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 11) 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?
Which of the following is an example of a reversible process?
All natural processes are:
A quasi-static process carried out with friction between the piston and cylinder walls is:
For a given expansion of an ideal gas from volume V₁ to volume V₂, the work done by the gas is:
During a reversible process, the entropy change of the universe is:
Which of the following statements about irreversible processes is INCORRECT?
An ideal gas is compressed from volume V to V/2. In which case does the surroundings do minimum work on the gas?
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 11).
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.) ---
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, p. 11Relations holding during reversible adiabatic process. gamma = Cp/Cv.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | pressure | Pa |
| V | volume | m^3 |
| T | temperature | K |
| gamma | adiabatic index | - |
Change in internal energy = heat ADDED minus work DONE BY the system. Energy conservation including thermal energy.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Delta_U | change in internal energy | J |
| Q | heat added to system | J |
| W | work done BY system | J |
Work done by ideal gas during isothermal expansion. Q = W (since Delta_U = 0). Reverse for compression.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | moles | mol |
| R | gas constant 8.314 | J/mol/K |
| T | temperature | K |
| V_i, V_f | initial/final volume | m^3 |
For ideal gas: difference of molar specific heats equals gas constant R. Useful for converting between Cp and Cv.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Cp | molar specific heat at const P | J/mol/K |
| Cv | molar specific heat at const V | J/mol/K |
| R | 8.314 | J/mol/K |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Graph Interpretation
Student misidentifies which P-V curve is adiabatic (steeper) vs isothermal.
P-V graph showing one or more processes; question asks for process type.
Adiabatic curve is STEEPER than isothermal at the same point (slope ratio = γ). Adiabat: PV^γ; isotherm: PV = const.
Root cause: graph misread
Adiabatic curve is STEEPER than isothermal at the same P-V point. Slope ratio at same point: adiabatic / isothermal = γ. Memorise: 'adiabatic angles down sharper'.
forgets net zero cycle
Computes only one segment's W
swaps adiabatic isothermal curves
Confuses which curve is steeper
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