Re: [Phys-l] T dS versus dQ

[Brian Whatcott <betwys1@sbcglobal.net>]

Bob Sciamanda wrote:
> ...when you just drop the weighted piston, it will fall and oscillate
> momentarily.  But when it has finally comes to rest,  where has all of the 
> mgh energy gone?  Under the stated constraints it can only have gone to the gas.
>
>   
> /snip/
> In the reversible case, you slowly lower the weight by pulling up with the variable force F so that at every instant the net external downward pressure (mg-F)/A only infinitesimally exceeds the gas pressure P.  The gas pressure P slowly rises in value and you continuously reduce the strength of F until finally F=0 and mg/A=P.  The total work done on the gas is then the area under the (mg-F) curve.  This will be less than mgh.
>
> Bob Sciamanda
It helps me to examine the 'dropped' piston case with the slowly lowered 
piston case in a somewhat realistic way. If Bob holds the piston back 
from falling  fast, he applies an up force to balance most of the 
weight, but he allows the piston to drop.

This amounts to a f. -dx work increment - the piston is returning energy 
to Bob's hand which could be set towards some other potential energy 
reservoir.
The quickly dropped piston oscillates to a higher rest position 
supported on the warmer gas - which would let it down as the warm gas cools
back to ambient.

In my always concrete modeling conceptualizations, I want to put a 
number on these woolly concepts of 'insulated','isolated' and the other 
apparatus of the isothermal/adiabatic/isentropic/enthalpy conserving
models.

Half filling a test cylinder with warmed gas, and monitoring its cooling 
rate
can easily serve me as an indicator, without which any discussion seems 
to some extent unsatisfactory.

Brian W