Describe changes of state in terms of melting, boiling, evaporating, freezing and condensing

States of Matter – Solids, Liquids and Gases (IGCSE 0620)

1. Core properties of the three states

2. Diffusion (Core)

• Definition: The spontaneous spreading of particles from an area of higher concentration to an area of lower concentration.
• Everyday example: A few drops of perfume eventually fill an entire room.
• Rate of diffusion (core ordering): gases > liquids > solids.
• Effect of relative molecular mass (core requirement): Heavier gases diffuse more slowly than lighter ones because the average speed of the particles is lower (e.g. Cl₂ diffuses slower than H₂).

3. Changes of State – Core

All changes are explained with the kinetic particle theory: heating adds kinetic energy to the particles, cooling removes it.

3.1 List of changes

• Melting – solid → liquid
• Freezing – liquid → solid
• Boiling – liquid → gas (formation of bubbles throughout the liquid)
• Evaporation – liquid → gas (surface phenomenon, occurs below the boiling point)
• Condensation – gas → liquid

3.2 Particle‑model explanations

1. Melting

   Heating supplies enough kinetic energy for particles to break the ordered lattice. At the melting point the solid becomes a liquid. Energy is absorbed (endothermic).

2. Freezing

   Cooling removes kinetic energy. When the temperature reaches the freezing point (identical to the melting point for a pure substance) the particles arrange into a regular lattice. Energy is released (exothermic).

3. Boiling

   Boiling occurs when the vapour pressure of the liquid equals the external atmospheric pressure. Bubbles of vapour form throughout the liquid and rise to the surface. Raising the external pressure raises the boiling point. Continuous heat supply is required; energy is absorbed.

4. Evaporation

   Even below the boiling point, a few surface molecules have sufficient kinetic energy to escape into the air. This surface‑only process can occur at any temperature, but the rate increases with higher temperature, larger surface area, lower humidity, and stronger wind. Energy is absorbed, producing a cooling effect.

5. Condensation

   When a gas loses kinetic energy (by cooling or compression) its particles come together to form a liquid. The process releases heat to the surroundings (exothermic).

4. Relationship between temperature, pressure and gas volume (Supplement)

For a fixed amount of gas:

• At constant pressure, volume ∝ temperature (Charles’s law, \(V \propto T\)).
• At constant temperature, volume ∝ 1/pressure (Boyle’s law, \(V \propto 1/P\)).
• Combined, these give the ideal‑gas equation \(PV = nRT\).

Legend: \(P\) = pressure (kPa), \(V\) = volume (dm³), \(n\) = amount of substance (mol), \(R\) = universal gas constant (8.314 kPa·dm³ mol⁻¹ K⁻¹), \(T\) = absolute temperature (K).

5. Energy changes summary

6. Key points for exam answers

• State clearly whether heat is absorbed (endothermic) or released (exothermic) for each change.
• Differentiate:
  • Boiling – bubbles form throughout the liquid; occurs when vapour pressure = external pressure.
  • Evaporation – only surface molecules escape; can occur at any temperature.
• For a pure substance, the melting point is identical to the freezing point, and the boiling point is identical to the condensation point.
• Use the terms endothermic and exothermic accurately.
• When drawing diagrams, show particle spacing and motion appropriate to each state and indicate the direction of energy flow with arrows.
• Remember the qualitative pressure‑volume relationship for gases (higher pressure → smaller volume) and the quantitative laws (Charles’s, Boyle’s, ideal‑gas equation) when a question involves temperature or pressure changes.

7. Sample question & answer

Question: A sample of water is heated from 20 °C to 120 °C at 1 atm pressure. List each change of state that occurs and state whether heat is absorbed or released.

Answer:

1. Evaporation (20 °C → 100 °C) – heat absorbed (endothermic).
2. Boiling at 100 °C – heat absorbed (latent heat of vapourisation).
3. Super‑heated steam (100 °C → 120 °C) – heat absorbed (sensible heat).

8. Formulae for quantitative problems

Latent‑heat calculations

\(q = m \times L\)

where q = heat energy (J), m = mass (kg), L = latent heat (J kg⁻¹) for the specific change of state.

Sensible‑heat calculations

\(q = m \times c \times \Delta T\)

where c = specific heat capacity (J kg⁻¹ K⁻¹) and \(\Delta T\) = temperature change (K).

9. Suggested revision diagram