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What is the difference between reversible and irreversible adiabatic expansion?

Is it true that the work done by the gas is the same but the pressure applied externally differ between two process?

If $P_{ext}$ = work done by gas, then how will there be an expansion, if they are pushing against each other at the same force?

Roger V.
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PhysC
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2 Answers2

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The difference is that one expansion is quasi-static (the reversible one) while the other is spontaneous because of a dramatic change of the external constraints (the irreversible one).

  • In the quasi-static case, you start off indeed in the state where gas pressure equates external pressure. An external operator then slightly decreases the outside pressure so that the gas expands a little bit before reaching very fast a new equilibrium state. You then repeat this procedure as many times as necessary to reach the sought pressure.

  • If instead you decrease instantaneously and dramatically the outside pressure (you decrease it by, say, a factor 3), then the gas will expand until it reaches equilibrium but in a fashion very very different from that of sum of minutes changes that would be done quasi-statically. As an example, for a start, quantities like internal pressure, temperature etc.. are not even well defined during an irreversible expansion.

At the end of the day, these two very different thermodynamic "trajectories" for your system will result in leading to two different final states (in your adiabatic case). And that's the reason why, the entropy change is not the same in the two cases.

gatsu
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  • I heard somewhere that the entropy change for adiabatic reversible reaction is constant but couldn't fathom why? – Abhinav May 18 '18 at 10:38
  • For a simple compressible system, the entropy change during a quasi-static (or, equivalently, reversible) adiabatic process (be it compression or expansion) is zero. – gatsu May 19 '18 at 12:36
  • @gatsu Suppose that a system undergoes adiabatic reversible and irreversible expansion. When undergoes adiabatic irreversible expansions the change in internal energy is: $$\Delta U = -P_{ex} \cdot \Delta V$$ For adiabatic reversible expansion the change in internal energy would be: $$\Delta U = -nR\ln\left(\frac{V_f}{V_i}\right)$$ which is bigger than the work for irreversible. But where that extra work goes? I mean what if instead of pressure above the gas we had an object with a variable mass? The object in boths cases would have the same change in potential energy. Thanks in advance. – Antonios Sarikas Mar 15 '20 at 19:05
  • @adosar I would say that the temperature of the gas is not the same at the end of the process and neither is the final volume. Since the only thing that matters is reaching mechanical equilibrium, depending on how fast the temperature cools down as expansion occurs leads to different final volumes. So in the case you mention the mass won't have the same altitude at the end depending on whether this is reversible or irreversible. – gatsu Mar 26 '20 at 15:20
  • @gatsu "An external operator then slightly decreases outside pressure" - Doesn't this mean that some work has been done on surroundings, by from somewhere? Won't it make ot irreversible since it cannot returns to the same state? – Aurelius Oct 03 '23 at 10:55
  • @Aurelius You are absolutely right. The reasoning I gave serves only the purpose of determining the contribution to the entropy production of the universe stemming from the transformation (expansion) undergone by the gas itself. If the mechanisms by which the nobs controlling the expansion necessitate some entropy production, then it will contribute to the overall entropy increase of the universe. – gatsu Oct 06 '23 at 09:44
  • @Aurelius For example, when people talk of the reversibility of a thermodynamic process for a thermal engine cycle, they don't mention the fact that the burning of the fuel to achieve the heating of the gas will obviously be irreversible. But that is inconsequential with regards to what has been 'experienced' so to speak by the gas undergoing the cycle and conveying heat and work between different heat reservoirs. – gatsu Oct 06 '23 at 09:48
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There are 3 questions in your post:

1) What is the difference between reversible and irreversible adiabatic expansion?

Answer: Reversible means actually the time-reversible. When system pressure is applied against the external surrounding (resistant), the external pressure reacts back at the next instant. The system pressure is always slightly higher than the external pressure. The other aspect of reversible is the external pressure gradually rises; whereas in the irreversible condition, the external pressure is constant.

2) Is it true that the work done by the gas is the same but the pressure applied externally differ between two process?

Answer: No. Workdone in reversible is always more than the workdone in irreversible. The reason is the system has to work against the gradual higher external pressure in reversible condition. Whereas the external pressure is constant, no rising in irreversible condition, hence work is less. The average external pressure in reversible is in-between P-initial and P-final whereas the external pressure is P-final in irreversible process.

3) If Pext = work done by gas, then how will there be an expansion if they are pushing against each other at the same force?

Answer: The pressure of gas during expansion is not equal but near-extra, a slightly higher. The pressure energy (mechanical/potential) is very sensitive compare to temperature energy (thermal). The work is performed with slight pressure difference. It's a quasi-equilibrium phenomena.

Soumen Maiti
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