Entropy — Clausius definition dS = dQ_rev - T: Worked examples

This page is the "flight simulator" for the Clausius definition $d S = \frac{d Q _{r e v}}{T}$ from the parent note Entropy — Clausius definition. Before you touch an exam problem, you should have flown through every kind of case entropy can throw at you: heat in, heat out, temperature constant, temperature changing, adiabatic-and-reversible, adiabatic-and-irreversible, two bodies meeting, and the sneaky "the answer is zero" traps.

Every symbol used here was built in the parent note. As a one-line refresher:

The scenario matrix

Every entropy problem you will meet is one (or a combination) of the cells below. The worked examples that follow are tagged with the cell they cover. Note that Cell J (system vs surroundings vs universe) is a bookkeeping layer that can sit on top of any process — Example 5 shows it for a spontaneous two-body flow and Example 8 shows it for a gas-plus-reservoir expansion, so the two are different applications, not a repeat.

Cell	Situation	Sign of $Δ S_{sy s}$	Key move
A	Heat in at constant $T$ (isothermal expansion)	$+$	$Q_{r e v} / T$ , $T$ pulls out
B	Heat out at constant $T$ (isothermal compression)	$-$	same, $Q < 0$
C	Temperature rising (heating, variable $T$ )	$+$	integrate $\int m c d T / T$
D	Temperature falling (cooling)	$-$	same, $T_{2} < T_{1}$
E	Reversible adiabatic ( $Q = 0$ , reversible)	$0$	isentropic — trap-buster
F	Irreversible adiabatic (free expansion, $Q = 0$ )	$+$	reversible substitute path
G	Two bodies to thermal equilibrium (irreversible mixing)	$+$ overall	sum two variable- $T$ integrals
H	General ideal-gas state change (both $T$ and $V$ change)	any sign	two-log master formula
I	Real-world word problem (ice melting)	$+$	latent heat at constant $T$
J	Bookkeeping layer: system vs surroundings vs universe	check each	compute $Δ S_{u ni v}$ (Ex 5 = spontaneous flow, Ex 8 = gas + reservoir)

We now cover all ten cells with eight examples (some examples hit two cells at once).

Example 1 — Isothermal expansion and compression (Cells A + B)

Forecast: guess the signs before reading. Expansion → gas gets more room → more places for molecules → bigger mess. Compression should undo it exactly.

$Δ U = 0$ on each leg. Why this step? Internal energy of an ideal gas depends only on $T$ , and $T$ is fixed, so $Δ U = 0$ .
First Law → all heat becomes work: With $Δ U = 0$ the First Law $d U = d Q_{r e v} - P d V$ gives $d Q_{r e v} = P d V$ . Why does the integral become a log? Substitute the ideal-gas law $P = \frac{n R T}{V}$ , so $\int P d V = \int \frac{n R T}{V} d V = n R T ln \frac{V _{2}}{V _{1}}$ (with $T$ constant), giving $Q_{r e v} = W = n R T ln \frac{V _{2}}{V _{1}}$ .
Entropy: $Δ S = \frac{Q _{r e v}}{T} = n R ln \frac{V _{2}}{V _{1}}$ . Why can $T$ come out? It is constant, so it factors straight out of the integral.
(a) expansion: $Δ S_{a} = 2 (8.314) ln 3 = + 18.26 J K^{- 1}$ .
(b) compression: $V_{2} / V_{1} = 1/3$ , so $Δ S_{b} = 2 (8.314) ln \frac{1}{3} = - 18.26 J K^{- 1}$ .

Verify: signs match the forecast (expand $+$ , compress $-$ ). The round trip gives $Δ S_{a} + Δ S_{b} = 0$ — correct, since $S$ is a state function and we returned to the start. Units: $mol \cdot J K^{- 1} mol^{- 1} = J K^{- 1}$ . ✔

Example 2 — Heating and cooling water (Cells C + D)

Forecast: heating should give $Δ S > 0$ , cooling $Δ S < 0$ , and equal magnitudes.

$d Q_{r e v} = m c d T$ . Why "rev"? Imagine an infinite ladder of reservoirs, each only $d T$ hotter than the water, so every slice moves reversibly.
$d S = \frac{m c d T}{T}$ . Why not just $Q / T$ ? Because $T$ is not constant — you must use the local temperature of each slice, so you integrate.
$Δ S = m c ln \frac{T _{2}}{T _{1}}$ . Why a log? $\int d T / T = ln T$ ; the ratio $T_{2} / T_{1}$ is what survives.
(a): $Δ S_{a} = 0.5 (4186) ln \frac{350}{290} = + 394.9 J K^{- 1}$ . Note on rounding: with $c = 4186 J kg^{- 1} K^{- 1}$ the log of $350/290$ is $ln (1.2069) = 0.18814$ , so $0.5 (4186) (0.18814) = 394.9 J K^{- 1}$ (carry the full log, do not approximate $Δ T / T$ ).
(b): $Δ S_{b} = 0.5 (4186) ln \frac{290}{350} = - 394.9 J K^{- 1}$ .

Verify: heating $+$ , cooling $-$ , magnitudes equal (state function, same endpoints reversed). Units $kg \cdot J kg^{- 1} K^{- 1} = J K^{- 1}$ . ✔

Example 3 — Reversible adiabatic (Cell E, the "answer is zero" case)

Forecast: $Q = 0$ makes many people say " $Δ S = 0$ trivially." Careful — is it zero for the right reason?

Adiabatic means $d Q_{r e v} = 0$ at every instant. Why this matters: the numerator of $d Q_{r e v} / T$ is zero all along the path.
$Δ S = \int \frac{0}{T} = 0$ . Why this is legitimate here: the process is reversible, so the definition applies directly along the actual path — no substitute path needed.

Verify: cross-check with the master ideal-gas formula (Example 6). Using the two facts about $γ$ above and plugging $\frac{T _{2}}{T _{1}} = (\frac{V _{1}}{V _{2}})^{γ - 1}$ into $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}}$ gives exactly $0$ (verified numerically in the checks). The zero is earned, not assumed. ✔

Example 4 — Free expansion into vacuum (Cell F)

Forecast: $Q = 0$ , $W = 0$ (nothing pushed back), so $Δ U = 0$ and $T$ stays constant. Tempting to write $Δ S = 0$ . Is that right?

Identify the endpoints: same $T$ , volume $V_{1} \to 2 V_{1}$ . Why endpoints only? $S$ is a state function — the actual violent path is irrelevant.
Build a reversible substitute: a reversible isothermal expansion between the same endpoints. Why allowed? Any reversible path between the same states gives the true $Δ S$ .
Apply Example 1's result: $Δ S = n R ln \frac{V _{2}}{V _{1}} = (1) (8.314) ln 2 = + 5.76 J K^{- 1}$ .

Verify: $Δ S > 0$ as it must be for an irreversible process in an isolated system. The actual heat $Q_{a c t u a l} = 0$ does not enter — that is exactly the parent note's headline trap. ✔

Example 5 — Two bodies reaching equilibrium (Cells G + J, spontaneous flow)

Figure — Entropy — Clausius definition dS = dQ_rev - T

Forecast: the metal cools ( $Δ S < 0$ ), the water warms ( $Δ S > 0$ ). Since the process (heat flowing across a finite temperature gap) is irreversible, guess $Δ S_{u ni v} > 0$ . The figure above shows why the final temperature sits close to the water's value: follow the two curves — the aluminium (orange) plunges a long way while the water (teal) barely rises, because the water's heat capacity dwarfs the metal's, and they meet at the dashed plum line $T_{f}$ .

Energy balance for $T_{f}$ : heat lost by Al = heat gained by water. $m_{A l} c_{A l} (400 - T_{f}) = m_{w} c_{w} (T_{f} - 300) .$ Why? No heat leaves the insulated system, so First Law $\Rightarrow$ conserved internal energy. Read this on the figure as "how far each curve travels vertically."
Solve: $0.30 (900) (400 - T_{f}) = 1.0 (4186) (T_{f} - 300)$ . Numerically $T_{f} = 306.06 K$ . Why sensible? The water's heat capacity ( $4186 J K^{- 1}$ ) hugely outweighs the metal's ( $270 J K^{- 1}$ ), so equilibrium sits close to the water's start — exactly what the near-flat teal curve in the figure shows.
$Δ S$ of each body — each cools/warms over a range, so use the Cell-C/D integral: $Δ S_{A l} = m_{A l} c_{A l} ln \frac{T _{f}}{400}, Δ S_{w} = m_{w} c_{w} ln \frac{T _{f}}{300} .$ Why local $T$ ? Both bodies change temperature, so integrate $m c d T / T$ along an imagined ladder of reservoirs.
Numbers: $Δ S_{A l} = 270 ln \frac{306.06}{400} = - 72.42 J K^{- 1},$ $Δ S_{w} = 4186 ln \frac{306.06}{300} = + 83.73 J K^{- 1} .$ Why these signs and sizes follow: $T_{f} < 400$ so the aluminium's log is negative (it cooled); $T_{f} > 300$ so the water's log is positive (it warmed). The water's magnitude is larger because its $m c$ (the multiplier in front of the log) is roughly 15 times the metal's, which more than makes up for its smaller temperature swing.
Universe: $Δ S_{u ni v} = Δ S_{A l} + Δ S_{w} = + 11.31 J K^{- 1}$ .

Verify: metal negative, water positive (forecast ✔). Sum is positive, as the Second Law demands for a spontaneous, irreversible heat flow across a finite gap. If we had let heat flow through infinitely many intermediate reservoirs (reversible), the sum would tend to $0$ . ✔

Example 6 — General ideal-gas state change (Cell H, master formula)

Forecast: temperature doubles ( $+$ from the $T$ term) and volume doubles ( $+$ from the $V$ term). Both push entropy up, so expect a solidly positive answer.

Master formula (derived in parent): $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}}$ . Why two terms? After dividing the First Law by $T$ , one exact piece came from $d U = n C_{V} d T$ and one from $P d V = n R T d V / V$ .
Plug in with $C_{V} = \frac{3}{2} (8.314) = 12.471$ : $Δ S = 0.5 (12.471) ln 2 + 0.5 (8.314) ln 2.$
Compute: $= 4.322 + 2.882 = + 7.20 J K^{- 1}$ .

Verify: both terms positive (forecast ✔). Sanity check on the tool: setting $T_{2} = T_{1}$ recovers Example 1's isothermal result; setting $V_{2} = V_{1}$ recovers Example 2's heating result. Reversible-adiabatic (Example 3) makes the two terms cancel exactly. ✔

Example 7 — Melting ice, real-world word problem (Cell I)

Forecast: solid → liquid means molecules gain freedom to wander, so entropy should rise.

Heat absorbed: $Q_{r e v} = m L_{f}$ . Why "rev"? Melting happens at a fixed $273 K$ with the surroundings only infinitesimally warmer — genuinely reversible.
Constant $T$ → $T$ pulls out: $Δ S = \frac{Q _{r e v}}{T} = \frac{m L _{f}}{T}$ . Why not integrate? Temperature does not change during a phase change.
Compute: $Δ S = \frac{0.20 ( 3.34 \times 1 0 ^{5} )}{273} = + 244.7 J K^{- 1}$ .

Verify: positive (solid → liquid, more disorder ✔). Units: $\frac{kg \cdot J kg ^{- 1}}{K} = J K^{- 1}$ . ✔

Example 8 — Exam twist: gas plus reservoir (Cell J, reversible vs irreversible)

Forecast: reversible should give $Δ S_{u ni v} = 0$ exactly; irreversible should give $Δ S_{u ni v} > 0$ .

Reversible case:

System: $Δ S_{sy s} = n R ln 2 = 8.314 ln 2 = + 5.76 J K^{- 1}$ . Why? Cell A result.
Surroundings: the reservoir loses the same heat $Q = n R T ln 2$ at the same $T$ , so $Δ S_{s u r r} = - \frac{Q}{T} = - 5.76 J K^{- 1}$ . Why negative? Heat leaving lowers the reservoir's entropy.
Universe: $Δ S_{u ni v} = 5.76 - 5.76 = 0$ . Why zero? Reversible ⇒ no entropy generated.

Irreversible (free) case: 4. System: same endpoints ⇒ same $Δ S_{sy s} = + 5.76 J K^{- 1}$ . 5. Surroundings: $Q_{a c t u a l} = 0$ , reservoir untouched ⇒ $Δ S_{s u r r} = 0$ . Why zero? In a free expansion the gas pushes on nothing and the insulated walls pass no heat, so no heat is exchanged with the reservoir — with zero heat crossing at temperature $T$ , $Δ S_{s u r r} = Q / T = 0$ . 6. Universe: $Δ S_{u ni v} = + 5.76 J K^{- 1} > 0$ .

Verify: reversible → $Δ S_{u ni v} = 0$ ; irreversible → $Δ S_{u ni v} > 0$ (both forecasts ✔). Same $Δ S_{sy s}$ in both cases confirms $S$ is a state function; only the surroundings' entropy tells the two processes apart. This is the operational content of the Clausius Inequality and the Second Law of Thermodynamics. ✔

Active recall

Recall Which cell? Match the scenario to its move (hide answers)

"Ice melts at 0 °C" ::: Cell I — constant- $T$ latent heat, $Δ S = m L_{f} / T$ .
"Gas doubles volume into vacuum, insulated" ::: Cell F — reversible isothermal substitute, $Δ S = n R ln 2 > 0$ .
"Reversible adiabatic compression" ::: Cell E — $Δ S = 0$ (isentropic).
"Hot metal into cold water" ::: Cell G — sum two $m c ln (T_{f} / T_{i})$ terms, universe $> 0$ .
"Both $T$ and $V$ change for a gas" ::: Cell H — master formula $n C_{V} ln (T_{2} / T_{1}) + n R ln (V_{2} / V_{1})$ .

Connections

Parent · Clausius definition
Carnot Cycle and Efficiency — why $Q_{H} / T_{H} = Q_{C} / T_{C}$ underpins these sums
First Law of Thermodynamics — energy balance in Examples 5 and 6
Second Law of Thermodynamics — $Δ S_{u ni v} \geq 0$ in Examples 5 and 8
Reversible vs Irreversible Processes — Cells E vs F, reversible vs free expansion
Exact and Inexact Differentials — why the substitute-path trick works
Statistical Entropy — Boltzmann S = k ln W — "more room ⇒ more microstates"
Clausius Inequality — Example 8's irreversible universe increase