Reversible and irreversible thermodynamic processes
Distinguish reversible and irreversible processes and explain why reversible paths set ideal performance limits.
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ContinueReturning a gas to its initial volume is not enough to undo a process. A reversible process can be reversed while restoring both the system and its surroundings, without a lasting net effect. This ideal limit makes thermodynamic paths calculable and sets performance bounds; a slow process can still be irreversible.
Build on heat and work as path transfers and isothermal expansion. Use Q > 0 and Wₒₙ > 0 into the system, with Δ U = Q + Wₒₙ. A fixed amount of ideal gas and negligible bulk energy changes are assumed in the gas examples.
Distinguish quasistatic motion from complete reversibility
A quasistatic process passes through states arbitrarily close to thermodynamic equilibrium, allowing state variables to describe the system along the path. Complete reversibility additionally excludes dissipative effects and finite driving differences: friction, mixing, free expansion and heat transfer across a finite temperature difference leave changes that an infinitesimal reversal does not undo.
A slowly moving piston with dry friction illustrates the distinction. Changing its motion reverses the friction force, but does not retrieve the mechanical energy dissipated as heat. Likewise, a body can warm slowly and remain nearly uniform inside while receiving heat from a much hotter reservoir. Internal near-equilibrium does not make the external heat transfer reversible.
For a mechanically quasistatic piston, a justified pressure at its boundary can be used in ∫ P dV. That alone does not certify thermal reversibility. For rapid nonequilibrium gas motion, there may be no single equilibrium gas pressure along the route; do not replace the actual work model with an imagined smooth state curve.
Use entropy to separate state change from generation
Entropy S is a state function. Between equilibrium endpoints, calculate its change along a suitable reversible reference path: Δ S = ∫ᵢ^f(δ Qᵣₑᵥ)/T. This reference can calculate the endpoint change even if the actual path is irreversible. It does not mean the actual heat transfer equals T dS everywhere. Total entropy generation in the system and surroundings is nonnegative; it is zero in the reversible limit.
For fixed-temperature reservoirs, heat Q received by one changes its entropy by Q/T. If heat of magnitude Q moves from a hot reservoir at Tₕ to a cold one at Tc < Tₕ, their combined entropy change is Δ Sₜₒₜₐₗ = Q(1/Tc-1/Tₕ) > 0. Heat can be conserved while entropy is generated. Making the transfer slower does not change this result for the same Q and reservoir temperatures. Approaching reversibility requires an infinitesimal thermal difference; that limit is not a guarantee of finite transfer rate at exactly equal temperatures.
Guided practice 1
Account for heat transfer that happens slowly
Problem
A connection insulated from its surroundings transfers 600 J from a reservoir at 500 K to one at 300 K, with no work or mass transfer. The connection returns to its initial thermodynamic state, and the transfer is made very slow. Find the entropy change of each reservoir and their combined change. Does slowing it make the process reversible?
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
The reservoirs stay at different fixed temperatures. Transfer rate and total transfer are different quantities.Show solution step by step
Assign the reservoir entropy changes
Method
Use signed heat into each reservoir divided by its temperature.Reason
The same energy transfer removes and adds different amounts of entropy.Working
Answer
Δ Sₕ = -600/500 = -1.20 J/K and Δ Sc = +600/300 = +2.00 J/K.
Judge the combined process
Method
Add the changes and inspect the source of irreversibility.Reason
A system’s negative entropy change is allowed; total entropy generation is the relevant test.Working
Answer
Δ Sₜₒₜₐₗ = -1.20 + 2.00 = +0.80 J/K. There is no external heat or mass transfer for the combined isolated account, so this is its generated entropy. The finite temperature difference makes it irreversible regardless of how slowly the stated 600 J is transferred.
Compare actual and reference paths for a gas
For an ideal gas with constant molar CV, a reversible reference path gives Δ S = nCV ln Tf/Tᵢ + nR ln Vf/Vᵢ. The ratios are dimensionless and temperatures are absolute. One way to obtain the expression is reversible constant-volume heating, with δ Qᵣₑᵥ = nCV dT, followed by reversible isothermal expansion, with δ Qᵣₑᵥ = nRT dV/V. The sum depends only on the endpoints.
In an insulated free expansion of ideal gas into vacuum, the equilibrium endpoint temperature is unchanged, but volume increases. Actual heat and work are zero while Δ S = nR ln(Vf/Vᵢ) > 0. Using actual Q/T = 0 as the entropy change would ignore generation. By contrast, a reversible isothermal expansion between those endpoints receives heat and transfers the corresponding entropy from its reservoir; total generation is zero.
A gas can also lose entropy during reversible compression while its reservoir gains the same amount. “Entropy never decreases” applies to the isolated combined account, not to every subsystem. Similarly, adiabatic means no heat transfer; only a reversible adiabatic process is isentropic in this closed-system model.
Comparing reversible and free expansion
Worked example 2
Same endpoints, different work
Problem
Show full solution
Identify the pressure that resists expansion
Method
Use internal equilibrium pressure for the reversible path and zero external pressure for free expansion.Reason
The reversible gas does work against an almost balancing load; expansion into vacuum has no load to receive work.Working
Answer
- Path A: reversible isothermal expansion W_(out,A) = nRT ln V₂/V₁ > 0
- Path B: free expansion into vacuum W_(out,B) = 0
Compare the outputs
Working
Answer
The reversible output is nRT ln(V₂/V₁) > 0, while free expansion gives zero work. The final ideal-gas temperatures can agree despite different work, so endpoints alone do not determine work.
Test a quasistatic heating process with a finite temperature difference
Check your understanding 1
A closed ideal-gas sample at fixed volume has constant total heat capacity C = 100 J/K. It is warmed slowly from 300 K to 400 K by a single 400 K reservoir, remaining internally near equilibrium. There is no work, mass transfer or heat exchange elsewhere, and connecting hardware returns to its initial thermodynamic state. Find the heat received by the gas, its entropy change, the reservoir’s entropy change and the total change. Is internal quasistatic behaviour sufficient for reversibility? How would the thermal surroundings need to change to approach reversible warming between the same gas endpoints?
Show answer
The gas receives Q = C(400-300) = 10000 J. Its endpoint entropy change, calculated with a reversible constant-volume reference, is Δ Sg = C ln(400/300) = 28.7682… J/K. The reservoir loses 10000/400 = 25.0 J/K, giving Δ Sₜₒₜₐₗ = 3.7682… J/K > 0.
The actual heat crosses a finite thermal difference for much of the warming. Slow heating can keep the gas internally near equilibrium but does not remove that external irreversibility. Using Q/400 as the gas entropy change would substitute the reservoir temperature for the varying gas temperature and miss the generation.
To approach reversible warming, use a succession of reservoirs, each infinitesimally warmer than the gas at that stage. A single fixed 400 K reservoir does not do this. The same constant-volume gas endpoints still require 10000 J of heat and give the same Δ Sg, but the reversible surroundings lose ∫δ Qᵣₑᵥ/T = Δ Sg of entropy, rather than 25 J/K. The combined generation then tends to zero. Check that your comparison separates the gas’s endpoint state change from the surroundings and actual transfer conditions.
Practice set
Check your understanding 2
Is a quasistatic process with dry friction reversible? Explain.
Show hint
Slow motion removes large departures from equilibrium, but does it remove dissipation?
Show answer
No. Friction dissipates mechanical energy. Reversing the motion does not undo that dissipation and restore both system and surroundings, so quasistatic motion alone is insufficient.
Check your understanding 3
Why is free expansion into vacuum irreversible?
Show hint
Consider how the gas could be returned to its original volume.
Show answer
The gas passes through nonequilibrium states and does not return by an infinitesimal reversal of the driving conditions. Compressing it back requires work and changes the surroundings; the original process cannot be retraced without a net effect.
Check your understanding 4
Why study reversible cycles when real cycles have irreversible effects?
Show hint
Think of a benchmark between the same reservoirs.
Show answer
They give an ideal efficiency ceiling and a calculable reference. Comparing a real cycle with that ceiling helps identify losses from irreversibility; it does not imply a real engine can attain the ideal limit.
What to keep
Reversibility requires restoring system and surroundings, not merely slow motion or matching endpoints. Use a reversible reference to calculate endpoint entropy changes, then inspect total generation for the actual path. Heat transfer across a finite temperature difference remains irreversible even when it is slow.
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Syllabus and review details
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