[FRIAM] Entropy Redux

[Pieter Steenekamp]

I'm not sure if you deliberately left out any mention of the initial
pressures — perhaps as part of the “catch,” which is common in trick
questions like this. As I mentioned in my initial answer, it’s common to
assume the initial pressures are equal, in which case the slider would move
to the tight.
However, if the twist is that the right-hand cylinder was initially filled
with high-pressure air, then of course the slider would move to the left.

On Wed, 28 May 2025 at 07:14, Nicholas Thompson <thompnickson2 at gmail.com>
wrote:

> Thanks everybody.   Anybody else?   Remember, please no peeking.  Your
> efforts have already helped me identify an ambiguity in how I stated the
> problem.  Here is an edited version.  See if that clarification changes any
> answers.
> Remember, no peeking before you kick in your answers.  No herding
> allowed.
>
> Side by side, butt-to-butt, two cylinders of air, of equal volume and
> equal temperature, but different histories. They are separated by a
> frictionless sliding divider which is now pegged.  Before compression or
> heating, both started with air at 10 degrees C, and after compression or
> heating, both arrive at 20 degrees C. The left one has arrived at its
> present temperature and volume by being compressed, the right-hand one
> by having its volume set to the same value as the left, BEFORE BEING FILLED
> WITH AIR AT 20 DEGREES C, and ONLY  then having contents  heated.  All
> compartments insulated, all manipulations quasi-static.  Now, remove the
> peg, allowing the partition between the two parts to slide, one way or the
> other.  Which way would it slide and why. What would your explanation
> be?  Please don't read others' answers until you have submitted your own.
> This is enormously helpful to me.
>
> On Tue, May 27, 2025 at 10:19 PM Pieter Steenekamp <
> pieters at randcontrols.co.za> wrote:
>
>> Assuming the pressures were equal at the start, it will slide from the
>> left (having been compressed) to the right.
>> From the ideal gas law PV = nRT, both have equal nRT;s so the one with
>> smaller V must have higher P and it would then push to the right.
>>
>> On Wed, 28 May 2025 at 04:59, Nicholas Thompson <thompnickson2 at gmail.com>
>> wrote:
>>
>>> Side by side, butt-to-butt, two cylinders of air, of equal volume and
>>> equal temperature, but different histories. They are separated by a
>>> frictionless sliding divider which is now pegged.  Before compression
>>> or heating, both started with air at 10 degrees C, and after compression or
>>> heating, both arrive at 20 degrees C. The left one has arrived at its
>>> present temperature and volume by being compressed, the right-hand one
>>> by having its volume set to the same value as the left and then having
>>> contents  heated.  All compartments insulated, all manipulations
>>> quasi-static.  Now, remove the peg, allowing the partition between the two
>>> parts to slide, one way or the other.  Which way would it slide and
>>> why. What would your explanation be?  Please don't read others' answers
>>> until you have submitted your own.
>>> --
>>> Nicholas S. Thompson
>>> Emeritus Professor of Psychology and Ethology
>>> Clark University
>>> nthompson at clarku.edu
>>> https://wordpress.clarku.edu/nthompson
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>
>
> --
> Nicholas S. Thompson
> Emeritus Professor of Psychology and Ethology
> Clark University
> nthompson at clarku.edu
> https://wordpress.clarku.edu/nthompson
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> ... --- -- . / .- .-. . / ..- ... . ..-. ..- .-..
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