Physics of a chemistry experiment

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A chemistry experiment demonstrates how to find the fraction of air that is oxygen by using rusting steel wool in a graduated cylinder. As the steel wool rusts, oxygen is consumed, causing the water level to rise. The discussion revolves around understanding the pressure relationships in the setup, specifically the pressures exerted by the trapped gas, atmospheric pressure, and the water column. Clarifications were made regarding terminology, particularly the definition of "reservoir" and its relevance to the pressure calculations. The educators plan to revisit the experimental results to align them with the correct physics principles.
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There is a simple chemistry experiment for finding the fraction of air that is oxygen. You push some steel wool to the bottom of a graduated cylinder. Partly fill the cylinder with water, and invert the tube into a water reservoir such that the steel wool is now above a pocket of air that is above a raised column of water. (Here is an online account with a http://www.practicalchemistry.org/experiments/intermediate/oxidation-and-reduction/how-much-air-is-used-during-rusting,208,EX.html" ). As the steel wool rusts, oxygen is depleted and the water level rises.

The chem class I teach is wrestling through problems with this and variations on the experiment. We are having trouble with one student question: what is the relationship involving the pressure exerted down by the gas trapped in the tube (Pi), the atmospheric pressure acting on the water in the reservoir,(PA) and the \rhogh pressure of the water column in the cylinder? Is it simply that the three sum to zero?
 
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For the sake of the experiment you can safely assume \rhogh=0 because the height of the water column represents a tiny fraction of atmospheric pressure. But there is no reason why you can't calculate what it actually is if you want to. Preservoir + Pwatercolum = Patmospheric
 
russ_watters said:
Preservoir + Pwatercolum = Patmospheric

I'm not sure of your meaning, so let me ask to clarify. I used "reservoir" to mean the water in the outer container, whose surface is exposed to the atmosphere. Is that what you mean? If so, why isn't the pressure of the gas trapped in the tube in the calculation?
 
That was a poor choice of word - you are correct about what it should be.
 
Thanks, Russ. That is very helpful. What you tell is (satisfyingly) what I expected the relationship to be. I was, though, bugged that the surface area of that outer reservoir of water didn't come into play. (I now see my mistake). When the experimental result the student got was inconsistent with my physics, I assumed my physics were wrong.

We'll go back and look at the experimental results again.
 
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