Review - Exam Prep question 10.3 - manometer question

[Zellmer, Robert]

I know it's late but imagine some of you may be up studying.  You should hit the
sack soon so you're not too tired for the exam tomorrow.

Near the end of the review, I went over Exam Prep question 10.3.  It's at the end
of the chapter and in Mastering.  This is the one dealing with a gas in a manometer
open to the atmosphere.  It started out with the gas having a pressure greater than
the atmospheric pressure (the mercury in the U-tube of the manometer was lower
in the arm connected to the flask with the gas than in the arm open to the atmosphere).
Then the gas was cooled and the pressure of the gas was lowered so it was less than
that of the atmosphere (the mercury in the U-tube of the manometer was higher in the
arm connected to the flask with the gas than in the arm open to the atmosphere).

The question asked for the height of the mercury in the arm open to the atmosphere
from the bottom of the U-tube.  I showed the answer and the explanation.  It gave a
hint: the sum of the heights of both arms must remain constant.  You also have to
relate the pressure of the gas and the atmosphere to the difference in height
between the two arms in the final state after the gas has cooled.  The sum of the
heights of the arms in before the gas was cooled was 240.2 mm Hg based on the
numbers given in the problem.  That meant the sum of the heights after it was
cooled also had to equal 240.2 mm Hg.  They came up (h +x) + x is the sum of
the heights of mercury in the two arms after the gas cooled.  There really wasn't
an explanation of where they got this.

I made a video explanation of the problem and I've put it on Carmen.  I put it in
the "Exam Reviews - Zoom Recordings" submodule in the Lecture module.
I've got pictures to go along with the explanation.  Hopefully, the helps to better
explain things.

Dr. Zellmer
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