Answers · The particle model

Temperature and absolute zero — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. PQ22 · Practice

Model answer. On average the molecules in Q are faster, though some in P are faster than some in Q. The temperature of a gas is a measure of the average kinetic energy of its molecules: at 70 °C the average is greater than at 30 °C, so on average the molecules in Q move faster. But in each flask the molecules do not all move at the same speed: collisions keep changing each one’s speed, so at any moment some are fast, some slow and most in between. So a few slow molecules in Q are slower than a few fast ones in P. The molecules are the same size in both flasks, and both flasks hold the same number of them.

A full answer includes

I chose: on average Q’s are faster, though some in P are faster than some in Q.
I wrote that a higher temperature means a greater average kinetic energy.
I wrote that the molecules in each flask have a range of speeds, so a slow molecule in Q can be slower than a fast one in P.
I did not choose that the molecules in Q have grown bigger: heating does not change the molecules.

Also accept

b: a higher temperature means a greater average kinetic energy, but each flask has a range of speeds, so not every molecule in Q is faster.

Watch for

“Every single molecule in Q is moving faster than every one in P.” — Only the average is greater in Q. In each flask, collisions keep changing the speeds, so some molecules in Q are slow and some in P are fast.
“The molecules in Q have grown bigger, so they carry more kinetic energy.” — Heating does not change the size of the molecules. They have more kinetic energy on average because they move faster.
“Q has more molecules in it, so each one has more kinetic energy.” — Both flasks hold the same mass of nitrogen, so the same number of molecules. The difference is how fast they move, on average.

2. PQ23 · Practice

Model answer. Row A. The copper is hotter, so its particles move faster on average and their average kinetic energy is greater. It is still a solid, so the particles stay in their fixed positions and only vibrate, faster than before. Heating does not change the particles themselves: each one is the same size. (The block gets very slightly bigger because the particles, vibrating harder, take up a little more room, not because they grow.)

A full answer includes

I chose row A.
I identified that the particles vibrate faster about their fixed positions.
I identified that their average kinetic energy is greater.
I identified that each particle stays the same size.
I did not choose row C: the copper is still a solid, so its particles stay in fixed positions.

Also accept

A: the particles vibrate faster about fixed positions, with a greater average kinetic energy, and stay the same size.

Watch for

“Row B: the particles get larger when the copper is heated.” — The particles stay the same size. A hotter block is very slightly bigger because its particles vibrate harder and take up a little more room.
“Row C: hotter particles move about, sliding past each other.” — That happens only when the copper melts. At 80 °C it is still a solid, so its particles vibrate faster about fixed positions.
“Row D: the particles of a solid cannot move faster, because they are fixed.” — They are fixed in position, but they vibrate. When the solid is hotter, they vibrate faster, so their average kinetic energy is greater.

3. PQ24 · Practice

Model answer. Carried on with a ruler, the straight line reaches zero kinetic energy at about −273 °C. It cannot be carried on below that: it would give a negative kinetic energy, and a particle cannot move more slowly than not moving. So −273 °C, absolute zero, is the lowest possible temperature, where the particles have the least kinetic energy.

Checked automatically

Temperature where the line reaches zero kinetic energy: -273 °C (within ±12 °C)
- Common wrong answer 273 : The line reaches zero far to the left of 0 °C, at a temperature below zero. Did you leave out the minus sign?
- Common wrong answer 0 : At 0 °C the line is still at about 5.7 × 10⁻²¹ J. Carry it on much further to the left.
- Common wrong answer -150 : That is the coldest point drawn, where the energy is still 2.55 × 10⁻²¹ J. Carry the line on further.
- Common wrong answer -300 : That is the end of the axis. The line reaches zero before that: look where it meets the axis.

A full answer includes

I carried the straight line on, with a ruler, below the last point drawn.
I read about −273 °C where it reaches zero kinetic energy (anything from −285 °C to −261 °C from a careful drawing).
I wrote that below this temperature the line would give a negative kinetic energy, which is impossible.
I did not give the temperature of the coldest point drawn.

Also accept

About −273 °C (from −285 °C to −261 °C); below that the kinetic energy would be negative, which is impossible.

Watch for

“The line reaches zero at 0 °C.” — At 0 °C the line is still high on the energy scale. Zero kinetic energy is much further to the left.
“The line could be carried on for ever, to any colder temperature.” — Below zero kinetic energy, the line would give a negative energy of motion, which is impossible. Nothing can be colder than absolute zero.

4. PQ25 · Practice

Options as numbered on the pupil sheet: 1 — 0 m/s; 2 — 438 m/s; 3 — 485 m/s; 4 — 515 m/s; 5 — 582 m/s.

Model answer. The hotter a gas, the faster its molecules move on average. So the coldest temperature, −50 °C, goes with the slowest of the four speeds, 438 m/s; 0 °C with 485 m/s; 35 °C with 515 m/s; and the hottest, 120 °C, with the fastest, 582 m/s. 0 m/s is not used. 0 °C is only the temperature at which ice melts: air molecules at 0 °C still move at hundreds of metres per second.

A full answer includes

I matched −50 °C to 438 m/s.
I matched 0 °C to 485 m/s.
I matched 35 °C to 515 m/s.
I matched 120 °C to 582 m/s.
I did not match 0 °C to 0 m/s.

Also accept

−50 °C – 438 m/s; 0 °C – 485 m/s; 35 °C – 515 m/s; 120 °C – 582 m/s; 0 m/s not used.

Watch for

“0 °C goes with 0 m/s, because the temperature is zero.” — 0 °C is only the temperature at which ice melts. Air molecules at 0 °C still move fast, at hundreds of metres per second.
“The coldest temperature goes with the fastest speed.” — It is the other way round: the colder the gas, the more slowly its molecules move on average.

5. PQ26 · Practice

Options as numbered on the pupil sheet: 1 — They hit each speck harder; 2 — The speck’s jumps are bigger: it jiggles more; 3 — The water is warmed; 4 — The uneven push on a speck at each moment is bigger; 5 — Its molecules move faster, with more kinetic energy on average.

Model answer. When the water is warmed, its molecules move faster, so their average kinetic energy is greater. They hit each speck harder. The hits are still uneven, from all sides, but the uneven push on a speck at each moment is bigger. So the speck’s jumps are bigger: it jiggles more.

A full answer includes

Step 1: I put “The water is warmed”.
Step 2: I put that its molecules move faster, with more kinetic energy on average.
Step 3: I put that they hit each speck harder.
Step 4: I put that the uneven push on a speck at each moment is bigger.
Step 5: I put that the speck’s jumps are bigger: it jiggles more.
I did not put “the uneven push is bigger” before the molecules hit harder.

Also accept

warmed; molecules faster, more kinetic energy; hit harder; uneven push bigger; bigger jumps, jiggles more.

Watch for

“The specks jiggle more because the warm specks have more energy of their own.” — The specks are pushed about by the water molecules. They jiggle more because the molecules hit them harder.
“Warming makes the specks smaller, so they are easier to push.” — The specks are the same size at 40 °C as at 10 °C. Only the molecules that hit them have changed: they move faster.

6. PQ27 · Practice

Model answer. (a) The temperatures are 60 °C apart. The fall between neighbouring values is 7.521 − 6.278 = 1.243, 6.278 − 5.036 = 1.242 and 5.036 − 3.793 = 1.243, all in units of 10⁻²¹ J. So the average kinetic energy falls by 1.24 × 10⁻²¹ J for every 60 °C colder, to 3 significant figures. (b) At −90 °C there is 3.793 × 10⁻²¹ J left. That runs out after 3.793 ÷ 1.24 ≈ 3.06 more steps of 60 °C, which is about 3.06 × 60 ≈ 183 °C. So the average kinetic energy reaches zero at about −90 − 183 = −273 °C.

Checked automatically

(a) Fall in average kinetic energy for every 60 °C colder: 1.24×10^-21 J (3 significant figures)
- Common wrong answer 3.73e-21 : That is the fall over the whole 180 °C, from 90 °C to −90 °C. Divide by 3 for each 60 °C.
- Common wrong answer 1.24 : The energies in the table are in units of 10⁻²¹ J. Give the answer in joules, with the power of ten (type it as 1.24x10⁻²¹).
- Common wrong answer 2.07e-21 : That is the fall for every 100 °C. The question asks for every 60 °C, the step between the temperatures in the table.
- Common wrong answer 2.07e-23 : That is the fall for each 1 °C. The question asks for every 60 °C.
(b) Temperature where the average kinetic energy reaches zero: -273 °C (within ±4 °C)
- Common wrong answer -90 : At −90 °C there is still 3.793 × 10⁻²¹ J left. Work out how much further down it runs out.
- Common wrong answer 183 : That is how far below −90 °C the energy runs out. Now find the temperature: −90 − 183.
- Common wrong answer 93 : You went the wrong way. The energy runs out 183 °C colder than −90 °C: −90 − 183.
- Common wrong answer 273 : The temperature is below zero. Did you leave out the minus sign?

A full answer includes

(a) I found the fall for one 60 °C step: for example 7.521 − 6.278, or (7.521 − 3.793) ÷ 3.
(a) I wrote 1.24 × 10⁻²¹ J, to 3 significant figures.
(a) I did not leave out the power of ten.
(b) I divided the energy left at −90 °C by the fall per step: 3.793 ÷ 1.24 ≈ 3.06 steps of 60 °C, about 183 °C.
(b) I wrote about −273 °C.
(b) I did not give 183 °C, the distance below −90 °C, as the temperature.

Also accept

(a) 1.24 × 10⁻²¹ J (3 s.f.); (b) about −273 °C.

Watch for

“(a) The fall is 3.73 × 10⁻²¹ J.” — That is the fall over the whole table, 180 °C. Divide by 3 for each 60 °C step.
“(b) The energy reaches zero at −90 °C, the coldest temperature in the table.” — At −90 °C there is still 3.793 × 10⁻²¹ J left. The line keeps falling below the table.

7. PQ28 · Practice

Model answer. Correct: the hydrogen molecules are still moving, so they have kinetic energy; on average they move more slowly than at −200 °C, because the gas is colder; and −250 °C is about 23 °C above the lowest possible temperature, −273 °C. Incorrect: the molecules have not stopped, because −250 °C is warmer than absolute zero, and even there real particles keep a tiny motion. No cooler, however good, could reach −280 °C: at absolute zero the particles already have the least kinetic energy they can have, so nothing can be colder. And −250 °C is not colder than absolute zero: −273 °C is colder still.

A full answer includes

I chose: its molecules are still moving, so they have kinetic energy.
I chose: on average, its molecules move more slowly than at −200 °C.
I chose: it is about 23 °C above the lowest possible temperature.
I did not choose that the molecules have stopped: they still move above absolute zero.
I did not choose that a better cooler could reach −280 °C: nothing can be colder than absolute zero.
I did not choose that it is colder than absolute zero: −250 °C is warmer than −273 °C.

Also accept

a, b and c.

Watch for

“Its molecules have stopped moving completely, as it is so cold.” — −250 °C is still 23 °C above absolute zero, so the molecules still move. Even at absolute zero, real particles keep a tiny motion.
“A better cooler could take the hydrogen down to −280 °C.” — The limit is not the cooler. At −273 °C the particles already have the least kinetic energy they can have, so nothing can be colder.
“It is colder than absolute zero, so it has no kinetic energy.” — Absolute zero, about −273 °C, is colder than −250 °C. Nothing can be colder than absolute zero.

8. PQ29 · Practice

Model answer. The pupil is incorrect. If the molecules had stopped, their speed and their kinetic energy would both be zero. At −180 °C the model molecules still move at a typical speed of 283 m/s, faster than a racing car, and their average kinetic energy, 1.93 × 10⁻²¹ J, is well above zero. They have slowed down compared with room temperature, but they are still moving fast, in random directions. Cooling makes particles slower, not still.

A full answer includes

I wrote that the molecules are still moving, at a typical speed of 283 m/s.
I wrote that their average kinetic energy, 1.93 × 10⁻²¹ J, is well above zero.
I wrote that the molecules have only slowed down: they move more slowly than at room temperature.
I did not write that cold air has no kinetic energy.

Also accept

They still move at a typical 283 m/s, and their average kinetic energy, 1.93 × 10⁻²¹ J, is not zero: they are slower, not stopped.

Watch for

“Below 0 °C, particles cannot move.” — Particles move at every temperature above absolute zero. In the model, air at −180 °C still has molecules moving at hundreds of metres per second.
“Very cold air has no kinetic energy left.” — The model shows 1.93 × 10⁻²¹ J for each molecule at −180 °C: less than at room temperature, but well above zero.

9. PQ30 · Practice

Model answer. Helium. It boils at −269 °C, so all the way down to −260 °C it is still a gas, and its molecules can be measured as a gas. Argon (−186 °C), neon (−246 °C) and hydrogen (−253 °C) all boil at temperatures above −260 °C, so each one would turn into a liquid, then a solid, before she got there. That is why the model keeps its air a gas all the way down: the cold end of its straight line is idealised. Real gases turn into liquids before absolute zero.

A full answer includes

I chose helium.
I wrote that helium boils at −269 °C, so at −260 °C it is still a gas.
I wrote that argon, neon and hydrogen boil above −260 °C, so they would be liquids, or solids, before she got there.
I did not choose hydrogen: −253 °C is warmer than −260 °C.

Also accept

Helium: it boils at −269 °C, below −260 °C; the other three boil above −260 °C, so they would be liquids by then.

Watch for

“Argon, because it has the highest boiling point, so it stays a gas the longest.” — It is the other way round. A gas with a high boiling point turns into a liquid early, as it is cooled. Argon is a liquid below −186 °C.
“Neon, because −246 °C is very cold.” — Neon is a liquid below −246 °C, so it is no longer a gas long before −260 °C.
“Hydrogen, because −253 °C is close enough to −260 °C.” — −253 °C is warmer than −260 °C, so hydrogen is already a liquid at −260 °C.

10. PT3 · Apply

Model answer. (a) The temperature of a gas is a measure of the average kinetic energy of its molecules. At −20 °C the average is smaller than at 20 °C, so on average the molecules move more slowly. But at both temperatures the molecules have a range of speeds, because collisions keep changing each one’s speed. So some molecules at −20 °C are faster than some were at 20 °C: only the average is lower. (b) A better headline: “Xenon atoms cooled almost to absolute zero: slowest motion ever”. The atoms in the crystal still vibrate about fixed positions. They are not quite at absolute zero, and even there real particles have the least kinetic energy they can have, not none: “stop dead” is incorrect. (c) The specks jiggle more at 50 °C. There the water molecules move faster on average, with a greater average kinetic energy, so they hit the specks harder. The hits are still uneven, from all sides, but each uneven push is bigger, so the jumps are bigger. At 10 °C the specks still jiggle, only less.

A full answer includes

(a) I wrote that the temperature tells us only about the average kinetic energy of the molecules.
(a) I wrote that at both temperatures the molecules have a range of speeds, so some at −20 °C are faster than some were at 20 °C.
(a) I wrote that, on average, the molecules at −20 °C are slower.
(b) I suggested a headline that says the atoms are nearly at absolute zero, with almost the least kinetic energy, not stopped.
(b) I wrote that the atoms still vibrate: they have not reached absolute zero, and even there real particles keep a tiny motion.
(c) I predicted that the specks jiggle more at 50 °C.
(c) I wrote that at 50 °C the water molecules move faster on average, with more kinetic energy.
(c) I wrote that they hit the specks harder, so the uneven pushes are bigger.
(c) I did not write that the specks at 10 °C stop jiggling.

Also accept

(a) only the average speed falls, and each sample has a range of speeds; (b) e.g. “Xenon cooled almost to absolute zero”: least kinetic energy, not stopped; (c) at 50 °C: faster molecules, harder uneven hits.

Watch for

“(a) In the cold store, every molecule slows down by the same amount.” — Collisions keep changing each molecule’s speed, at any temperature. Only the average kinetic energy falls; some molecules are still fast.
“(b) At absolute zero the atoms have no energy at all, so “stop dead” is right.” — At absolute zero the particles have the least kinetic energy they can have, but real particles keep a tiny motion. And the crystal is still 0.01 °C above it.
“(c) At 10 °C the specks do not jiggle at all.” — At 10 °C the water molecules still move fast, so the specks still jiggle, only a little less than at 50 °C.