States of Matter

Physical Chemistry Weightage: AS · Papers 1 and 2 (gas calculations also in Paper 3 and Paper 4) Topic 4

“Welcome! This topic splits into two very different skills. The gas half is a calculation you can practise until it never fails: convert the units, substitute, and state the answer with its unit. The solid half is an explanation skill: name the structure, name the particles, name what has to be overcome, and the marks follow. Both halves reward exactly the same habit of showing every step.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in this topic

Gas calculations: one mark for converting units (T in K, V in m³, p in Pa), one for rearranging and substituting correctly, one for the answer with its unit and sensible significant figures.
Ideal-gas explanations: the two assumptions are separate marks: negligible volume and no intermolecular forces. A deviation answer needs the condition and the reason.
Structure questions: name the structure type, name the particles, then name what must be overcome. “Strong bonds” without saying which bonds loses the mark.
Graphite and diamond appear in almost every session in some form. Learn each of them completely, including the reason for conduction.

1. The Three States

Going from solid to liquid to gas, the particles gain kinetic energy and move further apart. Melting and boiling of a simple molecular substance overcome intermolecular forces only; for ionic, metallic and giant covalent substances the strong bonds themselves must be overcome.

State	Arrangement	Movement	Fixed shape / volume?
Solid	Regular, closely packed	Vibrate about fixed positions	Fixed shape and volume
Liquid	Close but irregular	Move past each other	Fixed volume, takes shape of container
Gas	Far apart	Fast, random, in straight lines between collisions	Neither: fills the container

Melting and boiling need energy to overcome the forces holding the particles together. Freezing and condensation release the same energy.
Gas pressure is caused by gas molecules colliding with the walls of the container. More frequent or more forceful collisions give a higher pressure.

2. Ideal and Real Gases

An ideal gas is one that obeys pV = nRT exactly. It behaves as if:

the gas molecules have zero (negligible) volume compared with the container;
there are no intermolecular forces between the molecules;
collisions with each other and the walls are perfectly elastic (no loss of kinetic energy).

Real gases are not ideal because molecules do have volume and do attract one another.

A real gas deviates from ideal behaviour at high pressure (the molecules are close together, so their own volume is not negligible) and at low temperature (molecules move slowly, so intermolecular forces matter). Ideal behaviour is closest at low pressure and high temperature, and for small non-polar molecules such as He and H₂.

High pressure: the molecules are squeezed close together, so their own volume is no longer negligible compared with the volume of the container.
Low temperature: the molecules move slowly, so intermolecular forces are large enough to pull molecules together (and cause the gas to liquefy).
A real gas is most ideal at low pressure and high temperature. Small, non-polar molecules with very weak forces, such as He and H₂, are the closest to ideal. Polar gases such as NH₃ deviate more because of hydrogen bonding.

⚠️ Examiner Trap: Low pressure means what? At low pressure the molecules are far apart, so both problems (their volume and their attraction) become negligible. Writing “high pressure makes the gas ideal” is a classic slip. Always pair the condition with its reason.

For a fixed amount of an ideal gas: V ∝ 1/p at constant temperature, and V ∝ T at constant pressure. Always use kelvin: 27 °C is 300 K, and doubling the kelvin temperature doubles the volume.

3. The Ideal Gas Equation

pV = nRT where p is pressure (Pa), V is volume (m³), n is amount (mol), R = 8.31 J K⁻¹ mol⁻¹ and T is temperature (K).

pV = nRT. Put every quantity into SI units before substituting, unless you are using the kPa with dm³ shortcut. A wrong unit conversion is the most common way to lose the final mark.

Worked example 1: volume. Calculate the volume of 0.320 mol of oxygen at 200 kPa and 32 °C.

Convert: p = 200 000 Pa; T = 32 + 273 = 305 K.
V = nRT ÷ p = (0.320 × 8.31 × 305) ÷ 200 000 = 4.06 × 10⁻³ m³.
In dm³: 4.06 dm³ (1 m³ = 1000 dm³).

Worked example 2: pressure. 0.0500 mol of a gas occupies 1.00 dm³ at 300 K. Calculate the pressure.

V = 1.00 dm³ = 1.00 × 10⁻³ m³.
p = nRT ÷ V = (0.0500 × 8.31 × 300) ÷ (1.00 × 10⁻³) = 1.25 × 10⁵ Pa = 125 kPa.

Changing conditions without n: for a fixed amount, p₁V₁ ÷ T₁ = p₂V₂ ÷ T₂. For example, 2.00 dm³ of gas at 27 °C (300 K) heated to 127 °C (400 K) at constant pressure becomes 2.00 × 400 ÷ 300 = 2.67 dm³.

⚠️ Examiner Trap: Celsius in gas laws. Never substitute °C. Volume is proportional to the temperature in kelvin: 27 °C to 127 °C is 300 K to 400 K, an increase of one third, not a factor of 127 ÷ 27.

4. Finding Relative Molecular Mass

Since n = m ÷ M_r, the equation becomes pV = (m ÷ M_r)RT, so

M_r = mRT ÷ pV

The method for a volatile liquid uses a gas syringe in an oven: inject a known mass of liquid (found by weighing a syringe before and after), let it vaporise completely at a stated temperature above its boiling point, and record the volume of vapour and the pressure.

Worked example 3. 0.126 g of a volatile liquid is injected into a gas syringe held at 100 °C. It forms 45.0 cm³ of vapour at 101 kPa. Find M_r.

T = 373 K; V = 45.0 cm³ = 4.50 × 10⁻⁵ m³; p = 1.01 × 10⁵ Pa.
n = pV ÷ RT = (1.01 × 10⁵ × 4.50 × 10⁻⁵) ÷ (8.31 × 373) = 1.47 × 10⁻³ mol.
M_r = 0.126 ÷ 1.47 × 10⁻³ = 85.9 (a hydrocarbon with M_r = 86 is C₆H₁₄).

This method is less accurate than a mass spectrum: some vapour may condense, the liquid may not vaporise completely, and the vapour is not perfectly ideal.

5. Liquids and Vapour Pressure

Evaporation happens below the boiling point: the fastest molecules at the surface escape. The liquid cools because the most energetic particles leave.
In a closed container, evaporation and condensation reach a dynamic equilibrium. The pressure of the vapour in equilibrium with its liquid is the vapour pressure, and it rises as the temperature rises.
The boiling point is the temperature at which the vapour pressure equals the external pressure. At lower external pressure a liquid boils at a lower temperature.

The vapour pressure of a liquid rises steeply with temperature because more molecules have enough energy to escape. The boiling point is the temperature at which the vapour pressure equals the external pressure, so a lower external pressure (for example on a mountain) gives a lower boiling point.

6. Lattice Structures

Four lattice types. Ionic: strong attraction throughout. Simple molecular: only weak intermolecular forces to overcome (low melting point). Diamond: a rigid 3-D network of covalent bonds (very hard, very high melting point, no conduction). Graphite: strong layers, weak forces between layers (soft, conducts along the layers).

Structure	Particles / bonding	Melting point	Conducts?	Examples
Giant ionic	Ions in a lattice; strong ionic bonding	High	Molten or aqueous only	NaCl, MgO
Giant metallic	Positive ions in a sea of delocalised electrons	Usually high	Solid and liquid	Cu, Mg
Giant molecular (covalent)	Atoms joined by covalent bonds throughout	Very high	No (graphite yes)	Diamond, graphite, SiO₂
Simple molecular	Small molecules; weak intermolecular forces	Low	No	I₂, C₆₀, ice, CO₂

Ionic: dissolves in water (polar molecules surround the ions), brittle, high melting point because of the strong attraction between oppositely charged ions in all directions.
Simple molecular: low melting and boiling points because only weak intermolecular forces (id–id, pd–pd or hydrogen bonding) break; poor conductor because there are no mobile charged particles; usually soluble in non-polar solvents.
Iodine is a simple molecular solid with weak id–id forces, so it sublimes easily. Ice is simple molecular too, held in an open lattice by hydrogen bonds.
Metallic: conducts as a solid because of mobile delocalised electrons, and is malleable. Note that melting points vary: Group 1 metals melt at low temperatures.

⚠️ Examiner Trap: Simple versus giant. Silicon dioxide (SiO₂) is giant molecular; carbon dioxide (CO₂) is simple molecular. Both have covalent bonds, but only in the simple structure is the melting point set by weak intermolecular forces. Explain melting of SiO₂ by breaking strong covalent bonds, not weak forces.

7. Carbon and Unfamiliar Structures

Allotrope	Bonding	Key properties and reasons
Diamond	Each C bonded to 4 others, tetrahedral, giant	Very hard, very high mp (many strong covalent bonds); no mobile electrons so does not conduct
Graphite	Each C bonded to 3 others in layers, giant	High mp (strong bonds in layers); soft and slippery (weak id–id between layers); conducts along the layers (fourth electron is delocalised)
Fullerene C₆₀	3 bonds per C, 12 pentagons and 20 hexagons, simple molecular	Sublimes at low temperature (weak forces between balls); soft; poor conductor
Graphene	Single layer of graphite	Excellent conductor, very strong for its mass
Nanotube	Graphene rolled into a tube	High tensile strength and conductivity along the tube

Allotropes and related structures built from rings of carbon atoms. Nanotubes and graphene are not named in the syllabus, but Cambridge may give you an unfamiliar structure and ask you to deduce its properties from the bonding: strong covalent bonds and delocalised electrons mean high melting point and electrical conductivity; weak forces between separate molecules (C₆₀) mean low sublimation point.

Nanotubes and graphene are not named in the syllabus, but Cambridge may show you a diagram of an unfamiliar structure. Use the same reasoning every time: count bonds per atom, decide if the structure is giant or simple, and ask whether any electrons are free to move.

8. Deducing Structure from Data

Low melting point (below about 500 K) and does not conduct → simple molecular.
High melting point, conducts only when molten or dissolved → giant ionic.
High melting point, conducts as a solid → giant metallic (or graphite if it is a non-metal).
Very high melting point, does not conduct, insoluble → giant molecular.

⚠️ Examiner Trap: Explain the conductivity properly. “Ionic solids do not conduct because the ions cannot move” is the standard answer. “There are no electrons” is wrong: the electrons are there, but they are not delocalised and the ions are fixed.

9. Quick Sheet and Checklist

Ideal gas assumptions and why real gases deviate (high p, low T).
pV = nRT with SI units; the kPa with dm³ shortcut; kelvin only.
M_r = mRT ÷ pV and the gas syringe method.
Vapour pressure, dynamic equilibrium and boiling at lower pressure.
Four lattice types, with examples, properties and reasons.
Diamond, graphite, C₆₀: bonds per carbon and properties.
Data table → structure type.

Before you leave this topic, can you:

State the two assumptions of the ideal gas and why a real gas deviates at high pressure?
Calculate the pressure of 0.0500 mol of gas in 1.00 dm³ at 300 K?
Explain why graphite conducts but diamond does not?
Say why SiO₂ melts at a much higher temperature than CO₂?

If yes to all four, attempt the ten exam-style questions with the mark schemes covered, then open the flashcards.