IGCSE Physics Gases and the Absolute Scale of Temperature Questions and Worked Solutions (2.1.3) | 0625

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IGCSE Physics 0625 · Topic 2.1.3

IGCSE Physics Gases and the Absolute Scale of Temperature Questions and Worked Solutions

50 topic-wise practice questions on Gases and the Absolute Scale of Temperature (Unit 2: Thermal Physics), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions22 Core18 Extended10 Challenge130 marksTextbook §2.1 · pp. 86–99
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Gases and the Absolute Scale of Temperature

The pressure of a gas depends on its temperature and volume. On the kelvin scale, zero is absolute zero. (Extended) For a fixed mass at constant temperature, pressure is inversely proportional to volume (Boyle's law).

Key formulas

T (K) = θ (°C) + 273temperature conversionCore
p V = constantfixed mass of gas at constant temperatureExtended
p₁ V₁ = p₂ V₂before and after a change at constant temperatureExtended

Prerequisite knowledge

  • Particle model of gases (2.1.2)
  • Inverse proportion
  • Rearranging equations

Exam tips & common mistakes

  • Units of p and V only need to be the same on both sides of p₁V₁ = p₂V₂
  • Halving the volume doubles the pressure (at constant temperature)
  • Explain with particles — 'more frequent collisions' is the key phrase

Key facts from the syllabus (2.1.3)

  • Increasing temperature at constant volume increases pressure: particles move faster and hit the walls harder and more often
  • Decreasing volume at constant temperature increases pressure: particles hit the walls more often
  • A temperature change of 1 °C equals a change of 1 K
  • Absolute zero is 0 K = −273 °C
  • Extended a graph of p against V is a curve; p against 1/V is a straight line through the origin

Textbook: §2.1 Kinetic particle model of matter, pp. 86–99

1
Core[2 marks]

Convert 27 °C into kelvin.

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Method: T = θ + 273.

T = 27 + 273 = 300 K
2
Core[2 marks]

Convert 373 K into °C.

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Method: θ = T − 273.

θ = 373 − 273 = 100 °C
3
Core[2 marks]

Convert −40 °C into kelvin.

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Method: T = θ + 273.

T = −40 + 273 = 233 K
4
Core[2 marks]

A planet has a surface temperature of 93 K. Calculate this in °C.

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Method: θ = T − 273.

θ = 93 − 273 = -180 °C
5
Core[1 mark]

State the temperature of absolute zero in kelvin.

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Method: 0 K.

0 K
6
Core[3 marks]

Describe, in terms of particles, what happens to the pressure of a gas when it is heated at constant volume.

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Method: Faster particles hit walls more often and harder.

Pressure increases because particles move faster, colliding with the walls more frequently and with more force
7
Core[3 marks]

Describe, in terms of particles, what happens to the pressure of a gas when its volume is reduced at constant temperature.

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Method: Same speed, more collisions per area.

Pressure increases because particles hit the walls more frequently (smaller volume, same speed)
8
Core[2 marks]

A tyre is pumped up on a cold morning and later becomes hot on the road. State and explain what happens to the tyre pressure.

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Method: Temperature rises at roughly constant volume.

Pressure increases: hotter air particles move faster and hit the walls more often and harder
9
Core[2 marks]

An aerosol can carries a warning not to heat it. Explain why.

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Method: Heating at constant volume raises pressure.

Heating increases the gas pressure, which could make the can explode
10
Core[2 marks]

Convert 0 °C and 100 °C into kelvin.

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Method: Add 273.

273 K and 373 K
11
Core[2 marks]

State the temperature change in kelvin when water warms from 20 °C to 35 °C.

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Method: A change of 1 °C = a change of 1 K.

15 K
12
Core[2 marks]

Convert 15 °C into kelvin.

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Method: Add 273.

288 K
13
Core[2 marks]

Convert 77 K into °C.

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Method: Subtract 273.

-196 °C
14
Core[2 marks]

Convert −273 °C into kelvin.

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Method: Add 273.

0 K
15
Core[2 marks]

Convert 500 K into °C.

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Method: Subtract 273.

227 °C
16
Core[2 marks]

The temperature of a gas rises from 300 K to 350 K. State the rise in °C.

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Method: Same size degree.

50 °C
17
Core[2 marks]

Explain why there are no negative temperatures on the kelvin scale.

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Method: Starts at absolute zero.

The kelvin scale starts at absolute zero, the lowest possible temperature
18
Core[3 marks]

A sealed container of gas is cooled at constant volume. Describe and explain the change in pressure.

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Method: Slower particles.

Pressure decreases: particles move more slowly, hitting the walls less often and with less force
19
Core[2 marks]

A football is pumped up indoors and taken outside on a cold day. Explain why it feels softer.

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Method: Pressure drops.

The air cools, its particles slow down and the pressure falls
20
Core[2 marks]

A gas is compressed slowly so its temperature does not change. Explain why its pressure increases.

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Method: More frequent collisions.

Particles hit the walls more often because the volume is smaller
21
Core[2 marks]

A sealed bag of crisps swells when taken up a mountain. Explain why.

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Method: Lower outside pressure.

The outside air pressure is lower, so the gas inside expands
22
Core[1 mark]

State the temperature in °C at which particles have the least possible kinetic energy.

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Method: Absolute zero.

−273 °C
23
Extended[1 mark]

State the relationship between pressure and volume for a fixed mass of gas at constant temperature.

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Method: Boyle's law.

pV = constant (pressure is inversely proportional to volume)
24
Extended[3 marks]

A gas at 100 kPa occupies 6.0 dm³. It is compressed to 4.0 dm³ at constant temperature. Calculate the new pressure.

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Method: p₁V₁ = p₂V₂.

p₂ = 100 × 6.0 ÷ 4.0 = 150 kPa
25
Extended[3 marks]

A gas at 2.0 × 10⁵ Pa has a volume of 30 cm³. Calculate its volume at 5.0 × 10⁵ Pa (constant temperature).

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Method: V₂ = p₁V₁ ÷ p₂.

V₂ = 2.0×10⁵ × 30 ÷ 5.0×10⁵ = 12.0 cm³
26
Extended[3 marks]

A bubble of volume 2.0 cm³ forms at a pressure of 300 kPa. It rises to where the pressure is 100 kPa. Calculate its new volume (constant temperature).

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Method: pV = constant.

V = 300 × 2.0 ÷ 100 = 6.00 cm³
27
Extended[2 marks]

The pressure of a fixed mass of gas is doubled at constant temperature. State what happens to its volume.

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Method: Inverse proportion.

The volume halves
28
Extended[2 marks]

Sketch and describe the shape of a graph of pressure against volume for a gas at constant temperature.

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Method: Inverse curve.

A curve that falls steeply then levels out (an inverse-proportion curve), never touching the axes
29
Extended[2 marks]

Describe the graph of pressure against 1/volume for a gas at constant temperature.

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Method: Linear through origin.

A straight line through the origin
30
Extended[3 marks]

A gas of volume 500 cm³ at 1.2 × 10⁵ Pa is compressed until its pressure is 3.0 × 10⁵ Pa. Calculate the new volume.

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Method: V₂ = p₁V₁ ÷ p₂.

V₂ = 1.2×10⁵ × 500 ÷ 3.0×10⁵ = 200 cm³
31
Extended[3 marks]

A syringe holds 40 cm³ of air at 100 kPa. The outlet is sealed and the volume is reduced to 25 cm³ at constant temperature. Calculate the new pressure.

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Method: p₂ = p₁V₁ ÷ V₂.

p₂ = 100 × 40 ÷ 25 = 160 kPa
32
Extended[3 marks]

A gas at 120 kPa has a volume of 250 cm³. Calculate its volume at 80 kPa (constant temperature).

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Method: pV const.

V = 120 × 250 ÷ 80 = 375 cm³
33
Extended[3 marks]

A gas cylinder holds 0.050 m³ of gas at 1.2 × 10⁶ Pa. Calculate the volume this gas would occupy at 1.0 × 10⁵ Pa at the same temperature.

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Method: V₂ = p₁V₁/p₂.

V₂ = 1.2×10⁶ × 0.050 ÷ 1.0×10⁵ = 0.600 m³
34
Extended[3 marks]

A gas's volume is reduced from 800 cm³ to 200 cm³ at constant temperature. The initial pressure was 100 kPa. Calculate the final pressure.

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Method: p₂ = p₁V₁/V₂.

p₂ = 100 × 800 ÷ 200 = 400 kPa
35
Extended[2 marks]

A gas's volume is reduced to one third at constant temperature. State the factor by which its pressure changes.

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Method: Inverse.

It triples
36
Extended[3 marks]

An air bubble of volume 1.5 cm³ is released by a diver where the pressure is 250 kPa. Calculate its volume at the surface, where pressure is 100 kPa.

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Method: pV const.

V = 250 × 1.5 ÷ 100 = 3.75 cm³
37
Extended[3 marks]

Table: p = 100 kPa, V = 60 cm³; p = 150 kPa, V = 40 cm³; p = 200 kPa, V = 30 cm³. Show that these obey pV = constant.

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Method: Calculate pV.

pV = 6000 kPa cm³ in each case, so pV is constant
38
Extended[2 marks]

State the conditions under which pV = constant applies.

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Method: Fixed mass, constant temperature.

For a fixed mass of gas at constant temperature
39
Extended[3 marks]

A tyre pump contains 300 cm³ of air at 100 kPa. The outlet is blocked and the air is compressed to 75 cm³. Calculate the pressure.

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Method: p₂ = p₁V₁/V₂.

p₂ = 100 × 300 ÷ 75 = 400 kPa
40
Extended[3 marks]

Explain in terms of particles why pV = constant for a gas at constant temperature.

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Method: Halve V, double collision rate.

At constant temperature the particles' average speed is unchanged; halving the volume doubles the rate of collisions with each area of wall, doubling the pressure

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[3 marks]

A gas is cooled from 127 °C to −73 °C. Convert both temperatures to kelvin and state the factor by which the average kinetic energy of the particles changes.

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Method: KE ∝ T(K).

400 K to 200 K; the average kinetic energy halves
42
ChallengeCore[4 marks]

Explain, using particles, why the pressure in a car tyre increases after a long drive, and suggest why drivers are advised to check pressures when the tyres are cold.

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Method: Temperature rise.

The air warms, particles move faster and hit the walls harder and more often; hot readings would be falsely high
43
ChallengeCore[3 marks]

Explain why aerosol cans can explode if thrown into a fire.

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Method: Pressure rises with T.

The gas inside heats up at constant volume, increasing the pressure until the can bursts
44
ChallengeExtended[4 marks]

A gas occupies 2.4 dm³ at 100 kPa. It is compressed at constant temperature to 0.60 dm³. Calculate the new pressure and explain the change using particles.

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Method: pV const.

p = 100 × 2.4 ÷ 0.60 = 400 kPa; particles hit the walls four times as often in the smaller volume
45
ChallengeExtended[5 marks]

A diver at 30 m depth (total pressure 400 kPa) breathes out a 0.50 dm³ bubble. Calculate its volume at depths where the pressure is 200 kPa and at the surface (100 kPa), assuming constant temperature.

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Method: pV const.

1.00 dm³ and 2.00 dm³
46
ChallengeExtended[4 marks]

A cylinder contains 0.040 m³ of gas at 3.0 × 10⁶ Pa. How many 0.0050 m³ balloons can be filled to 1.2 × 10⁵ Pa at constant temperature? (The cylinder keeps 0.040 m³ at 1.2 × 10⁵ Pa.)

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Method: Total volume at lower p.

Total volume = 3.0×10⁶ × 0.040 ÷ 1.2×10⁵ = 1.00 m³; available = 0.96 m³; balloons = 192
47
ChallengeExtended[4 marks]

A syringe of air at 100 kPa has volume 60 cm³. The end is sealed and the plunger is pushed until the pressure is 250 kPa. Calculate the new volume and the distance the plunger moved if the cross-sectional area is 3.0 cm².

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Method: V₂, then ΔV/A.

V₂ = 24.0 cm³; ΔV = 36 cm³; distance = 12.0 cm
48
ChallengeExtended[3 marks]

A student plots p against 1/V for a gas and gets a straight line through the origin. State what this shows and the conditions required.

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Method: Inverse proportion.

p is inversely proportional to V (pV = constant), for a fixed mass of gas at constant temperature
49
ChallengeExtended[4 marks]

Data: V = 20, 25, 40 cm³; p = 200, 160, 100 kPa. Check whether pV is constant and predict p at V = 50 cm³.

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Method: Calculate pV.

pV = 4000 kPa cm³ each time; p = 80 kPa
50
ChallengeExtended[4 marks]

Explain why compressing a gas quickly makes it warmer, so that pV = constant no longer applies exactly.

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Method: Work done on gas.

Work is done on the gas, increasing the particles' kinetic energy and temperature; Boyle's law only holds at constant temperature
All IGCSE Physics topics← 2.1.2 Particle Model2.2.1 Thermal Expansion of Solids, Liquids and Gases →

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