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

AI Thread Summary
To determine the pressure inside a bubble 4.0 m below the surface of ethyl alcohol, the correct equation is P = P₀ + (density)(gravity)(height). The initial pressure, P₀, is 1.1 atm, and the density of ethyl alcohol is 806 kg/m³. The calculation should convert the pressure from atm to pascals for consistency in units. The confusion arises from incorrect calculations and unit conversions, leading to incorrect pressure values. Properly applying the equation with consistent units will yield the correct pressure inside the bubble.
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?
 
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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 P_0*\rho gh
Its P_0 + \rho gh
 
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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