Reversible vs Irreversible Processes

Push a piston in one violent shove, and a gas wastes some of your effort as heat and turbulence. Push it in a million tiny, patient increments, and every joule of work is recovered as usable energy. This is the difference between irreversible and reversible processes. The distinction matters because work is not a state function. It depends on the path, and the path depends on how carefully you control the surroundings at every instant. Chemists care because reversible paths give the maximum work obtainable, setting a theoretical ceiling used throughout thermodynamics. This article builds the quasi-static reversible path from first principles. You will see why irreversible work is always smaller in magnitude. Then you will compute both cases numerically for the same gas. Finally you will see how this idea foreshadows entropy and the Second Law.

Key takeaways

A reversible process moves through a continuous sequence of near-equilibrium states, reversible by an infinitesimal change in conditions.
An irreversible process is finite and abrupt; the system lags behind the imposed external pressure.
For expansion, reversible work equals the maximum possible work extractable between two states.
Irreversible work uses a constant external pressure, giving a smaller magnitude than the reversible integral.
Reversible paths take infinite time in the idealized limit; real "reversible" processes only approximate this.
Key idea: Work depends on the path because $P_{ext}$ can differ from the system's own pressure at every step, and only the quasi-static limit closes that gap.

The core insight

A reversible process keeps the system infinitesimally close to internal equilibrium at every stage, so $P_{ext} \approx P$ throughout, and the work done equals the integral of the system's own pressure over volume — the largest magnitude of work exchange possible between two given states.

The physical picture

Consider a gas confined by a piston, expanding from volume $V_1$ to $V_2$ . As covered under pressure–volume work, the work done by the system is

w = -\int_{V_1}^{V_2} P_{ext}\, dV

The entire question of reversibility hinges on what $P_{ext}$ is doing during the expansion.

Irreversible (free or single-stage) expansion. Suppose the external pressure is suddenly dropped to some fixed value $P_{ext} = P_f$ and held there. The gas rushes to fill the available volume. Molecules near the piston face a pressure gradient, gas flows turbulently, and the system is far from internal equilibrium during the process. Only at the very end does the gas re-equilibrate to a uniform pressure. Because $P_{ext}$ is constant and set below the gas's initial pressure, the work is

w_{irrev} = -P_{ext}(V_2 - V_1)

Reversible (quasi-static) expansion. Now imagine removing sand grains from the piston one at a time, an infinite sequence of infinitesimal steps. At each stage, $P_{ext}$ is adjusted to be only infinitesimally less than the system's own pressure $P$ . The gas is always essentially in internal equilibrium; a tiny nudge in either direction would reverse the process direction. In this limit $P_{ext} \to P$ continuously, so

w_{rev} = -\int_{V_1}^{V_2} P\, dV

with $P$ given by the system's own equation of state along the path. Because the reversible path always pushes against the largest possible restraining pressure consistent with not exceeding the system's own pressure, it extracts the most work the system can give up (or requires the least work to compress it). Any real, finite-rate process cuts corners relative to this ideal path, so irreversible work is always smaller in magnitude than reversible work for an expansion, and irreversible work required for compression is always larger in magnitude.

Figure

Reversible vs irreversible expansion paths on a P–V diagram

The blue curve traces $P_{ext}=P$ at every step; the shaded area under it is the reversible work magnitude. The red rectangle is smaller because the constant $P_f$ used in the single irreversible step is lower than the gas pressure for most of the expansion.

Figure

Multistage approach to the reversible limit

The core relations

w = -\int_{V_1}^{V_2} P_{ext}\, dV

where: the general definition of expansion work; $P_{ext}$ is the pressure opposing the system at the moving boundary.

w_{irrev} = -P_{ext}(V_2 - V_1), \qquad P_{ext} = \text{constant}

where: single-stage irreversible expansion or compression against a fixed external pressure.

w_{rev} = -\int_{V_1}^{V_2} P\, dV

where: $P$ is the system's own pressure, tracked continuously as it changes with $V$ along the quasi-static path.

w_{rev,\,isothermal,\,ideal\,gas} = -nRT \ln\!\left(\frac{V_2}{V_1}\right)

where: valid for an ideal gas at constant $T$ along the reversible path; foreshadows the isothermal expansion topic.

|w_{rev}| \ge |w_{irrev}| \quad \text{(expansion, same } V_1 \to V_2\text{)}

where: the reversible path always yields the maximum work magnitude between two given states.

Derivation AWhy the reversible path maximizes work magnitude

Goal. Show that for an expansion between fixed $V_1$ and $V_2$ , no irreversible path can extract more work than the reversible path, and generally extracts less.

Step 1. Set up the comparison. Both processes start at $(P_1, V_1)$ and end at $(P_2, V_2)$ for the same gas sample and the same $T_1, T_2$ (same state function values). The only freedom is the choice of $P_{ext}(V)$ along the way, since work is path-dependent while internal energy is not.

Step 2. Apply the mechanical constraint. For the piston to move outward (expansion) at all, the system must push harder than the resistance, requiring

P(V) \ge P_{ext}(V)

at essentially every point of any real process. Equality holds only in the reversible limit, where the imbalance driving motion is kept infinitesimal.

Step 3. Compare the work integrals. Since $dV > 0$ during expansion, integrating the inequality term by term gives

\int_{V_1}^{V_2} P\, dV \ge \int_{V_1}^{V_2} P_{ext}\, dV

Multiplying by $-1$ reverses the inequality:

-\int_{V_1}^{V_2} P\, dV \le -\int_{V_1}^{V_2} P_{ext}\, dV

Step 4. Interpret the signs. Recall $w_{rev} = -\int P\,dV$ and $w_{irrev} = -\int P_{ext}\,dV$ for the actual process; both are negative numbers for an expansion (system does work on surroundings, so $w<0$ under the "work done on the system" sign convention). The inequality above states

w_{rev} \le w_{irrev} \le 0

Since both are negative, this means $|w_{rev}| \ge |w_{irrev}|$ : the reversible path releases work of the greatest magnitude.

Step 5. Confirm the compression case. Running the argument in reverse, for compression $dV<0$ and now $P_{ext} \ge P$ is required to push the piston in. Integrating gives $w_{irrev} \ge w_{rev} \ge 0$ : the reversible path requires the least work input, again the optimum.

\boxed{|w_{rev}| = \max |w| \text{ over all paths between the same two states}}

Worked examples

Example 1 — Foundational: single irreversible step vs. reversible isothermal expansion. One mole of ideal gas at $T = 298\ \text{K}$ occupies $V_1 = 10.0\ \text{L}$ at high pressure, held by a piston against $P_1$ . It expands isothermally to $V_2 = 20.0\ \text{L}$ . First find the initial and final pressures using $PV = nRT$ , with $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} = 0.08206\ \text{L atm K}^{-1}\text{mol}^{-1}$ .

P_1 = \frac{nRT}{V_1} = \frac{(1.00)(0.08206)(298)}{10.0} = 2.446\ \text{atm}

P_2 = \frac{nRT}{V_2} = \frac{(1.00)(0.08206)(298)}{20.0} = 1.223\ \text{atm}

Irreversible case: the external pressure drops suddenly and stays fixed at $P_2 = 1.223\ \text{atm}$ throughout the expansion.

w_{irrev} = -P_{ext}(V_2 - V_1) = -(1.223\ \text{atm})(20.0 - 10.0\ \text{L})

w_{irrev} = -12.23\ \text{L atm}

Converting with $1\ \text{L atm} = 101.325\ \text{J}$ :

w_{irrev} = -12.23 \times 101.325 = -1239\ \text{J} \approx -1.24\ \text{kJ}

Reversible case: $P_{ext}$ tracks $P = nRT/V$ continuously.

w_{rev} = -nRT \ln\!\left(\frac{V_2}{V_1}\right) = -(1.00)(8.314)(298)\ln\!\left(\frac{20.0}{10.0}\right)

w_{rev} = -(2477.6)(0.6931) = -1717\ \text{J} \approx -1.72\ \text{kJ}

Sanity check. $|w_{rev}| = 1.72\ \text{kJ} > |w_{irrev}| = 1.24\ \text{kJ}$ , confirming the reversible path extracts more work, exactly as the derivation predicts. Both are less than the (unphysical) work obtained by using the initial pressure $P_1$ the whole way, $|w| = P_1\Delta V = 2.446 \times 10.0 \times 101.325 = 2478\ \text{J}$ , which would violate $P_{ext} \le P$ throughout and can never occur for a real expansion.

Example 2 — Applied: two-stage irreversible expansion approaching the reversible limit. Same gas, same $V_1 = 10.0\ \text{L} \to V_2 = 20.0\ \text{L}$ at $T = 298\ \text{K}$ , but now the expansion happens in two irreversible steps instead of one, to see how work rises toward the reversible value as steps increase. Stage A: $P_{ext}$ drops immediately to the pressure the gas has at an intermediate volume $V_m = 15.0\ \text{L}$ :

P_m = \frac{nRT}{V_m} = \frac{(1.00)(8.314)(298)}{15.0 \times 10^{-3}\ \text{m}^3} = 165{,}170\ \text{Pa} = 1.630\ \text{atm}

Work for stage A ( $V_1 \to V_m$ , constant $P_{ext} = P_m$ ):

w_A = -P_m (V_m - V_1) = -(1.630\ \text{atm})(5.0\ \text{L}) = -8.150\ \text{L atm}

w_A = -8.150 \times 101.325 = -826\ \text{J}

Stage B: $P_{ext}$ drops again to $P_2 = 1.223\ \text{atm}$ (computed above) and the gas expands from $V_m$ to $V_2$ :

w_B = -P_2(V_2 - V_m) = -(1.223\ \text{atm})(5.0\ \text{L}) = -6.115\ \text{L atm}

w_B = -6.115 \times 101.325 = -620\ \text{J}

Total two-stage work:

w_{2-stage} = w_A + w_B = -826 - 620 = -1446\ \text{J} \approx -1.45\ \text{kJ}

Sanity check. Compare all three results: one-stage $|w| = 1.24\ \text{kJ}$ , two-stage $|w| = 1.45\ \text{kJ}$ , reversible (infinite-stage) $|w| = 1.72\ \text{kJ}$ . Work magnitude climbs monotonically toward the reversible ceiling as the number of stages increases, exactly the staircase-to-smooth-curve picture in the second figure. This trend is the numerical seed for the idea, developed later, that the reversible limit gives the maximum work extractable between two fixed states.

Limiting cases & common traps

Situation	What happens to work
Single-stage expansion against vacuum ( $P_{ext}=0$ )	$w = 0$ ; no work is done regardless of reversibility language, since there is no opposing force
Infinite stages, quasi-static limit	$w \to w_{rev}$ , the theoretical maximum magnitude
Compression instead of expansion	Reversible path needs the least work input, not the most
Constant-volume process	$w = 0$ always, reversible or not — reversibility only matters when $V$ changes

Reversible does not mean "slow" in a colloquial sense; it means the driving force is always infinitesimal, which in practice does require an experiment to proceed slowly, but slowness alone does not guarantee reversibility if friction or turbulence is present.
$w_{rev}$ and $w_{irrev}$ refer to the same endpoints $V_1, V_2$ ; comparing work along different volume ranges is not a valid test of the inequality.
$\Delta U$ is identical for both paths if $T_1$ and $T_2$ match (state function), but $q$ differs to compensate, since $\Delta U = q + w$ for each path separately.

The chemist behind it

The formal machinery of reversibility as an idealized limiting process was sharpened through the nineteenth-century development of thermodynamics, most notably by Rudolf Clausius, who used the concept to define entropy and state the Second Law. Clausius recognized that the quasi-static, reversible path was not merely a convenient mathematical fiction but the very construction needed to make heat, like work, expressible as an integral of a state-independent quantity divided by temperature. His insistence on precise reversible cycles, building on Sadi Carnot's earlier engine analysis, transformed reversibility from an engineering ideal into the conceptual foundation of chemical thermodynamics.

Origins · 1 minRudolf Clausius and how this was discovered

Rudolf Clausius

1822–1888 · Formalized reversible cycles and introduced entropy via reversible heat exchange

Ask the tutor: Can you walk me through how this same reversible-work integral becomes the isothermal expansion formula step by step?

Is a reversible process ever achieved exactly in the laboratory?

No. A truly reversible process requires infinitesimal steps and infinite time, an idealized limit rather than a physical event. Real experiments approximate it by using many small, slow steps, such as gradually bleeding gas out of a cylinder, letting the system re-equilibrate between each tiny change. The closer the approach, the nearer the measured work comes to the theoretical $w_{rev}$ from the integral of $P\,dV$ , but perfect reversibility remains a limiting construction useful for setting bounds on real performance.

Why does reversible work represent a maximum rather than some arbitrary reference value?

The mechanical requirement for spontaneous expansion is $P \ge P_{ext}$ at every instant; equality is the most demanding condition a real process can approach without stalling or reversing direction. Any looser condition, $P_{ext} < P$ , wastes potential work by letting the gas expand against less resistance than it could sustain. The reversible path is therefore not one choice among many but the extremal case, proven rigorously by the integral inequality in the derivation above.

Does reversibility depend on whether the process is isothermal, adiabatic, or something else?

No — reversibility describes how a path is traversed, not which variable is held fixed. You can have a reversible isothermal expansion, a reversible adiabatic expansion, or a reversible isobaric process, each defined by keeping $P_{ext}$ infinitesimally close to the system's own pressure throughout. The isothermal case is especially convenient because $P(V)$ has the simple hyperbolic form $nRT/V$ , which is why it is the standard first example, covered in detail in the next topic.

How does this connect to the idea of maximum work discussed later?

This article establishes the mechanical inequality $|w_{rev}| \ge |w_{irrev}|$ for expansion at constant temperature. The maximum-work theorem generalizes this: among all possible processes connecting two states, including those exchanging heat with multiple reservoirs, the reversible path yields the greatest work output and the smallest necessary work input. That broader theorem underlies the definitions of Gibbs and Helmholtz free energy as bounds on non-expansion work, extending the P–V argument you have just verified numerically.