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Re: [Phys-l] T dS versus dQ

The "moment" here is when the sound waves have dissipated and the molecular motions of the gas are back to random, i.e., a Boltzman distribution. The work done is still mgh. I assume h is the same whether the process was done fast or slow - the waves have lost their identity.

The difference between Bob Sciamanda's example and the others appears to be the knowledge of the energy input - mgh. After equilibrium is attained shouldn't the states be the same, fast or slow? Fast here being a case where one starts with a piston of mass m1 held up by the pressure of the gas amd then we add a mass m2 all at once to compress the gas, slow being a gradual adding of small masses until a mass m2 has been added, m being m1 + m2 in both cases.

The example with the enclosed speaker requires speed dependent energy input because of the radiation reaction - making fast and slow have different end states - and helping clear a big mental block I am having on this. In Bob's scenario the input energy is the same, unless I am missing something about the final equilibrium value of h.

Bob at PC

From: [] On Behalf Of John Denker []
Sent: Sunday, January 17, 2010 12:20 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] T dS versus dQ

On 01/17/2010 09:44 AM, LaMontagne, Bob wrote:
... follows the same P-V adiabatic curve
during the oscillation as it does in the slow adiabat (neglecting for
the moment the sound wave, etc., points that have been raised on this

In other words, fast is the same as slow (if we neglect for
a moment the difference between fast and slow).

This «moment» has gone on far too long.

"Other than that, Mrs. Lincoln, how did you enjoy the play?"

There are some things that ought not to be neglected.

Forum for Physics Educators