Entropy — Clausius definition dS = dQ_rev - T: Exercises

This page is a ladder. Each rung is harder than the last, and every problem has a complete solution hidden inside a collapsible callout so you can test yourself first. Everything here builds on the parent note Entropy — Clausius definition.

Before you start, keep these three tools within reach — we will earn each one as we use it.

A quick reminder of what each symbol means, because we will never use one without it being clear:

$n$ = number of moles (how much gas).
$T$ = absolute temperature in kelvin (K) — never celsius, because $1/ T$ must never blow up or go negative.
$V$ = volume; $P$ = pressure; $Q$ = heat; $c$ = specific heat (energy to warm 1 kg by 1 K); $C_{V}$ = molar heat capacity at constant volume.
$ln$ = the natural logarithm — the "undo" button for the number $e$ . It appears because $\int \frac{d x}{x} = ln x$ , and dividing heat by temperature always leaves us integrating something of the form $\frac{1}{x} d x$ .

Level 1 — Recognition

Goal: can you spot which formula applies and read off a sign?

Problem 1.1

State the Clausius definition of an entropy change and its SI unit.

Recall Solution 1.1

$d S = \frac{d Q _{r e v}}{T}, Δ S = \int_{A}^{B} \frac{d Q _{r e v}}{T} .$ The subscript rev means the heat must be imagined as flowing along a reversible path. Unit: $J K^{- 1}$ (joules per kelvin), because we divide energy (J) by temperature (K).

Problem 1.2

A gas expands. Without any numbers, does its entropy rise or fall, and why?

Recall Solution 1.2

It rises. More volume means the molecules have more places to be — more microscopic arrangements (microstates). In the formula $n R ln \frac{V _{2}}{V _{1}}$ , expansion means $V_{2} > V_{1}$ , so $\frac{V _{2}}{V _{1}} > 1$ , and $ln$ of a number bigger than 1 is positive. See the figure: a wider box, more dots spread out.

Figure — Entropy — Clausius definition dS = dQ_rev - T

Problem 1.3

A reversible adiabatic process has $d Q_{r e v} = 0$ throughout. What is $Δ S$ ?

Recall Solution 1.3

$Δ S = \int \frac{0}{T} = 0.$ Zero heat exchanged reversibly ⇒ zero entropy change. This special case is called isentropic ("same entropy").

Level 2 — Application

Goal: plug numbers into one formula correctly.

Problem 2.1

$2 mol$ of an ideal gas expands isothermally and reversibly to three times its original volume. Find $Δ S$ .

Recall Solution 2.1

Isothermal ⇒ only the volume term survives: $Δ S = n R ln \frac{V _{2}}{V _{1}} = 2 \times 8.314 \times ln 3.$ $ln 3 = 1.0986$ , so $Δ S = 2 \times 8.314 \times 1.0986 = 18.27 J K^{- 1} .$ Positive, as expected for expansion. ✔

Problem 2.2

Heat $0.5 kg$ of water ( $c = 4186 J kg^{- 1} K^{- 1}$ ) from $290 K$ to $340 K$ . Find $Δ S$ of the water.

Recall Solution 2.2

The temperature changes, so we cannot use a single $Q / T$ ; we integrate with the local $T$ : $Δ S = m c ln \frac{T _{2}}{T _{1}} = 0.5 \times 4186 \times ln \frac{340}{290} .$ $ln (340/290) = ln (1.1724) = 0.15906$ , so $Δ S = 2093 \times 0.15906 = 332.9 J K^{- 1} .$

Problem 2.3

$1 mol$ of ideal gas ( $C_{V} = \frac{3}{2} R$ ) is heated from $300 K$ to $600 K$ at constant volume. Find $Δ S$ .

Recall Solution 2.3

Constant volume ⇒ the volume term vanishes ( $ln 1 = 0$ ): $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} = 1 \times \frac{3}{2} \times 8.314 \times ln 2.$ $\frac{3}{2} \times 8.314 = 12.471$ , and $ln 2 = 0.6931$ , so $Δ S = 12.471 \times 0.6931 = 8.644 J K^{- 1} .$

Level 3 — Analysis

Goal: two effects at once, or an irreversible process needing a reversible stand-in.

Problem 3.1

$1 mol$ of ideal gas ( $C_{V} = \frac{3}{2} R$ ) goes from $(T_{1}, V_{1}) = (300 K, 1 L)$ to $(T_{2}, V_{2}) = (600 K, 4 L)$ . Find $Δ S$ .

Recall Solution 3.1

Both temperature and volume change, so use the full ideal-gas formula: $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}} .$ Term 1: $\frac{3}{2} \times 8.314 \times ln 2 = 12.471 \times 0.6931 = 8.644 J K^{- 1}$ . Term 2: $8.314 \times ln 4 = 8.314 \times 1.3863 = 11.526 J K^{- 1}$ . $Δ S = 8.644 + 11.526 = 20.17 J K^{- 1} .$ Because $S$ is a state function, this answer does not care whether the gas was heated first then expanded, or expanded first then heated — only the endpoints matter.

Problem 3.2 (Free expansion)

$3 mol$ of ideal gas double their volume by rushing into an evacuated, insulated chamber. Find the entropy change of the gas.

Recall Solution 3.2

The real process is irreversible, so we may not use the actual heat ( $Q_{a c t u a l} = 0$ ). Instead:

Insulated + no external pressure ⇒ $Q = 0$ , $W = 0$ , so $Δ U = 0$ ⇒ $T$ unchanged (ideal gas $U$ depends only on $T$ ).
Same endpoints as a reversible isothermal doubling. Use that substitute path: $Δ S = n R ln \frac{V _{2}}{V _{1}} = 3 \times 8.314 \times ln 2 = 24.94 \times 0.6931 = 17.29 J K^{- 1} .$ Positive — the irreversibility shows up as created entropy. The figure shows the trick: the crooked real path is replaced by a smooth reversible one joining the same two dots.

Problem 3.3 (Both directions of heat in a cycle)

A Carnot engine absorbs $Q_{H} = 500 J$ at $T_{H} = 400 K$ and rejects heat at $T_{C} = 300 K$ . Verify that $\oint d Q_{r e v} / T = 0$ by finding $Q_{C}$ and the two entropy pieces.

Recall Solution 3.3

For a reversible Carnot cycle, $\frac{Q _{H}}{T _{H}} = \frac{Q _{C}}{T _{C}}$ , so $Q_{C} = Q_{H} \frac{T _{C}}{T _{H}} = 500 \times \frac{300}{400} = 375 J .$ Entropy gained by gas taking heat in: $+ \frac{500}{400} = + 1.25 J K^{- 1}$ . Entropy lost when heat is rejected: $- \frac{375}{300} = - 1.25 J K^{- 1}$ . Sum around the loop: $1.25 - 1.25 = 0$ . ✔ This is the Clausius theorem in one concrete cycle.

Level 4 — Synthesis

Goal: combine several bodies; track entropy of system and surroundings.

Problem 4.1 (Mixing / thermal contact)

A $1 kg$ block of copper ( $c = 385 J kg^{- 1} K^{- 1}$ ) at $400 K$ is dropped into a huge reservoir at $300 K$ . Find $Δ S_{b l oc k}$ , $Δ S_{r eser v o i r}$ , and $Δ S_{u ni v er se}$ .

Recall Solution 4.1

Block (its $T$ falls $400 \to 300$ , integrate): $Δ S_{b l oc k} = m c ln \frac{T _{2}}{T _{1}} = 385 \times ln \frac{300}{400} = 385 \times (- 0.28768) = - 110.8 J K^{- 1} .$ Negative — cooling loses entropy, correct.

Reservoir (so large its $T$ stays at $300 K$ ; it absorbs the heat the block gives up): $Q_{in t o r es} = m c (T_{1} - T_{2}) = 385 \times (400 - 300) = 38500 J, Δ S_{r es} = \frac{+ 38500}{300} = + 128.3 J K^{- 1} .$ (Here we divide by the single reservoir temperature because the reservoir never changes temperature.)

Universe: $Δ S_{u ni v} = - 110.8 + 128.3 = + 17.5 J K^{- 1} > 0.$ The heat flowing from hot to cold is irreversible, so the universe's entropy rises. ✔

Problem 4.2 (Two finite blocks reaching a common temperature)

Two identical blocks, each mass $m$ , specific heat $c$ , at $T_{1} = 400 K$ and $T_{2} = 200 K$ , are placed in contact and isolated. Show the final temperature is the arithmetic mean and find $Δ S_{u ni v}$ for $m c = 100 J K^{- 1}$ .

Recall Solution 4.2

Final temperature: equal masses/heat capacities, energy conserved: $m c (T_{f} - T_{1}) + m c (T_{f} - T_{2}) = 0 \Rightarrow T_{f} = \frac{T _{1} + T _{2}}{2} = \frac{400 + 200}{2} = 300 K .$ Entropy of each block (integrate, its $T$ varies): $Δ S = m c ln \frac{T _{f}}{T _{1}} + m c ln \frac{T _{f}}{T _{2}} = m c ln \frac{T _{f}^{2}}{T _{1} T _{2}} .$ $= 100 \times ln \frac{30 0 ^{2}}{400 \times 200} = 100 \times ln \frac{90000}{80000} = 100 \times ln 1.125 = 100 \times 0.11778 = 11.78 J K^{- 1} .$ Positive again. Notice $T_{f}^{2} = 90000 > T_{1} T_{2} = 80000$ : the arithmetic mean always beats the geometric mean, which is why $Δ S_{u ni v} > 0$ for any two unequal temperatures. That inequality is the second law hiding in plain algebra.

Level 5 — Mastery

Goal: prove a general result, or engineer a process with zero entropy change.

Problem 5.1 (Isentropic condition — deriving a well-known law)

For a reversible adiabatic ideal-gas process, $Δ S = 0$ . Starting from the full formula, derive the relation between $T$ and $V$ . (Take $γ = C_{P} / C_{V}$ and use $C_{P} - C_{V} = R$ .)

Recall Solution 5.1

Set the total entropy change to zero: $n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}} = 0.$ Divide by $n C_{V}$ and use $R / C_{V} = (C_{P} - C_{V}) / C_{V} = γ - 1$ : $ln \frac{T _{2}}{T _{1}} + (γ - 1) ln \frac{V _{2}}{V _{1}} = 0 \Rightarrow ln [\frac{T _{2}}{T _{1}} (\frac{V _{2}}{V _{1}})^{γ - 1}] = 0.$ Exponentiate (undo the $ln$ ): $\frac{T _{2}}{T _{1}} (\frac{V _{2}}{V _{1}})^{γ - 1} = 1 \Rightarrow T V^{γ - 1} = const .$ So the familiar adiabatic law is exactly the statement "entropy stays constant." Numerically, for $γ = 5/3$ , doubling the volume ( $V_{2} / V_{1} = 2$ ) forces $\frac{T _{2}}{T _{1}} = 2^{- (γ - 1)} = 2^{- 2/3} = 0.6300,$ i.e. the gas cools to $63.0%$ of its start temperature. Check: term 1 $= C_{V} ln 0.6300$ and term 2 $= R ln 2$ cancel.

Problem 5.2 (Designing a reversible heat exchange)

You must heat $1 mol$ of gas ( $C_{V} = \frac{5}{2} R$ ) from $300 K$ to $600 K$ at constant volume, but you want the universe entropy change to be as small as possible. Compare (a) using a single $600 K$ reservoir versus (b) using an infinite ladder of reservoirs each infinitesimally hotter. Find $Δ S_{u ni v}$ in each case.

Recall Solution 5.2

The gas entropy change is the same both ways (state function): $Δ S_{g a s} = n C_{V} ln \frac{T _{2}}{T _{1}} = \frac{5}{2} (8.314) ln 2 = 20.785 \times 0.6931 = 14.41 J K^{- 1} .$ Heat needed: $Q = n C_{V} Δ T = 20.785 \times 300 = 6235.5 J$ .

(a) Single hot reservoir at $600 K$ gives up all $6235.5 J$ at $600 K$ : $Δ S_{r es} = - \frac{6235.5}{600} = - 10.39 J K^{- 1}, Δ S_{u ni v} = 14.41 - 10.39 = + 4.02 J K^{- 1} .$

(b) Ladder of reservoirs, each matching the gas temperature $T$ as it rises, gives up $d Q = n C_{V} d T$ at that same $T$ : $Δ S_{r es} = - \int_{300}^{600} \frac{n C _{V} d T}{T} = - n C_{V} ln 2 = - 14.41 J K^{- 1}, Δ S_{u ni v} = 14.41 - 14.41 = 0.$ The ladder is the reversible limit: heat always crosses between bodies at (nearly) equal temperature, so no entropy is created. This is the physical meaning of "reversible" — matching temperatures at every step.

Connections

Carnot Cycle and Efficiency — Problem 3.3 uses $Q_{H} / T_{H} = Q_{C} / T_{C}$ .
First Law of Thermodynamics — energy conservation behind every $Q$ we tracked.
Second Law of Thermodynamics — every $Δ S_{u ni v} > 0$ result is it in action.
Reversible vs Irreversible Processes — the L5 ladder is the reversible limit.
Exact and Inexact Differentials — why $1/ T$ turns $d Q$ into a state function.
Clausius Inequality — the L3 free-expansion trap.
Statistical Entropy — Boltzmann S = k ln W — the microstate picture behind Problem 1.2.