Reversible Processes

Reversible thermodynamic processes: reversibility conditions, zero universe entropy change, maximum Carnot efficiency and ideal operating limit.

Conceptual context

The second law of thermodynamics establishes that natural processes generate entropy in the universe: $\Delta S_{\mathrm{univ}}$ > 0 for every real process. However, there exists an ideal limit where the process generates no net entropy: $\Delta S_{\mathrm{univ}}$ = 0. Such processes are called reversible and represent the theoretical reference against which every real process is compared.

A reversible process is one that, once completed, can be undone exactly along the same path, returning both system and surroundings to their initial states without leaving any trace in the universe. It destroys no exergy, degrades no energy quality, and defines the maximum achievable efficiency of any heat engine.

Understanding reversible processes is essential because they provide the theoretical ceiling: the Carnot efficiency $\eta_{\mathrm{Carnot}}$, the maximum extractable work, and the minimum heat that must be released to the cold reservoir. Every second-law analysis is based on comparing the real with the reversible.

🟢 Essential level

A reversible process is an ideal process that generates no entropy in the universe. The defining condition is that the universe entropy balance is zero: system and surroundings compensate each other exactly.

This means that whatever entropy the system gains, the surroundings lose exactly, and vice versa. There is no net energy degradation: all energy entering as heat can, in principle, be fully recovered as work.

[!TIP] Think of a reversible process as one that can be "rewound" without anything changing in the universe. If you play a video forward and then backward, everything returns exactly as it was. In an irreversible process, rewinding would leave something permanently changed.

The conditions for a process to be reversible are:

• No friction: no mechanical dissipation at any point.
• No finite temperature gradients: heat is transferred with infinitesimal T differences between system and surroundings.
• No finite pressure gradients: expansion or compression occurs with infinitesimal P differences.
• No spontaneous mixing: no free diffusion of substances.
• Quasi-static process: slow enough for the system to be in internal equilibrium at every instant.

In practice, no real process meets all these conditions. The reversible process is an ideal limit useful for calculating maximum efficiencies and entropies.

🔵 Formal level

Basic formulas

The reversibility condition is formalized as:

Reversibility condition (universe entropy)

For a reversible process with heat exchange with a reservoir at temperature $T_{\mathrm{caliente}}$ or $T_{\mathrm{frío}}$, the reversible heat $Q_{\mathrm{rev}}$ and entropy change are related by Clausius's definition: $\Delta S_{\mathrm{sis}}$ = $Q_{\mathrm{rev}}$ / T (for isothermal processes) or $\Delta S_{\mathrm{sis}}$ = integral of dQ_rev / T (for non-isothermal processes).

[!NOTE] The condition $\Delta S_{\mathrm{univ}}$ = 0 does not imply $\Delta S_{\mathrm{sis}}$ = 0. The system can gain or lose entropy; what matters is that the surroundings compensate exactly: $\Delta S_{\mathrm{ent}}$ = −$\Delta S_{\mathrm{sis}}$.

Real student questions

If a reversible process doesn't exist in reality, why study it? It serves as a maximum reference. Just as the speed of light is an unattainable limit but essential for relativistic physics, the reversible process is an unattainable limit but essential for evaluating the efficiency of any real machine. Without that reference, we could not quantify how much work is lost to irreversibilities.

Is a quasi-static process always reversible? No. A quasi-static process with friction is slow but irreversible: friction dissipates energy as internal heat and generates entropy. Quasi-static is a necessary but not sufficient condition for reversibility. Additionally, there must be no friction, no spontaneous mixing, and no finite gradients.

Why does Carnot efficiency depend only on temperatures? Because the condition $\Delta S_{\mathrm{univ}}$ = 0 imposes a fixed relationship between absorbed and released heat: Q_C/Q_H = $T_{\mathrm{frío}}$/$T_{\mathrm{caliente}}$. This relationship is derived from the equality $\Delta S_{\mathrm{sis}}$(hot isotherm) + $\Delta S_{\mathrm{sis}}$(cold isotherm) = 0 in the cycle. The working fluid does not appear in the derivation.

Can a heat pump have COP > 1? Yes, and it violates no law. The COP of a heat pump is Q_H/W, not an efficiency in the strict sense. The Carnot COP for a heat pump is $T_{\mathrm{caliente}}$/($T_{\mathrm{caliente}}$ − $T_{\mathrm{frío}}$), which can be much greater than 1. This is possible because the heat pump "moves" heat, it does not "create" it.

Final synthesis

A reversible process is defined by $\Delta S_{\mathrm{univ}}$ = 0: it generates no net entropy in the universe. It requires ideal conditions (no friction, no finite gradients, quasi-static) that are unattainable in practice. Carnot efficiency $\eta_{\mathrm{Carnot}}$ = 1 − $T_{\mathrm{frío}}$/$T_{\mathrm{caliente}}$ is the direct consequence for cyclic heat engines and constitutes the universal theoretical performance ceiling. Every real process has $\Delta S_{\mathrm{univ}}$ > 0 and efficiency less than Carnot.