In article ,
says...

Sure, there's less potential energy in the
final, stored gas of that two-tank system, than
in the original, one-tank configuration.
Don's point: energy does *not* appear as
heat "in the regulator" during the gas transfer.

- He's correct.

The gas regulator does not get hot.

Well, he's correct in the observation that the regulator
does not get hot, not in the conclusion he draws from that
fact--that there is thus no loss in the regulator.




It's a tolerably complicated subject; a proper
treatment has all kinds of equations and funny symbols
littering the discussion. Some here have complained
that putting stuff like that in a post is needlessly
complicated. I agree. But to *really* understand what
goes on when somthing as blastedly simple as 'one
tank of compressed air gets hooked up to an empty
tank' the rigorous treatment is the ONLY way to get
the correct answer.


Yes, without the thermo the best you can do is illustrate a
point with a narrowly defined ideal case. That's what I was
trying to do with the reservoir/regulator/piston example,
which if treated isothermally can be solved with Boyle's
Law. But a thorough analysis and complete understanding
requires thermo with all its ugly abstract concepts.

Ned Simmons