/hydrochloric acid tube investigation is normally done inside a fume cupboard with the sash (glass screen) mostly closed, rather than in the open room. Explain why this is important for getting a reliable result, in addition to safety.
[3]
Show complete worked solution
(a)
A larger crystal contains more potassium permanganate, so it releases more coloured particles into the water and creates a stronger starting concentration gradient than a smaller crystal. This would likely make the colour appear to spread the fixed distance faster, for a reason unrelated to temperature, making it impossible to fairly compare that student's timing result with the others'.
(b)
Use crystals of a measured, consistent mass (e.g. weighed on an electronic balance to $0.05\,\text{g}$ each) rather than crystals simply judged "about the same size" by eye, so that every test genuinely starts with the same amount of potassium permanganate and only temperature is different between tests.
(c)
Air currents or draughts, whether from an open room or from the fume cupboard's own extraction fan if the sash is fully open, could push the gases along the tube in one direction or disturb them unevenly. This would mean the white ring's position reflects air movement rather than showing where the two gases would have met by diffusion alone. Keeping the tube still and undisturbed (with the sash mostly closed to minimise draughts) ensures the result reflects diffusion, not air currents.
QUESTION 11 3 marks Criterion C
Easy

Time (s)010203040
Distance colour has spread (cm)01234

a. Describe the pattern shown by the data.
[1]
b. Calculate the rate of diffusion shown by this data, in cm per second.
[2]
Show complete worked solution
(a)
The distance increases steadily at a constant rate — it increases by $1\,\text{cm}$ every $10$ seconds.
(b)
$$ \text{rate} = \frac{4\,\text{cm}}{40\,\text{s}} = 0.1\,\text{cm/s} $$
QUESTION 12 5 marks Criterion C
Medium

Time (min)02468
Distance spread at 20°C (cm)00.81.62.43.2
Distance spread at 50°C (cm)02.04.06.08.0

a. Calculate the rate of diffusion at $20\,^\circ\text{C}$ and at $50\,^\circ\text{C}$, in cm/min.
[2]
b. How many times faster is diffusion at $50\,^\circ\text{C}$ compared with $20\,^\circ\text{C}$?
[1]
c. Explain this result using the particle model.
[2]
Show complete worked solution
(a)
$20\,^\circ\text{C}$: $\dfrac{3.2}{8}=0.4\,\text{cm/min}$. $\quad 50\,^\circ\text{C}$: $\dfrac{8.0}{8}=1.0\,\text{cm/min}$.
(b)
$$ \frac{1.0}{0.4} = 2.5 \text{ times faster} $$
(c)
At the higher temperature, particles have more kinetic energy and move faster on average, so they spread out (diffuse) through the water more quickly, covering a greater distance in the same amount of time.
QUESTION 13 5 marks Criterion C
Medium

Trial1234
Distance from ammonia end to ring (cm)32333132
Distance from HCl end to ring (cm)18171918
This data is from an ammonia/hydrochloric acid diffusion tube, $50\,\text{cm}$ long, repeated four times.

a. Calculate the mean distance from each end of the tube to the ring.
[2]
b. Which gas diffused further along the tube, and therefore diffuses faster?
[1]
c. Suggest why ammonia particles diffuse faster than hydrogen chloride particles, in terms of the particle model.
[2]
Show complete worked solution
(a)
Ammonia end: $\dfrac{32+33+31+32}{4}=\dfrac{128}{4}=32\,\text{cm}$. HCl end: $\dfrac{18+17+19+18}{4}=\dfrac{72}{4}=18\,\text{cm}$.
(b)
Ammonia gas diffused further ($32\,\text{cm}$ vs $18\,\text{cm}$), so ammonia diffuses faster.
(c)
Ammonia particles ($\text{NH}_3$) have a smaller mass than hydrogen chloride particles ($\text{HCl}$). For a given amount of kinetic energy at the same temperature, lighter particles move at higher speeds, so ammonia diffuses (spreads/travels) faster along the tube than the heavier hydrogen chloride gas.
QUESTION 14 5 marks Criterion C
Medium

Gas% in air breathed in% in air breathed out
Oxygen21%16%
Carbon dioxide0.04%4%

a. Calculate the change in percentage of oxygen and of carbon dioxide between the air breathed in and the air breathed out.
[2]
b. Explain, using the idea of diffusion and concentration gradients, why oxygen decreases and carbon dioxide increases between breathing in and breathing out.
[3]
Show complete worked solution
(a)
Oxygen: decreases by $21-16=5$ percentage points. Carbon dioxide: increases by $4-0.04=3.96 \approx 4$ percentage points.
(b)
In the lungs (alveoli), the concentration of oxygen is higher in the air than in the blood, so oxygen diffuses (net movement) from the air into the blood, down its concentration gradient — meaning the air breathed back out contains less oxygen than was breathed in. At the same time, the concentration of carbon dioxide is higher in the blood than in the air, so carbon dioxide diffuses from the blood into the air in the alveoli, down its own concentration gradient — meaning the air breathed out contains more carbon dioxide than was breathed in.
QUESTION 15 5 marks Criterion C
Medium
05101520253001020304050 Time (min) Concentration at point (%)

A dye is released at one end of a tank of still water. The graph shows the concentration of dye measured at a fixed point elsewhere in the tank over time.

a. Describe how the concentration at this point changes over the 30 minutes.
[1]
b. Calculate the rate of change of concentration between $t=0$ and $t=5$ minutes, and between $t=20$ and $t=25$ minutes.
[2]
c. Explain, using the idea of diffusion, why the rate of change slows down over time.
[2]
Show complete worked solution
(a)
It rises quickly at first, then rises more and more slowly, before levelling off (becoming almost constant) at around $42\%$ after about $20$–$25$ minutes.
(b)
$t=0$ to $5\,\text{min}$: $\dfrac{20-0}{5}=4\%/\text{min}$. $\quad t=20$ to $25\,\text{min}$: $\dfrac{42-41}{5}=0.2\%/\text{min}$.
(c)
Diffusion happens faster when there is a bigger difference (concentration gradient) between two regions. As particles spread out and the concentration becomes more even throughout the water, the concentration gradient gets smaller, so the net rate of diffusion — and therefore the rate at which the concentration at this point changes — slows down, eventually levelling off once the particles are evenly spread (equilibrium) throughout the water.
QUESTION 16 7 marks Criterion C
Hard

Trial12345
Time for colour to reach 5 cm mark, at 25°C (s)11812211595120

a. Identify the anomalous result and explain how you identified it.
[2]
b. Calculate the mean time, excluding the anomalous result.
[2]
c. Suggest a plausible reason for the anomalous trial 4 result, and calculate the rate of diffusion (cm/s) using the corrected mean time.
[3]
Show complete worked solution
(a)
Trial $4$ ($95\,\text{s}$) is anomalous — it is noticeably faster (lower) than the other four fairly consistent readings ($115$–$122\,\text{s}$).
(b)
$$ \text{mean} = \frac{118+122+115+120}{4} = \frac{475}{4} = 118.75\,\text{s} $$
(c)
Trial $4$ may have used a slightly larger or already-crushed crystal, been affected by an accidental knock that stirred the water slightly, or involved a misread stopwatch — any of these could make the colour appear to reach $5\,\text{cm}$ faster without diffusion genuinely being quicker. Using the corrected mean: $$ \text{rate} = \frac{5}{118.75} = 0.0421\,\text{cm/s}\ (\text{3 s.f.}) $$
QUESTION 17 7 marks Criterion C
Hard

GasDistance travelled in 60 s (cm)
Gas P22
Gas Q11
This is a single trial for each gas. A student concludes: "Because Gas P travelled exactly twice as far as Gas Q in this data, Gas P definitely diffuses exactly twice as fast as Gas Q, and this must always be true."

a. Evaluate whether this single set of data is enough evidence to support the student's conclusion.
[2]
b. Explain why the phrase "must always be true" is not fully justified, even if the ratio were confirmed by repeats, using ideas about the conditions of the test.
[2]
c. Describe what additional data collection would make this conclusion much more reliable.
[3]
Show complete worked solution
(a)
No — this is based on only one trial for each gas, with no repeats. A single measurement could easily be affected by a random error (e.g. a slight draught, an inconsistent starting release of gas) and doesn't show whether the result is reliable or repeatable. Without repeating the test multiple times and finding a similarly consistent ratio, it is not possible to confidently conclude the ratio is "exactly" two.
(b)
The relative diffusion rates measured here were only tested under one set of conditions (e.g. one particular temperature and concentration). Diffusion rate depends on temperature, concentration gradient and particle properties, so the ratio between Gas P and Gas Q's rates could be different under other conditions (e.g. a different temperature). A conclusion based on one test at one temperature cannot safely be generalised to "always."
(c)
Repeat the measurement for both Gas P and Gas Q multiple times (e.g. $5$ repeats each) under the exact same conditions, and calculate a mean distance travelled for each gas, to check whether the roughly $2{:}1$ ratio is consistent and not due to chance. Ideally, also repeat the whole comparison at a few different temperatures, to check whether the same ratio between the two gases' rates holds generally, before concluding anything about how the gases compare "always."
QUESTION 18 5 marks Criterion D
Medium

Fresh food such as salad, sliced meat or bread is often sold in sealed plastic packets filled with a special mixture of gases (low in oxygen, high in carbon dioxide or nitrogen) instead of ordinary air, known as Modified Atmosphere Packaging (MAP). This slows the diffusion of oxygen into the food, which slows the growth of bacteria and the chemical reactions that cause food to spoil or turn brown.

Discuss one benefit and one drawback of using this packaging method.

Show complete worked solution

Benefit: By reducing the concentration of oxygen inside the sealed packet, the concentration gradient driving oxygen diffusion into the food (and supporting bacterial growth and browning reactions) is much smaller, so food stays fresher for significantly longer. This reduces food waste for both shops and households, and allows fresh food to be transported further or stored longer before it needs to be eaten.

Drawback: This packaging typically uses more plastic material (to make an airtight seal that stops outside air diffusing back in) than simpler packaging, adding to plastic waste. The mixed layers of different plastics used for a good seal can also be difficult or impossible to recycle, so the environmental benefit of less food waste has to be weighed against the extra packaging waste produced.

QUESTION 19 5 marks Criterion D
Medium

Understanding how gases diffuse and spread through the air helps scientists predict how far pollutant gases released by factories, power stations or traffic will travel, and how concentrated they will be at a given distance. This is used to set safe locations for schools and housing, and to design tall chimneys ("smokestacks") that release pollution high above ground level.

Discuss one benefit and one drawback related to how pollutant gases diffuse through the air.

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Benefit: Because diffusion spreads pollutant gases from a smokestack outward and upward through a very large volume of air, releasing them from a tall chimney means the gas is diluted (spread out and mixed with far more air) by the time it reaches ground level near people, reducing the concentration people actually breathe in near the source. This predictable spreading behaviour also lets planners keep new housing or schools a safe distance from major pollution sources.

Drawback: Diffusion means pollutant gases inevitably spread beyond the immediate area of the factory or road, so people and ecosystems far from the original source can still be affected over time as the gases gradually diffuse and get carried by wind over long distances. Pollution cannot simply be "contained" at its source, which is why controlling emissions at the source, rather than just relying on dilution and diffusion, is still necessary.

QUESTION 20 6 marks Criterion D
Hard

During surgery, patients are often given anaesthetic gases to breathe, which diffuse from the lungs into the bloodstream and then to the brain, allowing complex, pain-free operations to be carried out safely. After the operation, exhaled anaesthetic gas is normally extracted by the hospital's ventilation system and released into the outside air. Some commonly used anaesthetic gases, such as nitrous oxide, are also very potent greenhouse gases — much more effective at trapping heat in the atmosphere than the same mass of carbon dioxide.

Evaluate the impact of using inhaled anaesthetic gases in medicine, discussing both a benefit and a concern.

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Benefit: Because these gases diffuse rapidly from the lungs into the blood and reach the brain, they allow doctors to safely and reliably keep a patient unconscious and free of pain during long or complex operations that would otherwise be impossible to carry out. This diffusion-based delivery method is fast-acting and its effects can also be reversed relatively quickly once the gas supply stops and it diffuses back out of the body, making it a controllable and essential tool in modern medicine, saving countless lives.

Concern: Because these gases inevitably diffuse away from the patient (some remains in exhaled breath and some escapes from equipment), and hospitals worldwide use large volumes of them every day, gases like nitrous oxide that leak or vent into the atmosphere contribute a meaningful amount to global greenhouse gas emissions and climate change, given how much more strongly they trap heat than carbon dioxide. Hospitals are increasingly investing in gas-capture and recycling systems to reduce how much anaesthetic gas is released into the atmosphere, balancing patient care against environmental impact.