What Is the Pressure Inside a Bubble 4.0 m Below the Surface of Ethyl Alcohol?

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Homework Help Overview

The problem involves determining the pressure inside a bubble located 4.0 m below the surface of ethyl alcohol, given the air pressure above the liquid and the density of the alcohol. The context is fluid mechanics, specifically dealing with pressure calculations in a liquid medium.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the pressure equation, questioning the consistency of units and the correctness of the density value. There are attempts to clarify the equation used and its components.

Discussion Status

Some participants have provided guidance on unit conversion and have pointed out potential errors in the equation format. Multiple interpretations of the pressure equation are being explored, and there is ongoing confusion regarding the calculations presented.

Contextual Notes

There is mention of needing to convert pressure units from atmospheres to pascals for consistency. Participants are also questioning the accuracy of the density value used in the calculations.

davidatwayne
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Homework Statement



Air is trapped above liquid ethyl alcohol in a rigid container. If the air pressure above the liquid is 1.1 atm, determine the pressure inside a bubble 4.0 m below the surface of the liquid.

Homework Equations



P= P(nought) + (density)(gravity)(height)
density of alcohol .806 x 10^3

The Attempt at a Solution



Pnought would be the 1.1 atm
P= 1.1 + (806 kg/m^3)(9.8 m/s^2)(4 m)
= 31596

That isn't the right answer, so I'm very confused. Do I need to change atmospheres to kPas? Is that density correct? Am i even using the right equation?
 
Last edited:
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I would definitely convert the 1.1 atm to Pa to see if that's your problem. Your units aren't consistent the way they are.
 
It isn't [tex]P_0*\rho gh[/tex]
Its [tex]P_0 + \rho gh[/tex]
 
davidatwayne said:
P= P(nought) x (density)(gravity)(height)
density of alcohol .806 x 10^3

The Attempt at a Solution



Pnought would be the 1.1 atm
P= 1.1 + (806 kg/m^3)(9.8 m/s^2)(4 m)
= 338.22

That isn't the right answer, so I'm very confused. Do I need to change atmospheres to kPas? Is that density correct? Am i even using the right equation?


You've got two versions of the equation written down here. The one you used in your calculation is the right one, but I can't figure out how you got the 338.22 even if it is wrong.
 

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