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a common example for irreversible process given is the rapid expansion of a piston due to sudden reduction in external pressure(as against very slow reduction in a reversible case). In this case, it is stated that maximum useful work derived is less, because piston does less useful work (as illustrated via area under the curve in P-V work).But this is mathematical explanation more or less. What I am trying to understand is intuitively, why does this happen? why is work done less? where does the "non-useful " work end up as? assume it is a frictionless system. Describe at a microscopic level, keeping it less mathematical.

My guess: the molecules of gas within piston chamber end up doing compression-expansion work of some sort on themselves which doesn't show up as useful work. But I am not able to visualize this. also in this example has the entropy of system increased as well as surroundings? or only system? or both?

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  • $\begingroup$ Where is it stated that useful work is less? Can you post an image of the text, or provide a link? $\endgroup$
    – sammy gerbil
    May 8, 2017 at 12:43
  • $\begingroup$ Part of the potential to do work is lost as a result of viscous dissipation associated with the rapid deformation of the gas within the cylinder. The lost work is converted to internal energy of the gas (i.e., higher temperature). Think of a spring and damper in parallel. In a rapid deformation, some of the energy stored in the spring is lost by viscous damping, but in a slow deformation, most of the energy stored in the spring converts to useful work. A gas behaves in an analogous way. $\endgroup$
    – Chet Miller
    May 8, 2017 at 14:16
  • $\begingroup$ Sammy, it is irreversible process, so you don't derive maximum possible useful work out of it, that's what I meant $\endgroup$
    – daraj
    May 9, 2017 at 5:15
  • $\begingroup$ Thanks Chester, yes "viscous dissipation" is sort of what I was looking for to visualize, that helps a lot. So the rise in internal energy of the gas can lead to entropy increase of the system? In this case piston does work on surroundings as it pushes out on the system. Without doing any calculations can we say entropy of both system(gas within chamber) and surroundings(because of work done on it)increases ? consider both isolated as well as a closed system(which permits heat transfer) $\endgroup$
    – daraj
    May 9, 2017 at 5:20
  • $\begingroup$ in a closed but not isolated system this increase in internal energy of the system can lead to heat transfer out to the surroundings, so why is it still considered "non-useful"? Is only P-V work considered useful? $\endgroup$
    – daraj
    May 9, 2017 at 5:22

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If the piston moves outwards faster than any of the molecules in the cylinder are moving, clearly no work will be done on the piston by the gas, nor, in the case of an ideal gas, will there be any any energy change. If the piston is moving out fast, but not as fast as before, clearly there will be some work done by the gas on the piston, but less than if it were moving out at a speed many times less than the rms speed of the molecules (quasi-statically).

It's not, then, a case of non-useful work taking the place of useful work – simply of not as much (useful) work being done as in the quasi-static case.

As for entropy, consider the case of an ideal gas expanding adiabatically. The entropy of the gas (assuming constant $C_V$) is $$S= C_V \ ln T + nR\ ln V+S_0.$$ It's easy to show that if the expansion is quasi-static, so that the gas does the maximum possible amount of work, the fall in entropy due to the fall in temperature exactly compensates for the rise in entropy due to the rise in volume. But if the expansion is too quick (see first paragraph) not as much work is done, so the fall in entropy due to the temperature fall does not compensate for the rise due to rise in volume; that is the entropy goes up. This is only one of many ways of understanding what's going on.

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  • $\begingroup$ Philip, you talk about "fall in temperature" whereas Chester talks about rise in temperature due to viscous dissipation, so a bit confused. which effect dominates more? also, what about entropy of surroundings in this irreversible process? do they go up? you can assume 2 scenarios-closed(no mass transfer but energy transfer is there) and isolated(no type of transfer takes place) $\endgroup$
    – daraj
    May 9, 2017 at 5:26
  • $\begingroup$ I considered an ideal gas, to make life simple. When a gas expands quasi-statically, pushing a piston, it does work, so, according to the First Law of Thermod, its internal energy falls, if the conditions are adiabatic (no heat exchange with surroundings). For an ideal gas, for which $\Delta U=nC_v \Delta T$, the temperature must fall. If the expansion is very rapid, less work will be done and the temperature will fall less! Entropy of surroundings won't change in this adiabatic process. $\endgroup$
    – Philip Wood
    May 9, 2017 at 10:11
  • $\begingroup$ To leave no room for doubt as to what I'm saying: if a gas expands, without any heat flow into or out of it, its temperature falls. It falls most if it expands slowly, pushing a piston. If the expansion is very rapid, less work will be done by the gas, and its temperature will fall less. In the limit of the piston moving out at a speed many times greater than the rms speed of the molecules (or the gas expanding into a vacuum), there will be no fall in temperature for an ideal gas. $\endgroup$
    – Philip Wood
    May 18, 2017 at 14:18
  • $\begingroup$ If the gas is not ideal, there will still be a fall in temperature in this extreme case for this reason... Although there is no change in the gas's total internal energy, it gains potential energy due to the molecules moving further apart (on average) against attractive forces, and therefore it loses (random) kinetic energy. $\endgroup$
    – Philip Wood
    May 18, 2017 at 14:18
  • $\begingroup$ Thanks Philip, I guess I should imagine a real gas as molecules with more potential interactions as compared to an ideal gas. Mathematically, how is this difference in internal energy drop treated between the 2 types of gases?(I have avoided math so far , but just for the sake of closure) also, what about viscous dissipation leading to increase in temp. as Chester pointed out? under what circumstance does that happen? you have talked about only cooling vs less cooling so far for the 2 gases. But can there be heating due to viscous dissipation in real gases? $\endgroup$
    – daraj
    May 18, 2017 at 19:12

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