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Mercury manometer

by Bengo
Tags: manometer, mercury
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Bengo
#1
Apr23-14, 06:50 AM
P: 44
In my book it talks about a mercury manometer where one side of the U tube is open to air and the other side is connected to a system where a reaction takes place. It then derives ΔP=ρgΔh by setting the force pushing the mercury up equal to the force pushing the mercury down. The force pushing the mercury up is the pressure of the gas from the system multiplied by the cross sectional area of the mercury which is F=PA. The force pushing the mercury down is gravity, F=mg. My question is why don't they consider the atmospheric pressure pushing down on the mercury since one side of the manometer is open to the air?

Thank you! You guys have been so helpful!
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paisiello2
#2
Apr23-14, 02:24 PM
P: 558
You are interested in the change in pressure caused by the reaction.

Presumably the gas was at some pressure before the reaction and you calibrated the manometer before the reaction took place. Therefore the atmospheric pressure was already balanced by the pressure that the gas exerted before the reaction.

Once the reaction occurs you get the change in pressure caused by the reaction only, which is all you are interested in. The atmospheric pressure has already cancelled out with the gas pressure before the reaction.

Now if you wanted to know the absolute pressure in the system then I agree you would have to add atmospheric pressure to the equation. But you would also have to calibrate your manometer to read -1atm when the system was a vacuum (i.e. P=0).


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