Re: [Phys-l] T dS versus dQ

["LaMontagne, Bob" <RLAMONT@providence.edu>]

If you would take the time to read what I said - I assumed none of the items you mention. I am simply assuming that insulated implies no energy of any kind leaves the inside of the cylinder.

Adiabatic and ideal gas gives specific relationships between P, V, and T that you can look up in any general physics text. 

My conclusion was simply that the final temperature after the compression would be the same regardless of how fast it was done and that in the end the sound and shock waves didn't matter.. How do you get isothermal out of that?

Bob at PC

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of John Mallinckrodt [ajm@csupomona.edu]
Sent: Tuesday, January 12, 2010 3:11 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ


To which Bob LaMontagne wrote:

> Since you specified a thermally insulated cylinder and piston, it
> doesn't really seem to matter if you do the compression quickly or
> slowly - the final temperature will be the same.  Since a
> reversible adiabat will get you from the same initial state and to
> the same final state, it appears dS = 0 as well as dq = 0.

Unless, I'm misunderstanding this, I think Bob might be laboring
under some combination of the following points of confusion:

1. Thermal insulation insures isothermality
2. All adiabatic processes are reversible
3. Adiabatic processes are isothermal processes.

None of the above are true.

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Bob at PC: And none of the above were assumed by me.