Heating a gas at a fixed volume — answer key

1. GQ11 · Practice

Model answer. They move faster, so they hit the walls more often and harder. Heating makes the particles move faster. Faster particles reach the walls more often, and each hit is harder. Both give a bigger force on each unit of area of the walls, so the pressure rises. The can is sealed, so there are no more particles than before; they do not grow; and the volume is fixed, so they are no closer together on average.

A full answer includes

• I chose “They move faster”, “They hit the walls more often” and “They hit the walls harder”.
• I explained that faster particles hit the walls more often and harder, so the force on each unit of area of the walls, the pressure, rises.

Also accept

• a, b, c

Watch for

• “The particles get bigger when they are heated.” — The particles keep their size. Heating makes them move faster, and that alone changes the hits on the walls.
• “Heating makes more particles.” — The can is sealed, so the number of particles is fixed. The pressure rises because the same particles move faster.

2. GQ12 · Practice

Model answer. (a) T = θ + 273 = 27 + 273 = 300 K. (b) θ = T − 273 = 400 − 273 = 127 °C. Check: the number in kelvin is the bigger one each time.

Checked automatically

• (a) 27 °C in kelvin: 300 K (exact value)
  • Common wrong answer -246 : You subtracted 273. From °C to K, add 273; there are no negative temperatures in kelvin.
  • Common wrong answer 246 : You subtracted 273. From °C to K, add 273: 27 + 273.
• (b) 400 K in °C: 127 °C (exact value)
  • Common wrong answer 673 : You added 273. From K to °C, subtract 273: 400 − 273.

A full answer includes

• (a) T = 27 + 273 = 300 K.
• (b) θ = T − 273 = 400 − 273 = 127 °C.

Also accept

• (a) 300 K; (b) 127 °C

Watch for

• “(a) 27 − 273 = −246 K.” — There are no negative temperatures in kelvin. From °C to K you add 273.
• “(b) 400 + 273 = 673 °C.” — From K to °C you subtract 273: the temperature in °C is always the smaller number.

3. GQ13 · Practice

Model answer. (1) slowly: colder particles move more slowly. (2) often: slower particles reach the walls less often. (3) force: fewer and gentler hits give a smaller force on each unit of area. (4) falls: a smaller force on each unit of area is a lower pressure.

A full answer includes

• (1) slowly
• (2) often (or frequently)
• (3) force
• (4) falls (or decreases)

Also accept

• slowly; often; force; falls.

Watch for

• “(1) closer together” — The bottle is rigid, so the particles have the same space as before. Cooling changes their speed.
• “(4) rises” — Fewer and gentler hits on each unit of area mean a smaller force on it, so a lower pressure.

4. GQ14 · Practice

Model answer. The pressure rises by about the same amount for each 25 °C rise. The rises are 8, 9, 8 and 8 kPa for each 25 °C: about the same each time, so the points lie on a straight line. It is not proportional to the temperature in °C, because at 0 °C it is 91 kPa, not zero. It does not double from 50 °C to 100 °C: it rises from 108 to 124 kPa. The pressure rises because the particles of a warmer gas move faster, so they hit the walls more often and harder, giving a bigger force on each unit of area.

A full answer includes

• I chose “The pressure rises by about the same amount for each 25 °C rise in temperature”.
• I explained that warmer particles move faster, so they hit the walls more often and harder: a bigger force on each unit of area, a higher pressure.

Also accept

• b; faster particles hit the walls more often and harder.

Watch for

• “The pressure doubles from 50 °C to 100 °C.” — It rises from 108 to 124 kPa. Doubling a temperature in °C does not double the pressure, because 0 °C is not where the pressure is zero.
• “The volume is fixed, so the pressure stays the same.” — A fixed volume keeps the space the same, but the particles still speed up as the gas warms, so the pressure rises.

5. GQ15 · Practice

Model answer. The pressure rises by about 44 kPa for every 100 °C. Going down from 120 kPa at 0 °C, it falls by 44 kPa for each 100 °C colder, so it reaches zero after about 120 ÷ 44 × 100 ≈ 273 °C: at about −273 °C. On the Kelvin scale this is 0 K, because T = θ + 273 = −273 + 273 = 0. A real gas would turn to liquid before it got that cold; this is an ideal-gas extrapolation.

Checked automatically

• Temperature at which the line reaches zero pressure: -273 °C (within ±15 °C)
  • Common wrong answer 0 : At 0 °C the graph shows 120 kPa, not zero. Carry the line on to colder temperatures.
  • Common wrong answer -100 : At −100 °C the line is still at about 76 kPa. Keep going.
  • Common wrong answer 273 : The line goes down towards colder temperatures, so the answer is below 0 °C: about −273 °C.

A full answer includes

• I carried the straight line on and found that it reaches zero pressure at about −273 °C.
• I stated that −273 °C is 0 K, the start of the Kelvin scale.

Also accept

• about −273 °C; 0 K

Watch for

• “The pressure is zero at 0 °C.” — At 0 °C the graph shows 120 kPa. 0 °C is only where water freezes; the gas particles are still moving fast there.
• “The line reaches zero at about −100 °C.” — At −100 °C the line is still at about 76 kPa. Keep going: it falls by only about 44 kPa for each 100 °C.

6. GQ16 · Practice

Options as numbered on the pupil sheet: 1 — 273 K; 2 — 373 K; 3 — 0 K; 4 — 310 K; 5 — 37 K; 6 — −546 K.

Model answer. Add 273 each time. 0 °C: 0 + 273 = 273 K (a). 100 °C: 100 + 273 = 373 K (b). −273 °C: −273 + 273 = 0 K (c). 37 °C, body temperature: 37 + 273 = 310 K (d).

A full answer includes

• 0 °C = 273 K (a).
• 100 °C = 373 K (b).
• −273 °C = 0 K (c).
• 37 °C = 310 K (d).

Also accept

• 0 °C – a; 100 °C – b; −273 °C – c; 37 °C – d

Watch for

• “−273 °C is −546 K.” — Add 273: −273 + 273 = 0 K. The Kelvin scale starts at −273 °C, so it has no negative temperatures.
• “0 °C is 0 K.” — The two scales have the same size of step but different zeros: 0 °C is 273 K.

7. GQ17 · Practice

Model answer. Overnight the air in the tyres cooled down, so its particles move more slowly. Slower particles reach the walls of the tyre less often, and each hit is gentler. The force on each unit of area of the walls is smaller, so the pressure is lower. No air has leaked out and the volume is about the same: only the speed of the particles has changed.

A full answer includes

• I said the air in the tyres is colder in the morning, so its particles move more slowly.
• I said slower particles hit the walls of the tyre less often and less hard, so the force on each unit of area, the pressure, is smaller; the number of particles and the volume have not changed.

Also accept

• Colder air, slower particles, fewer and gentler hits on each unit of area, so a lower pressure.

Watch for

• “Some air leaked out overnight.” — The tyres do not leak. The same particles are there; they move more slowly because the air is colder.
• “The cold makes the particles smaller, so they press less.” — The particles keep their size. Cooling slows them down, so they hit the walls less often and less hard.

8. GQ18 · Practice

Model answer. The pressure rises by 7 kPa for each 20 °C rise: 100, 107, 114 kPa. From 60 °C to 100 °C is two more 20 °C steps, so the pressure at 100 °C is about 114 + 2 × 7 = 128 kPa. A pupil who thinks the pressure is proportional to the temperature in °C would get 114 × 100 ÷ 60 = 190 kPa, which is far too high, because 0 °C is not where the pressure is zero.

Checked automatically

• Pressure at 100 °C: 128 kPa (within ±1.5 kPa)
  • Common wrong answer 500 : You made the pressure proportional to the temperature in °C. Add 7 kPa for each 20 °C instead.
  • Common wrong answer 190 : You scaled 114 kPa by 100 ÷ 60, as if the pressure were proportional to the temperature in °C. Add 7 kPa for each 20 °C instead.
  • Common wrong answer 121 : That is the pressure at 80 °C. One more 20 °C step takes you to 100 °C.

A full answer includes

• I found that the pressure rises by 7 kPa for each 20 °C.
• I added two more steps: 114 + 7 + 7 = 128 kPa at 100 °C (127 to 128 kPa).

Also accept

• about 128 kPa (127 to 128 kPa)

Watch for

• “100 °C is five times 20 °C, so the pressure is 5 × 100 = 500 kPa.” — The pressure is not proportional to the temperature in °C: at 0 °C it is still about 93 kPa. Use equal rises for equal steps of temperature.
• “100 °C is not far above 60 °C, so the pressure is still about 114 kPa.” — Every 20 °C rise adds about 7 kPa, and 60 °C to 100 °C is two more steps.

9. GT2 · Apply

Model answer. (a) The ball is sealed, so it holds a fixed mass of air. In the hot water the air warms and its particles move faster. They hit the inside of the ball more often and harder, so the force on each unit of area of the inside, the pressure, rises. Once it pushes harder than the air and water outside, the soft dent is pushed back out. (b) 20 °C = 20 + 273 = 293 K; 70 °C = 70 + 273 = 343 K. (c) The Celsius scale does not start where the pressure of a gas is zero: at 0 °C the air still has a pressure of over 90 kPa. So 3.5 times the number in °C is not 3.5 times anything for the gas. Equal rises in temperature give equal rises in pressure, and warming from 20 °C to 70 °C adds only about a sixth. (d) From 20 °C to 70 °C is 50 °C, which is 2.5 steps of 20 °C. The pressure rises by about 2.5 × 7 = 17.5 kPa, to about 117 kPa.

A full answer includes

• (a) The warmed air particles move faster, so they hit the inside of the ball more often and harder; the force on each unit of area rises above that of the air outside and pushes the dent out.
• (b) 20 + 273 = 293 K; 70 + 273 = 343 K.
• (c) 0 °C is not where the pressure is zero, so a temperature in °C cannot be used as a ratio; equal temperature rises give equal pressure rises, which here is far less than 3.5 times.
• (d) 50 °C is 2.5 steps of 20 °C: 100 + 2.5 × 7 ≈ 117 kPa (117 to 118 kPa).

Also accept

• (a) faster particles, more often and harder hits, higher pressure; (b) 293 K, 343 K; (c) 0 °C is not zero pressure; (d) about 117 kPa

Watch for

• “(a) The heat makes the air particles expand, so they take up more room and push the dent out.” — The particles do not get bigger. They move faster, so they hit the inside of the ball more often and harder.
• “(d) 100 × 70 ÷ 20 = 350 kPa.” — That treats the pressure as proportional to the temperature in °C. Add about 7 kPa for each 20 °C instead.