# Pressure of a bubble in Liquid

• davidatwayne
In summary, the problem involves determining the pressure inside a bubble located 4.0 m below the surface of a liquid ethyl alcohol in a rigid container. Using the equation P = P(nought) + (density)(gravity)(height), with the density of alcohol being .806 x 10^3 and the air pressure above the liquid being 1.1 atm, a calculation results in a pressure of 338.22, which may not be the correct answer. Converting the given air pressure to Pa may help in solving the problem.

## 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:
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.