# When is an adiabatic process reversible?

Throughout my physics education this far, I have always understood that adiabatic expansions and compressions involving ideal gases are reversible and follow $pV^\gamma =$ constant, provided the process is quick enough.

However, this year my thermodynamics notes contain the following:

"In rapid adiabatic expansion, the gas ends up moving fast as a fluid – i.e. some U is converted not to work on the piston, but KE of the bulk fluid. When the expansion stops, unless you are very ingenious (e.g. in a gas turbine), this KE converts back to U as the fluid slops around in the container. If you then compress the gas slowly back to its original volume it will be warmer than it was at the start – ie the 1st expansion was irreversible"

Can anybody help me get my head around this? Isn't the internal energy of an ideal gas already defined to be the kinetic energy? If energy is conserved in the process, how can we end up with a hotter gas in the end?

• Note that by definition, an adiabatic process is one for which $Q = 0$. The statement that a gas expansion is adiabatic if it is "quick enough" is usually a way of saying that the heat flow from the environment into the gas is negligible. However, a reversible process has to happen slow enough that the system maintains internal equilibrium throughout. In other words, just saying that an expansion is "quick" probably means that it's approximately adiabatic, but if it's "too quick", then it's also irreversible. – Michael Seifert Mar 10 '18 at 20:45

## 2 Answers

An adiabatic (zero heat exchanged with the surroundings) process is reversible if the process is slow enough that the system remains in equilibrium throughout the process.

Mathematically, the characteristic relaxation time has to be smaller than some characteristic time of action of the applied process: $\tau_{relax} << \tau_{process}$.
The system then equilibriates faster than the process is able to change it.

This $\tau_{process}$ can be linked with the speed $v$ of a changing system size $\Delta x$ (e.g. a moving piston) as $\tau_{process} \approx \frac{\Delta x}{v}$.

In your case this is violated due to the rapid expansion, thus rendering the process irreversible.

"Isn't the internal energy of an ideal gas already defined to be the kinetic energy?"

Correct.

"If energy is conserved in the process, how can we end up with a hotter gas in the end?"

Because more work was done on the gas by pushing the piston slowly back in than was done by the gas pushing the piston out fast.

I look at it like this… First consider the extreme case: a gas escapes into a vacuum ("Joule expansion") The cylinder containing the gas and the expansion space that was originally evacuated are both thermally insulated. So we have adiabatic expansion when no work at all is done! Now consider a less extreme case when a piston moves out very fast, at a speed that isn't negligible compared with the rms speed of the molecules. This means that the gas pressure on the piston is less than that in the bulk of the gas, and less work is done than in a slow expansion between the same initial and final volumes.