What is the water pressure ina lake?

Question

At the surface of a freshwater lake the air pressure is 0.9 atm. At what depth under water in the lake is the water pressure 3.5 atm?

Answer 1

Each 1033.23 centimeters of fresh water adds 1 atmosphere of pressure.

Here is why:

1 atmosphere of pressure is equal to 14.696 pounds per square inch, which is the same as saying 1033.23 grams per square centimeter.

Each CUBIC centimeter of water has a mass of 1 gram, because 1cm^3 = 1mL which is by definition 1g of pure water (and fresh water has little to no impurities). That means a 1033.23-cm tall column of water 1cm by 1cm will have a mass of 1033.23 grams, all pressing down on 1 square cm area at the bottom -- that column will be exerting 1 atmosphere of pressure on the bottom area of 1 square centimeter.

The area at the bottom does not matter, since as you increase the AREA you increase the VOLUME (thus the MASS) by the same ratio as long as the water column stays the same height -- but it is easier to imagine based on a 1cm x 1cm column.

1033.23 centimeters = 33.9 feet.

So what we need to do to solve your question, is figure out how much pressure IN ADDITION TO THE ATMOSPHEREIC PRESSURE, must be supplied by the water:

3.5 atm - 0.9atm = 2.6 atm

2.6 atm * 1033cm/atm = 2686.4 cm

So, at 2686.4 cm (26.864 meters, or 88.14 feet) the water will be adding 2.5 atmospheres of pressure to the 0.9 atmospheres pressure provided by the atmosphere itself, for a total of 3.5 atmospheres of pressure.

BTW, there is a way to work it out without all the "american" units (feet and pounds, just using the metric unit of force which is a Newton) but I'm so used to doing it in "american" units that this way is quicker for me...

Answer 2

Using this formula to calculate pressure in fluids:

P = hρg

P is pressure ( measured in Pa or N m⁻² )
h is depth
ρ is the object's density
g is the acceleration of objects under gravitional influence

Lets assume that the depth is x meter. So the pressure from the surface to x is:

P = 3.5 atm - 0.9 atm
P = 2.6 atm
1 atm = 10⁵ Pa
2.6 atm = 2.6 × 10⁵ Pa

The density of fresh water ( lake water ), ρ is 1000 kgm⁻³
The acceleration of objects under gravitional influence at earth surface is roughly 9.80665 ms⁻²

So:

P = hρg
2.6 × 10⁵ Pa = h × 1000 kgm⁻³ × 9.80665 ms⁻²
h = 26.51262154 m

Please correct me if I made mistake. Thanks! ^^

Answer 3

1 atm is equivalent to 34 feet of water.

The equivalent atm at the point where pressure is due to water pressure alone =3.5atm-0.9atm
=2.6atm

Convert 2.6atm to feet of water:

Depth or feet of water=2.6*34
=88.4feet

Answer 4

a) p(absolute) = p(atmospheric) + dgh d - density in kg/m^3 g = 9.8 m/s^2 H - depth in meters b) F = pA A - area of the window in square meters (A = pi*R^2) Plug and chug

Answer 5

1 atm =33.92' of water.
3.5-.9=2.6 atm
2.6*33.92=88.2'