# How does non-useful work manifest in this irreversible example?

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?

• Where is it stated that useful work is less? Can you post an image of the text, or provide a link? May 8, 2017 at 12:43
• 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. May 8, 2017 at 14:16
• Sammy, it is irreversible process, so you don't derive maximum possible useful work out of it, that's what I meant May 9, 2017 at 5:15
• 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) May 9, 2017 at 5:20
• 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? May 9, 2017 at 5:22

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.
• 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. May 9, 2017 at 10:11