Entropy — Clausius definition dS = dQ_rev - T: Question bank

Reminders of the notation used below, so nothing is unearned:

$Δ S$ = entropy change of a chosen body between two states, unit $J K^{- 1}$ .
$Q$ = heat that actually flowed; $Q_{r e v}$ = heat along an imagined reversible path with the same endpoints.
$T$ = absolute temperature (kelvin, always positive).
"System" = the stuff we track; "surroundings/reservoir" = everything it exchanges heat with; "universe" = system + surroundings.
$\oint$ = integral all the way around a closed cycle (back to the start state).

True or false — justify

Reversible adiabatic process has

Δ S = 0

True. Adiabatic means

d Q = 0

and reversible means we may use

d S = d Q_{r e v} / T = 0

, so no entropy is created — this is exactly an isentropic process.

Every adiabatic process has

Δ S = 0

False. Free expansion is adiabatic (

Q = 0

) yet

Δ S = n R ln (V_{2} / V_{1}) > 0

;

d Q = 0

only kills entropy change when the path is also reversible, so you must not plug

Q = 0

into an irreversible process.

Entropy is a form of stored heat inside a body.

False. Heat is not stored (it is an inexact differential — you cannot say "5 J of heat sits in this gas"); entropy is a state function counting how spread-out the energy is, measured in

J K^{- 1}

, not joules.

For any real (irreversible) process the entropy of the universe increases.

True. The Second Law of Thermodynamics says

Δ S_{u ni v er se} \geq 0

, with strict inequality for irreversible processes — that inequality is what "irreversible" means.

If a gas is compressed, its entropy must fall.

False. Compression lowers the

n R ln (V_{2} / V_{1})

term, but if the temperature rises the

n C_{V} ln (T_{2} / T_{1})

term can more than compensate; the sign of

Δ S

depends on both endpoints, not on volume alone.

Along a reversible cycle,

\oint d Q_{r e v} / T = 0

but

\oint d Q_{r e v}

need not be zero.

True.

\oint d Q_{r e v} / T = 0

because

S

is a state function, yet

\oint d Q_{r e v} =

net heat

=

net work

\neq = 0

in general — that non-zero loop integral is precisely why raw heat is inexact and needs the

1/ T

integrating factor.

The system's entropy can decrease during a real process.

True. Only the universe's entropy is forced up; a system can dump entropy to its surroundings (e.g. a fridge cooling water), as long as the surroundings gain more.

Two different reversible paths between the same states give different

Δ S

False. Because

S

is a state function, every reversible path between fixed endpoints yields the identical

Δ S

— that path-independence is the whole point of dividing by

T

Spot the error

"Gas expands freely into vacuum,

Q = 0

, therefore

Δ S = 0

The error is using the actual heat of an irreversible path in

d S = d Q_{r e v} / T

. Replace the path by a reversible isothermal one with the same endpoints:

Δ S = n R ln (V_{2} / V_{1}) > 0

"An irreversible engine's entropy change is

\oint d Q / T

, which we set to zero."

For an irreversible cycle the Clausius Inequality gives

\oint d Q / T < 0

, not

= 0

; only reversible cycles reach equality. Setting it to zero silently assumes reversibility.

"Heating water, I'll just use

Δ S = Q / T

with

T

the final temperature."

The system temperature changes throughout, so a single

T

is wrong; you must integrate

d S = m c d T / T

using the local

T

, giving

Δ S = m c ln (T_{2} / T_{1})

"The

T

d Q_{r e v} / T

is the reservoir temperature, so for system entropy I use the reservoir's

T

For system entropy you use the system's temperature. In a reversible exchange the two temperatures differ only infinitesimally so it does not matter, but for system bookkeeping always take the system's

T

"Since

Δ S_{u ni v er se} \geq 0

, no process can ever lower any entropy."

The bound is on the universe, not on parts. A local region (system) can cool and lose entropy provided the surroundings gain at least as much.

"For a Carnot cycle

Q_{H} = Q_{C}

because it returns to its start."

What returns to start is

U

(so net

Q =

net

W

), not the individual heats. What actually balances is

Q_{H} / T_{H} = Q_{C} / T_{C}

, i.e. the entropy exchanged, giving

\oint d Q_{r e v} / T = 0

"Entropy has units of joules because it comes from heat."

Entropy is heat divided by temperature, so its unit is

J K^{- 1}

, not J — the division by

T

is exactly what changes the dimensions.

Why questions

Why must the heat be reversible in

d S = d Q_{r e v} / T

Only along a reversible path does

d Q / T

become an exact (path-independent) differential equal to the true state change; irreversible

Q / T

falls short of

Δ S

Why is

1/ T

the "integrating factor" for heat and not some other function?

Because the Carnot Cycle and Efficiency forces

Q_{H} / T_{H} = Q_{C} / T_{C}

, so multiplying

d Q_{r e v}

1/ T

makes every reversible loop integrate to zero — the defining property of an exact differential. See Exact and Inexact Differentials.

Why does dividing by

T

turn a path-dependent quantity into a path-independent one?

Raw heat

\oint d Q_{r e v} \neq = 0

around a loop, but

\oint d Q_{r e v} / T = 0

(Clausius theorem, from tiling with tiny Carnot cycles); a quantity with zero loop integral is the differential of a state function.

Why does the same heat cause a bigger entropy jump in a cold body than a hot one?

Because

d S = d Q / T

has

T

in the denominator: a small

T

makes

d Q / T

large. Physically, adding energy to something already "quiet" opens up proportionally more new microstates.

Why can we compute

Δ S

of an irreversible process at all, if the definition needs reversibility?

Because

S

is a state function, its change depends only on endpoints. We invent any reversible path with the same endpoints and integrate along it; the answer is the true

Δ S

regardless of what really happened.

Why does entropy give time a direction ("arrow of time")?

Real processes only ever raise

Δ S_{u ni v er se}

; the reverse (spontaneously un-mixing, un-spreading) would lower it and never occurs, so the increasing-entropy direction labels "future." Microscopically this is Statistical Entropy — Boltzmann S = k ln W.

Why isn't

Δ S

for the surroundings just

- Δ S_{sy s t e m}

Only for reversible processes do they cancel exactly. In irreversible processes the surroundings gain more than the system loses, leaving a positive

Δ S_{u ni v er se}

Edge cases

A reversible cycle: what is the system's total

Δ S

over one full loop?

Zero. The system returns to its start state and

S

is a state function, so

\oint d S = 0

— regardless of how much heat crossed the boundary.

Isothermal expansion at

T = 0 K

: is

Δ S = n R ln (V_{2} / V_{1})

still finite?

The formula is finite, but

T = 0

is unreachable (third law); more importantly, at

T \to 0

any real heat exchange

d Q / T

blows up, signalling why absolute zero cannot be attained in finite steps.

Mixing two identical gases at the same

T

and

P

: what is

Δ S

Zero. There is no distinguishable change of state (the "before" and "after" are physically identical), so no new microstates open — this is the resolution of the Gibbs paradox.

A process with

Q > 0

(heat absorbed) but

Δ S_{sy s t e m} < 0

: possible?

Not for a single reversible input, since

d S = d Q_{r e v} / T

shares the sign of

d Q_{r e v}

. It can happen only if simultaneous work/other paths reshape the state; for a pure reversible heat exchange,

Q > 0 \Rightarrow Δ S_{sy s t e m} > 0

Reversible isothermal compression (

V_{2} < V_{1}

): sign of

Δ S

Negative, since

ln (V_{2} / V_{1}) < 0

. Squeezing removes accessible microstates — but the surroundings absorb the expelled heat and gain at least as much, keeping

Δ S_{u ni v er se} \geq 0

Perfect insulation and perfectly reversible: what is

Δ S_{sy s t e m}

Zero — this is a reversible adiabat (isentropic). Both conditions are needed: reversibility alone or insulation alone does not force

Δ S = 0

Heat flowing between two bodies at equal temperature: what happens to entropy?

Nothing net — at equal

T

there is no spontaneous flow, and any infinitesimal reversible exchange has

+ d Q / T

on one side cancelling

- d Q / T

on the other, so

Δ S_{u ni v er se} = 0

. Unequal

T

is what makes flow irreversible and

Δ S_{u ni v er se} > 0

Connections

Clausius Inequality — the $\oint d Q / T \leq 0$ that powers most "spot the error" items
Reversible vs Irreversible Processes — the line every trap here straddles
Second Law of Thermodynamics — source of $Δ S_{u ni v er se} \geq 0$
Carnot Cycle and Efficiency — where $Q_{H} / T_{H} = Q_{C} / T_{C}$ comes from
Exact and Inexact Differentials — why $1/ T$ integrates heat
Statistical Entropy — Boltzmann S = k ln W — microstate meaning behind "more spread out"
First Law of Thermodynamics — energy accounting underneath $Q$ and $W$