States of matter — solid, liquid, gas, plasma; macroscopic vs particulate view: Exercises

This page is a self-testing ladder. Each problem builds on the parent topic. Try each one before opening its solution. The levels climb from recognising states to synthesising the whole particle picture.

Two constants you will reuse everywhere:

Figure — States of matter — solid, liquid, gas, plasma; macroscopic vs particulate view

Level 1 — Recognition

Exercise 1.1 — Name the state from behaviour

A sample keeps its own volume but flows to take the shape of its container. It can barely be squeezed smaller. Which state is it, and what are its particles doing?

Recall Solution

What we see (macroscopic): definite volume + indefinite shape + almost incompressible.

Definite volume ⇒ particles are still in contact (not a gas).
Indefinite shape ⇒ particles can slide past each other (not a solid).

Answer: liquid.

Particulate view: particles are close-packed (so volume is fixed and it resists squeezing, just like a solid), but $K E \approx F_{I M F}$ , so a particle can escape its cage of neighbours and drift — giving flow. Motion is translational + rotational + vibrational. See Kinetic Molecular Theory.

Exercise 1.2 — Match property to state

For each property, name the state (solid / liquid / gas): (a) Expands to fill any container. (b) Vibrates about fixed lattice sites only. (c) ~1000× less dense than the other two condensed states.

Recall Solution

(a) Gas — indefinite shape and indefinite volume; particles fly freely, $K E ≫ F_{I M F}$ .
(b) Solid — only vibrational motion, particles locked in a lattice.
(c) Gas — particles are ~10 diameters apart, so mostly empty space ⇒ very low density.

Level 2 — Application

Exercise 2.1 — Thermal energy per particle

Compute the average kinetic energy of a single particle at $T = 298 K$ (room temperature). Report in joules and convert to kJ/mol.

Recall Solution

Per particle: $⟨ K E ⟩ = \frac{3}{2} k_{B} T = \frac{3}{2} (1.381 \times 1 0^{- 23}) (298) = 6.17 \times 1 0^{- 21} J .$ Per mole (multiply by $N_{A}$ ): $⟨ K E ⟩_{mol} = (6.17 \times 1 0^{- 21}) (6.022 \times 1 0^{23}) = 3.72 \times 1 0^{3} J/mol = 3.72 kJ/mol .$ Why this step? This single number is the "opponent" that every intermolecular force must beat to hold a state together. Keep it handy: room-temperature thermal energy ≈ 3.7 kJ/mol per mole.

Exercise 2.2 — Molar volume of an ideal gas

Using $P V = n R T$ , find the volume of $1 mol$ of an ideal gas at $T = 273.15 K$ and $P = 101325 Pa$ .

Recall Solution

$V = \frac{n R T}{P} = \frac{( 1 ) ( 8.314 ) ( 273.15 )}{101325} = 0.02241 m^{3} = 22.41 L .$ Why $P V = n R T$ and not something else? The parent note derived this from particle collisions with walls — pressure is nothing but momentum delivered by billions of impacts. So this law is the natural tool whenever $K E ≫ F_{I M F}$ (particles ignore each other). See Gas Laws.

Level 3 — Analysis

Exercise 3.1 — Will it be solid or liquid?

At $T = 298 K$ , a certain liquid has intermolecular bonds of strength $E_{I M F} = 10 kJ/mol$ . Estimate the Boltzmann factor $e^{- E_{I M F} / R T}$ for a particle escaping its cage, and interpret it.

Recall Solution

Work per mole, so use $R$ (not $k_{B}$ ): $\frac{E _{I M F}}{R T} = \frac{10000}{( 8.314 ) ( 298 )} = 4.036.$ $P_{escape} \propto e^{- 4.036} = 0.0177 \approx 1.8 \times 1 0^{- 2} .$ Interpretation: roughly 1 in 56 attempts has enough energy to break the local cage — rare but constant. That steady trickle of cage-escapes is exactly what makes a liquid flow rather than freeze. If $E_{I M F}$ were far larger (small exponential), escapes would essentially never happen ⇒ solid. See Intermolecular Forces.

Exercise 3.2 — Compare IMF to thermal energy for ice

An H-bond in ice is about $20 kJ/mol$ ; each water molecule holds ~4 of them. At $0^{\circ} C$ , is thermal energy able to break the network? Justify with a ratio.

Recall Solution

Thermal energy per mole at $T = 273.15 K$ : $R T = (8.314) (273.15) = 2271 J/mol = 2.27 kJ/mol .$ Total H-bond anchoring per molecule $\approx 4 \times 20 = 80 kJ/mol$ . $ratio = \frac{80}{2.27} \approx 35.$ The binding is ~35× the thermal energy, so molecules cannot escape — they only vibrate. That is precisely why ice is a solid at $0^{\circ} C$ : $K E ≪ F_{I M F}$ .

Level 4 — Synthesis

Exercise 4.1 — From gas back to particle speed

For $1 mol$ of helium ( $M = 4.00 \times 1 0^{- 3} kg/mol$ ) at $273 K$ , find the root-mean-square speed $v_{r m s} = ⟨ v^{2} ⟩$ .

Recall Solution

Start from the KMT bridge $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ , then scale to a mole by multiplying inside by $N_{A} / N_{A}$ : $v_{r m s} = \frac{3 R T}{M} = \frac{3 ( 8.314 ) ( 273 )}{4.00 \times 1 0 ^{- 3}} .$ Numerator: $3 (8.314) (273) = 6809.2$ . Divide by $M$ : $6809.2/4.00 \times 1 0^{- 3} = 1.7023 \times 1 0^{6}$ . $v_{r m s} = 1.7023 \times 1 0^{6} = 1305 m/s .$ Why this matters: ~1.3 km/s explains why helium leaks and diffuses fast — the particles are genuinely moving faster than a jet plane, colliding and spreading. The same $⟨ K E ⟩ = \frac{3}{2} k_{B} T$ that governed ice's stability now hands us gas speeds. One idea, two states. See Temperature Heat.

Exercise 4.2 — Predict the phase transition direction

Water at $25^{\circ} C$ is heated toward boiling. Using $P_{escape} \propto e^{- E_{I M F} / R T}$ , explain (with the ratio at $25^{\circ} C$ vs $100^{\circ} C$ ) why boiling happens, treating $E_{I M F} \approx 10 kJ/mol$ for a single transient H-bond.

Recall Solution

At $T_{1} = 298 K$ : $E / R T = 10000/ (8.314 \cdot 298) = 4.036$ , factor $e^{- 4.036} = 0.0177$ . At $T_{2} = 373 K$ : $E / R T = 10000/ (8.314 \cdot 373) = 3.225$ , factor $e^{- 3.225} = 0.0397$ . Escape probability more than doubles ( $0.0397/0.0177 = 2.24$ ). More molecules cross the IMF barrier per second ⇒ H-bond network breaks faster than it reforms ⇒ liquid → gas: boiling. The direction is dictated entirely by $K E$ climbing relative to fixed $F_{I M F}$ . See Evaporation and Boiling and Phase Diagrams.

Level 5 — Mastery

Exercise 5.1 — The full macro↔particulate audit of plasma

The Sun's core reaches $\sim 1.5 \times 1 0^{7} K$ . (a) Compute $⟨ K E ⟩$ per particle there. (b) Hydrogen's ionization energy is $13.6 eV = 2.18 \times 1 0^{- 18} J$ . Compare and explain why matter there is plasma, not gas.

Recall Solution

(a) $⟨ K E ⟩ = \frac{3}{2} k_{B} T = \frac{3}{2} (1.381 \times 1 0^{- 23}) (1.5 \times 1 0^{7}) = 3.11 \times 1 0^{- 16} J .$ (b) Ratio to ionization energy: $\frac{⟨ K E ⟩}{E _{ion}} = \frac{3.11 \times 1 0 ^{- 16}}{2.18 \times 1 0 ^{- 18}} = 143.$ Average thermal energy is ~140× the energy needed to rip an electron off a hydrogen atom. So collisions routinely ionize atoms into free nuclei + electrons — the defining feature of the fourth state. In the parent note's language: $K E$ is now so large it doesn't just separate molecules, it dismantles the atoms themselves. See Plasma in Stars. Macro↔particulate audit:

Macroscopic: glowing, electrically conducting, responds to magnetic fields.
Particulate: a soup of unbound charged particles, $K E ≫ E_{ion}$ .

Exercise 5.2 — One number governs everything

Explain, using a single organising ratio, why the same substance (water) can be solid, liquid, or gas, and place ice ( $T = 273 K$ , $E_{I M F} \approx 80 kJ/mol$ network) and steam ( $T = 373 K$ ) on that scale with numbers.

Recall Solution

The organising ratio is $\frac{K E}{F _{I M F}}$ , measured cleanly as $\frac{R T}{E _{I M F}}$ (both per mole).

Ice, whole cage ( $E \approx 80 kJ/mol$ ): $R T / E = 2271/80000 = 0.028$ . Far below 1 ⇒ locked ⇒ solid.
Liquid water ( $E \approx 10 kJ/mol$ single bond, $T = 298$ ): $R T / E = 2478/10000 = 0.248$ . Approaching 1 ⇒ bonds break and reform ⇒ liquid flow.
Steam ( $T = 373$ , IMF essentially unbound): $R T / E$ climbs toward and past 1 ⇒ particles free ⇒ gas. The synthesis: identity of the molecule fixes $E_{I M F}$ ; temperature sets $K E$ ; their ratio alone picks the state. Change temperature, slide along the ratio, and matter marches solid → liquid → gas → (at extreme $K E$ ) plasma. That is the entire chapter in one fraction.

Recall Self-test checklist

I can name a state from macro clues ::: Yes — shape + volume + compressibility (Ex 1.1–1.2) I can compute thermal energy per particle and per mole ::: Yes — $\frac{3}{2} k_{B} T$ and $\frac{3}{2} R T$ (Ex 2.1) I can get molar volume from the ideal gas law ::: Yes — $V = n R T / P = 22.41$ L at STP (Ex 2.2) I compare IMF to thermal energy in matching units ::: Yes — kJ/mol with $R T$ , J/particle with $k_{B} T$ (Ex 3.1–3.2) I can find gas particle speed from temperature ::: Yes — $v_{r m s} = 3 R T / M$ (Ex 4.1) I can predict transition direction from Boltzmann factors ::: Yes — rising $T$ raises escape probability (Ex 4.2) I can explain plasma by comparing $K E$ to ionization energy ::: Yes — ratio $\approx 143$ in the solar core (Ex 5.1) I can reduce all states to one ratio $K E / F_{I M F}$ ::: Yes (Ex 5.2)