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Particle models of gases liquids and solids

Learning objectives

  • Outcome

    Describe the arrangement, mean spacing and type of motion of particles in the solid, liquid and gaseous states.

  • Outcome

    Compare the three states by compressibility, retention of shape and volume, explaining the differences through interparticle spacing and interaction forces.

  • Outcome

    Estimate the mean interparticle spacing using d≈(V/N)1/3d \approx (V/N)^{1/3} and compare it with the size of a molecule.

  • Outcome

    Predict how the gas pressure and the liquid volume change when a sealed rigid vessel is heated, using the particle model.

Understand the idea

Particle models of gases, liquids and solids

Text

All matter consists of particles — atoms, ions or molecules — that are in continuous thermal motion, and temperature is a measure of their mean kinetic energy: ⟨Ek⟩∝T\langle E_k \rangle \propto T, and for an ideal gas ⟨Ek⟩=32kBT\langle E_k \rangle = \frac{3}{2}k_B T. Thermal motion is opposed by attractive forces between particles, and it is the balance between thermal energy and binding energy that determines the state of the substance. In a solid, particles are arranged in an ordered crystal lattice, the distances between them are comparable to the particle size, and they only vibrate about equilibrium positions — so a solid keeps both its shape and its volume and is almost incompressible. In a liquid, particles are also close together (spacings of the order of the particle size), but the order is only short-range and particles can jump from place to place, so a liquid keeps its volume, takes the shape of the container and is practically incompressible. In a gas, the mean distance between particles is many times their size — under normal conditions d≈10 dmold \approx 10\,d_{\text{mol}} — attractive forces are almost negligible, particles travel in straight lines until they collide, so a gas fills any available volume and is easily compressed. For quantitative estimates the mean interparticle spacing is conveniently calculated as d≈(V/N)1/3d \approx (V/N)^{1/3}, where VV is the volume and NN is the number of particles.

Worked examples

Direct

Problem

From the description of observations, identify the state of matter and justify your answer through the interparticle spacing and the freedom of particle motion:

(a) particles form a regular lattice and vibrate about fixed equilibrium positions;

(b) particles are close together and disordered, jumping past one another;

(c) particles are far apart and move in straight lines until they collide.

Solution

(a) Solid. Particles are held by attractive forces at lattice sites, the mean spacing is comparable to the particle size, and the thermal energy is sufficient only for vibration about an equilibrium position, not for hopping. Hence the shape and volume are retained and compressibility is small.

(b) Liquid. The mean spacing is as small as in a solid, but the order is only short-range: a particle can jump into a neighbouring site. Therefore the volume is retained while the shape is not, and compressibility remains very small.

(c) Gas. The mean spacing d≈(V/N)1/3d \approx (V/N)^{1/3} is many times the particle size, attractive forces hardly affect the motion, and particles fly freely until they collide. Therefore a gas fills the whole volume and is easily compressed.

Application

Problem

A sample of gas contains N=6.02×1023N = 6.02\times10^{23} particles and occupies a volume V=24 dm3V = 24\ \mathrm{dm^3}. Estimate the mean interparticle spacing using d≈(V/N)1/3d \approx (V/N)^{1/3} and compare it with the molecular diameter dmol≈3.5×10−10 md_{\text{mol}} \approx 3.5\times10^{-10}\ \mathrm{m}. What conclusion can be drawn about the fraction of volume occupied by the particles themselves?

Solution

Convert the volume to SI: V=24 dm3=24×10−3 m3V = 24\ \mathrm{dm^3} = 24\times10^{-3}\ \mathrm{m^3}.

Find the volume per particle:

VN=24×10−36.02×1023≈3.99×10−26 m3.\frac{V}{N} = \frac{24\times10^{-3}}{6.02\times10^{23}} \approx 3.99\times10^{-26}\ \mathrm{m^3}.

Hence

d≈(3.99×10−26)1/3≈3.4×10−9 m.d \approx (3.99\times10^{-26})^{1/3} \approx 3.4\times10^{-9}\ \mathrm{m}.

The ratio to the molecular size:

ddmol≈3.4×10−93.5×10−10≈10.\frac{d}{d_{\text{mol}}} \approx \frac{3.4\times10^{-9}}{3.5\times10^{-10}} \approx 10.

Conclusion: in a gas under normal conditions the mean spacing is about 10 times the molecular diameter. Along a line a molecule occupies about 10% of the length (dmol/d≈0.10d_{\text{mol}}/d \approx 0.10), and the volume fraction filled by the particles themselves is (dmol/d)3≈1.0×10−3(d_{\text{mol}}/d)^3 \approx 1.0\times10^{-3} — as a dimensionless fraction this is about 0.1%. This is exactly why a gas can be strongly compressed: there is plenty of free space between the particles.

Transfer

Problem

A sealed rigid steel vessel contains water and air above it. The vessel is heated from 20 ∘C20\ ^\circ\mathrm{C} to 80 ∘C80\ ^\circ\mathrm{C}. Using the particle model, predict and justify: (1) how the air pressure changes; (2) how the liquid volume changes; (3) whether the mean spacing between air particles changes. Separately explain why the gas in such a vessel is easily compressed by a piston while the liquid is almost not.

Solution

(1) The air pressure will rise. The temperature increases and ⟨Ek⟩∝T\langle E_k \rangle \propto T, so the mean speed of the molecules increases. At constant volume the molecules hit the walls more often and transfer a larger momentum in each impact, so the pressure rises: for an ideal gas at V=constV = \text{const} we have p∝Tp \propto T, so the pressure increases by about a factor 353293≈1.2\frac{353}{293} \approx 1.2.

(2) The liquid volume will hardly change. Liquid particles are already packed closely, with spacings of the order of the molecular size, and attractive forces keep them together. At 80 ∘C80\ ^\circ\mathrm{C} the thermal energy is still insufficient to break the bonds and increase the mean spacing appreciably; only a small thermal expansion (of the order of a few percent) is observed.

(3) The mean spacing between air particles will not change, because the vessel is rigid and the gas volume is fixed: d≈(V/N)1/3=constd \approx (V/N)^{1/3} = \text{const} at constant VV and NN.

Why the gas is compressed while the liquid is not. In a gas the particles are separated by large free spaces (d≈10 dmold \approx 10\,d_{\text{mol}}), so a piston can bring them closer, reducing VV and hence dd. In a liquid the particles are already essentially in contact, and bringing them closer requires overcoming the strong repulsion of the electron shells, so the compressibility of a liquid is negligible.

Practice

Text

Safe self-study plan in a simulation. Do not heat sealed vessels or pressurise anything at home — carry out all observations in a virtual laboratory.

  1. Open a states-of-matter simulation (for example PhET "States of Matter: Basics" or a similar molecular-dynamics model) and select the substance "water".
  2. Switch in turn to the "solid", "liquid" and "gas" modes. For each state record in a table: (a) the ordering of the particle arrangement, (b) the mean interparticle spacing in units of the molecular diameter, (c) the type of motion (vibration, hopping, free flight).
  3. Slowly raise and lower the temperature and note under what conditions the transitions between states occur and how the mean particle speed changes.
  4. Then switch on the "compression" mode: act with a "piston" on the gas and on the liquid and compare how easily the volume changes in each case. Explain the result using d≈(V/N)1/3d \approx (V/N)^{1/3}.
  5. Write a one-paragraph conclusion: how the interparticle spacing and the interaction forces explain the differences between the three states.

Visual reference

Purpose

To show, on a single common scale, how the three states differ in ordering, mean interparticle spacing and freedom of motion, and thereby to explain the differences in compressibility and retention of shape.

Representation

A sketch of three equally sized rectangles: "Solid | Liquid | Gas". In each rectangle the particles are drawn as circles: in the solid they form a regular lattice and carry short double-headed vibration arrows; in the liquid they are close together and disordered, with arrows showing hops between neighbouring sites; in the gas the sparse circles are far apart, with long straight velocity arrows and collision points. Below the panels is a common scale bar labelled d≈(V/N)1/3d \approx (V/N)^{1/3} and marking the molecular diameter dmold_{\text{mol}}, so that d≫dmold \gg d_{\text{mol}} is visible only in the gas. On the right are labels: "keeps shape and volume", "keeps volume, takes the shape of the container", "fills the whole volume".

Real-world examples

Air inside a bicycle tyre

Scenario

The inner volume of a road-bike tyre is 4×10−4 m34\times10^{-4}\ \text{m}^3 (0,4 L0{,}4\ \text{L}). A pump raises the pressure to 8×105 Pa8\times10^{5}\ \text{Pa} at 293 K293\ \text{K} — that is about 0,13 mol0{,}13\ \text{mol}, roughly 8×10228\times10^{22} nitrogen and oxygen molecules with a total mass of 3,8 g3{,}8\ \text{g}. Their mean speed at 293 K293\ \text{K} is 3RT/M≈510 m/s\sqrt{3RT/M}\approx 510\ \text{m/s}; each molecule covers hundreds of metres every second yet never leaves the tyre.

Connection

This is the gas model at work: the mean separation of the molecules is tens of times their size (∼0,3 nm\sim 0{,}3\ \text{nm}), there are no fixed sites, and the particles fill whatever volume is available. Two observable facts follow: a gas has neither its own shape nor its own volume, and the tyre wall feels a continuous hail of molecular impacts.

Water in a bottle and in a syringe

Scenario

1 L1\ \text{L} of water with a mass of 1 kg1\ \text{kg} (density 1000 kg/m31000\ \text{kg/m}^3) is poured into a two-litre bottle: the water occupies exactly 1 L1\ \text{L}, takes the shape of the lower part of the bottle, and its surface stays horizontal when the bottle is tilted. If the water is drawn into a syringe and the plunger of area 1 cm21\ \text{cm}^2 is pushed with a force of 100 N100\ \text{N}, the pressure rises by Δp=FS=106 Pa=1 MPa\Delta p=\frac{F}{S}=10^{6}\ \text{Pa}=1\ \text{MPa}, while the volume falls by only ΔpK=1062,2×109≈0,05 %\frac{\Delta p}{K}=\frac{10^{6}}{2{,}2\times10^{9}}\approx 0{,}05\ \%.

Connection

The liquid model applies here: the particles are packed almost touching, yet they do not form a rigid lattice and can slide past one another. Two measured properties follow directly — near-incompressibility (fixed volume: 0,05 %0{,}05\ \% instead of the 100 %100\ \% a gas would show) and fluidity (no fixed shape: the water copies the shape of the bottle).

Expansion gap at a rail joint

Scenario

A steel rail 25 m25\ \text{m} long is laid at 5 ∘C5\ ^\circ\text{C} and heats up to 35 ∘C35\ ^\circ\text{C} in summer, so ΔT=30 K\Delta T=30\ \text{K}. For steel α=12×10−6 K−1\alpha=12\times10^{-6}\ \text{K}^{-1}, hence ΔL=αLΔT=12×10−6×25×30≈9 mm\Delta L=\alpha L\Delta T=12\times10^{-6}\times25\times30\approx 9\ \text{mm}. That is exactly the gap left between rails: without it the ends press against each other and the track buckles sideways.

Connection

This is the solid-state model: the particles are held near fixed equilibrium positions — lattice sites — and only vibrate with an amplitude of about 0,01 nm0{,}01\ \text{nm}. Heating increases that amplitude, so the mean distance between neighbouring particles grows slightly: the body keeps both its shape and its volume, yet it still expands by 9 mm9\ \text{mm} in 25 m25\ \text{m}.