Question bank — Spontaneity — second law; entropy ΔS

False. Spontaneity is decided by $\Delta S_{univ}$, not by heat direction; endothermic processes like ${\text{NH}}_4{\text{NO}}_3$ dissolving absorb heat yet happen on their own because $\Delta S_{sys}$ is large and positive.

False. Spontaneity is about direction (which way is allowed), not rate; diamond → graphite is spontaneous but takes eons because rate belongs to kinetics, not thermodynamics.

False. Only $\Delta S_{univ}$ is forbidden from decreasing; the system may order itself (water freezing) as long as the surroundings gain more entropy from the dumped heat.

True. Reversible means the system stays infinitesimally close to equilibrium, so $\Delta S_{sys}$ and $\Delta S_{surr}$ are exactly equal and opposite; the total is zero.

False. The system's entropy can fall freely (condensation, freezing, gas compression); it is the entropy of the universe that cannot decrease.

False. It assumes constant pressure (so $q_P=\Delta H$) and constant $T$ so the surroundings act as a fixed-$T$ reservoir; the general form is $\Delta S=\int d{q}_{rev}/T$, which you must integrate when $T$ varies.

True. At $0\text{K}$ a perfect crystal has exactly one microstate ($W=1$), and $S=k_B\ln 1=0$.

False in general. Entropy is additive for independent systems, but mixing identical gases at the same conditions changes nothing measurable, whereas mixing different gases adds an entropy of mixing.

True. $\Delta S$ is a state function, so it depends only on endpoints; but you must compute it using $q_{rev}$ — the reversible heat — even when the real path releases the smaller $q_{irrev}$.

The error is confusing spontaneity with energy lowering. Melting absorbs heat ($\Delta H_{fus}>0$); it is spontaneous because $\Delta S_{sys}$ rises enough that $\Delta S_{univ}>0$.

Freezing orders the molecules, so $\Delta S_{sys}<0$. The second law is satisfied because the released heat raises $\Delta S_{surr}$ by more, keeping $\Delta S_{univ}\ge 0$.

Wrong heat. The entropy definition demands $q_{rev}$, not the smaller irreversible $q_{irrev}$; using the real fast-path heat gives a wrong, path-dependent number.

Not guaranteed. Spontaneity needs $\Delta S_{univ}=\Delta S_{sys}-\Delta H_{sys}/T>0$; at high $T$ the surroundings term shrinks and a negative $\Delta S_{sys}$ can dominate, making it non-spontaneous.

At exactly the boiling point liquid and vapour coexist at equilibrium, so $\Delta S_{univ}=0$. Spontaneous boiling needs a temperature just above the boiling point.

The real reason is additivity: entropies of independent systems add ($S_1+S_2$) while microstates multiply ($W_1{W}_2$), and only the logarithm converts a product into a sum. See the microstate picture.

Only when $T$ is constant. If $T$ changes along the path you must keep $T$ inside the integral: $\Delta S=\int d{q}_{rev}/T$, because each slice of heat is divided by the temperature at which it flows.

The system alone can lower its entropy by dumping heat outward; only the total, $\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}$, captures both effects and is the quantity the second law constrains.

A very exothermic reaction pours heat into the surroundings, and $\Delta S_{surr}=-\Delta H_{sys}/T$ becomes large and positive — the surroundings' disorder does the driving. This links directly to enthalpy and ΔG.

Because $\Delta S_{surr}=-\Delta H_{sys}/T$ has $T$ in the denominator; the same heat spread over hotter surroundings produces less relative disorder, so the enthalpy term matters less at high $T$.

Heat is a path function, so raw $q$ differs between paths; the reversible path is the unique "gentlest" one that gives a value depending only on the endpoints, making $q_{rev}/T$ a state function.

Gas molecules have vastly more accessible positions and speeds than atoms locked in a lattice, so $W$ jumps enormously, and since $S=k_B\ln W$, the entropy rises steeply.

For isothermal expansion $\Delta U=0$, so the first law gives $q_{rev}=-w_{rev}$; that reversible heat, divided by $T$, is exactly $\Delta S=nR\ln (V_2/V_1)$.

$\ln W$ is a pure number (a count), but entropy is measured in ${\text{J K}}^{-1}$; $k_B$ is the fixed conversion factor that gives the count physical energy-per-temperature units.

$S=0$, because a perfect crystal has a single microstate ($W=1$) and $\ln 1=0$ — the basis of the third law.

Zero. With $d{q}_{rev}=0$ everywhere, $\Delta S=\int d{q}_{rev}/T=0$; such a process is called isentropic, and note it is generally not isothermal — the temperature usually changes.

A reversible process at equilibrium — for example ice and water coexisting at $273\text{K}$, where neither direction is favoured.

$\Delta S=nR\ln (V_2/V_1)\to nR\ln 1=0$; no volume change means no extra positions gained, so zero entropy change is exactly right.

Negative, since $V_2/V_1<1$ makes $\ln (V_2/V_1)<0$; molecules lose accessible positions, so $W$ and $S$ fall — allowed as long as surroundings compensate.

It is zero over the whole cycle, because every state function returns to its initial value and the reversible path never generates net entropy.

Yes — with no enthalpy term, $\Delta S_{univ}=\Delta S_{sys}>0$, so the system's own disorder increase drives it, exactly like free expansion of an ideal gas into vacuum.

No — you must integrate $\Delta S=\int d{q}_{rev}/T$ because $T$ is not a single constant; the $q_{rev}/T$ shortcut only applies at fixed $T$.