Under Pressure Finding the height of compressed air

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Homework Help Overview

The problem involves a cylinder containing air at a specific pressure and a piston, with mercury being added until the cylinder overflows. The goal is to find the height of the compressed air using the ideal gas law and principles of fluid statics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the ideal gas law and fluid statics to derive the height of the compressed air but expresses confusion regarding the results, particularly with the quadratic equation yielding complex solutions.
  • Some participants question whether the atmospheric pressure has been adequately considered in the calculations.
  • Others suggest a revised equation that incorporates atmospheric pressure alongside the weight of the mercury.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to adjust the original approach to account for atmospheric pressure. There is a sense of collaboration as participants seek to clarify assumptions and refine the calculations.

Contextual Notes

The original poster indicates that multiple individuals, including a professor and a lab instructor, have reviewed their work, suggesting that there may be a common misunderstanding or oversight in the application of the principles involved.

IslandHead
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Homework Statement


A 1.0--tall cylinder contains air at a pressure of 1 . A very thin, frictionless piston of negligible mass is placed at the top of the cylinder, to prevent any air from escaping, then mercury is slowly poured into the cylinder until no more can be added without the cylinder overflowing. Find the height of the compressed air. It recommends P1V1=P2V2 and that temperature is constant.


Homework Equations


P1V1=P2V2
F=mg
P=F/A



The Attempt at a Solution


P1V1=P2V2
P1A1m=P2hA
(P1*1M)/h=P2

The system is at rest so the force gravity is pulling on the Hg is balanced by the air pressure
PA=F=mg
A(P1*1M)/h=ρ(1-h)Ag
(P1*1M)/pg=h(1-h)
.76=-h^2 + h
0=-h^2 +h -.76
The answers is 76cm ( think the book is wrong) and that quadratic has answers with i in it. I'm unsure what I'm doing wrong! I asked my prof and he couldn't help me. I asked my lab instructor and he says what I did looked right. So 3 people (including me) have missed what is wrong with my work. Please help me :'(
 
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Am I on the right track for this?
 
Welcome to PF, IslandHead! :smile:

I believe you are forgetting that the outside force is not just the weight of the mercury, but also the atmospheric pressure.
 
so then it will be:

A(P1*1M)/h = 101300A + ρ(1-h)Ag

I'll try solving this
 
It works :D i got thanks!
 
Cheers! :smile:
 

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