Third law of thermodynamics — S → 0 as T → 0: Worked examples

Everything here rests on two tools from the parent topic:

The scenario matrix

Every Third-Law question falls into one of these cells. The examples that follow are tagged with the cell they cover so nothing is left out.

Cell	What varies	The question type	Example
A	$W = 1$ (perfect order)	entropy is exactly zero	Ex 1
B	$W > 1$ frozen-in, binary	residual entropy $R ln g$	Ex 2
C	$W > 1$ non-binary ( $g = 1.5$ , ice)	fractional-log residual	Ex 3
D	$C_{p} = a T^{3}$ (Debye limit)	absolute entropy by integration	Ex 4
E	$C_{p}$ = sum of pieces + phase change	full calorimetric ladder	Ex 5
F	Two states merge at $T = 0$	unattainability / cooling steps	Ex 6
G	Real-world word problem	glass vs crystal, measured mismatch	Ex 7
H	Exam twist / trap	"is $C_{p} / T$ divergent?" limiting check	Ex 8

The three "degenerate/limiting" edges — $W = 1$ (no disorder), $C_{p} \to 0$ (integrand tamed), and $T \to 0$ unreachable — are each given their own example so you never meet an untested boundary.

Cell A — Perfect order, $W = 1$

Worked example Example 1 — Entropy of a flawless silicon crystal at 0 K (Cell A)

Q: A perfect crystal of $N = 5.0 \times 1 0^{22}$ silicon atoms is cooled to exactly 0 K, every atom locked on its lattice site. Find $S$ .

Forecast: Does the answer depend on $N$ ? Guess before reading.

Count arrangements. Perfect crystal → each atom has exactly one site → there is only one way to build the whole thing, so $W = 1$ . Why this step? Entropy counts distinguishable arrangements; perfect order leaves no choice anywhere.
Feed the counting engine. $S = k_{B} ln W = k_{B} ln 1.$ Why this step? We are at $T = 0$ , so the counting engine (not the integral) is the right tool — there is no heat to integrate.
Evaluate the log. $ln 1 = 0$ , so $S = k_{B} \cdot 0 = 0$ . Why this step? $ln 1 = 0$ because $e^{0} = 1$ — "one way" means zero information, zero entropy.

Answer: $S = 0 J K^{- 1}$ , independent of $N$ . The $5 \times 1 0^{22}$ was a distractor.

Verify: Multiply the count by any factor — $W = 1$ stays $1$ , $ln 1 = 0$ stays $0$ . Units: $k_{B}$ is $J K^{- 1}$ , times a pure number, so $S$ has $J K^{- 1}$ . Consistent.

Cell B — Binary frozen disorder

Worked example Example 2 — Residual entropy of solid CO (Cell B)

Q: In solid carbon monoxide each molecule freezes pointing either CO or OC, chosen at random. Find $S (0)$ for one mole.

Forecast: Two options per molecule and $6 \times 1 0^{23}$ molecules — will the answer be astronomically huge, or a modest number like $6 J K^{- 1}$ ?

Count total arrangements. Each molecule independently has $g = 2$ choices; for $N$ molecules $W = 2^{N}$ . Why this step? Independent choices multiply: $2 \times 2 \times \dots = 2^{N}$ .
Apply the counting engine. $S = k_{B} ln 2^{N} = N k_{B} ln 2.$ Why this step? $ln (x^{N}) = N ln x$ turns the giant power into a manageable product.
Go per mole. For one mole $N = N_{A}$ , and $N_{A} k_{B} = R$ : $S (0) = R ln 2 = 8.314 \times 0.6931.$ Why this step? Replacing $N_{A} k_{B}$ by $R$ converts to per-mole units.

Answer: $S (0) = R ln 2 \approx 5.76 J mo l^{- 1} K^{- 1}$ .

Verify: The number is modest (not $1 0^{23}$ ) because the log crushed the exponent — good sanity check. Units: $R$ in $J mol^{- 1} K^{- 1}$ times a pure number → correct.

Cell C — Non-binary / fractional log (ice)

Worked example Example 3 — Residual entropy of ice (

$g = 3/2$ ) (Cell C) Q: In water ice each oxygen has 4 hydrogen bonds; Pauling's counting gives an effective $g = 3/2$ orientations per molecule. Find $S (0)$ per mole and compare with CO.

Forecast: Ice's $g = 1.5$ is smaller than CO's $g = 2$ . Should ice's residual entropy be bigger or smaller than CO's $5.76$ ?

Effective count. $W = (3/2)^{N}$ — Pauling showed the ice-rules leave, on average, $3/2$ allowed configurations per molecule. Why this step? The number need not be a whole integer; it's an average multiplicity per molecule, and the log handles fractions fine.
Counting engine per mole. $S (0) = R ln (\frac{3}{2}) = 8.314 \times ln 1.5.$ Why this step? Same $ln (x^{N}) = N ln x$ move; per mole gives $R ln (3/2)$ .
Number. $ln 1.5 = 0.4055$ , so $S (0) = 8.314 \times 0.4055$ . Why this step? Plug the log value.

Answer: $S (0) \approx 3.37 J mo l^{- 1} K^{- 1}$ — indeed smaller than CO's $5.76$ , as forecast, because $3/2 < 2$ .

Verify: Monotonic check — a smaller $g$ must give smaller entropy: $ln 1.5 < ln 2$ ✓. This matches the experimentally measured ice residual entropy ( $\approx 3.4 J mol^{- 1} K^{- 1}$ ). See Residual entropy of ice and CO.

Cell D — Absolute entropy from the Debye tail

Worked example Example 4 — Entropy of a solid obeying

$C_{p} = a T^{3}$ (Cell D) Q: Near 0 K a crystal obeys the Debye law $C_{p} = a T^{3}$ with $a = 4.8 \times 1 0^{- 4} J mol^{- 1} K^{- 4}$ . Find $S$ at $T = 10 K$ .

Forecast: Will $S (10)$ be roughly equal to $C_{p} (10)$ , one-third of it, or three times it? (Look at the figure below first.)

Set up the integral from zero. Because $S (0) = 0$ (Third Law), start the ladder at absolute zero: $S (T) = \int_{0}^{T} \frac{C _{p} ( T ^{'} )}{T ^{'}} d T^{'} = \int_{0}^{T} \frac{a T ^{'3}}{T ^{'}} d T^{'} .$ Why this step? The Third Law is what lets us start at $0$ ; otherwise we'd only get an entropy difference.
Cancel one power of $T^{'}$ . $\frac{T ^{'3}}{T ^{'}} = T^{'2}$ , so the integrand is $a T^{'2}$ . Why this step? This is exactly why the integral is safe — the dangerous $1/ T^{'}$ is eaten by the $T^{'3}$ , leaving a tame $T^{'2}$ (this is Cell H's worry, pre-empted).
Integrate. $\int_{0}^{T} a T^{'2} d T^{'} = a \frac{T ^{3}}{3}$ . Why this step? Power rule $\int T^{'2} d T^{'} = T^{'3} /3$ ; the lower limit $0$ contributes nothing.
Recognise the shortcut. Since $a T^{3} = C_{p} (T)$ , $S (T) = \frac{a T ^{3}}{3} = \frac{C _{p} ( T )}{3} .$ Why this step? A neat Debye rule: entropy is one-third of the heat capacity in the $T^{3}$ region.
Number. $C_{p} (10) = 4.8 \times 1 0^{- 4} \times 1 0^{3} = 0.48 J mol^{- 1} K^{- 1}$ , so $S (10) = 0.48/3$ .

Answer: $S (10) = 0.16 J mo l^{- 1} K^{- 1}$ (and $S (10) = C_{p} (10) /3$ , matching the forecast "one-third").

Verify: Units: $a [J mol^{- 1} K^{- 4}] \times T^{3} [K^{3}] /3 = J mol^{- 1} K^{- 1}$ ✓. As $T \to 0$ , $S \to 0$ ✓ (Third Law self-consistent). See Heat capacity and Debye T-cubed law.

Cell E — Full calorimetric ladder with a phase change

Worked example Example 5 — Absolute entropy up through a melting transition (Cell E)

Q: A substance is warmed from 0 K. Below the Debye region assume $C_{p} = a T^{3}$ ( $a$ as in Ex 4) up to $T_{1} = 10 K$ . Between $10 K$ and its melting point $T_{m} = 200 K$ the solid has a constant $C_{p}^{solid} = 25 J mol^{- 1} K^{- 1}$ . At $T_{m}$ it melts with latent heat $L_{f u s} = 5000 J mol^{- 1}$ . Find the total entropy $S (T_{m}^{+})$ just after melting.

Forecast: Three contributions add up (Debye tail + heating + melting). Which do you think dominates — guess before computing.

Debye piece (0→10 K). From Ex 4, $S_{1} = C_{p} (10) /3 = 0.16 J mol^{- 1} K^{- 1}$ . Why this step? Reuse the converged Debye integral — the only safe way through $T = 0$ .
Constant- $C_{p}$ heating (10→200 K). $S_{2} = \int_{10}^{200} \frac{C _{p}^{solid}}{T ^{'}} d T^{'} = C_{p}^{solid} ln \frac{200}{10} = 25 ln 20.$ Why this step? When $C_{p}$ is constant, $\int d T^{'} / T^{'} = ln (T_{hi} / T_{lo})$ — the logarithm is the tool for " $C_{p}$ constant" segments.
Number for $S_{2}$ . $ln 20 = 2.996$ , so $S_{2} = 25 \times 2.996 = 74.9 J mol^{- 1} K^{- 1}$ .
Melting (phase change at fixed $T_{m}$ ). $S_{3} = \frac{L _{f u s}}{T _{m}} = \frac{5000}{200} = 25.0 J mol^{- 1} K^{- 1} .$ Why this step? At a phase change $T$ is constant, so $Δ S = Q / T = L / T_{m}$ — no integral needed.
Add the ladder. $S (T_{m}^{+}) = S_{1} + S_{2} + S_{3} = 0.16 + 74.9 + 25.0$ . Why this step? Entropy is a state function — contributions along the reversible path simply sum.

Answer: $S (T_{m}^{+}) \approx 100.1 J mo l^{- 1} K^{- 1}$ ; the heating segment $S_{2}$ dominates, as forecast if you noticed the wide temperature span.

Verify: Units of each term $J mol^{- 1} K^{- 1}$ ✓. The Debye tail is tiny (0.16) — sensible, since little heat flows near 0 K. This is exactly how Absolute entropy and standard molar entropy tables are built.

Cell F — Unattainability / cooling steps

Worked example Example 6 — Why each cooling step gives less (Cell F)

Q: In adiabatic demagnetization, one magnetize–demagnetize cycle drops the temperature by a factor, not a fixed amount. Suppose each cycle multiplies the temperature by $r = 0.5$ (halves it), starting from $T_{0} = 1 K$ . How many cycles to reach exactly $T = 0$ ?

Forecast: A finite number, or never?

Model the steps. After $n$ cycles, $T_{n} = T_{0} r^{n} = (0.5)^{n} K$ . Why this step? Each step scales rather than subtracts, because the two entropy curves merge as $T \to 0$ (Third Law), so gains shrink geometrically.
Set $T_{n} = 0$ and solve. $(0.5)^{n} = 0$ requires $n \to \infty$ . Why this step? A positive number to any finite power is still positive — geometric decay never hits zero.
Check the trend numerically. $T_{1} = 0.5, T_{2} = 0.25, T_{3} = 0.125, \dots$ — always positive, halving forever. Why this step? Confirms the "diminishing returns" picture in the figure: the curves squeeze together but never touch.

Answer: Infinitely many steps — absolute zero is unreachable in any finite number of operations (the unattainability form of the Third Law).

Verify: $lim_{n \to \infty} (0.5)^{n} = 0$ but no finite $n$ gives $0$ ✓. Physically consistent with Adiabatic demagnetization and reaching low temperatures.

Cell G — Real-world word problem

Worked example Example 7 — Glassy glycerol: measured vs. crystal entropy (Cell G)

Q: For glycerol, the crystal's calorimetric entropy at 298 K integrates (with $S (0) = 0$ ) to a spectroscopic value. But if you cool the glass and integrate its $C_{p}$ , you get a value that is too low by $\approx 19 J mol^{- 1} K^{- 1}$ . Interpret this deficit using the Third Law, and estimate the effective number of frozen configurations $g$ per molecule.

Forecast: Is $g$ near 1 (nearly ordered), near 2 (binary), or much larger (very disordered)?

Diagnose the deficit. The glass freezes into many random configurations, so its true $S (0) = k_{B} ln g^{N} = R ln g$ per mole is not zero. Integrating its $C_{p}$ from zero assumes $S (0) = 0$ , so it misses exactly $R ln g$ . Why this step? Only a perfect crystal has $S (0) = 0$ ; a glass is Cell B/C in disguise — residual entropy hides in the missing $R ln g$ .
Match to the measured deficit. $R ln g = 19 ⟹ ln g = \frac{19}{8.314} = 2.285.$ Why this step? The deficit is the residual entropy.
Solve for $g$ . $g = e^{2.285} = 9.8 \approx 10$ . Why this step? Exponentiate to undo the log.

Answer: $g \approx 10$ frozen configurations per molecule — "much larger than 2," a strongly disordered glass.

Verify: $R ln 10 = 8.314 \times 2.303 = 19.1 J mol^{- 1} K^{- 1}$ ✓ — reproduces the deficit. This is precisely how chemists detect residual entropy in the lab.

Cell H — Exam twist: the "diverging integrand" trap

Worked example Example 8 — Does

$\int_{0}^{T} C_{p} / T^{'} d T^{'}$ blow up? (Cell H) Q: A student panics: " $C_{p} / T^{'}$ has a $1/ T^{'}$ that diverges at $T^{'} = 0$ , so absolute entropy is undefined." Resolve it for the Debye tail $C_{p} = a T^{'3}$ by checking the limit of the integrand.

Forecast: Does the integrand go to $\infty$ , a finite constant, or $0$ as $T^{'} \to 0$ ?

Write the integrand. $f (T^{'}) = \frac{C _{p}}{T ^{'}} = \frac{a T ^{'3}}{T ^{'}} = a T^{'2}$ . Why this step? The naive fear ignores that $C_{p}$ itself vanishes; you must substitute before judging.
Take the limit. $T^{'} \to 0 lim a T^{'2} = 0$ . Why this step? $T^{'2} \to 0$ — the integrand not only stays finite, it goes to zero. No divergence.
Contrast the bad case. If $C_{p}$ were a constant $C_{0}$ (which the Third Law forbids near 0 K), then $f = C_{0} / T^{'} \to \infty$ and the integral would truly diverge. Why this step? Shows the Third Law's requirement $C_{p} \to 0$ is exactly what saves the integral — it isn't luck.

Answer: The integrand $\to 0$ ; the integral converges. The trap dissolves once you remember $C_{p} \to 0$ .

Verify: Numerically at $T^{'} = 0.01 K$ : $f = 4.8 \times 1 0^{- 4} \times (0.01)^{2} = 4.8 \times 1 0^{- 8}$ — vanishingly small ✓, confirming convergence.

Cell coverage recap

Recall Did we hit every cell?

A (Ex1 $W = 1$ ) ::: exactly zero entropy B (Ex2 binary CO) ::: $R ln 2 \approx 5.76$ C (Ex3 ice $g = 3/2$ ) ::: $R ln 1.5 \approx 3.37$ D (Ex4 Debye) ::: $S = C_{p} /3 = 0.16$ at 10 K E (Ex5 ladder) ::: $\approx 100.1$ J mol⁻¹K⁻¹ with a melt F (Ex6 steps) ::: infinite steps, unattainable G (Ex7 glass) ::: $g \approx 10$ from a $19$ J deficit H (Ex8 trap) ::: integrand $\to 0$ , converges

Connections

Boltzmann entropy S = k ln W — the counting engine (Cells A, B, C, G).
Second law of thermodynamics — the $d S = δ Q / T$ used in Cells D, E, H.
Heat capacity and Debye T-cubed law — the $C_{p} = a T^{3}$ tail (Cells D, E, H).
Absolute entropy and standard molar entropy — the payoff table built by Cell E.
Adiabatic demagnetization and reaching low temperatures — Cell F in the lab.
Residual entropy of ice and CO — Cells B, C, G exceptions.
Back to parent: Third Law topic note.

Worked examples — Third law of thermodynamics — S → 0 as T → 0