Exercises — Third law of thermodynamics — S → 0 as T → 0

Statement: For a perfect crystalline substance, the entropy approaches zero as the temperature approaches absolute zero. ${\lim }_{T\to 0}S=0\text{(perfect crystal)}$ The word "perfect" is doing all the work — it guarantees a single ground arrangement.

• $S$ = entropy, a number measuring "how many ways the system can secretly be arranged while looking the same from outside."
• $k_B$ = the conversion factor from "count of arrangements" to physical units of J/K.
• $W$ = the count of microstates — the number of distinct arrangements consistent with what we measure. See Boltzmann entropy S = k ln W.

"Non-degenerate" means exactly one lowest-energy arrangement, so $W=1$. $S=k_B\ln 1=k_B\cdot 0=0.$ $\ln 1=0$ because $e^0=1$: no factor of $e$ is needed to reach $1$, so the "logarithm-power" is zero.

Perfect order $\Rightarrow W=1$ regardless of how many atoms there are. $S=k_B\ln 1=0.$ Answer: $S=0$, and it does not depend on $N$. The count of atoms never enters because there is still only one whole-crystal arrangement.

Each of $N_A$ molecules has $2$ independent choices, so the whole crystal has $W=2^{N_A}.$ $S=k_B\ln \left(2^{N_A}\right)=N_A{k}_B\ln 2=R\ln 2.$ $S=8.314\times 0.6931\approx 5.76\mathbf{J}\mathbf{mo}{\mathbf{l}}^{-1}{\mathbf{K}}^{-1}.$ Answer: $\approx 5.76{\text{J mol}}^{-1}{\text{K}}^{-1}$.

Because the Third Law fixes $S(0)=0$, integrate the calorimetric ladder from zero (see Absolute entropy and standard molar entropy): $S(T)={\int }_0^T\frac{C_p}{T^{\prime }}d{T}^{\prime }={\int }_0^T\frac{a{T}^{\prime 3}}{T^{\prime }}d{T}^{\prime }=a{\int }_0^T{T}^{\prime 2}d{T}^{\prime }=\frac{a{T}^3}{3}=\frac{C_p(T)}{3}.$ One power of $T^{\prime }$ cancels against the $T^{\prime }$ in the denominator — that cancellation is why the integral converges (integrand $\to 0$). $S(10)=\frac{4.8\times 10^{-4}\times 10^3}{3}=\frac{0.48}{3}=0.16\mathbf{J}\mathbf{mo}{\mathbf{l}}^{-1}{\mathbf{K}}^{-1}.$

Let the ground state be $g$-fold degenerate: $g$ distinct lowest-energy arrangements, so $W=g$ at $0\text{K}$. $S(0)=k_B\ln g.$ This vanishes iff $\ln g=0$, i.e. $g=1$. A perfect crystal is precisely the case $g=1$. So Planck's "$S_0=0$" is the special case, and $k_B\ln g$ is the general truth. Degeneracy $g>1$ is residual entropy.

If $C_p$ were a nonzero constant $c$ near $0$, the integrand would be $c/T^{\prime }$, whose integral is $c\ln T^{\prime }$ — which $\to -\infty$ as $T^{\prime }\to 0$. Diverges. But the Third Law forces $C_p\to 0$; Debye gives $C_p=a{T}^{\prime 3}$, so the integrand is $a{T}^{\prime 2}$, which $\to 0$ smoothly. Then ${\int }_0^{T}a{T}^{\prime 2}d{T}^{\prime }=\frac{a{T}^3}{3},\text{finite.}$ Convergence is not luck — the law that pins $S(0)=0$ is the same law that kills $C_p$. See Heat capacity and Debye T-cubed law.

$S(T)-S(ϵ)={\int }_{ϵ}^T\frac{c}{T^{\prime }}d{T}^{\prime }=c[\ln T{]}_{ϵ}^T=c\ln \frac{T}{ϵ}.$ As the lower limit $ϵ\to 0$, $\ln (T/ϵ)\to +\infty$, so $S(ϵ)\to -\infty$. Entropy would run off to $-\infty$ rather than to a finite constant $0$. The Third Law demands a finite limit; a constant $C_p$ is therefore forbidden as $T\to 0$. Hence $C_p$ must vanish.

• Adiabatic step (horizontal, green): no heat exchange means $S$ stays constant, so you move left along a flat line from curve $B$ down onto curve $A$ at a lower $T$.
• Isothermal step (vertical, orange): at fixed $T$ you re-magnetize, jumping from curve $A$ up to curve $B$, dumping entropy to the surroundings.
• Because both curves merge at $T=0$ (a direct consequence of $S\to 0$ for both states), each horizontal step lands on a $T$ that is closer to zero but by an ever-smaller amount — the gap between the curves shrinks to nothing.
• You get an infinite staircase whose steps shrink geometrically; you approach $0$ but never arrive in finitely many steps. This is the unattainability form of the Third Law. See Adiabatic demagnetization and reaching low temperatures.
• Crucial link: if the curves did not merge (i.e. if $S(0)$ differed between states), a single finite step could bridge the gap — so unattainability is equivalent to $S\to$ the same limit.

Per mole $W=(3/2{)}^{N_A}$, so $S(0)=k_B\ln [(3/2{)}^{N_A}]=R\ln \frac{3}{2}.$ $S(0)=8.314\times \ln 1.5=8.314\times 0.4055\approx 3.37\mathbf{J}\mathbf{mo}{\mathbf{l}}^{-1}{\mathbf{K}}^{-1}.$ This is less than CO's $R\ln 2\approx 5.76$ because ice has fewer effective orientations per molecule ($1.5<2$). Both are genuine violations of the naive "$S=0$" but perfectly consistent with $S(0)=k_B\ln g$. See Residual entropy of ice and CO.

(a) As $T\to 0^{+}$, the exponent $-\epsilon /k_{B}T\to -\infty$ (since $\epsilon >0$), so $e^{-\epsilon /k_{B}T}\to 0$. Then $p\to \frac{0}{1+0}=0.$ Every spin sits in the lower level. Why the exponential? The Boltzmann factor $e^{-\epsilon /k_{B}T}$ measures how likely thermal energy $k_{B}T$ is to kick a spin up; when $k_{B}T$ dies, that likelihood dies. (b) All spins down is one arrangement: $W=1$, so $S=k_B\ln 1=0$. The system obeys the Third Law with $g=1$. (If $\epsilon =0$ — degenerate levels — both orientations are equally allowed even at $T=0$, giving $W=2^N$ and residual entropy $N{k}_B\ln 2$; a clean illustration of the $g>1$ exception.)

Let $x=\epsilon /(k_{B}T)$. As $T\to 0$, $x\to +\infty$, and $C\propto x^2{e}^{-x}.$ The exponential decay $e^{-x}$ beats the polynomial growth $x^2$: for large $x$, $e^{-x}$ shrinks faster than any power of $x$ grows, so $x^2{e}^{-x}\to 0$. Therefore $C\to 0$. Hence $C/T\to 0$ too, keeping the entropy integral finite — exactly the self-consistency the Third Law promises. Why does exponential beat polynomial? Each unit increase in $x$ multiplies $e^{-x}$ by the fixed factor $1/e\approx 0.37$ while only adding a bounded relative amount to $x^2$; repeated multiplication by $<1$ crushes any power law.

At $x=20$: $x^2{e}^{-x}=400\times e^{-20}\approx 400\times 2.06\times 10^{-9}\approx 8.2\times 10^{-7}$ — already essentially zero, confirming $C\to 0$.