Exercises — Entropy and disorder — Boltzmann S = k·ln(W)

$k=1.38\times 10^{-23}\text{J/K}$ (joules per kelvin). For $W=1$: $S=k\ln 1=k\cdot 0=0\text{J/K}$. A unique state (only one way to build it) carries zero entropy — this is the seed of the Third Law of Thermodynamics.

$S=k\ln 6=(1.38\times 10^{-23})(1.7918)=2.47\times 10^{-23}\text{J/K}.$ Here $\ln 6=1.7918$ because $e^{1.7918}\approx 6$. This is the "2 heads out of 4 coins" macrostate from the parent note.

The $W=100$ one is higher (more $\ln W$). The difference is $\Delta S=k\ln 100-k\ln 10=k\ln \frac{100}{10}=k\ln 10=1.38\times 10^{-23}\times 2.3026=3.18\times 10^{-23}\text{J/K}.$ Note: $W$ jumped by a factor of $10$, but $S$ only rose by $k\ln 10$ — a tiny additive bump, not a tenfold one.

$W=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!2!}=\frac{120}{6\cdot 2}=10.$ $S=k\ln 10=(1.38\times 10^{-23})(2.3026)=3.18\times 10^{-23}\text{J/K}.$ We used $\left(\genfrac{}{}{0ex}{}{5}{3}\right)$ because we are choosing which 3 of the 5 coins show heads — an unordered selection.

Each molecule independently gains twice as many position-states, so $W_f=2^N{W}_i$ (position choices multiply across independent molecules — see Free Expansion of an Ideal Gas). $\Delta S=k\ln \frac{W_f}{W_i}=k\ln 2^{100}=100k\ln 2.$ Numerically $\Delta S=100\times 1.38\times 10^{-23}\times 0.6931=9.57\times 10^{-22}\text{J/K}$.

$\frac{\Delta S}{N}=\frac{Nk\ln 2}{N}=k\ln 2=9.57\times 10^{-24}\text{J/K per molecule}.$ It is constant because each molecule contributes the same factor of 2 to $W$, and $\ln$ turns that product into a sum of equal terms $k\ln 2$. This is why entropy is extensive (scales with size $N$).

$\left(\genfrac{}{}{0ex}{}{4}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4,\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6,\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4,\left(\genfrac{}{}{0ex}{}{4}{4}\right)=1.$ Total microstates $=1+4+6+4+1=16=2^4$. ✓ Most probable macrostate: $n=2$ (two heads), with $W=6$. Fraction $=6/16=0.375=37.5\%$. The peaked middle wins because it has the most ways — the seed of the Second Law re-read as "drift to the biggest pile."

Multiplicities multiply: $W_{AB}=W_A{W}_B=8\times 32=256$. $S_{AB}=k\ln 256.$ Separately: $S_A+S_B=k\ln 8+k\ln 32=k(\ln 8+\ln 32)=k\ln (8\cdot 32)=k\ln 256.$ ✓ Numerically $\ln 256=5.545$, so $S_{AB}=7.65\times 10^{-23}\text{J/K}$. The additivity is exactly the $\ln$ property $\ln a+\ln b=\ln (ab)$ — this is why the logarithm was chosen (see Microstates and Macrostates).

$W_f/W_i=50^{N_A}$, so $\Delta S=k\ln (50^{N_A})=N_{A}k\ln 50=R\ln 50,$ using $R=N_{A}k=8.314\text{J/(mol⋅K)}$. $\Delta S=8.314\times \ln 50=8.314\times 3.912=32.5\text{J/(mol⋅K)}.$ Positive, as it must be: liquid has far more accessible microstates than the locked-in crystal.

Statistical: $\Delta S=N_{A}k\ln 2=R\ln 2$. $\Delta S=8.314\times 0.6931=5.76\text{J/(mol⋅K)}.$ This is exactly the value obtained from Clausius Entropy dS = dQ_rev / T by integrating $dS=d{Q}_{\text{rev}}/T$ along a reversible isothermal expansion from $V$ to $2V$ ($\Delta S=nR\ln (V_f/V_i)=R\ln 2$). Statistical and thermodynamic entropy agree — the whole point of $S=k\ln W$.

All heads: $W=\left(\genfrac{}{}{0ex}{}{10}{\mathrm{0...}}\right)$ — careful, "all heads" is $n=10$: $W=\left(\genfrac{}{}{0ex}{}{10}{10}\right)=1$, so $S=k\ln 1=0$. Five heads: $W=\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, so $S=k\ln 252=(1.38\times 10^{-23})(5.529)=7.63\times 10^{-23}\text{J/K}$. The system spends almost all its time near "5 heads": that macrostate owns $252$ of the $2^{10}=1024$ microstates ($24.6\%$), while "all heads" owns just $1$ ($0.098\%$). Higher $W$ ⇒ higher $S$ ⇒ where you find the system — the counting statement of the Second Law of Thermodynamics.

$\Delta H={\log }_2{2}^{100}=100\text{bits}.$ Each molecule's left/right choice is exactly 1 bit of missing information, and there are 100 molecules. The thermodynamic entropy is the same count re-scaled by $k\ln 2$ per bit: $\Delta S=(k\ln 2)\times 100=9.57\times 10^{-22}\text{J/K},$ matching L2.2. Same $\ln (\text{count})$ idea — joules vs. bits is only a choice of the constant out front.

One per cell: the particles are distinguishable and the cells are distinct, so this is the number of ways to assign 3 labelled particles to 3 labelled cells one-to-one $=3!=6$. Entropy $S=k\ln 6=2.47\times 10^{-23}\text{J/K}$. All in one cell: there is exactly $1$ arrangement, so $W=1$ and $S=k\ln 1=0$. The spread-out macrostate has strictly higher entropy — more distinct orderings. (This is the microscopic logic behind the Boltzmann Distribution: energy populates many accessible cells because doing so maximises $W$.)

Ground state is unique ⇒ $W=1$ ⇒ $S=k\ln 1=0$. This is the Third Law of Thermodynamics: entropy tends to zero as $T\to 0$ for a perfectly ordered pure crystal. Caveat: if the crystal contains randomly-placed isotopes, there are many ways to arrange which lattice site holds which isotope even at $T=0$. Then $W>1$ and a residual entropy $S=k\ln W_{\text{isotope}}>0$ survives. "Order you can see" is not the same as "$W=1$" — always count microstates, never eyeball tidiness.

(a) Two independent free expansions, each of $50$ molecules doubling volume: $\Delta S=2\times (50k\ln 2)=100k\ln 2=9.57\times 10^{-22}\text{J/K}.$ (b) All $100$ molecules each double their position-states: $W_f=2^{100}{W}_i$, so $\Delta S=100k\ln 2=9.57\times 10^{-22}\text{J/K}.$ Same answer. Why: entropy is additive (Step 2 of the parent's derivation). Whether you count the two 50-molecule contributions separately and add ($50k\ln 2+50k\ln 2$), or count all 100 at once ($100k\ln 2$), the $\ln$ turns the multiplied microstate counts into the same sum. Additivity guarantees the bookkeeping is consistent.