- #1
Bengo
- 46
- 0
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!
Thank you! You guys have been so helpful!