1.7.23 · D5Thermodynamics
Question bank — Entropy change in irreversible processes — always - 0
True or false — justify
True or false: In every real process the entropy of the system increases.
False — only the universe's entropy must rise; a system can lose entropy (freezing gives ) provided the surroundings gain at least as much.
True or false: A process with is physically impossible.
False — setting the equation to its floor gives exactly the reversible limit; it is the idealized boundary reached as every driving gradient , not a forbidden state. See Reversible vs irreversible processes.
True or false: If a gas is compressed and its entropy drops, the second law is violated.
False — compression work pushes heat out to the surroundings, so rises by at least as much as falls, keeping .
True or false: For an adiabatic () process, .
False — only a reversible adiabat has ; an irreversible adiabat (e.g. free expansion, the dashed path in Figure s03) has yet .
True or false: A slow, quasi-static process is always reversible.
False — slowness only shrinks pressure/temperature gradients to zero; friction still injects each step, so a quasi-static-with-friction process is irreversible.
True or false: Entropy production is a state function like .
False — depends only on the endpoints, but (the entropy manufactured along the way) depends on the path, exactly as the two different loop-integrals in the Figure s02 proof show.
True or false: for a Carnot cycle.
True — the Carnot cycle is reversible, so heat crosses at the body's own () and the Clausius inequality holds with equality.
True or false: If over some interval, the process just needs more time to correct itself.
False — is exact and instantaneous ( at every step), so a net decrease never occurs for a real isolated system; there is nothing to correct.
True or false: Heat flowing from cold to hot always violates the second law.
False — it is fine if driven (a fridge); the driving work dumps enough extra entropy elsewhere that overall. Only spontaneous cold→hot flow, with no input, is forbidden.
Spot the error
"Free expansion has , so ." — find the flaw.
The formula requires reversible heat; free expansion is irreversible (dashed path, Figure s03), so you route through an imagined reversible isothermal path (solid curve) giving .
"For heat flowing across a finite , use one temperature in for both bodies." — find the flaw.
Each body is scored at its own temperature: the hot loses , the cold gains ; the mismatch is exactly the plotted in Figure s01.
" exactly." — find the flaw.
It should be , not ; the gap is the entropy production (Step 4 of the Figure s02 proof), and equality holds only in the reversible limit — this is the Clausius inequality.
"Since entropy always increases, a refrigerator can't lower the entropy of its contents." — find the flaw.
The contents (system) can lose entropy; the compressor's work dumps heat plus extra entropy to the room (), so the total still rises.
"Ice melting cools the room, so could be negative." — find the flaw.
The same heat enters the ice at the lower melting temperature (gain ) while leaving the room at higher (loss ); dividing by the smaller makes the gain bigger, so .
"A reversible process has ." — find the flaw.
Reversible only forces ; the system's entropy can change freely (reversible isothermal expansion has ) as long as the surroundings change by exactly the opposite amount.
Why questions
Why must we evaluate along a reversible path even for an irreversible process?
Because is a state function — its change depends only on endpoints, and the definition is only valid on reversible paths, so we borrow a convenient one (e.g. the solid isotherm in Figure s03) with the same endpoints.
Why does irreversibility force strict inequality rather than ?
Because a real gradient makes measurable entropy: heat crossing a finite gap yields (Figure s01), which is strictly positive whenever and only vanishes in the infinitesimal-gap limit.
Why is the surroundings scored at and not the system's temperature?
The surroundings is a large reservoir that exchanges heat internally reversibly at its own fixed , so ; the system sits at a different , and the finite gap between the two is the source of the extra .
Why does the Clausius inequality use a cycle to prove ?
Because is stated for closed loops only; closing an irreversible (orange) with a reversible (teal) in Figure s02 lets the reversible leg supply (Step 3) while the whole loop stays (Step 2), giving .
Why does the second law give time a direction?
Only is allowed, so the mixed-up direction is the future; a reversed movie (unmixing, milk un-spilling) would require , which never happens — this is the Arrow of time.
Why can two identical macroscopic states be reached by both reversible and irreversible paths yet share the same ?
Because depends only on endpoints; what differs is and hence , which is larger for the irreversible path by exactly the extra .
Why does the statistical picture agree that mixing raises entropy?
A mixed/spread configuration corresponds to vastly more microstates , so its (and thus ) is larger — see Statistical interpretation of entropy (Boltzmann S = k ln W).
Edge cases
Edge case: heat flows between two bodies at equal temperature . What is ?
Zero — the bracket vanishes (right edge of Figure s01), matching the reversible limit with no driving gradient.
Edge case: a gas expands into vacuum but the container is only half emptied (). Is still positive?
Yes — with surroundings unchanged, so any real expansion into vacuum, however small, raises universe entropy.
Edge case: what happens to in the limit of an infinitely slow, frictionless (reversible) process?
It approaches exactly from above; reversibility is the mathematical infimum of the allowed entropy production, never actually beaten.
Edge case: an isolated system already at maximum entropy (equilibrium). What further change is possible?
None spontaneously — cannot rise further and cannot fall, so the state is stable and no net macroscopic process occurs.
Edge case: a finite-temperature body dropped into an infinite reservoir at the same . Reversible or not?
Reversible in the limit — with no temperature gap, heat exchange produces no entropy; any nonzero gap makes it irreversible with .
Edge case: can while ?
Yes — exactly the free-expansion case: the surroundings exchange no heat () so , yet the system's entropy rises, keeping .
Edge case: a chemical reaction runs spontaneously at fixed and while absorbing heat (endothermic). Does that break ?
No — endothermic spontaneity means the system's own entropy rise is large enough to overpower the surroundings' loss , so is still (equivalently the Gibbs free energy ).
Edge case: a "negative absolute temperature" spin system ( on the kelvin scale) gives heat to a normal positive- body. Which way does heat flow, and is still obeyed?
Heat flows out of the negative- system into the positive- one, and still holds because a negative- state actually sits above all positive- states in energy/entropy ordering, so the transfer still increases total entropy.
Recall One-line summary to lock in
The law is ", strict for real processes" — never " always rises," and never " works on any path." Match the temperature to the body, keep = into the system, and always route through an imagined reversible path.