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Atoms molecules and particle scale

Learning objectives

  • Outcome

    Compare by order of magnitude the sizes of an atom, a molecule and a visible particle, and record them in metres and nanometres.

  • Outcome

    Explain diffusion and Brownian motion as observable consequences of the chaotic thermal motion of molecules.

  • Outcome

    Estimate the number of particles in a sample from its mass or volume using the Avogadro constant.

  • Outcome

    Explain the difference in compressibility of a gas and a liquid through the distances between particles in the particle model.

Understand the idea

Atoms, Molecules and Particle Scale: How Matter Is Built

Text

Physics studies phenomena through models: a model replaces a real object with a simplified image while keeping its essential properties. One of the basic models is the discrete structure of matter: every substance consists of an enormous number of tiny particles — atoms and molecules — with gaps between them. An atom is the smallest particle of a chemical element that retains its properties; a molecule consists of two or more atoms bonded together. Particle scales are conveniently expressed in nanometres: 1 nm=10−9 m1\ \text{nm}=10^{-9}\ \text{m}; an atom's diameter is about 10−10 m10^{-10}\ \text{m}, a small molecule about 10−9 m10^{-9}\ \text{m}, and a visible grain of sand about 10−3 m10^{-3}\ \text{m} — a million times larger than a molecule. The number of particles is described by the Avogadro constant: one mole of a substance contains NA≈6,02⋅1023N_A \approx 6{,}02\cdot 10^{23} particles. Particles cannot be seen by eye, yet they reveal themselves in Brownian motion and diffusion — both phenomena are explained by the continuous chaotic thermal motion of molecules and confirm the model.

Worked examples

Direct

Problem

The diameter of an atom is approximately d≈10−10 md \approx 10^{-10}\ \text{m}. Estimate how many atoms would fit in a row of length L=1 mmL = 1\ \text{mm}, and write the answer as a power of 10.

Solution

Convert the length to metres: L=1 mm=10−3 mL = 1\ \text{mm} = 10^{-3}\ \text{m}.
In the model in which atoms are spheres packed side by side, the number of particles in the row is

N=Ld=10−310−10=107.N = \frac{L}{d} = \frac{10^{-3}}{10^{-10}} = 10^{7}.

This is 10710^{7}, that is, ten million atoms. Conclusion: the nanometre world is seven orders of magnitude smaller than everyday sizes, so individual atoms cannot be seen by eye.

Application

Problem

A drop of water has volume V=1 mm3V = 1\ \text{mm}^3. The density of water is ρ=1000 kg/m3\rho = 1000\ \text{kg}/\text{m}^3, and its molar mass is M=18 g/molM = 18\ \text{g}/\text{mol}. Estimate the number of molecules in the drop using NA≈6,02⋅1023 mol−1N_A \approx 6{,}02\cdot 10^{23}\ \text{mol}^{-1}.

Solution
  1. Volume in SI units: V=1 mm3=10−9 m3V = 1\ \text{mm}^3 = 10^{-9}\ \text{m}^3.
  2. Mass of the drop: m=ρV=1000⋅10−9=10−6 kg=10−3 gm = \rho V = 1000 \cdot 10^{-9} = 10^{-6}\ \text{kg} = 10^{-3}\ \text{g}.
  3. Amount of substance: ν=m/M=10−3/18≈5,6⋅10−5 mol\nu = m/M = 10^{-3}/18 \approx 5{,}6\cdot 10^{-5}\ \text{mol}.
  4. Number of molecules: N=νNA≈5,6⋅10−5⋅6,02⋅1023≈3,4⋅1019.N = \nu N_A \approx 5{,}6\cdot 10^{-5} \cdot 6{,}02\cdot 10^{23} \approx 3{,}4\cdot 10^{19}.
    Answer: about 3⋅10193\cdot 10^{19} molecules. This result shows why matter is conveniently described statistically rather than particle by particle.

Transfer

Problem

Using the particle model, explain why air in a syringe can be compressed easily while water can hardly be compressed. Estimate the average distance between air molecules under normal conditions (Vm=22,4 L/molV_m = 22{,}4\ \text{L}/\text{mol}) and compare it with the molecular diameter d≈3⋅10−10 md \approx 3\cdot 10^{-10}\ \text{m}.

Solution
  1. Number of molecules per unit volume: n=NAVm=6,02⋅102322,4⋅10−3≈2,7⋅1025 m−3.n = \frac{N_A}{V_m} = \frac{6{,}02\cdot 10^{23}}{22{,}4\cdot 10^{-3}} \approx 2{,}7\cdot 10^{25}\ \text{m}^{-3}.
  2. Average distance between molecules: a≈n−1/3=(2,7⋅1025)−1/3≈3,3⋅10−9 m=3,3 nm.a \approx n^{-1/3} = (2{,}7\cdot 10^{25})^{-1/3} \approx 3{,}3\cdot 10^{-9}\ \text{m} = 3{,}3\ \text{nm}.
  3. Comparison with the diameter: a/d≈3,3⋅10−9/(3⋅10−10)≈11a/d \approx 3{,}3\cdot 10^{-9}/(3\cdot 10^{-10}) \approx 11.
    In a gas the molecules are separated by gaps of about ten diameters, so compressing the gas reduces the volume without bringing the molecules into contact. In a liquid or a solid the molecules are nearly touching (a≈da \approx d), and further compression is opposed by the repulsion of electron shells. Thus the particle model explains the difference in compressibility without introducing new entities.

Practice

Text

Goal: see how the particle model explains the observed behaviour of matter.

  1. Open the PhET States of Matter: Basics simulation (or a similar one) on a computer or phone.
  2. Choose the gas state, turn on the particle display and observe the motion for one minute; sketch 5–6 trajectories.
  3. Raise the temperature and note how the particle speed changes; then lower it and compare the picture.
  4. Switch to the liquid and solid states; compare the distances between particles and their degree of ordering.
  5. State a conclusion: how the distances between particles are related to the compressibility of matter.
    Safety: direct observation of atoms is impossible, so we use a simulation; do not heat closed vessels and do not experiment with unknown substances.

Visual reference

Purpose

Make the enormous range of scales visible: show that an atom is smaller than a molecule, that a molecule is far smaller than a visible grain, and link these sizes to the Avogadro constant.

Representation

A staircase-of-scales diagram on a logarithmic axis: from left to right it marks an atom (10−10 m10^{-10}\ \text{m}), a small molecule (10−9 m10^{-9}\ \text{m}), a nanoparticle (10−810^{-8}–10−7 m10^{-7}\ \text{m}), a cell (10−5 m10^{-5}\ \text{m}) and a grain of sand (10−3 m10^{-3}\ \text{m}); under each mark the corresponding power of ten is written and a small schematic picture is given (a sphere, a two-atom molecular model, a cell outline, a grain). An inset shows the water molecule model: two hydrogen atoms and one oxygen atom bonded together, illustrating the link atom → molecule.

Real-world examples

Bicycle tyre puncture

Scenario

A tyre is inflated to 4 bar4\ \mathrm{bar}; a puncture hole has diameter 1 mm=10−3 m1\ \mathrm{mm}=10^{-3}\ \mathrm{m}. An air molecule is about 0.3 nm=3×10−10 m0.3\ \mathrm{nm}=3\times10^{-10}\ \mathrm{m}. Across the hole, 10−3/(3×10−10)≈3×10610^{-3}/(3\times10^{-10})\approx3\times10^6 molecules fit in a row.

Connection

The example uses the scale ratio: the hole is macroscopic, but its width is millions of molecular diameters, so escaping air is a stream of particles, not a continuous sheet.

Phone lithium-ion battery

Scenario

A phone battery is 5 mm=5×10−3 m5\ \mathrm{mm}=5\times10^{-3}\ \mathrm{m} thick. In graphite, lithium ions sit between layers separated by 0.335 nm=3.35×10−10 m0.335\ \mathrm{nm}=3.35\times10^{-10}\ \mathrm{m}. Along the thickness, 5×10−3/(3.35×10−10)≈1.5×1075\times10^{-3}/(3.35\times10^{-10})\approx1.5\times10^7 interlayer gaps fit.

Connection

The scenario exercises the atomic/molecular scale of layered solids: charge is stored because ions enter gaps comparable to atomic distances.

Sugar cube in tea

Scenario

A sugar cube has edge 1 cm=10−2 m1\ \mathrm{cm}=10^{-2}\ \mathrm{m}. A sucrose molecule is about 1 nm=10−9 m1\ \mathrm{nm}=10^{-9}\ \mathrm{m}. Along one edge, 10−2/10−9=10710^{-2}/10^{-9}=10^7 molecules fit; in the whole cube, about (107)3=1021(10^7)^3=10^{21} molecules. When stirred, they leave the crystal and spread through the water.

Connection

This uses the enormous number and tiny size of molecules: a visible solid cube is a dense assembly of molecules that can separate without visible fragments.