A reversible process is an idealised thermodynamic process that can be retraced exactly in reverse, restoring both the system and its surroundings to their original states with no net change anywhere in the universe. NCERT Class 11 Physics Chapter 11 (page 236) defines it as a process carried out infinitely slowly through a continuous sequence of equilibrium states — a "quasi-static" process with no dissipative effects.
Why NEET cares: Questions on this topic test whether you can distinguish reversible from irreversible processes, identify which real-world processes fall into which category, and recall the thermodynamic consequences of irreversibility. The concept appears at medium weight within the thermodynamics chapter, typically as recall or conceptual-application MCQs worth 4 marks.
The core distinction:
A reversible process requires: (1) the system passes through continuous equilibrium states (quasi-static), (2) no friction, viscosity, or other dissipative forces, (3) no finite temperature difference between system and surroundings, (4) the process can be exactly reversed, leaving zero net change in the universe.
An irreversible process violates one or more of these conditions. All natural (spontaneous) processes are irreversible — free expansion of a gas, heat flow from hot to cold, mixing of gases, combustion, friction. The system may be restored to its initial state, but the surroundings cannot be simultaneously restored.
Key consequences: In a reversible process, work done is maximum for expansion and minimum for compression. Irreversible processes always produce less useful work (expansion) or require more work (compression) than their reversible counterparts. Entropy of the universe increases in every irreversible process and remains unchanged in a reversible one.
Watch out: "Quasi-static" is necessary but not sufficient for reversibility — a quasi-static process with friction is still irreversible. This distinction is a common source of wrong answers.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following is a necessary condition for a thermodynamic process to be reversible?
Answer: C. A reversible process requires the system to pass through continuous equilibrium states (quasi-static) with no dissipative forces such as friction or viscosity (NCERT Class 11 Physics, Chapter 11, page 236).
Why A is wrong: A is wrong because reversible processes require infinitely slow (quasi-static) execution, not rapid changes — rapid processes drive the system away from equilibrium, making them irreversible.
Why B is wrong: B is wrong because heat exchange is not a necessary condition for reversibility. A reversible adiabatic process involves zero heat exchange yet is still reversible.
Why D is wrong: D is wrong because reversibility is not restricted to constant-volume processes. Reversible processes can occur at constant pressure, constant temperature, or with all variables changing, provided the quasi-static and no-dissipation conditions are met.
Which of the following is an example of a reversible process?
Answer: B. An infinitely slow (quasi-static) isothermal compression with a frictionless piston satisfies both conditions for reversibility: continuous equilibrium states and no dissipative effects (NCERT Class 11 Physics, Chapter 11, page 236).
Why A is wrong: A is wrong because free expansion is a rapid, uncontrolled process where the gas rushes into vacuum — the system passes through non-equilibrium states and cannot be reversed without external work.
Why C is wrong: C is wrong because spontaneous heat transfer across a finite temperature difference is irreversible. To reverse it, external work would be needed, changing the surroundings permanently.
Why D is wrong: D is wrong because mixing of gases is a spontaneous, entropy-increasing process. The gases cannot unmix on their own — the process is inherently irreversible.
Two processes take the same ideal gas between the same initial state (P1, V1) and the same final state (P2, V2): Process X is reversible, Process Y is irreversible. Since entropy is a state function, the system's entropy change, delta-S_system, is identical for X and Y. Which statement correctly compares the entropy change of the surroundings, delta-S_surroundings, for the two processes?
Answer: A. Entropy is a state function, so delta-S_system is fixed by the endpoints alone and is the same for X and Y. The second law requires the TOTAL entropy change of the universe to be exactly zero for the reversible process X and strictly positive for the irreversible process Y. Since delta-S_system is fixed and identical for both, delta-S_surroundings(Y) must exceed delta-S_surroundings(X) — i.e. delta-S_surroundings is more positive (or less negative) for Y than for X — so that delta-S_system + delta-S_surroundings equals zero for X but is strictly greater than zero for Y (NCERT Class 11 Physics, Chapter 11, pages 236-237).
Why B is wrong: B wrongly extends the state-function property of entropy to the surroundings: delta-S_system is path-independent because it depends only on the fixed initial and final states of the SYSTEM, but delta-S_surroundings is exchanged via heat transfer, which is path-dependent — this difference is exactly why total entropy change differs between reversible and irreversible processes between the same two system states.
Why C is wrong: C gets the direction of delta-S_surroundings right but the total wrong: the second law requires the total entropy change to be exactly zero only for the reversible process X and strictly positive for the irreversible process Y, not zero for both — claiming the total stays zero for Y directly contradicts the definition of irreversibility.
Why D is wrong: D overclaims: delta-S_system needs no path information (fixed by the endpoints), and the second law lets us conclude how delta-S_surroundings must compare between the two NAMED processes (reversible vs irreversible) without needing their exact paths, since reversibility alone fixes whether the total is zero or positive.
A quasi-static process carried out with friction between the piston and cylinder walls is:
Answer: A. Quasi-static is a necessary but not sufficient condition for reversibility. The presence of friction (a dissipative force) makes the process irreversible even though it proceeds through equilibrium states (NCERT Class 11 Physics, Chapter 11, page 236).
Why B is wrong: B is wrong because being quasi-static alone does not guarantee reversibility. A process must ALSO be free from all dissipative effects (friction, viscosity, electrical resistance) to qualify as reversible.
Why C is wrong: C is wrong for the same reason as A — equilibrium states are maintained in a quasi-static process, but friction dissipates energy as heat that cannot be recovered, making the process irreversible.
Why D is wrong: D is wrong because every thermodynamic process is classified as either reversible or irreversible. A process with friction has a definite classification: irreversible.
For a given expansion of an ideal gas from volume V₁ to volume V₂, the work done by the gas is:
Answer: B. In a reversible expansion, the external pressure is infinitesimally less than the gas pressure at every instant, so the gas does maximum work. Any irreversible expansion involves a larger pressure drop, and the gas does less work against a lower external pressure (NCERT Class 11 Physics, Chapter 11, page 236).
Why A is wrong: A is wrong because irreversible expansion involves the gas pushing against a lower external pressure (often a sudden drop), yielding less work than the continuous, maximum-pressure path of a reversible process.
Why C is wrong: C is wrong because work is path-dependent in thermodynamics. The reversible path extracts maximum work; any irreversible path between the same states extracts less.
Why D is wrong: D is wrong because a reversible expansion does positive work — in fact, it produces the maximum possible work for a given volume change. Zero work occurs only in free expansion (irreversible, against zero external pressure).
During a reversible process, the entropy change of the universe is:
Answer: D. In a reversible process, the entropy gained by the system equals the entropy lost by the surroundings (or vice versa), so the total entropy change of the universe is zero. This is a defining thermodynamic characteristic of reversibility (NCERT Class 11 Physics, Chapter 11, page 236).
Why A is wrong: A is wrong because a positive entropy change of the universe is the hallmark of an irreversible process. If ΔS_universe > 0, the process cannot be reversed to restore both system and surroundings.
Why B is wrong: B is wrong because the second law of thermodynamics prohibits the total entropy of the universe from decreasing. A negative ΔS_universe would violate the second law.
Why C is wrong: C is wrong because entropy change is well-defined for any process between equilibrium states. For a reversible process connecting two equilibrium states, ΔS_universe = 0 exactly.
Which of the following statements about irreversible processes is INCORRECT?
Answer: C. By definition, an irreversible process cannot be exactly reversed to restore both the system and its surroundings to their original states simultaneously. Option C describes a reversible process, not an irreversible one (NCERT Class 11 Physics, Chapter 11, page 236).
Why A is wrong: A is a correct statement about irreversible processes — they proceed through non-equilibrium intermediate states, unlike quasi-static reversible processes.
Why B is wrong: B is a correct statement — the second law requires ΔS_universe > 0 for all irreversible processes.
Why D is wrong: D is a correct statement — free expansion is the textbook example of irreversibility: the gas expands against zero external pressure, doing no work, and the process cannot be undone spontaneously.
An ideal gas is compressed from volume V to V/2. In which case does the surroundings do minimum work on the gas?
Answer: C. For compression, the reversible path requires the minimum work input from surroundings. In a reversible isothermal compression, the external pressure is only infinitesimally greater than the gas pressure at each step, minimising the total work done on the gas. Irreversible compression with a sudden pressure increase requires more work (NCERT Class 11 Physics, Chapter 11, page 236).
Why A is wrong: A is wrong because work is path-dependent. The reversible path minimises work input for compression, while irreversible paths require more work for the same final state.
Why B is wrong: B is wrong because sudden (irreversible) compression applies a constant high external pressure throughout, compressing the gas along a non-equilibrium path. This always requires MORE work than the gradual, quasi-static reversible path.
Why D is wrong: D is wrong because adiabatic compression (no heat removal) results in a temperature rise, and the gas pressure increases more steeply during compression. This requires more work than isothermal compression (where heat is continuously removed, keeping temperature and pressure lower).
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Given
An ideal gas is confined in a cylinder fitted with a frictionless piston. The gas is expanded from volume V₁ to volume V₂ = 2V₁ by two different methods:
• Process A: The external pressure is reduced infinitesimally at each step (quasi-static, no dissipation).• Process B: The external pressure is suddenly dropped to a constant value equal to the final equilibrium pressure, and the gas expands rapidly.
Both processes start and end at the same equilibrium states.
Required
Determine which process is reversible and which is irreversible. Explain which process produces more work.
Concept
A reversible process requires the system to pass through continuous equilibrium states with no dissipative effects. Work done by the gas in expansion is the area under the P-V curve. A reversible path traces the maximum-area curve; an irreversible path encloses less area (NCERT Class 11 Physics, Chapter 11, page 236).
Formula
Work done by gas: W = ∫P_ext dV. For a reversible process, P_ext = P_gas at every instant (the integral follows the equilibrium P-V curve). For an irreversible process, P_ext < P_gas and is often constant.
Substitution
• Process A (reversible): W_rev = ∫(from V₁ to 2V₁) P_gas dV — the full area under the equilibrium curve.• Process B (irreversible): W_irr = P_final × (2V₁ − V₁) = P_final × V₁ — a rectangle of smaller area.
Calculation
Since the equilibrium P-V curve for an ideal gas is a smooth, decreasing function (P = nRT/V for isothermal, or steeper for adiabatic), the area under the curve (W_rev) is always greater than the rectangle P_final × ΔV (W_irr).
Therefore: W_rev > W_irr.
Final answer
• Process A is reversible (quasi-static, frictionless, continuous equilibrium).• Process B is irreversible (sudden pressure drop, non-equilibrium intermediate states).• The reversible expansion produces more work than the irreversible expansion between the same initial and final states.
Common trap
Confusing "quasi-static" with "reversible." A process can be quasi-static yet irreversible if dissipative forces (friction) are present. Always check both conditions: quasi-static AND no dissipation.
Similar NEET-style question
"An ideal gas expands from state (P₁, V₁) to state (P₂, V₂) via a reversible process and also via a free expansion. Compare the work done in each case." (Answer: reversible expansion does positive work; free expansion does zero work since P_ext = 0.)
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A reversible process passes through a continuous sequence of equilibrium states; can be reversed by infinitesimal change. Real processes are irreversible (involve friction, finite temp gradients, non-equilibrium expansion).
-- NCERT Class 11 Physics, Ch. 11, p. 236More in Thermodynamics: 2 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
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