| 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 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. 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.
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.
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.
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