What Is an Irreversible Process in Thermodynamics?

Chemistry · Thermodynamics · NEET

An irreversible process is any real change that happens at a normal, finite speed, so the system is NOT in equilibrium with its surroundings while it happens. NCERT says it plainly: any process that is not reversible is irreversible. You cannot undo it by just a tiny (infinitesimal) change, and it always does LESS work than a reversible process between the same two states. Memory hook: "Irreversible = In a hurry" — fast, one-shot, against a fixed external pressure, and it wastes energy.
Work: Reversible (max) vs Irreversible (less) expansionVolumePressureinitialfinalReversible pathsmooth curve, big area= MAXIMUM workIrreversible pathfixed p_ext, small areaw = -p_ext(V_f - V_i), LESS work
Both paths join the same start and end states, but the irreversible path pushes against a fixed low external pressure, so its area (work) is a small rectangle, while the reversible curve encloses a larger area and gives the maximum work.

Your doubts, answered

What exactly makes a process irreversible in simple words?

A process is irreversible when it happens at a real, finite speed, so the system and surroundings are NOT in balance (not in near-equilibrium) at every moment. In an irreversible gas expansion the gas pushes against a FIXED external pressure (p_ext) that is much lower than the gas pressure. Because the two pressures are very different, the change rushes forward and cannot be undone by just a tiny change. NCERT's definition is short: any process that is not reversible is irreversible. Almost every real-life process (a gas bursting into a bigger space, ice melting in a warm room) is irreversible.

How is an irreversible process different from a reversible one?

A reversible process is imaginary and infinitely slow. It moves through a long chain of equilibrium states, with the gas pressure and external pressure almost equal at every step, so a tiny change can reverse it. An irreversible process is real and finite-speed. The external pressure stays fixed and is clearly less than the gas pressure, so equilibrium does NOT hold during the change. Key result for NEET: for the SAME expansion, the reversible path gives the MAXIMUM work; the irreversible path gives less work.

Why does an irreversible expansion do less work than a reversible one?

Work in expansion is the area under the curve on a p-V graph. In an irreversible expansion the gas pushes only against a low, constant external pressure, so the area (work) is a small rectangle: w = -p_ext(V_f - V_i). In a reversible expansion the gas pushes against a pressure that stays almost equal to its own (much higher) pressure the whole time, so the area under the smooth curve is larger. More area = more work. That is why reversible work is the maximum possible.

Is free expansion of a gas into vacuum reversible or irreversible?

Free expansion (gas expanding into vacuum) is irreversible. Because the external pressure is zero, the work done is w = -p_ext x deltaV = 0. It is a sudden, one-way process that cannot undo itself by a tiny change. Note: even though work is zero and (for an ideal gas at constant T) internal energy does not change, the total entropy of the universe still INCREASES — that increase is the fingerprint of any irreversible process.

How do I know which formula to use for irreversible work in a NEET numerical?

If the question says 'against a constant external pressure' or gives you a single p_ext value, the process is IRREVERSIBLE, so use w = -p_ext(V_f - V_i). If it says 'reversible isothermal', use w = -nRT ln(V_f/V_i) = -2.303 nRT log(V_f/V_i). Reading the wording is half the marks. Watch units: 1 L bar = 100 J.

Does an irreversible process reach the same final state as a reversible one?

Yes, it can. Both paths can start and end at the same state, so state functions like deltaU, deltaH and deltaS(system) are the SAME for both (they depend only on start and end, not the path). But path functions like work (w) and heat (q) are DIFFERENT: the reversible path gives more work. Only the reversible path also keeps the surroundings in balance, so it is the only one where deltaS(universe) = 0.

⚠️ The NEET trap
For an isothermal irreversible expansion of an ideal gas, students pick deltaU not equal to 0, or think total entropy stays zero like a reversible process.
For ANY isothermal ideal-gas process deltaU = nCv deltaT = 0 (because deltaT = 0). Because the expansion is IRREVERSIBLE, the entropy of the universe increases, so deltaS(total) is NOT zero. Correct answer: deltaU = 0 and deltaS(total) not equal to 0.
🧠 Isothermal - deltaU is always 0. Irreversible - deltaS(universe) is always positive. Two separate facts; do not mix them.

Real NEET questions

2021

For the irreversible expansion of an ideal gas under isothermal conditions, the correct option is:

A · deltaU = 0, deltaS(total) not equal to 0
B · deltaU not equal to 0, deltaS(total) = 0
C · deltaU = 0, deltaS(total) = 0
D · deltaU not equal to 0, deltaS(total) not equal to 0
Solution: Isothermal means deltaT = 0, so for an ideal gas deltaU = nCv deltaT = 0. Because the expansion is irreversible, the process is not in equilibrium and the total entropy of the universe increases, so deltaS(total) = deltaS(system) + deltaS(surroundings) > 0, i.e. not equal to 0. Hence option A.
2019

Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is (1 L bar = 100 J):

A · -30 J
B · 5 kJ
C · 25 J
D · 30 J
Solution: 'Against a constant external pressure' signals an irreversible expansion, so w = -p_ext(V_f - V_i) = -2 bar x (0.25 - 0.10) L = -0.30 L bar = -0.30 x 100 = -30 J. The minus sign means the gas does work on the surroundings. Answer A.
2020

The correct option for free expansion of an ideal gas under adiabatic condition is:

A · q < 0, deltaT = 0 and w = 0
B · q > 0, deltaT > 0 and w > 0
C · q = 0, deltaT = 0 and w = 0
D · q = 0, deltaT < 0 and w > 0
Solution: Free expansion is irreversible expansion into vacuum, so p_ext = 0 and w = -p_ext deltaV = 0. Adiabatic means q = 0. First law: deltaU = q + w = 0, and for an ideal gas deltaU = nCv deltaT = 0 gives deltaT = 0. Hence q = 0, deltaT = 0, w = 0 (option C).

Solved Thermodynamics NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

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Frequently asked

Give one everyday example of an irreversible process.

Ice melting in a warm room, a gas rushing out into a larger empty space, or a hot cup of coffee cooling down. All of these happen on their own at a finite speed and cannot undo themselves by a tiny change, so they are irreversible.

Is every real process irreversible?

Yes. Every actual process in nature is irreversible because it happens at a finite rate and is never perfectly in equilibrium. The reversible process is an idealised limit that we use only as a benchmark for maximum work and for calculations.

Which does more work, reversible or irreversible expansion?

Reversible expansion does the maximum work between two given states. An irreversible expansion (against a fixed lower external pressure) always does less, because the area under its p-V path is smaller.

Why does NEET keep asking about irreversible processes?

Because they hide two easy traps: the correct work formula (w = -p_ext deltaV, not the reversible log formula) and the entropy fact (deltaS(universe) > 0). Knowing which formula the wording demands earns quick, sure marks.