Attraction and repulsion of particles

Particles of matter (atoms, molecules, ions) interact with one another: at large distances attraction dominates, at very small distances repulsion dominates. Between these regions there is a distance $r_0$ at which the forces balance and the interaction potential energy is minimal. Taking the force $F$ as positive when it is repulsive:

$$r < r_0 \;\Rightarrow\; F > 0, \qquad r > r_0 \;\Rightarrow\; F < 0.$$

At $r \gg r_0$ the interaction weakens quickly and practically vanishes. The properties of the states of matter follow from this balance: in gases the mean separation of particles is much greater than $r_0$, the forces are weak, and the particles fly almost freely; in liquids the particles sit near $r_0$, so attraction holds them together while still allowing them to jump from place to place; in solids the particles vibrate about equilibrium positions at $r_0$. Thermal motion continuously pushes particles away from equilibrium, and the interaction pulls them back.

Two molecules are separated by $r = 4.5\cdot 10^{-10}$ m. The equilibrium separation for this substance is $r_0 = 3.0\cdot 10^{-10}$ m. Taking the force as positive when it is repulsive, determine the sign of the net force and state whether the molecules attract or repel each other.

Estimate the mean separation between molecules in liquid water ($\rho = 1000$ kg/m$^3$, $M = 0.018$ kg/mol) and in water vapour under normal conditions (molar volume $V_m = 22.4$ L/mol). Compare both results with $r_0 \approx 0.3$ nm and explain why a liquid holds together as a drop while vapour does not gather into a drop by itself.

A solid is heated, and its particles vibrate about $r_0$ with increasing amplitude. Using the distance dependence of the force

$$F_r(r) = C\left[\left(\frac{r_0}{r}\right)^{13} - \left(\frac{r_0}{r}\right)^{7}\right],$$

explain why the body expands on heating and why a solid is harder to compress than a gas.

Direct observation of individual molecules at home is impossible, so work with a computer model (for example, PhET "Atomic Interactions" or "States of Matter: Basics").

1. In a spreadsheet, plot $F_r(r) = C\left[\left(\frac{r_0}{r}\right)^{13} - \left(\frac{r_0}{r}\right)^{7}\right]$ for $C = 1$, $r_0 = 1$ with $r$ from $0.8$ to $3$. Mark the repulsion region, the point $F_r = 0$, and the attractive tail on the graph.
2. In the simulation, place two particles and smoothly change their separation, recording the sign of the force in a table (at least six points).
3. State three conclusions: where the force is zero, where the particles attract, where they repel, and how this relates to the stability of a solid and a liquid.

Safety: use the virtual model only; do not heat or compress closed vessels and do not try to reproduce gas compression with real containers.

Show that the interaction force changes sign at $r = r_0$, and connect this transition with particle equilibrium and the stability of the states of matter.

A graph of the radial force $F_r$ versus distance $r$: a positive branch for $r < r_0$ (repulsion), zero at $r_0$, and a negative branch for $r > r_0$ (attraction) tending to zero as $r \to \infty$. Beside it, a sketch of two particles joined by a segment marked $r_0$, with force arrows.

A road bike tyre is inflated with a hand pump to $7\,\text{bar}$ ($700\,\text{kPa}$). The pump barrel is $250\,\text{mm}$ long with an inner diameter of $28\,\text{mm}$ and holds about $154\,\text{cm}^3$ of air at $1\,\text{bar}$; when the piston is pushed down and the volume shrinks by a factor of about $7$, the pressure in the barrel rises to $7\,\text{bar}$.

At $1\,\text{bar}$ and $20\,^\circ\text{C}$ each air molecule has an average volume of about $(3\,\text{nm})^3$; at $7\,\text{bar}$ the mean spacing falls to roughly $1.8\,\text{nm}$. $\text{N}_2$ and $\text{O}_2$ molecules start to repel each other as soon as their electron shells come closer than about $0.3\,\text{nm}$, and this short-range repulsion is exactly the force the hand overcomes on every pump stroke. The weak attraction between molecules explains why air can be liquefied at all: nitrogen boils at $77\,\text{K}$.

A $25\,\text{g}$ ice cube at $0\,^\circ\text{C}$ is dropped into a glass holding $250\,\text{g}$ of lemonade at $20\,^\circ\text{C}$. Melting the cube takes $25\,\text{g}\times 334\,\text{J/g}=8.35\,\text{kJ}$ drawn from the drink, so the lemonade cools by about $8\,^\circ\text{C}$, down to $12\,^\circ\text{C}$.

In an ice crystal each water molecule is held by hydrogen bonds to four neighbours about $0.28\,\text{nm}$ away, so the molecules only vibrate about fixed positions. Melting one mole of ice takes $6.0\,\text{kJ}$, about $0.06\,\text{eV}$ per molecule — the energy needed for attraction to stop pinning those neighbours in place, even though the molecules stay in contact: water keeps a density of about $1000\,\text{kg/m}^3$ and the molecules slide past one another.

A balloon $30\,\text{cm}$ across holds about $14\,\text{L}$ of helium, roughly $3.5\times 10^{23}$ atoms at $20\,^\circ\text{C}$ and $1\,\text{atm}$. After two days it no longer floats and sinks to the floor.

A helium atom has a filled electron shell, and the attraction between He atoms is the weakest of all substances: liquid helium exists only below $4.2\,\text{K}$ at $1\,\text{atm}$, whereas nitrogen boils at $77\,\text{K}$ and oxygen at $90\,\text{K}$ precisely because their molecules attract one another much more strongly. At $20\,^\circ\text{C}$ a helium atom has a mean speed $\sqrt{3RT/M}\approx 1.35\,\text{km/s}$, and these atoms, which barely interact with each other, gradually leak out through the latex walls.