Third law of thermodynamics — S → 0 as T → 0: Question bank

Before we start, three words you must be sure of (the parent note built them — this is the quick reminder):

See the parent Third Law topic note and its Boltzmann entropy S = k ln W bridge whenever a symbol feels unearned.

True or false — justify

Every one of these is a statement someone would write on an exam. Decide true/false and say why — the "why" is the whole point.

Entropy of every substance is exactly zero at 0 K.

False. Only a perfect crystal (

W = 1

) reaches

S = 0

. Glasses, CO, ice and mixed crystals freeze into

g > 1

patterns and keep residual entropy

S_{0} = k_{B} ln g > 0

The Planck form of the Third Law fixes the absolute zero-point of entropy.

True. Nernst only claimed

Δ S \to 0

(a change). Planck strengthened it to

S_{0} = 0

itself, which is why we can quote an absolute molar entropy rather than only differences.

The Nernst form (

Δ S \to 0

) already implies

S_{0} = 0

False.

Δ S \to 0

only says all states share the same limiting entropy; it does not tell you that shared value is zero. Planck's extra assertion (

S_{0} = 0

) is genuinely stronger.

Heat capacity

C_{p}

approaches a positive constant as

T \to 0

False. The Third Law forces

C_{p} \to 0

(Debye's

C_{V} \propto T^{3}

confirms it). A nonzero constant

C_{p}

would make

C_{p} / T

diverge and entropy would not be finite.

The entropy integral

\int_{0}^{T} \frac{C _{p}}{T ^{'}} d T^{'}

diverges because we divide by

T^{'} \to 0

False. Because

C_{p} \to 0

faster than

T^{'}

(like

T^{'3}

), the ratio

C_{p} / T^{'} \to 0

, so the integrand is finite. The law guarantees convergence, not coincidence.

We can reach exactly

T = 0

K if our refrigerator is powerful enough.

False. The unattainability principle (an equivalent form of the Third Law) says

T = 0

needs infinitely many steps. The entropy curves of the two working states merge at

T = 0

, so each cooling step removes less and less.

Residual entropy violates the Third Law.

False. It obeys the general law "

S \to

a constant." That constant is only zero when

g = 1

. For

g > 1

the constant is

k_{B} ln g

— fully consistent with the statistical statement.

Boltzmann's

S = k_{B} ln W

and the Third Law are two unrelated statements.

False. The Third Law is a consequence of

S = k_{B} ln W

: cool to the ground state, count its microstates, and if

W = 1

then

S = k_{B} ln 1 = 0

Adding more atoms

N

to a perfect crystal raises its 0 K entropy.

False.

W = 1

regardless of

N

(still one arrangement), so

S = k_{B} ln 1 = 0

independent of

N

. Residual entropy, by contrast, does scale with

N

(it's

N k_{B} ln g

Spot the error

Each line contains a plausible-sounding chain of reasoning with one broken link. Name the broken link.

S = k_{B} ln W

, and at 0 K there are zero microstates, so

S = k_{B} ln 0 = - \infty

The error is "zero microstates." The system is always in some state, so

W \geq 1

. For a perfect crystal

W = 1

, giving

S = k_{B} ln 1 = 0

, not

- \infty

"CO freezes with 2 orientations per molecule, so its residual entropy is

S_{0} = k_{B} ln 2

Missing the

N

. Each of

N

molecules chooses independently, so

W = 2^{N}

and

S_{0} = N k_{B} ln 2 = R ln 2

per mole — not

k_{B} ln 2

"Since

S (0) = 0

for a perfect crystal,

Δ S = 0

for any reaction between such crystals at low

T

The conclusion is right but the stated reason is loose: it holds because each reactant and product tends to its own zero, so the difference tends to zero — the Nernst form. It is not because entropy is somehow conserved.

C_{p} \to 0

, therefore

\int_{0}^{T} \frac{C _{p}}{T ^{'}} d T^{'} = 0

Wrong.

C_{p} \to 0

only at the lower limit. Over the range

0 \to T

the integrand is positive, so the integral is a positive number (e.g.

\frac{1}{3} a T^{3}

for the Debye case), not zero.

"Glass is a solid, and solids reach

S = 0

at 0 K, so glass has zero entropy at 0 K."

"Solids reach

S = 0

" is only true for perfect crystalline solids. Glass is amorphous — frozen disorder — so it keeps a positive residual entropy.

"Temperature is thermal energy, so at

T = 0

the atoms have zero total energy and stop moving completely."

Two slips:

T

measures available energy to populate excited levels, not total energy; and quantum zero-point motion remains even in the ground state. What vanishes is thermal excitation, leaving one ground state, hence

S \to 0

"If the ground state were 3-fold degenerate,

S (0) = k_{B} \cdot 3

The formula is

S = k_{B} ln W

, so a 3-fold degenerate state gives

S_{0} = k_{B} ln 3

, not

k_{B} \cdot 3

. The logarithm is essential — it is what makes

W = 1

give exactly zero.

Why questions

State the deep reason, not the headline.

Why does the Third Law let us define an absolute entropy, while the First and Second Laws only give differences?

The Second Law defines

d S = δ Q_{r e v} / T

, fixing only changes in

S

. The Third Law pins the constant of integration:

S (0) = 0

, so integrating

C_{p} / T

from 0 gives an absolute number. See Absolute entropy and standard molar entropy.

Why must

C_{p} \to 0

T \to 0

for the Third Law to be self-consistent?

C_{p}

stayed finite,

\int_{0}^{T} C_{p} / T^{'} d T^{'}

would diverge, so

S

could not be finite and pinned at zero. The law needs

C_{p} / T \to 0

, which requires

C_{p} \to 0

— delivered by Heat capacity and Debye T-cubed law.

Why is "perfect crystal" the essential clause, not decoration?

Only a defect-free, order-locked lattice has a non-degenerate ground state (

W = 1

). Drop that clause and the ground state can be degenerate (

g > 1

), leaving residual entropy

k_{B} ln g

. See Residual entropy of ice and CO.

Why can't absolute zero be reached in finitely many cooling steps?

Each step (e.g. an adiabatic demagnetization) drops

T

by exploiting the entropy gap between two states. As

T \to 0

both entropy curves converge to the same value, so the gap — and the cooling per step — shrinks to zero. See Adiabatic demagnetization and reaching low temperatures.

Why does thermal energy

\sim k_{B} T \to 0

force the system into its ground state?

With no thermal energy to lift particles into higher-energy levels, the Boltzmann weight of every excited state vanishes; only the lowest level stays occupied, so the accessible microstate count collapses.

Why does residual entropy scale with

N

but perfect-crystal entropy does not?

Residual entropy comes from independent choices per particle:

W = g^{N}

, so

S = N k_{B} ln g

grows with

N

. A perfect crystal has one global arrangement (

W = 1

) regardless of

N

, so

S = 0

always.

Why is the log in

S = k_{B} ln W

exactly what makes

W = 1

give zero?

ln 1 = 0

: one arrangement means no choice, no missing information, no disorder. The log converts "one way" into "zero entropy" — a linear function of

W

could not do this cleanly.

Edge cases

The scenarios the headline statement quietly skips.

What is

S (0)

if the ground state is

g

-fold degenerate?

S (0) = k_{B} ln g > 0

— residual entropy. The strict "

S \to 0

" only holds for

g = 1

; the general law is "

S \to

a constant."

What happens to the Third Law for an ideal gas as

T \to 0

A true ideal gas has no interactions, so it can't crystallize; the classical entropy formula would go to

- \infty

. Reality steps in: real gases condense/solidify before 0 K, and quantum statistics take over, keeping

S

finite and

\to 0

for the ordered solid.

What is the 0 K entropy of a substance with a unique lowest state but low-lying excited states very close above it?

Still

S (0) = 0

— degeneracy is about the ground state only. Nearby excited states affect entropy at small nonzero

T

, but as

T \to 0

their occupation vanishes and

S \to k_{B} ln 1 = 0

Two perfect crystals of different elements — does

Δ S \to 0

for their reaction at 0 K?

Yes. Each pure perfect crystal tends to

S = 0

, so reactants and products both approach zero and

Δ S \to 0

(Nernst form) — the basis of low-temperature reaction-entropy calculations.

A crystal with a single frozen-in point defect vs. one with

1 0^{20}

random defects — 0 K entropy?

The single fixed defect (one specified arrangement) still gives

W = 1

S = 0

. Randomly-placed defects give many arrangements,

W ≫ 1

, and a positive residual entropy — disorder, not mere imperfection, is what matters.

Does zero entropy at 0 K mean the atoms are perfectly still?

No. Quantum zero-point motion persists even in the ground state.

S = 0

means only one quantum state is occupied (

W = 1

), not that motion has ceased.

Is entropy exactly zero at 0 K, or does it only approach zero?

The law is a limit:

lim_{T \to 0} S = 0

. Since

T = 0

is unattainable, the crystal never literally sits at

S = 0

— it approaches it as closely as we cool.

Recall One-line summary of the traps

The three killers are: forgetting the "perfect crystal" clause (residual entropy exists), thinking $T = 0$ is reachable (it isn't), and fearing the entropy integral diverges (it converges because $C_{p} \to 0$ ).

Connections

Boltzmann entropy S = k ln W — the counting engine every answer leans on.
Second law of thermodynamics — supplies $d S = δ Q_{r e v} / T$ .
Heat capacity and Debye T-cubed law — why $C_{p} \to 0$ and the integral converges.
Absolute entropy and standard molar entropy — the payoff of $S (0) = 0$ .
Adiabatic demagnetization and reaching low temperatures — unattainability in practice.
Residual entropy of ice and CO — the exceptions behind half these traps.