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

True. This is the fundamental postulate of statistical mechanics; equilibrium is not a preferred microstate but a macrostate that happens to own the most of these equally-likely microstates.

Almost, but say it right. Nature drifts to the macrostate with the largest $W$ because it holds nearly all microstates; for a system of about $N\sim 10^{23}$ particles that peak is so sharp it is effectively certain — but it is a statement of overwhelming probability, not a law of force.

False. $S=k\ln W$, so doubling $W$ adds only $k\ln 2$; entropy grows with the logarithm of $W$, not $W$ itself.

True. $S_{AB}=S_A+S_B$ because $W_{AB}=W_A{W}_B$ and $\ln$ turns that product into a sum — this additivity is the very reason the logarithm appears.

True. $S=k\ln 1=0$; a single unique arrangement (like a perfect flawless crystal cooled to $T=0$, where all jiggling stops) has nothing to count.

False. The entropy of an isolated system never decreases; a sub-system (a fridge interior, a freezing puddle) can lose entropy as long as the surroundings gain more, so the total still rises.

False. "Disorder" is a fuzzy visual label; entropy is precisely $k\ln (\text{accessible microstates})$. A neat-looking crystal made of mixed isotopes can carry large entropy from the countless isotope arrangements you cannot see.

False. $k$ only sets units (J/K) so that $S=k\ln W$ matches the older thermodynamic (Clausius) entropy; the ranking of macrostates depends only on $W$, which $k$ never reorders.

False (subtle). Free expansion into vacuum is irreversible with $Q=0$; the relation $Q=T\Delta S$ (from $dS=d{Q}_{\text{rev}}/T$) holds only along a reversible path. Entropy still rises because it is a state function — you compute it along an imaginary reversible route, not the actual one.

False in general. For ordinary systems yes, but a system with a bounded, finite energy ladder (e.g. spins that can only point up or down) can reach a state where adding energy forces more spins to the higher level, actually reducing $W$. There $S$ falls as energy rises — this is precisely what "negative absolute temperature" means (hotter-than-infinity, not colder-than-zero).

The error is inventing a force. Nature is blind — every microstate is equally likely; the disordered macrostate simply contains vastly more of them, so you land there by counting alone.

Wrong operation. $\ln (3W)=\ln 3+\ln W$, so $S$ goes up by $k\ln 3$ (an addition), not up by a factor of 3.

The microstate counts multiply ($W_{AB}=W_A{W}_B$), but entropies add: $S_{AB}=S_A+S_B=2S$. You multiply $W$, you add $S$ — that is exactly what the $\ln$ enforces.

You must count which coins are heads: $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ different choices. $W$ is a count of arrangements, not the number of heads.

Zero entropy means $W=1$, i.e. exactly one microstate — that is perfectly possible, just very improbable compared to the many-way middle macrostates. Rare $\ne$ forbidden.

The heat enables new configurations, but the entropy rise is from $W_{\text{liquid}}\gg W_{\text{solid}}$ — molecules gain access to far more positions/orientations. The count is the cause; heat is the enabler.

Same core idea. Shannon entropy is $\ln (\text{count})$ of possibilities measured in bits; Boltzmann's is the same log-of-multiplicity measured in joules per kelvin via the factor $k$.

Error — this is the Gibbs paradox. If the molecules on both sides are identical and indistinguishable, swapping a left molecule with a right one produces no new microstate, so $W$ does not actually grow and $\Delta S=0$. The $Nk\ln 2$ jump only appears when the two gases are different (distinguishable) species.

Only $f(xy)=f(x)+f(y)$ makes entropy additive while multiplicities multiply, and the unique continuous solution to that equation is $f=\text{const}\cdot \ln$. A square root does not turn products into sums.

Because $W$ grows exponentially in $N$ (roughly $W\sim a^N$), so $\ln W=N\ln a$ — the log of an exponential is linear. The two effects cancel to give a bulk, extensive quantity.

For $N$ two-sided choices the counts follow the binomial $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$, whose peak sits at $n=N/2$. Its width (the spread of appreciable probability) is about $\sqrt{N}$ particles, while the peak position is $N/2$, so the relative width is $\sqrt{N}/N=1/\sqrt{N}$ — the Central-Limit shrinking of fluctuations. At $N\sim 10^{23}$, $1/\sqrt{N}\sim 10^{-11}$, an unimaginably thin spike, so essentially every microstate belongs to the central macrostate.

At absolute zero a system settles into its unique ground configuration, so $W\to 1$ and $S=k\ln 1=0$ — this is the statistical content of the third law of thermodynamics.

"Preferred" refers to the macrostate, not any single microstate. The equilibrium macrostate wins because it owns an overwhelming share of the equally-weighted microstates.

Each species suddenly accesses the whole doubled volume — position choices multiply just like a free expansion, so $W$ jumps and $S$ rises purely from counting, no energy transfer required. (Crucially, this needs the gases to be distinguishable; see the Gibbs-paradox trap above.)

A single low-energy state is more probable per state, but there are often many more high-energy states; the observed population balances "probability per state" against "number of states" — again a multiplicity argument.

$W=1$ and $S=k\ln 1=0$ — the degenerate floor of entropy, the reference point the third law fixes.

No physically realizable macrostate can; $W=0$ would mean it corresponds to no microstate, i.e. it simply does not exist. Every observable macrostate has $W\ge 1$.

Entropy is a property of the macrostate, not one microstate. After expansion the accessible position-space is doubled, so $W$ (and $S$) rose regardless of the specific microstate you happen to inspect.

Formally yes, $\Delta S=1\cdot k\ln 2$, but with one molecule fluctuations are enormous and the "Second Law" is only statistical — the molecule can wander entirely back into the original half quite often. Thermodynamic certainty needs large $N$.

Yes — that is the point of the Gibbs correction. Because permuting indistinguishable particles yields the same microstate, the honest count divides out those $N!$ fake rearrangements, lowering $W$ and keeping entropy properly extensive.

Yes, for a system with a capped energy ladder (spin systems, certain lasers). Once more than half the particles are in the top level, extra energy shoves the population toward the single "all-up" configuration, so $W$ shrinks and $S$ falls — the regime of negative absolute temperature, which is actually hotter than any positive temperature.

Yes, locally — it looks tidier. But it dumps enough entropy (usually as heat) into its surroundings that the total $W$ of system-plus-surroundings still increases, honouring the second law.

Yes — the central macrostate ($n\approx N/2$) has the largest $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$, hence the largest $\ln W$, hence the maximum entropy available to that fixed set of coins.