Skip to lesson content

Lesson 1 of 6

Kinetic Theory · Lesson 1 of 6

Molecules, Moles and Molecular Sizes

“Build the molecular picture of matter, count particles using moles, and estimate molecular volume and spacing.”

Learning Objectives

• Explain the molecular picture of solids, liquids and gases and the meaning of dynamic equilibrium. • Connect definite proportions, multiple proportions and gaseous combining volumes to atoms and molecules. • Convert between sample mass, molar mass, mole number and molecular count. • Estimate molecular occupied-volume fractions, molecular size and centre-to-centre spacing. • Distinguish molecular diameter, intermolecular spacing and mean free path.

How can an apparently still gas be full of motion?

A sealed bottle looks quiet, but the gas inside is a collection of rapidly moving molecules. They continually change direction in collisions with one another and the walls. Kinetic theory connects this unseen motion to measurable pressure, temperature and heat capacity. Boyle and Newton explored particle ideas about gases; Maxwell, Boltzmann and others developed the quantitative theory in the nineteenth century. Its success comes from making numerical predictions that can be tested, rather than merely saying that particles exist.

The atomic picture treats matter as built from small constituents in continual motion. Particles interact: attraction can hold nearby particles together, while very strong repulsion prevents them from being squeezed into the same space. Feynman emphasised how much of matter’s behaviour can be recovered from this simple starting idea. For dilute gases, particles spend most of their time far enough apart that intermolecular forces are small except during close encounters.

An old idea became a testable scientific theory

Kanada’s Vaiseshika tradition in India and Democritus in Greece proposed particulate descriptions long before modern experiments. The Indian account discussed fundamental particles, their combinations and characteristic attributes; it associated kinds of matter with earth, water, fire and air while treating space differently. The Greek account imagined different particle shapes to explain properties, such as smooth particles for flowing water and rough particles for hard matter. Historical Indian writings also contain estimates of extremely small particle sizes, of order 10⁻¹⁰ m. These are historical conjectures, not the experimental basis of today’s atomic model. Their lesson is that a useful scientific explanation must be tested and revised quantitatively.

Dalton’s atomic theory supplied a way to account for measured chemical proportions: elements consist of atoms, and compounds form when atoms combine in simple numerical ratios. Modern science has refined that picture. Atoms of an element can have isotopes; atoms have electrons and a nucleus; nuclei contain protons and neutrons, which themselves have quark structure. Atomic indivisibility was therefore a historical proposal, not a modern fact. More speculative ideas about deeper structure should not be confused with established evidence. For the gas physics here, a molecule or a monatomic gas atom can usually be treated as one moving particle.

Chemical proportions reveal discrete building blocks

The law of definite proportions says that a particular pure compound has a fixed mass ratio of its constituent elements. Water, for example, has hydrogen and oxygen in approximately the mass ratio 1:8 when ordinary rounded atomic masses are used. A fixed molecular composition explains why choosing a larger sample changes both masses together without changing the ratio.

The law of multiple proportions compares different compounds made from the same two elements. For a fixed mass of one element, the masses of the other that combine with it occur in small integer ratios. With 12 g of carbon, CO contains about 16 g of oxygen and CO₂ about 32 g. The oxygen masses are in the ratio 1:2 because the molecules contain one or two oxygen atoms per carbon atom. This is a relation between chemical amounts, not a claim that all molecules have the same mass.

Gay-Lussac’s law of combining volumes concerns gases measured at the same temperature and pressure: reacting gas volumes, and gaseous product volumes where applicable, occur in small integer ratios. Avogadro’s hypothesis explains the connection: equal gas volumes at the same temperature and pressure contain equal numbers of molecules in the ideal-gas limit. Thus two gas-volume units of hydrogen reacting with one of oxygen to form two of water vapour correspond to 2 H₂ + O₂ → 2 H₂O. The products must be gaseous and compared under suitable common conditions; condensing the water would remove the direct gas-volume comparison.

The same particle picture gives different states of matter

Typical atomic dimensions are about an ångström, where 1 Å = 10⁻¹⁰ m. In solids and liquids, neighbouring constituent particles are separated by only a few ångströms. A solid’s particles vibrate around relatively fixed equilibrium positions. In a liquid they remain close but can rearrange, allowing flow. In an ordinary dilute gas, molecular centres are separated by several molecular diameters, often a few tens of ångströms, and the gas spreads to fill its container. Interactions matter strongly in closely packed solids and liquids; the long intervals of nearly free motion make the dilute gas model simpler.

One molecular picture, different bulk statesSolid: vibrations about sitesLiquid: close but rearrangingGas: dilute random motionSchematic panels; molecular sizes are not on one shared scaleDiameter ≠ centre spacing ≠ distance travelled between collisions
Particle arrangements and molecular length scales— Positions and arrows are schematic. Solid particles vibrate about sites; liquid particles remain close while rearranging; gas molecules are more widely separated. No common geometric scale is implied between panels.

Thermal equilibrium is dynamic. Individual molecules collide and exchange kinetic energy, so a particular molecule need not keep the same velocity or speed. The measured bulk averages can nevertheless remain constant. Modern microscopy and many independent measurements support the atomic picture; the fact that one cannot track every molecule with the naked eye does not make the model mere speculation.

Moles connect a visible sample to a huge number of particles

Use N for the number of molecules, ν for the amount of substance in moles, M for molar mass in kg mol⁻¹, and msample for total sample mass. Avogadro’s constant NA is approximately 6.022×10²³ mol⁻¹. A mole contains NA specified particles. For neon the particle is an atom; for oxygen gas it is an O₂ molecule. Molar mass is mass per mole, whereas m₀ below is the mass of one molecule.

Particle-count and mass relationsLaTeX
Use kg and kg mol⁻¹ consistently. The familiar molecular mass in u has the same numerical value as molar mass in g mol⁻¹; convert to kg mol⁻¹ for SI calculations.

Number density n means N/V, the number of particles per unit volume, measured in m⁻³. It is not the mole number ν. Mass density ρ means msample/V and equals nm₀ for one molecular species. Equal number densities need not imply equal mass densities, because the masses of individual particles can differ.

Two different densitiesLaTeX
n is a particle count per volume; ρ is mass per volume in kg m⁻³. Number count N is dimensionless, while ν has unit mol.

Molecular occupied volume is much smaller than gas volume

The volume of a gas sample includes the large spaces between molecules. A molecular occupied-volume estimate adds the small volumes assigned to the individual molecules themselves. Those two volumes must not be confused. Treating a liquid as closely packed provides a rough way to estimate molecular volume from mass density; it is an order-of-magnitude model, not a precise determination of a molecule’s boundary.

How much of steam’s volume is molecular material?

Problem
Liquid water has density 1000 kg m⁻³, while water vapour at 100 °C and 1 atm has density 0.6 kg m⁻³. Estimate the fraction of vapour volume occupied by the molecules, treating the liquid volume of the same mass as the occupied molecular volume.

  1. 1.Choose the same water mass m in both phases. Its liquid volume is m/1000 and vapour volume is m/0.6.
  2. 2.The ratio is (m/1000)/(m/0.6) = 0.6/1000 = 6.0×10⁻⁴.
  3. 3.Only about 0.06% of the vapour volume is assigned to molecular material by this model. The vapour volume is about 1667 times the corresponding liquid volume.
  4. 4.This estimate assumes that the densely packed liquid volume represents the sum of molecular volumes. It does not mean real liquid molecules are perfectly space-filling hard spheres.

Estimate one molecule’s mass, volume and radius

An approximate size for a water molecule

Problem
Use water molar mass 0.018 kg mol⁻¹, NA ≈ 6.0×10²³ mol⁻¹ and liquid density 1000 kg m⁻³. Estimate the molecular volume, then the radius of a sphere with that volume.

  1. 1.One molecule has mass m₀ = 0.018/(6.0×10²³) = 3.0×10⁻²⁶ kg.
  2. 2.Using liquid density as the effective molecular density, vmolecule ≈ m₀/ρ = 3.0×10⁻²⁹ m³.
  3. 3.For a sphere vmolecule = 4πr³/3, so r = [3 vmolecule/(4π)]^(1/3) ≈ 1.93×10⁻¹⁰ m ≈ 2 Å.
  4. 4.The corresponding diameter is about 4 Å. This geometric estimate is rough; water is not literally a sphere, and an effective collision diameter need not equal a radius inferred from condensed-phase volume.
Sphere volume and hard-sphere diameterLaTeX
If a question gives a diameter, divide by two before using a radius formula. Lengths in Å must be converted to metres before cubing.

Particle spacing is a length, so it scales with a cube root

If the total volume assigned to each particle increases by a factor F, a characteristic centre spacing increases by F^(1/3), not by F. A cube with side a has volume a³; this is the geometric reason. For a number density n, a useful spacing estimate is a ≈ n^(−1/3). It assigns each particle a representative cubic cell. In a real gas positions are random, so this is a characteristic scale rather than an exact nearest-neighbour distance.

Spacing in liquid water and in its vapour

Problem
Estimate water-molecule spacing using the preceding mass and density data. Compare the liquid and vapour scales and explain the rough estimate of about 40 Å for vapour.

  1. 1.The liquid volume per molecule is about 3.0×10⁻²⁹ m³, so its representative cell spacing is (3.0×10⁻²⁹)^(1/3) ≈ 3.11 Å.
  2. 2.In vapour the volume per molecule is m₀/ρvapour = (3.0×10⁻²⁶)/0.6 = 5.0×10⁻²⁶ m³. Its cube root is about 3.68×10⁻⁹ m = 36.8 Å.
  3. 3.The ratio of spacing scales is (1000/0.6)^(1/3) ≈ 11.9. Thus “a few Å in liquid” becomes “a few tens of Å in vapour”.
  4. 4.A coarser estimate uses the roughly 4 Å molecular diameter as the starting length and rounds the volume increase to 10³, hence a length increase of about 10 and a spacing near 40 Å. Both are order-of-magnitude estimates, not exact equalities.
  5. 5.The centre-to-centre intermolecular spacing is not the distance between hydrogen and oxygen atoms bonded within a single water molecule.
Hard-sphere occupied fraction for oxygen at STP

Problem
Estimate the occupied molecular-volume fraction for one mole of oxygen at textbook STP, using molar gas volume 22.4 L, molecular diameter d = 3 Å and NA = 6.02×10²³ mol⁻¹.

  1. 1.Convert d = 3.0×10⁻¹⁰ m and gas volume V = 22.4×10⁻³ m³.
  2. 2.The volume assigned to one hard sphere is πd³/6 = 1.414×10⁻²⁹ m³.
  3. 3.The sum for one mole is NAπd³/6 ≈ 8.51×10⁻⁶ m³.
  4. 4.Dividing by 22.4×10⁻³ m³ gives about 3.80×10⁻⁴, or 0.038%. A rough one-significant-figure estimate is 4×10⁻⁴.
  5. 5.The 22.4 L is the container volume, not a claim that oxygen molecules themselves occupy that much space. The ideal-gas derivation of this molar volume follows in the next lesson.

Three lengths with different meanings

Molecular diameter characterises the size of one particle. Intermolecular spacing characterises how far apart particle centres are at a given density. Mean free path is the average total distance travelled between successive collisions. A fast molecule can pass through many intermolecular spacings before a collision because a collision requires a sufficiently close encounter, not merely passing the distance of one average cell. The mean-free-path calculation will make this distinction quantitative.

Quick Quiz

Quiz

Quick check

Which is a number density rather than a mole number?

Explanation

Number density is particles per volume, N/V. N/NA and msample/M are mole numbers.

Quick check

A gas has constant bulk pressure in equilibrium. What follows about each molecule?

Explanation

Equilibrium is dynamic: collisions redistribute velocities even when bulk averages remain constant.

Quick check

The ratio of vapour density 0.6 kg m⁻³ to liquid density 1000 kg m⁻³ gives an occupied-volume estimate of

Explanation

For equal mass, liquid volume divided by vapour volume equals ρvapour/ρliquid = 0.0006.

Quick check

If volume per particle increases by a factor 1000, a characteristic particle spacing increases by

Explanation

A linear scale varies as the cube root of volume: 1000^(1/3) = 10.

Quick check

In a hard-sphere calculation a molecule has diameter 3 Å. Its assigned radius is

Explanation

Radius is half the diameter. Confusing them multiplies the calculated sphere volume by eight.

Practice Problems

Practice Problems
  1. Explain definite proportions, multiple proportions and Avogadro’s hypothesis using water, carbon oxides and gaseous combining volumes. Distinguish mass ratios from molecule-number ratios.
  2. A 0.016 kg oxygen sample has molar mass 0.032 kg mol⁻¹. Using NA = 6.02×10²³ mol⁻¹, find its mole number and particle count. State whether the particles counted are oxygen atoms or O₂ molecules.
  3. Using liquid density 800 kg m⁻³ and vapour density 1.2 kg m⁻³ for the same substance, estimate the molecular occupied-volume fraction of the vapour under the close-packing approximation.
  4. A molecule has mass 4.0×10⁻²⁶ kg and an assigned condensed-phase density 1000 kg m⁻³. Estimate its volume and the radius of a sphere with that volume. State why this is not a precise molecular boundary measurement.
  5. For number density 8.0×10²⁷ m⁻³, estimate the representative cubic-cell spacing n^(−1/3). Find how it changes if the density becomes 1000 times smaller.
  6. For one mole of a hard-sphere gas of molecular diameter 2.0 Å occupying 22.4 L, estimate the occupied-volume fraction. Use NA = 6.02×10²³ mol⁻¹ and show all unit conversions.
  7. Compare molecular diameter, intermolecular centre spacing and mean free path. Explain why a molecule need not collide after travelling just one average intermolecular spacing.

Key Takeaways

Key Takeaways

• The molecular model becomes scientific through quantitative evidence and tested predictions. • Equilibrium leaves molecules in continual random motion; only bulk statistical properties are steady. • ν = msample/M = N/NA counts moles, while n = N/V counts particles per volume. • Molecular occupied volume can be a tiny fraction of gas volume. • Particle size, spacing and mean free path are distinct; spacing scales with the cube root of volume per particle.