# Why does slow compression of a gas doesn't change its entropy to increase?

This is quoted from Daniel V Schroeder's Thermal Physics:

It's interesting to think about why the slow compression of a gas doesn't change its entropy to increase. One way to think about it is to imagine that the molecules in the gas inhabit various quantum-mechanical wavefunctions, each filling the entire box with discrete energy levels. When you compress the gas, each wavefunction gets squeezed, so the energies of all the levels increase, and each molecule's energy increases accordingly. But if the compression is sufficiently slow, molecules will not be kicked up into higher energy levels; a molecule that starts in the $n$th level remains in the $n$th level (although the energy of the level increases). Thus the number of ways of arranging the molecules among various energy levels will remain the same, that is, the muliplicity and entropy do not change. On the other hand, if the compression is violent enough to kick molecules up in the higher levels, then the number of possible arrangements will increase and so will the entropy.

I've some queries on the above explanation of Schroeder:

$\bullet$ Why does squeezing the wavefunction increase the energies of each energy-level?

$\bullet$ The molecule was in $n$th state prior to compression; how did the molecule remain in the same $n$th state after compression also even if work was done on the system/

$\bullet$ Why is the molecule not kicked up to higher energy level when the compression was slow? Why did the violent compression work otherwise?

1. Squeezing the wavefunction means confining it to a smaller space. It takes more energy to confine something within a small space than within a big one.

2, 3: These are consequences of the quantum adiabatic theorem: if you take a system in state $n$ of some system, and act on the system sufficiently slowly, it ends up still in state $n$ of the new system.

Roughly speaking, if the compression is very slow, the system and the particle remain "in equilibrium" at all times. The energy being fed into the system is thus fed into the particle in just the right way to hold it at a given energy level. Slightly more correctly, the decay-timescale of the particle is fast compared to the compression timescale, so it is "perpetually decayed".

EDIT: The above is not really an adequate picture for isolated systems; see Are stationary quantum states attractors?.

In the opposite limit, when the perturbation is instantaneous, the system doesn't have time to decay at all, and its functional form doesn't change (until after the perturbation). For the particle in a box, the energy levels are $$E_n = \frac{n^2 h^2}{2mL^2}.$$ Instantaneously shrinking $L$ holds $E$ constant, so $n$ must increase. Thus the molecule is 'kicked up'.

Here's a classical analogy: you have a bunch of atoms bouncing around in a box. If you suddenly shrink the box, the atoms don't have time to notice right away. They at first have the same kinetic energy as before, but are out of equilibrium. While settling to equilibrium they will release entropy.

If you shrink the box really slowly, the atoms are continuously fed kinetic energy from the walls and remain always in equilibrium. No entropy is released.

• For more information, Wikipedia does a reasonable job at this with a nice animation of an adiabatically changing harmonic oscillator: en.wikipedia.org/wiki/Adiabatic_theorem Nov 11, 2015 at 21:21
• @AGML I think you meant instantaneously enlarging $L$? Otherwise it doesn't make sense why $n$ must increase. Nov 12, 2015 at 1:12