Cambridge Syllabus Notes

States of Matter – Solids, Liquids and Gases

Learning Objective

Explain, using the kinetic‑particle theory, how temperature and pressure affect the volume of a gas and describe the related concepts required by the Cambridge IGCSE Chemistry (0620) syllabus.

1. Distinguishing Properties of the Three States

Property	Solid	Liquid	Gas
Shape	Fixed	Adopts the shape of the container	Adopts the shape of the container
Volume	Fixed	Fixed (but not rigid)	Variable – expands to fill the container
Particle arrangement	Closely packed in a regular lattice	Close together, no regular pattern	Widely separated, no pattern
Particle motion	Vibrate in fixed positions	Vibrate and slide past one another	Move rapidly in all directions
Compressibility	Very low	Low	High

Link to the kinetic‑particle theory: The observed macroscopic properties arise because:

In a solid, particles are locked in a lattice; they can only vibrate, giving a fixed shape and volume and very low compressibility.
In a liquid, particles are still close together but are free to slide, so the liquid takes the shape of its container while retaining a roughly constant volume.
In a gas, particles are far apart and move independently; collisions with the container walls produce pressure, and the gas can be easily compressed or expanded.

2. Kinetic‑Particle Theory for Each State

Solid: Particles are in a regular, tightly packed arrangement. They possess only vibrational kinetic energy about fixed points.
Liquid: Particles remain close but are not in a fixed lattice. They have translational kinetic energy that allows them to slide past one another while still experiencing intermolecular attractions.
Gas: Particles are widely separated and move in straight lines until they collide with each other or the container walls. The average kinetic energy of the particles is directly proportional to the absolute temperature (K).

3. Changes of State (Core & Supplement)

Each change of state can be explained by the kinetic‑particle theory. The table gives a concise syllabus‑style statement and the underlying particle‑model reasoning.

Change of State	Syllabus‑style statement	Kinetic‑particle explanation
Melting (solid → liquid)	Heat supplied overcomes part of the intermolecular forces, allowing particles to move past each other.	Added kinetic energy breaks the rigid lattice; particles begin to slide while remaining close.
Freezing (liquid → solid)	Removal of heat allows particles to settle into a regular arrangement.	Loss of kinetic energy lets intermolecular attractions lock particles into a lattice.
Boiling / Evaporation (liquid → gas)	Particles acquire enough kinetic energy to overcome intermolecular forces and escape into the vapour phase.	When the vapour pressure equals the external pressure (boiling) or when surface particles have sufficient energy (evaporation), they break free and spread apart.
Condensation (gas → liquid)	Cooling reduces kinetic energy so that intermolecular attractions bring particles together.	Slower particles collide less violently, allowing attractive forces to dominate and form a liquid.
Sublimation (solid → gas)	Solid particles gain enough kinetic energy to bypass the liquid phase and become a gas.	Direct transition occurs when the kinetic energy supplied exceeds both lattice and liquid‑phase attractions.
Deposition (gas → solid)	Gas particles lose kinetic energy rapidly and arrange directly into a solid lattice.	Very low temperature causes particles to settle into the most ordered arrangement without forming a liquid.

Heating‑Cooling Curve (labelled sketch)

Horizontal plateaus at the melting point (solid–liquid) and boiling point (liquid–gas).
Sloping sections represent sensible heating where temperature rises with added heat.
Arrows indicating the direction of heat flow and the corresponding change in particle motion.

4. Diffusion

Diffusion is the spontaneous spreading of particles from an area of higher concentration to one of lower concentration.

Kinetic‑particle explanation: Fast‑moving particles collide randomly, gradually mixing throughout the available space.
Factors that increase the rate of diffusion:
- Higher temperature – particles move faster (greater kinetic energy).
- Lower molecular mass – at a given temperature, lighter molecules have higher average speeds (v ∝ 1/√M).
- Steeper concentration gradient – a larger difference in concentration provides a stronger driving force.

Quantitative example: At 298 K, the average speed of H₂ (M = 2 g mol⁻¹) is about 1.7 times that of CO₂ (M = 44 g mol⁻¹). Consequently, hydrogen diffuses roughly 1.7 times faster than carbon dioxide under the same conditions.

5. Effect of Temperature and Pressure on the Volume of a Gas

5.1 Qualitative description (kinetic‑particle justification)

Increasing temperature (pressure constant): Particles gain kinetic energy, move faster, collide with the container walls more frequently and with greater force. To maintain constant pressure, the gas expands – volume increases.
Increasing pressure (temperature constant): An external force pushes the particles closer together, reducing the space between them. With the same kinetic energy, the frequency of wall collisions rises, so the volume decreases.

5.2 Quantitative relationships (Gas Laws)

Law	Relationship	Formula	Typical graph
Boyle’s Law	V ↔︎ P (inverse, T constant)	\(P_{1}V_{1}=P_{2}V_{2}\)	Hyperbola (P on y‑axis, V on x‑axis)
Charles’s Law	V ↔︎ T (direct, P constant)	\(\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}\)	Straight line through the origin (V on y‑axis, T on x‑axis)
Gay‑Lussac’s Law	P ↔︎ T (direct, V constant)	\(\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}\)	Straight line through the origin (P on y‑axis, T on x‑axis)
Combined Gas Law	V, P, T together	\(\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}\)	Used for multi‑step calculations
Ideal Gas Equation	All variables plus amount of gas	\(PV=nRT\)	Linear relationship between PV and n (for a given T)

5.3 Important points for calculations

Temperatures must be in kelvin (K). Convert using \(K = °C + 273\).
Use consistent pressure units (kPa, atm or mm Hg). If you change units, apply the appropriate conversion factor (1 atm = 101.3 kPa = 760 mm Hg).
For the ideal‑gas equation, \(R = 8.314\;\text{kPa·L·mol}^{-1}\text{K}^{-1}\) when pressure is in kPa; use the matching value for other pressure units.

6. Practical Applications

Inflating a tyre – Pumping air raises the internal pressure; the tyre’s volume changes only slightly, but the higher pressure makes the tyre rigid enough to support a load.
Hot‑air balloon – Heating the air inside the envelope raises its temperature, causing expansion and a decrease in density; the balloon rises because the surrounding air is denser.
Respiratory system – During inhalation the chest cavity expands, lowering lung pressure; atmospheric pressure then forces air into the lungs (Boyle’s law in action).
Syringe operation – Pulling the plunger increases the gas volume inside the barrel, reducing its pressure and drawing fluid into the syringe (negative pressure).

7. Sample Questions (AO2)

7.1 Combined Gas Law – basic

Question: A 2.00 L sample of gas is at 300 K and 100 kPa. It is heated to 450 K while the pressure is increased to 150 kPa. What is the final volume?

Solution:

Write the combined gas law: \(\displaystyle\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\).
Insert the known values: \(\displaystyle\frac{100 \times 2.00}{300} = \frac{150 \times V_{2}}{450}\).
Calculate the left‑hand side: \(\displaystyle\frac{200}{300}=0.667\).
Re‑arrange for \(V_{2}\): \(\displaystyle0.667 = \frac{150V_{2}}{450}= \frac{V_{2}}{3}\).
Therefore \(V_{2}=0.667 \times 3 = 2.00\;\text{L}\).

Both temperature and pressure increase, but the proportional changes cancel, leaving the volume unchanged.

7.2 More challenging – unit conversion

Question: A 0.500 mol sample of nitrogen gas occupies 12.5 L at 25 °C and a pressure of 1.00 atm. The gas is compressed to a pressure of 750 mm Hg and heated to 80 °C. Calculate the final volume in millilitres (mL). (Take 1 atm = 760 mm Hg.)