Squeezing a gas — answer key

1. GQ1 · Practice

Model answer. Particles moving at random hit the walls and bounce off. The particles of a gas are far apart and move fast, at random, in all directions. Each time one hits a wall it bounces off and gives the wall a small push, and billions of billions of hits each second add up to the pressure. Because the motion is random, every wall gets the same hits whichever way it faces, so the gauge reads the same whichever way up the syringe is held. The weight of the air in a syringe is far too small to matter.

A full answer includes

• I chose “Particles moving at random hit the walls and bounce off”.
• I explained that the particles move at random, in all directions, so every wall is hit just as often whichever way it faces: the pressure is the same in all directions.

Also accept

• b; the particles move at random in all directions, so the hits are the same on every wall.

Watch for

• “The weight of the air presses down on the bottom of the syringe.” — The weight of 50 cm³ of air is tiny: from the top of the syringe to the bottom it changes the pressure by only about a thousandth of a kilopascal. The gauge reads 100 kPa whichever way up the syringe is held, and the air pushes just as hard on the side walls. The push comes from moving particles hitting in all directions.
• “The particles are packed tightly and push outwards on the walls.” — In a gas the particles are far apart, with empty space between them. They push on a wall only when they hit it and bounce off.

2. GQ2 · Practice

Model answer. (1) random: every direction is equally likely. (2) force: each hit is a small push on the wall. (3) area: the pressure is the force on each unit of area, such as each square centimetre. (4) often: more hits on each unit of area each second give a bigger force on it, so a bigger pressure.

A full answer includes

• (1) random (or all)
• (2) force (or push)
• (3) area
• (4) often (or frequently)

Also accept

• random; force; area; often.

Watch for

• “(3) volume” — Pressure is the force on each unit of area of a surface, such as each square centimetre of a wall, not on each unit of volume of the gas.
• “(4) slowly” — Slower particles hit the walls less often and less hard, so the pressure would fall, not rise.

3. GQ3 · Practice

Model answer. The force on the lid is bigger, but the pressure on both is the same. The particles move at random, so each square centimetre inside the box gets the same number of hits each second: the force on each unit of area is the same everywhere, so the pressure on both is 120 kPa. The lid has 10 times the area of the window, so it gets 10 times as many hits each second and 10 times the force. Bigger area, bigger force, same pressure.

A full answer includes

• I chose “The force on the lid is bigger, but the pressure on both is the same”.
• I explained that each square centimetre of the lid and of the window gets the same number of hits each second, so the force on each unit of area, the pressure, is the same; the lid has ten times the area, so it gets ten times as many hits and ten times the force.

Also accept

• b; same hits on each cm², so same pressure; ten times the area, ten times the force.

Watch for

• “The pressure on the lid is bigger, because it is bigger.” — The lid gets a bigger force, because it has more area. Pressure is the force on each unit of area, and each square centimetre of the lid is hit just as often as each square centimetre of the window.
• “The pressure on the window is bigger, because the force is concentrated on a small area.” — The small window gets fewer hits, so a smaller force, on a smaller area. The force on each square centimetre is the same.

4. GQ4 · Practice

Options as numbered on the pupil sheet: 1 — So the force on each unit of area of the walls increases: the pressure rises; 2 — The plunger moves in, so the same particles have less space; 3 — So each square centimetre of the walls is hit more often each second; 4 — The particles are more crowded, but they move at the same speed.

Model answer. Pushing the plunger in leaves the same particles less space, so they are more crowded. The temperature is the same, so they move at the same speed and each hit is just as hard; but each square centimetre of the walls is now hit more often each second. More hits each second mean a bigger force on each unit of area: a higher pressure.

A full answer includes

• Step 1: the plunger moves in, so the same particles have less space.
• Step 2: the particles are more crowded but move at the same speed, because the temperature has not changed.
• Step 3: each square centimetre of the walls is hit more often each second.
• Step 4: the force on each unit of area of the walls increases, so the pressure rises.

Also accept

• a, b, c, d

Watch for

• “The particles get smaller when the gas is squeezed.” — The particles keep their size. Only the space between them gets smaller.
• “The particles move faster because they are squeezed, so they hit harder.” — At a constant temperature the particles keep the same speed, so each hit is just as hard. What changes is how often they hit.

5. GQ5 · Practice

Model answer. Fixed mass and constant temperature, so p₁V₁ = p₂V₂. 100 kPa × 60 cm³ = 6000 kPa cm³ = p₂ × 20 cm³, so p₂ = 6000 ÷ 20 = 300 kPa. The volume fell to a third, so the pressure tripled.

Checked automatically

• New pressure of the air: 300 kPa (within ±2%)
  • Common wrong answer 33.3 : You divided the wrong way round: a smaller volume must give a bigger pressure. p₂ = 100 × 60 ÷ 20.
  • Common wrong answer 33 : You divided the wrong way round: a smaller volume must give a bigger pressure. p₂ = 100 × 60 ÷ 20.
  • Common wrong answer 140 : You added the change in volume to the pressure. It is p × V that stays the same.

A full answer includes

• I wrote p₁V₁ = p₂V₂: 100 kPa × 60 cm³ = p₂ × 20 cm³.
• I found p₂ = 6000 kPa cm³ ÷ 20 cm³ = 300 kPa: a third of the volume, three times the pressure.

Also accept

• 300 kPa

Watch for

• “p₂ = 100 × 20 ÷ 60 = 33 kPa.” — A smaller volume must give a bigger pressure. Keep p × V the same: 6000 ÷ 20.
• “The volume fell by 40 cm³, so the pressure rises by 40 kPa to 140 kPa.” — The product p × V stays the same, not the sum. Equal squeezes do not add equal amounts of pressure.

6. GQ6 · Practice

Model answer. p × V: 100 × 60 = 6000, 125 × 48 = 6000, 150 × 40 = 6000, 180 × 30 = 5400 and 250 × 24 = 6000 kPa cm³. For a fixed mass of gas at a constant temperature pV is constant, so the row at 30 cm³ is the wrong one. Its pressure should have been 6000 kPa cm³ ÷ 30 cm³ = 200 kPa.

Checked automatically

• Volume of the wrong row: 30 cm³ (exact value)
  • Common wrong answer 24 : At 24 cm³, p × V = 250 × 24 = 6000 kPa cm³, the same as the other rows: that row fits.
  • Common wrong answer 48 : At 48 cm³, p × V = 125 × 48 = 6000 kPa cm³, the same as the other rows: that row fits.
• Pressure that should have been recorded: 200 kPa (within ±2%)
  • Common wrong answer 180 : That is the reading that was written down wrongly. Divide 6000 kPa cm³ by 30 cm³.

A full answer includes

• I worked out p × V for each row: 6000 kPa cm³ every time except 180 × 30 = 5400 kPa cm³.
• I found the pressure that fits at 30 cm³: 6000 kPa cm³ ÷ 30 cm³ = 200 kPa.

Also accept

• 30 cm³; 200 kPa

Watch for

• “The pressure goes up by 25 kPa each time, so the 250 kPa row is the wrong one.” — Equal steps in pressure are not the rule; the product p × V is. 250 × 24 = 6000 kPa cm³, which fits.
• “A smaller volume makes p × V smaller, so every row fits.” — For a fixed mass of gas at a constant temperature p × V stays the same. Four rows give 6000 kPa cm³; one does not.

7. GQ7 · Practice

Model answer. The air in the packet is a fixed mass at a constant temperature, so p₁V₁ = p₂V₂. 100 kPa × 400 cm³ = 40 000 kPa cm³ = 80 kPa × V₂, so V₂ = 40 000 ÷ 80 = 500 cm³. The packet puffs up until the air inside, more spread out, hits each square centimetre of the packet only as often as the cabin air hits the outside.

Checked automatically

• Volume of the air on the plane: 500 cm³ (within ±2%)
  • Common wrong answer 320 : You multiplied by 80 ÷ 100. A lower pressure goes with a bigger volume: V₂ = 100 × 400 ÷ 80.
  • Common wrong answer 420 : You added the 20 kPa drop in pressure to the volume. It is p × V that stays the same.
  • Common wrong answer 400 : The packet is soft, so its volume can change: the mass of air is fixed, not the volume.

A full answer includes

• I used p₁V₁ = p₂V₂: 100 kPa × 400 cm³ = 80 kPa × V₂.
• I found V₂ = 40 000 kPa cm³ ÷ 80 kPa = 500 cm³: a lower pressure, a bigger volume, so the packet puffs up.

Also accept

• 500 cm³

Watch for

• “V₂ = 400 × 80 ÷ 100 = 320 cm³.” — A lower pressure must go with a bigger volume. Keep p × V the same: 40 000 ÷ 80.
• “The packet stays the same size, because it is sealed.” — Sealing keeps the mass of air fixed, not its volume. The soft packet swells until the pressure inside matches the cabin.

8. GQ8 · Practice

Options as numbered on the pupil sheet: 1 — A curve that falls steeply at small volumes and gets flatter; 2 — A straight line through the origin; 3 — A horizontal straight line; 4 — A straight line sloping down to the volume axis; 5 — A curve that rises more and more steeply.

Model answer. p against V: a (a curve that falls steeply at small volumes and gets flatter, never reaching the axes). p against 1/V: b (a straight line through the origin, because p is proportional to 1/V). p × V against V: c (a horizontal line, because p × V has the same value at every volume).

A full answer includes

• p against V: a curve that falls steeply at small volumes and gets flatter (a).
• p against 1/V: a straight line through the origin (b), because p is proportional to 1/V.
• p × V against V: a horizontal straight line (c), because pV is constant.

Also accept

• p–V: a; p–1/V: b; pV–V: c.

Watch for

• “p against V is a straight line sloping down.” — Equal squeezes do not add equal pressure: from 50 to 40 cm³ the pressure rises by 25 kPa, from 30 to 20 cm³ by 83 kPa. The points lie on a curve.
• “p against 1/V is a curve too.” — Doubling 1/V (halving V) doubles p, and 1/V = 0 would give p = 0: that is a straight line through the origin.

9. GQ9 · Practice

Model answer. The particles keep their size: squeezing a gas only reduces the space between them. The temperature is the same, so they keep the same speed and each hit is just as hard. But the same particles now have less space, so each square centimetre of the walls is hit more often each second. More hits each second give a bigger force on each unit of area: the pressure rises.

A full answer includes

• I said the particles do not change size, and at a constant temperature they keep the same speed; only the space between them gets smaller.
• I explained that the same particles in less space hit each square centimetre of the walls more often each second, so the force on each unit of area, the pressure, rises.

Also accept

• The particles stay the same size and speed; less space means more frequent hits on each unit of area, so a higher pressure.

Watch for

• “The particles move faster when they are squeezed, so they hit harder.” — At a constant temperature the particles keep the same speed, so each hit is just as hard. What changes is how often they hit.
• “The plunger pushes the particles against the walls.” — The particles are far apart and keep moving at random. The plunger only leaves them less space, so they reach the walls more often.

10. GQ10 · Practice

Model answer. With most of the air pumped out, there are far fewer particles inside the bottle, so each square centimetre of the inside of the wall is hit far less often each second: the pressure inside is much lower. The air outside is unchanged: its particles still hit each square centimetre of the outside of the wall as often as before, at 100 kPa. The force on each unit of area from outside is now much bigger than from inside, so the walls are pushed in until the bottle is crushed.

A full answer includes

• I said there are now far fewer particles inside, so each square centimetre of the inside of the wall is hit far less often each second: a much lower pressure inside.
• I said the air outside still hits each square centimetre as often as before, so the force on each unit of area from outside is bigger than from inside, and the walls are pushed in.

Also accept

• Fewer particles inside, fewer hits on each cm² from inside, lower pressure; the outside air pushes harder on each unit of area.

Watch for

• “The vacuum inside sucks the walls in.” — Nothing inside pulls on the walls: a vacuum has no particles to pull with. The walls are pushed in by the hits of the air particles outside.
• “The air outside gets heavier and presses down on the bottle.” — The outside air has not changed, and the bottle is crushed from every side, not only from above. What changed is the number of hits from inside.

11. GT1 · Apply

Model answer. (a) The air inside is at 70 kPa and the air outside at the beach is at 100 kPa. The outside air particles hit each square centimetre of the bottle more often each second than the particles inside do, so the force on each unit of area from outside is bigger, and the soft walls are pushed in. (b) Fixed mass, constant temperature: p₁V₁ = p₂V₂. 70 kPa × 1500 cm³ = 105 000 kPa cm³ = 100 kPa × V₂, so V₂ = 1050 cm³. (c) As the bottle is squeezed, the same particles have less space. At the same temperature they move at the same speed, but they hit each square centimetre of the inside more often each second: a bigger force on each unit of area, a higher pressure. The walls stop moving in once the pressure inside has risen to 100 kPa. (d) The mass of air must stay fixed, with no air getting in or out, and the temperature must be the same.

A full answer includes

• (a) At the beach the outside air hits each square centimetre of the bottle more often than the 70 kPa air inside does: a bigger force on each unit of area from outside, so the walls are pushed in.
• (b) 70 kPa × 1500 cm³ = 100 kPa × V₂, so V₂ = 105 000 ÷ 100 = 1050 cm³.
• (c) The same particles have less space, so at the same speed they hit each square centimetre of the inside more often each second: a bigger force on each unit of area, so a higher pressure.
• (d) The mass of air stays fixed (the cap is on, nothing leaks), and the temperature is the same.

Also accept

• (a) more hits per cm² outside, bigger force per unit area outside; (b) 1050 cm³; (c) less space, more frequent hits, higher pressure; (d) fixed mass, constant temperature

Watch for

• “(b) V₂ = 1500 × 100 ÷ 70 = 2140 cm³.” — The bottle is squeezed, so the volume must get smaller as the pressure rises. Keep p × V the same: 105 000 ÷ 100.
• “(a) The bottle is crushed because the air inside got colder.” — The temperature is the same in both places. What changed is the pressure of the air outside.