[FRIAM] Entropy Redux

Wed May 28 15:59:11 EDT 2025

<LOL/>

From: Friam <friam-bounces at redfish.com> On Behalf Of Santafe
Sent: Wednesday, May 28, 2025 12:53 PM
To: The Friday Morning Applied Complexity Coffee Group <friam at redfish.com>
Subject: Re: [FRIAM] Entropy Redux

No wonder it is, de facto, a trick question, at the same time Nick maintains that he does not intend it to be, or to believe it is, a trick question.  

On May 29, 2025, at 3:08, Nicholas Thompson <thompnickson2 at gmail.com <mailto:thompnickson2 at gmail.com> > wrote:

My reasons for doing this is that the distinction between an air temperature that is arrived at by compression and the same air temperature that is arrived at through heating is crucial to meteorology, about which I am trying to write. 

Temperature is a state variable.  That is inherent to its definition, so if the word is to mean anything, its meaning is to include that.  If meteorology is a discipline within physical science, that is true in meteorology as well as anywhere else.  

“State variable” means that its value tells you all it has to tell you about its role.  How it was arrived at historically is exactly _not_ distinguishing of anything.

Eric 

On Wed, May 28, 2025 at 8:38 AM Alexander Rasmus <alex.m.rasmus at gmail.com <mailto:alex.m.rasmus at gmail.com> > wrote:

Pieter got all of this, but I’ll pile on. I think Nick is intending us to think that the initial fill is the same on both sides, but he definitely hasn’t said this. The answer to the problem as stated is that the barrier will move left, right, or not at all. I think Nick is intending that the fill be the same, and that the answer to be that the barrier move to the right. The pressure in the right cell is lower as the density is lower (unless the EOS is real weird).

What’s air Nick? You also messed up your corrected version and set the second initial fill temp to be the post heating temperature 

Sent from my iPhone

On May 28, 2025, at 3:02 AM, Pieter Steenekamp <pieters at randcontrols.co.za <mailto:pieters at randcontrols.co.za> > wrote:

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 <mailto: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 <mailto: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 <mailto: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

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-- 

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