A reversible process in thermodynamics is an idealized transformation that can be reversed along its original path, restoring both a thermodynamic system and its surroundings exactly to their initial states. Restoring the system alone is insufficient: the reversal must leave no net change elsewhere. Reversible processes provide reference models for calculating thermodynamic properties and establishing performance limits, although actual processes generally involve some irreversibility. (openstax.org)
Reversibility and equilibrium
In the conventional equilibrium description, a reversible transformation proceeds through a continuous sequence of states arbitrarily close to thermodynamic equilibrium. The differences driving the transformation—such as differences in pressure or temperature—are infinitesimal. An infinitesimal adjustment of the external conditions can therefore reverse the direction of change. For example, a gas can expand against a frictionless piston when its pressure is infinitesimally greater than the opposing pressure, and compress when that difference is reversed. (openstax.org)
A quasistatic process proceeds sufficiently slowly for the system to remain approximately in internal equilibrium. This is necessary for reversibility in the conventional treatment, but it is not sufficient. Slow piston motion can still involve friction, and slow heating can still transfer heat across a finite temperature difference. Both introduce irreversibility. Slowness alone does not eliminate dissipative effects. (web.mit.edu)
Reversibility also depends on the surroundings. A system may undergo an internally reversible transformation while exchanging heat irreversibly with an external reservoir at a different temperature. Full reversibility requires the absence of irreversibility both within the system and in its interactions with the environment. (web.mit.edu)
Entropy criterion
The second law of thermodynamics gives a precise criterion: a fully reversible process generates no entropy. For the combined system and surroundings, treated as an isolated whole,
[ \Delta S_{\mathrm{total}} =\Delta S_{\mathrm{system}}+\Delta S_{\mathrm{surroundings}}=0. ]
An irreversible process instead produces a positive total entropy change. Zero entropy generation does not mean that the system’s own entropy remains constant; entropy can be transferred between the system and its surroundings by heat exchange. (ocw.mit.edu)
For an infinitesimal reversible step,
[ dS=\frac{\delta Q_{\mathrm{rev}}}{T}, ]
where (\delta Q_{\mathrm{rev}}) is heat entering the system and (T) is its absolute temperature, measured in kelvin. Entropy is a state function, so between equilibrium states (A) and (B),
[ S_B-S_A=\int_A^B\frac{\delta Q_{\mathrm{rev}}}{T}. ]
The integral may be evaluated along any suitable reversible path connecting those states, even when the actual transformation is irreversible. The hypothetical path is a calculation device, not a claim about how the real process occurred. (openstax.org)
Heat, work, and energy
Reversibility does not remove the requirement of energy conservation. With heat entering the system taken as positive and work done by the system taken as positive, the first law of thermodynamics is
[ dU=\delta Q-\delta W, ]
where (U) is internal energy. For a closed, simple compressible system undergoing reversible volume work,
[ \delta W_{\mathrm{rev}}=p,dV, \qquad dU=T,dS-p,dV. ]
Here (p) is the equilibrium system pressure. During an irreversible expansion, system pressure need not equal the external pressure opposing boundary motion; consequently, integrating system pressure over volume does not generally give the actual work. (web.mit.edu)
Under the same specified initial and final states and environmental constraints, reversible operation provides the limiting maximum work output, or minimum required work input. These qualifications matter: work depends on the path and available interactions, so endpoints alone do not establish a universal work maximum. (web.mit.edu)
Representative processes
For (n) moles of an ideal gas, a reversible isothermal expansion at temperature (T) obeys the equation of state (pV=nRT), where (R) is the molar gas constant. Expansion from (V_1) to (V_2) gives
[ W_{\mathrm{rev}}=nRT\ln!\left(\frac{V_2}{V_1}\right). ]
Because an ideal gas’s internal energy depends only on temperature, (\Delta U=0), so the heat absorbed equals the work delivered. Its entropy increases by (nR\ln(V_2/V_1)), while the reservoir’s entropy decreases by the same amount. (web.mit.edu)
By contrast, in free expansion into a vacuum inside an insulated enclosure, no boundary work is extracted and no heat enters. The ideal gas has the same initial and final temperatures, but its entropy increases by the same expression. There is no compensating reservoir entropy decrease. This illustrates why equal endpoints do not imply equal reversibility. (web.mit.edu)
A reversible adiabatic process exchanges no heat and therefore has constant entropy. Adiabatic conditions alone do not guarantee this: friction or free expansion can generate entropy without heat transfer through the boundary. (ocw.mit.edu)
Cycles and performance limits
The Carnot cycle consists of two reversible isothermal stages and two reversible adiabatic stages. For a heat engine operating between reservoirs at absolute temperatures (T_h) and (T_c), its efficiency is
[ \eta_{\mathrm{rev}}=1-\frac{T_c}{T_h}. ]
No engine operating solely between those reservoirs can exceed this efficiency; all reversible engines between them have the same efficiency, regardless of working substance. (openstax.org)
A cycle is not automatically reversible merely because the working substance returns to its starting state. That return ensures zero net change in its state functions, but irreversible processes during the cycle can still increase the surroundings’ entropy. Reversibility concerns the complete accounting of system and surroundings, not simply the closure of a path on a thermodynamic diagram. (ocw.mit.edu)