Reversible and irreversible processes

Push a syringe plunger in slowly and it slides back out the moment you let go, as if nothing happened. Pop a balloon, or watch a drop of ink bloom into a glass of water, and no amount of waiting ever undoes it. Both are changes of state. Only one of them is, even in principle, undoable. Thermodynamics gives this distinction a precise name: a reversible process is one that can be run backwards, retracing every intermediate state, leaving no permanent mark on the system or on anything around it; an irreversible process cannot. The ink never gathers itself back into a drop, the popped balloon never re-inflates itself, and a hot cup of tea never spontaneously pulls heat back out of the cooler room to reheat itself — even though none of these outcomes would violate the first law of thermodynamics. This chapter is where you finally get a physical feel for why the second law exists. You already know the first law is a strict accountant of energy, $\Delta Q = \Delta U + \Delta W$ , and it does not care which direction a process runs — energy books balance either way. Yet the everyday world has an unmistakable arrow: eggs break, they do not unbreak; gases spread out, they do not spontaneously bunch up. Reversibility and irreversibility are the concepts that let you classify any thermodynamic process — isothermal, adiabatic, free expansion, a stirred fluid, a cooling engine — along that arrow, and connect that classification directly to the Kelvin–Planck and Clausius statements you have just studied. By the end you should be able to look at a described process, list the physical features that make it fall short of ideal reversibility, and say precisely why it cannot be reversed without leaving the universe changed.

Key takeaways

A process is reversible if the system and its surroundings can both be restored to their exact initial states with no change anywhere else in the universe; a process is irreversible if this cannot be done.
A reversible process is carried out quasi-statically — infinitely slowly, through an unbroken chain of equilibrium states — with no friction, no turbulence, and no finite temperature or pressure difference driving the change.
Reversibility is an idealised limit, never fully achieved: every real process happens at a finite rate, so every real process is irreversible to some degree.
Common irreversible processes: free (unrestrained) expansion into vacuum, heat flow across a finite temperature difference, friction and viscous drag, combustion, sudden compression or expansion, diffusion/mixing of gases.
The idealised reversible limit: an infinitesimally slow isothermal or adiabatic compression/expansion, with the external driving pressure differing from the gas pressure by only an infinitesimal amount.
Irreversibility is the physical content behind the second law — a spontaneous process that could run backwards with nothing else changing would violate the Kelvin–Planck or Clausius statement, which is precisely why the second law forbids it.
Key idea: a process is reversible only if system and surroundings can both be restored to their exact starting states with zero net change anywhere else, which demands an idealised, infinitely slow, frictionless, quasi-static path — every real process falls short and is therefore irreversible.

The core insight

A process is reversible only when it can be undone so completely that the system, the surroundings, and the rest of the universe all return to their exact initial states with no trace left behind — a condition that real processes, always driven at a finite rate through friction, turbulence, or a finite temperature or pressure difference, can only approach and never fully reach.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Physics Thermodynamics unit, immediately after the first and second laws and alongside the classification of gas processes — isothermal, adiabatic, and cyclic — where $PV = \text{const}$ and $PV^\gamma = \text{const}$ are introduced. It is a qualitative, conceptual topic: no new formula is prescribed here beyond the first law you already know, and the goal is reasoning, not calculation. The 2025-26 CBSE prescription keeps the Carnot engine, heat-engine efficiency, and quantitative entropy ( $\Delta S = \int dQ/T$ ) out of the Class 11 syllabus — those are studied in Class 12 and beyond — so this article stays strictly at the level of classifying processes and connecting that classification to the second-law statements, without deriving any efficiency formula.

The physical picture

Picture a gas in a cylinder fitted with a frictionless piston, in contact with a large heat reservoir. If you push the piston in so slowly that the gas pressure inside is, at every instant, only infinitesimally greater than the external pressure, the gas passes through an unbroken sequence of equilibrium states. Reverse the push — release the external pressure by the same infinitesimal amount — and the gas retraces exactly the same sequence of states outward. This is the operational meaning of a quasi-static process, and it is the backbone of reversibility.

\Delta Q = \Delta U + \Delta W, \qquad \Delta W = P\,\Delta V \qquad (1)

where: $\Delta Q$ is heat supplied to the gas, $\Delta U$ is the change in internal energy, $\Delta W$ is work done by the gas, and $P\,\Delta V$ is that work for a quasi-static change at (near-)uniform pressure $P$ . The physical reading is that this bookkeeping equation is valid for any process, reversible or not — but $\Delta W = P\,\Delta V$ with a single well-defined $P$ only makes sense when the gas has a uniform pressure at every instant, which is exactly the quasi-static condition. In a rapid, irreversible expansion or compression the gas pressure is not uniform (it is turbulent, or has a spatial gradient), so there is no single $P$ to insert — you cannot even draw a proper $P$ – $V$ curve for the irreversible path, only mark its endpoints.

\text{no other change anywhere} \iff \text{Process is reversible} \qquad (2)

where: this is the defining criterion, not a formula to compute — a process is reversible only if running it forward and then finding some way to run it backward leaves the system back at its initial state and the surroundings back at their initial state and nothing else in the universe altered. Equation (2) is deliberately strict: it is not enough that the system returns to where it started (a compressed spring can always be released back to its natural length); the surroundings must be restored too, with no residual heating, no permanently displaced weight, no leftover disorder anywhere.

Figure

A cylinder with a piston being pushed in through a slow sequence of near-equilibrium states shown as closely spaced curve points forming a smooth reversible path, next to a second cylinder where a partition bursts and the gas rushes turbulently into vacuum, showing a jagged irreversible path that cannot be retraced

The second law of thermodynamics is where irreversibility gets its teeth. If a spontaneous process could somehow run in reverse with absolutely no other change anywhere, that reversed process would, in specific cases, transfer heat from cold to hot with no external work (violating the Clausius statement) or convert heat completely into work in a cycle with no other effect (violating the Kelvin–Planck statement). Since real processes never manage this, real processes are irreversible — and the reversible process is the theoretical limiting case in which neither statement is ever threatened, because nothing irretrievable ever happens.

Derivation AWhy free expansion cannot be a quasi-static, reversible process

The goal is to show, starting only from the first law and the physical description of free (unrestrained) expansion, that this standard example cannot satisfy the reversibility conditions — it fails the "equilibrium at every instant" test and the "no other change anywhere" test simultaneously, and the argument uses nothing beyond $\Delta Q = \Delta U + \Delta W$ and the definition of quasi-static.

Step 1. Set up the process. A gas occupies one half of a rigid, thermally insulated container; the other half is evacuated; a partition separates them. The partition is suddenly removed and the gas expands to fill the whole container. Because the container is rigid and insulated, no heat enters or leaves ( $\Delta Q = 0$ ), and because the gas expands into a vacuum there is no external pressure resisting it, so no work is done ( $\Delta W = 0$ ).

Step 2. Apply the first law. From $\Delta Q = \Delta U + \Delta W$ with $\Delta Q = 0$ and $\Delta W = 0$ , we get $\Delta U = 0$ : the internal energy of the gas is unchanged by free expansion. This already shows the process cannot be tracked on a normal $P$ – $V$ work diagram in the usual sense, since $\Delta W = 0$ throughout even though the volume has changed enormously.

Step 3. Check the quasi-static requirement. Immediately after the partition is removed, gas molecules stream into the vacuum at high, unequal local speeds; the pressure and density are not uniform across the container during the expansion — there is no single value of $P$ that describes the gas at that instant. Since a quasi-static process requires the system to pass through a continuous series of equilibrium states, and free expansion visibly does not (it is turbulent and non-uniform until the gas re-settles), it fails the first reversibility condition outright.

Step 4. Check whether it can be undone with no other change. To force the gas back into the original half, an external agent must now do work on it (compressing it back), and that work appears somewhere else in the universe as heat or as a permanently raised weight — a genuine, irreversible change in the surroundings. The gas itself can be squeezed back to its original volume, but the universe as a whole is left altered, so the "no other change anywhere" criterion of equation (2) fails.

Step 5. Conclude. Free expansion satisfies neither the quasi-static condition nor the strict reversibility criterion, so it is a textbook irreversible process — despite the deceptively simple energy bookkeeping $\Delta U = \Delta Q = \Delta W = 0$ . This is the standard trap: students who only check the first-law numbers wrongly conclude "nothing happened," when in fact reversibility is a statement about the path, not the energy balance.

Worked examples

Board level - classifying a slow, frictionless compression

A gas is compressed by pushing a piston in very slowly, remaining in equilibrium with its surroundings at every instant, with the external pressure differing from the gas pressure by only an infinitesimal amount, and with no friction between piston and cylinder wall. Classify this process and justify your answer.

Step 1. List the given conditions: the process is infinitesimally slow, the gas is in equilibrium throughout, the driving pressure difference is infinitesimal, and friction is absent.

Step 2. Test against the reversibility criteria. A continuous sequence of equilibrium states is maintained (quasi-static), and because the external pressure exceeds the gas pressure by only an infinitesimal amount, a tiny decrease in the external pressure at any stage would make the gas expand back along exactly the same states it just passed through — with no dissipative loss to friction anywhere along the way.

Step 3. Apply the definition. Since the compression can be exactly reversed at any point with an infinitesimal change of conditions, and neither the gas nor the surroundings retain any permanent alteration when it is reversed, both reversibility conditions of equation (2) are satisfied in the idealised limit.

Step 4. State the conclusion. This is an idealised reversible process — the theoretical limit that a real slow compression can only approximate, since no real piston push is perfectly frictionless or perfectly infinitesimal in its driving pressure difference. The process is reversible (idealised).

Competitive level - free expansion followed by a compression back to the original volume

A gas of internal energy $U_i$ , confined in one half of an insulated rigid container of total volume $2V$ , is allowed to expand freely (partition removed) to fill the whole container. It is then slowly and isothermally compressed, by an external agent doing work $W_c$ on it, back to its original volume $V$ . (a) Find $\Delta U$ for the free-expansion step. (b) State whether the overall two-step process (expansion, then compression) is reversible, and why.

Step 1. Analyse the free-expansion step using the first law. The container is insulated, so $\Delta Q = 0$ ; expansion into vacuum does no work, so $\Delta W = 0$ . From $\Delta Q = \Delta U + \Delta W$ , $\Delta U = \Delta Q - \Delta W = 0 - 0 = 0$ . $\Delta U = 0$ for the free-expansion step, so the internal energy — and, for an ideal gas, the temperature — is unchanged: $U_f = U_i$ .

Step 2. Recognise that the gas has been returned to its original volume by the second step, but ask whether the surroundings have also been restored. The isothermal compression required real work $W_c$ done on the gas by an external agent; that work was extracted from some outside source (a falling weight, a motor drawing electrical energy), and since the compression is isothermal, an equivalent amount of heat $Q_c = W_c$ must leave the gas into the surrounding reservoir to keep its temperature constant.

Step 3. Compare initial and final states of the universe, not just the gas. The gas is back at volume $V$ and its original temperature — so the system looks unchanged. But the surroundings are not: the external agent's energy source has been permanently depleted by $W_c$ , and the reservoir has absorbed extra heat $Q_c$ that it did not have before. This is a permanent change left behind outside the gas.

Step 4. Apply the strict definition. Because the surroundings retain a permanent change (energy has been irreversibly degraded, converted from ordered work into heat delivered to a reservoir, exactly as the first, spontaneously irreversible free-expansion step already made unavoidable), the overall two-step process fails the "no other change anywhere" test. The overall process is irreversible, even though the gas alone has returned to its starting state — a cyclic-looking outcome for the system is not the same as reversibility for the process. Timing note: In a board or competitive exam, do not spend time trying to compute $W_c$ or $Q_c$ numerically unless the question gives specific numbers (moles, temperature, volumes) — the marks here are for correctly identifying $\Delta U = 0$ in Step 1 and for the reasoning in Steps 2 to 4 that distinguishes "the system returned" from "the process is reversible." Budget under two minutes for the free-expansion part and the rest for the classification argument; a numeric slip in $W_c$ costs far less than missing the conceptual point.

Common traps

Treating "reversible" as "can be reversed by some means" — the correct, strict definition requires the system and the surroundings to both return to their initial states with no other change anywhere, which excludes many processes that merely look cyclic.
Believing an everyday slow process (a gently released piston, a slowly leaking valve) is truly reversible — real processes only approximate reversibility; genuine reversibility needs infinite slowness, zero friction, and an infinitesimal driving difference throughout.
Assuming a cyclic process (system back to its starting state) is automatically reversible — a cycle only guarantees the system's state variables have returned; if any leg involved friction, a finite temperature difference, or a sudden change, the overall process is still irreversible, exactly as in the free-expansion-then-compression example above.
Forgetting that free expansion has $\Delta W = 0$ and $\Delta Q = 0$ (so $\Delta U = 0$ ) even though it is a classic irreversible process — reversibility is a statement about the path through equilibrium states, not about the energy bookkeeping at the endpoints.
Confusing "quasi-static" with "reversible" — quasi-static (equilibrium at every instant) is a necessary condition for reversibility, but true reversibility additionally demands the complete absence of dissipative effects such as friction; a quasi-static process with friction is still irreversible.

You should be able to…

State the precise, strict definition of a reversible process, including the "no other change anywhere in the universe" condition.
List the conditions — quasi-static, frictionless, infinitesimal driving pressure or temperature difference — that define the idealised reversible limit.
Explain, using the first law, why free expansion has $\Delta U = 0$ even though it is a standard irreversible process.
Classify a described process (compression, expansion, heat flow, friction, mixing) as reversible or irreversible, with justification referencing equilibrium states and permanent change in the surroundings.
Distinguish a cyclic process (system returns to its initial state) from a reversible process (system and surroundings return, with no other change anywhere).
Connect irreversibility to the second law by naming which statement — Kelvin–Planck or Clausius — a spontaneous reversal would violate.

Ask the tutor: Want to check whether a specific process you have thought up — like a slowly deflating balloon, a stirred cup of tea, or a block sliding to rest on a rough floor — is reversible or irreversible? Ask the AI tutor to walk through the three-step test with you.

If a process can technically be undone by doing more work on it, why isn't that called reversible?

Being able to force a system back to its starting point is not the same as reversibility in the thermodynamic sense. The definition demands that nothing else in the universe changes when you undo it — no extra energy source gets used up, no reservoir ends up with leftover heat it did not have before. If undoing the process requires an external agent to spend work and dump heat somewhere, that spent work and dumped heat are a permanent mark left on the surroundings, so the process is still classified as irreversible even though the system itself looks fine again.

Is a very slow, real-world process like a piston pushed in over an hour actually reversible?

Not exactly — it is a very good approximation to reversibility, but not the real thing. Reversibility is an idealised limit that requires infinite slowness, zero friction between the piston and cylinder, and a driving pressure difference that never exceeds an infinitesimal amount. A real hour-long push still has some friction and moves at a finite (if small) rate, so a tiny amount of energy is always dissipated as heat that cannot be fully recovered. The slower and more frictionless the real process, the closer it gets to the reversible ideal, but it never reaches it exactly.

Why does the syllabus care about an idealisation that never actually happens?

Because the reversible process is the best-case benchmark against which every real process is measured. Just as you compare a real engine's efficiency against the theoretical best possible cycle (a topic you will meet formally when you study the Carnot engine in Class 12), thermodynamics uses the fully reversible process as the upper limit of performance and the cleanest case for deriving relationships between state variables — it is a tool for reasoning precisely, even though nature never quite delivers it.

Does every irreversible process have to involve heat flow across a temperature difference?

No — heat flow across a finite temperature difference is only one of several ways a process can be irreversible. Friction and viscous drag dissipate ordered mechanical energy as heat without necessarily involving two different temperatures at the start. Free expansion into a vacuum is irreversible even though, as shown above, no heat is exchanged at all ( $\Delta Q = 0$ ) — its irreversibility comes from the turbulent, non-equilibrium path the gas takes, not from any temperature mismatch. Any one of turbulence, friction, a finite driving pressure or temperature difference, or a sudden non-quasi-static change is enough on its own to make a process irreversible.