Skip to content

Kinetic theory of gases worksheet answers for GCSE physics (free PDF download)

  • by

The problems are on the kinetic theory of gases worksheet. Try them on your own first, then compare with the answers below.

Quick answer: Gas pressure is caused by particles colliding with the walls of their container. These worked answers correct the 20 problems of the Physics 9-10 worksheet on kinetic theory of gases, from ideas of kinetic theory to compress then heat: each solution shows the formula, the substitution and the final answer with its unit.

Kinetic theory of gases worksheet answers for GCSE physics (free PDF download)

Here are the kinetic theory of gases worksheet answers for physics grades 9-10, explained step by step. First try the worksheet on your own, then compare, and reread the lesson if a step is unclear.

Every answer states the law used, the substitution and the final value with its unit. Check that your temperatures were in kelvin where needed, and that both pressures and both volumes were in matching units. Small rounding differences in the last figure are acceptable.

Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: IGCSE 2.1, HS-PS3-2

Kinetic theory of gases worksheet answers: the answers explained

Problem 1 Ideas of kinetic theory

Any three: the particles are in constant random motion; they are very far apart compared with their size; they move in straight lines between collisions; there are almost no forces between them except at collisions; the average kinetic energy depends on the temperature.

Problem 2 Celsius to kelvin

1. 0 + 273 = 273 K
2. 25 + 273 = 298 K
3. −196 + 273 = 77 K
4. 300 − 273 = 27 °C

Problem 3 What raises the pressure?

(a) Increases: faster particles hit harder and more often.
(b) Increases: more collisions per second per unit area.
(c) Increases: more particles, so more collisions.

Problem 4 Halving the room

p2 = 100 × 4.0 / 2.0 = 200 kPa

Problem 5 Two boxes

1. Box B.
2. B has twice as many particles, so about twice as many collisions with the walls each second, giving about twice the pressure.

Problem 6 Squeezing a syringe

p2 = 100 × 40 / 25 = 160 kPa

Problem 7 Filling a tyre

V2 = p1V1 / p2 = 100 × 600 / 300 = 200 cm³

Problem 8 Heating a rigid cylinder

T1 = 293 K, T2 = 373 K
p2 = 150 × 373 / 293 = 190.96 = 191 kPa

Problem 9 Warning on an aerosol can

Heating makes the gas particles faster. They hit the walls harder and more often. The volume of the can is fixed, so the pressure rises and the can might burst, with the risk of injury.

Problem 10 Reading a pressure-volume graph

1. 250 kPa
2. 2 × 500 = 1000, 4 × 250 = 1000, 10 × 100 = 1000 kPa L. The product is constant, so pressure is inversely proportional to volume (Boyle’s law).
3. p = 1000 / 8 = 125 kPa

Problem 11 Weather balloon

V2 = 100 × 2.0 / 25 = 8.0 m³

Problem 12 Crisp packet on a hill

The air outside is at lower pressure at altitude. The pressure of the gas trapped inside is greater than outside, so the bag expands until the pressures balance. The mass of gas is the same, so the volume rises as the outside pressure falls.

Problem 13 Spot the error

The student used Celsius temperatures. T1 = 293 K, T2 = 353 K.
p2 = 120 × 353 / 293 = 144.6 = 145 kPa (3 s.f.)

Problem 14 Units first

1. 800 cm³ = 0.80 L
2. 100 kPa = 100 000 Pa = 1.0 × 10⁵ Pa
3. p2 = 100 × 0.80 / 0.50 = 160 kPa

Problem 15 Force on a piston

F = p A = 200 000 × 0.0030 = 600 N

Problem 16 Pumping a bicycle tyre

Total air at 100 kPa = 1.5 + 6 × 0.50 = 4.5 L. This is squeezed into 1.5 L.
p2 = 100 × 4.5 / 1.5 = 300 kPa

Problem 17 Rising bubble

1. V2 = 300 × 5.0 / 100 = 15 cm³
2. The air in the lungs expands as the outside pressure falls, here by a factor of three. If the breath is held, the lungs overstretch and can be damaged.

Problem 18 Extrapolating to absolute zero

1. At 0 °C, T = 273 K: 100 / 273 = 0.366 kPa/K. At 100 °C, T = 373 K: 136.6 / 373 = 0.366 kPa/K. The ratios are equal.
2. The dashed line reaches zero at −273 °C. This is absolute zero: the particles have minimum energy so they exert no pressure.

Problem 19 Jar in the freezer

1. T1 = 300 K, T2 = 255 K; p2 = 100 × 255 / 300 = 85 kPa
2. Pressure difference = 100 − 85 = 15 kPa = 15 000 Pa. F = p A = 15 000 × 6.0 × 10⁻⁴ = 9.0 N, pushing the lid inwards.

Problem 20 Compress then heat

Step 1 (Boyle): p = 100 × 3.0 / 1.0 = 300 kPa at 300 K.
Step 2 (constant volume): p2 = 300 × 450 / 300 = 450 kPa

← Back to the kinetic theory of gases worksheet

Physics and chemistry textbooks to download in PDF, from grade 6 to AP Physics and AP Chemistry

Follow us