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

This deep dive lives under the parent topic. The parent gave you the master rule:

$$\text{State}=f\left(\frac{KE}{F_{IMF}}\right)$$

where $KE$ is the kinetic energy (energy of motion) and $F_{IMF}$ is the strength of the intermolecular forces (the "stickiness" pulling particles together). Here we do the opposite of theory: we throw every possible scenario at that rule and watch it hold.

Before anything, one symbol we will lean on constantly — earn it now.

Look at the picture: the blue wave is $k_{B}T$, the coloured walls are the IMF energies of solids, liquids, gases. That single comparison decides the state.

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The scenario matrix

Every question this topic can ask falls into one of these case classes. Each worked example is tagged with the cell(s) it covers, and together they fill the whole grid.

#	Case class	What makes it tricky	Covered by
A	$KE\ll F_{IMF}$ (solid regime)	wall ≫ wave, must show why frozen	Ex 1
B	$KE\approx F_{IMF}$ (liquid regime)	the borderline, escape probability	Ex 2
C	$KE\gg F_{IMF}$ (gas regime)	numeric ideal-gas prediction	Ex 3
D	Ionization limit (plasma)	when $KE$ rips electrons off	Ex 4
E	Zero / degenerate input ($T\to 0$ K)	does motion truly stop?	Ex 5
F	Limiting behaviour ($T\to \infty$, or $P\to 0$)	when do laws break / hold perfectly	Ex 6
G	Real-world word problem	translate everyday scene → particulate	Ex 7
H	Exam-style twist (same substance, two states)	one number decides the state	Ex 8
I	Sign / direction of energy flow	melting vs freezing, $\pm Q$	Ex 9

Nine cells, nine examples. Nothing left uncovered.

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Example 1 — Cell A: the solid regime (wall ≫ wave)

Forecast: Guess first — will the thermal wave $k_{B}T$ be bigger or smaller than one H-bond? Which state does that predict?

1. Convert the wall to a per-molecule energy. $E_{IMF}=20$ kJ/mol $=20000$ J/mol. Divide by Avogadro's number $N_A=6.022\times 10^{23}$ per mol: $E_{IMF,\text{particle}}=\frac{20000}{6.022\times 10^{23}}=3.32\times 10^{-20}\text{J}.$ Why this step? The parent's rule compares energies per particle; you cannot compare a "per mole" wall to a "per particle" wave.

2. Compute the wave height. Thermal energy $k_{B}T=(1.381\times 10^{-23})(263)=3.63\times 10^{-21}$ J. Why this step? This is the actual jostling energy each molecule feels at that temperature.

3. Compare. Ratio $\frac{F_{IMF}}{KE}=\frac{3.32\times 10^{-20}}{3.63\times 10^{-21}}\approx 9.1$. The wall is $\approx 9\times$ taller than the wave. Why this step? $KE\ll F_{IMF}$ → the solid regime. Molecules can only vibrate, not escape.

Verify: Units cancel to a dimensionless ratio (J/J) — good. And the ratio $>1$ (in fact $\gg 1$) matches the known fact: ice is solid at $-10^{\circ }\text{C}$. ✓

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Example 2 — Cell B: the borderline liquid (escape probability)

Forecast: In a liquid $KE\approx F_{IMF}$. Do you expect the escape fraction to be almost zero (like a solid) or almost one (like a gas)?

1. Per-particle wall. $E=\frac{10000}{6.022\times 10^{23}}=1.66\times 10^{-20}$ J. Why this step? Same per-particle conversion as before, so the comparison is fair.

2. Wave height. $k_{B}T=(1.381\times 10^{-23})(298)=4.116\times 10^{-21}$ J.

3. Escape factor (Boltzmann). The parent gave $P_{\text{escape}}\propto e^{-E/k_{B}T}$: $e^{-E/k_{B}T}=\exp \left(-\frac{1.66\times 10^{-20}}{4.116\times 10^{-21}}\right)=e^{-4.03}\approx \mathrm{0.0178.}$ Why this step? The exponential is the tool that answers "what fraction of a random-motion crowd exceeds a threshold?" — no other simple function turns an energy ratio into a probability like this.

Verify: $\approx 1.8\%$ at any instant have enough kick to break free — small but not zero. That intermediate value is exactly the "borderline" signature of a liquid: molecules mostly stay in contact but constantly swap neighbours. ✓

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Example 3 — Cell C: the gas regime (ideal-gas prediction)

Forecast: Since $KE\gg F_{IMF}$ for helium (a noble gas — almost no stickiness), the ideal gas law should nail it. Guess: closer to $2$ L, $22$ L, or $220$ L?

1. Pick the law. Use $PV=nRT$. Why this step? Helium has negligible IMF and tiny atomic volume, so the two ideal-gas assumptions hold — the law is valid, not approximate.

2. Solve for $V$. $V=\frac{nRT}{P}=\frac{(1)(8.314)(273)}{101325}=0.02240{\text{m}}^3.$ Why this step? Rearrange for the unknown; keep SI units so $R=8.314$ J/(mol·K) fits.

3. Convert. $0.02240$ m³ $\times 1000=22.40$ L.

Verify: The textbook molar volume at STP is $22.4$ L — matched. Units: $\frac{\text{mol}\cdot \text{J/(mol⋅K)}\cdot \text{K}}{\text{Pa}}=\frac{\text{J}}{\text{Pa}}=\frac{\text{N⋅m}}{{\text{N/m}}^2}={\text{m}}^3$. ✓

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Example 4 — Cell D: the ionization limit (plasma)

Forecast: Plasma is the hottest state. Guess the order of magnitude: thousands, or tens of thousands, of kelvin?

1. Per-atom wall. $E=\frac{1312000}{6.022\times 10^{23}}=2.179\times 10^{-18}$ J. Why this step? Ionizing = ripping an electron off one atom, so we need the energy per atom.

2. Set wave = wall and solve for $T$. $T=\frac{E}{k_B}=\frac{2.179\times 10^{-18}}{1.381\times 10^{-23}}=1.58\times 10^5\text{K}.$ Why this step? At this temperature the average particle carries ionizing energy — atoms lose electrons wholesale, giving the charged soup we call plasma.

Verify: $\approx 158,000$ K — hotter than any liquid or gas on Earth, consistent with plasma living in stars (see Plasma in Stars). In reality partial ionization begins lower (thousands of K) because the tail of fast particles ionizes first, but the average-energy estimate correctly lands in the hundred-thousand-kelvin range. ✓

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Example 5 — Cell E: the zero/degenerate input ($T\to 0$ K)

Forecast: Plug $T=0$. The formula says... zero motion. True or false?

1. Plug in. $⟨KE⟩=\frac{3}{2}(1.381\times 10^{-23})(0)=0$ J. Why this step? This is the classical prediction — thermal jiggling vanishes.

2. Flag the degeneracy. Classically motion stops. But quantum mechanics forbids a particle from having both an exactly known position (its lattice site) and exactly zero momentum. So there remains zero-point energy — the particle still jiggles even at $0$ K. Why this step? This is exactly the degenerate case that breaks the simple formula; the parent note warned about it in the "particles don't move" mistake box.

Verify: The classical formula gives $0$ J (correct as a limit of thermal energy), but the substance is a solid with residual quantum vibration — so "0 K = frozen still" is a myth. Both facts stand without contradiction: thermal KE $=0$, total KE $>0$. ✓

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Example 6 — Cell F: limiting behaviour ($P\to 0$, $T\to \infty$)

Forecast: Push pressure toward zero. Does the gas get more or less ideal?

1. Reason about the limits. Low $P$ (and high $T$) → particles far apart → collisions rare → IMF negligible and the space they occupy is a vanishing fraction of $V$. That is precisely the regime where the two assumptions become exact. Why this step? The law's validity condition from the parent ("$KE\gg F_{IMF}$ and particle volume $\ll V$") is met best here.

2. Compute the (huge, nearly ideal) volume. $V=\frac{nRT}{P}=\frac{(1)(8.314)(1000)}{100}=83.14{\text{m}}^3.$ Why this step? An enormous volume means atoms are absurdly far apart — the textbook picture of an ideal gas.

Verify: $83.14$ m³ for one mole → density $\sim 10^{-5}$ of STP; at that separation IMF is truly negligible, so real ≈ ideal. As $P\to 0$, deviation → $0$. Conversely, at high $P$/low $T$ gases deviate and eventually condense (see Phase Diagrams). ✓

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Example 7 — Cell G: real-world word problem

Forecast: Same molecule, two behaviours. Which single quantity flips — the IMF, or the wave $k_{B}T$?

1. Puddle (liquid, $293$ K). $k_{B}T=(1.381\times 10^{-23})(293)=4.046\times 10^{-21}$ J. The fastest molecules in the tail of the distribution exceed the surface H-bond and escape as vapour — this is evaporation. Over hours, enough escape that the puddle empties. Why this step? Evaporation is the escape-probability of Ex 2 applied at the surface; a slow leak of the fastest molecules.

2. Ice (solid, $268$ K). $k_{B}T=(1.381\times 10^{-23})(268)=3.701\times 10^{-21}$ J — smaller wave, and now molecules are locked in a rigid lattice. Almost none reach escape energy, so the shape is held. Why this step? Same substance, lower $T$ → smaller $KE/F_{IMF}$ → the state (and behaviour) flips from flowing/evaporating to rigid.

Verify: The only number that changed meaningfully is $k_{B}T$ ($4.05$ vs $3.70\times 10^{-21}$ J) — the IMF (H-bond) is the same molecule's bond. Lower thermal wave ⇒ solid ⇒ shape held. Matches everyday observation. ✓

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Example 8 — Cell H: exam-style twist (one number decides the state)

Forecast: Two temperatures, same substance. Guess the two states before computing.

1. Cold case, $150$ K. $k_{B}T=(1.381\times 10^{-23})(150)=2.072\times 10^{-21}$ J. Ratio $\frac{KE}{F_{IMF}}=\frac{2.072\times 10^{-21}}{2.5\times 10^{-21}}=0.829$. Why this step? Ratio just under $1$ → $KE\approx F_{IMF}$, slightly below → liquid on the verge of freezing.

2. Hot case, $400$ K. $k_{B}T=(1.381\times 10^{-23})(400)=5.524\times 10^{-21}$ J. Ratio $=\frac{5.524\times 10^{-21}}{2.5\times 10^{-21}}=2.21$. Why this step? Ratio $\gg 1$ → $KE\gg F_{IMF}$ → gas.

Verify: One fixed IMF, two temperatures, ratios $0.83$ and $2.21$ straddling $1$ — the substance climbs from liquid to gas as $T$ rises. The wall (IMF) never moved; only the wave grew. Exactly the parent's central claim. ✓

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Example 9 — Cell I: sign / direction of energy flow

Forecast: To melt ice you must add energy to overcome IMF. So is $Q$ for melting positive or negative?

1. Melting (solid → liquid). You supply energy to break enough H-bonds so molecules can slide: $Q=+6.01$ kJ (endothermic, $Q>0$). Why this step? Raising $KE/F_{IMF}$ past the solid/liquid boundary costs energy — the sign encodes direction of flow into the system.

2. Freezing (liquid → solid). The reverse process releases the same energy: $Q=-6.01$ kJ (exothermic, $Q<0$). Why this step? Energy is a state property; the reverse path returns exactly what melting absorbed, with opposite sign.

Verify: $|Q_{melt}|=|Q_{freeze}|=6.01$ kJ and signs are opposite ($+6.01$ vs $-6.01$), so they sum to $0$ over a melt–freeze cycle — energy conservation holds. Relates to Temperature Heat and Kinetic Molecular Theory. ✓

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Recall Self-test

What single dimensionless comparison decides the state of matter? ::: The ratio $KE/F_{IMF}$ (equivalently, thermal wave $k_{B}T$ vs the IMF wall height). Why use the exponential $e^{-E/k_{B}T}$ for escape, not a simple ratio? ::: It converts an energy threshold into the fraction of a randomly-moving crowd that exceeds it — the natural probability weight for thermal motion. Does motion stop at $0$ K? ::: Thermal $KE\to 0$, but quantum zero-point vibration remains, so particles still jiggle. In which limit is a real gas most ideal? ::: Low pressure and high temperature (particles far apart, IMF and particle volume negligible).