# Pressure Underwater?

A 1-cm3 air bubble at a depth of 291 meters and at a temperature of 4 oC rises to the surface of the lake where the temperature is 10.4 oC, to the nearest tenth of a cm3, what is its new volume?

I guess my real question is: does the depth of the water make any difference or do i assume it's constant pressure? If i can't make that assumption, how do i calculate the pressure the water has on this?

The obvious answer is that if it's to the nearest tenth of a cm3, it's still coming out to 1.0 cm3 which is wrong.

OlderDan
Homework Helper
The depth of the water makes a huge difference. The pressure at depth is equal to the weight of a column of water of cross-sectional area A divided by that area A. The A divides out, leaving the pressure at any depth the same. You need to figure out how much a column of water 291 meters high with a given cross-sectional area would weigh, and divide by the area you used. That will give you the pressure.

I don't have cross-sectional area. How else can I go about getting the pressure... between surface and the depth given? I assume the initial pressure goes something along the lines of 9.81x291 (g*h).

Boyles law --> P.V/T = constant (pressure * volume / Temperature)
1 meter water = 0.0967841105 atmospheres

Just remember T is in Kelvin i.e 4 degrees = 277 degrees K

Surface pressure is 1, hence at 291 meters = 1+291*0.0967...

You have all the data, I have not got a calculator.

OlderDan