Work in expansion - reversible isothermal w = −nRT ln(V₂ - V₁), irreversible w = −P_ext ΔV: Question bank

Before we start, one reminder about the sign rule the whole page uses:

Recall The IUPAC sign convention (reveal if rusty)

Work done by the gas (expansion, gas pushes the world) is counted negative; work done on the gas (compression, the world pushes the gas) is positive. So a negative $w$ = the gas gave energy away.

True or false — justify

Work is a state function

False. Work depends on the path taken between two states — reversible and irreversible expansions between the same endpoints give different

w

, so it cannot depend on endpoints alone. See First Law of Thermodynamics.

For an isothermal expansion of an ideal gas,

Δ U = 0

True. Internal energy of an ideal gas depends only on temperature; constant

T

means constant

U

, so

Δ U = 0

and by the First Law of Thermodynamics

q = - w

Reversible isothermal expansion always releases more work than any irreversible path between the same two volumes

True. The reversible path pushes against the highest possible external pressure at every instant (

P_{ext} = P_{gas}

), so its area under the PV Diagram curve is maximal — every irreversible path pushes against something lower.

In a reversible expansion the gas is never actually in equilibrium

False. It is always in (quasi-)equilibrium — that is the whole point.

P_{ext}

trails

P_{gas}

by only an infinitesimal

d P

, so the system passes through a continuous chain of equilibrium states. See Isothermal Process.

w = - P_{ext} Δ V

is valid for any irreversible expansion

False. It holds only when

P_{ext}

is constant. For a general path you must integrate

- \int P_{ext} (V) d V

— the two-step example is exactly this in disguise.

During a reversible expansion, heat absorbed equals the magnitude of work done

True (for isothermal ideal gas). Since

Δ U = 0

, the first law gives

q = - w

, so all the heat drawn in leaves as work. This is why isothermal reversible expansion needs a heat reservoir.

The reversible work formula

w = - n R T ln (V_{2} / V_{1})

can be used for a non-isothermal reversible expansion

False. The derivation factored

T

out of the integral because it was constant. If

T

changes,

T

stays inside and the clean log form breaks.

If you compress a gas irreversibly and then expand it reversibly back to the start, the net work is zero

False. Different paths mean different work magnitudes; the loop encloses a nonzero area, so net work is nonzero — and it comes out as dissipated energy. This is the seed of Entropy and Irreversibility.

Spot the error

"The gas has pressure

P_{1} = n R T / V_{1}

, so for the irreversible burst

w = - P_{1} Δ V

Error: it uses

P_{gas}

where only

P_{ext}

belongs. During an irreversible expansion the gas pushes against the external pressure, which is finite and lower than

P_{gas}

"In reversible expansion

P_{ext} = P_{gas}

exactly, so there is no net force and the gas can't move."

Error: the difference is an infinitesimal

d P

, not exactly zero — enough to drive motion but small enough to vanish inside the integral. "Infinitely slow" is not "frozen."

"Both processes go

5 L \to 15 L

, so

Δ V

is the same, so the work is the same."

Error: equal

Δ V

does not mean equal work — work is the area under the pressure–volume path, and the two paths have very different shapes on the PV Diagram.

w_{irrev} = - P_{ext} Δ V

used

Δ V

in litres and

P

in kPa, answer in kJ."

Error: litre × kPa is not joules; you must convert litres to

m^{3}

(or use

1 L\cdotatm = 101.325 J

) so the units of energy are consistent.

"More expansion steps means more irreversibility, so less work is extracted."

Error: backwards. More steps let

P_{ext}

track

P_{gas}

more closely, so the process is closer to reversible and extracts more work — the reversible limit is infinitely many steps.

"Since

w

is negative for expansion, the gas is losing internal energy."

Error: negative

w

only means the gas did work on the surroundings. In an isothermal process that energy is replenished by heat (

q = - w

), so

Δ U = 0

— no net loss of internal energy.

Why questions

Why does the reversible path give the maximum work for expansion?

Because at every instant

P_{ext}

is as large as it possibly can be while still allowing expansion (just under

P_{gas}

), maximizing the integrand

P_{ext}

and hence the area

\int P_{ext} d V

Why does the same ratio

V_{2} / V_{1}

give the same reversible work regardless of the actual litre values?

Because the answer depends on

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

, and the logarithm only sees the ratio —

1 \to 2

and

10 \to 20

both give

ln 2

Why do we insist on "isothermal" before using

w = - n R T ln (V_{2} / V_{1})

Constant

T

lets us (a) apply

P V = n R T

throughout and (b) pull

T

out of the integral. Drop either and the derivation collapses. See Ideal Gas Law.

Why is the irreversible constant-

P_{ext}

work a simple rectangle on the PV diagram?

Because

P_{ext}

doesn't change with

V

, the "height" is flat — area = height × width =

P_{ext} (V_{2} - V_{1})

, a rectangle rather than the curved isotherm area.

Why can't we recover the "lost" work of an irreversible expansion by an infinitesimal nudge?

Because the gas was driven finitely far from equilibrium and dissipated the difference as heat/disorder; reversing needs a finite effort, and the dissipated energy is not retrievable — the hallmark of Entropy and Irreversibility.

Why does a reversible isothermal expansion require a heat reservoir at all?

Because the gas does work while

Δ U = 0

, so energy must flow in as heat to replace it exactly (

q = - w > 0

) and keep temperature constant.

Edge cases

V_{2} = V_{1}

(no volume change), what is the work?

Zero in both formulas:

ln (1) = 0

and

Δ V = 0

. No boundary moves, so no expansion work is done regardless of path.

A gas expands into a vacuum (

P_{ext} = 0

). What is the work?

Zero, because

w = - \int P_{ext} d V = 0

when there is nothing to push against. This is Free Expansion — irreversible yet does no work, even though volume increases.

For compression (

V_{2} < V_{1}

), what is the sign of reversible work?

Positive, since

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

makes

- n R T ln (V_{2} / V_{1}) > 0

— work is done on the gas, matching the convention.

In the limit of infinitely many infinitesimal irreversible steps, what does

w_{irrev}

approach?

The reversible value,

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

. The staircase of rectangles converges to the smooth area under the isotherm.

If the gas were not ideal, would

w = - n R T ln (V_{2} / V_{1})

still hold for reversible isothermal expansion?

No. That formula used

P_{gas} = n R T / V

; a real gas obeys a different equation of state, so the integrand and the result change.

At exactly

P_{ext} = P_{gas}

with truly zero difference, does the gas expand?

No net expansion — a strictly zero difference means mechanical equilibrium and no motion. Real reversible expansion needs the infinitesimal

d P

imbalance; the "

=

" is a limiting idealization.

Recall One-line summary of the whole trap-set

Reversible = max work, always in equilibrium, needs isothermal + ideal for the log formula; irreversible = less work, $P_{ext}$ rules (not $P_{gas}$ ), and $- P_{ext} Δ V$ only when $P_{ext}$ is constant.