[Physics] When a diving bell is submerged in water, why must I also consider the pressure of the atmosphere

fluid-statics

I understand how to solve these types of problems, but I'm not exactly sure why I need to do complete a certain step. Consider the problem:

A diving bell contains 10 cubic meters of air at sea level. It is lowered to a depth of 30m. What will the volume of air be in the tank after it is lowered? (Temperature remains constant)

The actual problem has more parts, but I'm simplifying it just to ask a specific question. In a previous part of the problem, it was found that the pressure pushing against the object due to the water at that depth would be 300 000 Pa (or 3 atm. Again, I'm simplifying the numbers just to get to my question).

As temperature remains constant, my first instinct was to simply use:

p1V1 = p2V2

Where:

  • p1 = the air pressure at sea level (100 000 Pa)

  • V1 = the volume of the air at sea level (10 cubic meters)

  • p2 = see below

  • V2 = the volume of the air at 30 m depth (what I am solving for)

My first instinct was to use "300 000 Pa" as the value for p2. I've seen (and can memorize) that what I should actually do is add the atmospheric pressure (100 000 Pa) to this value, which gives me 400 000 Pa.

I have two questions:

1) Can somebody explain why the pressure of the atmosphere needs to be included with the pressure of the water at this depth in order to solve this problem correctly? and,

2) If I include the pressure of the atmosphere in those types of problems, why do I not include it when I'm answering a question like this:

  • "What is the pressure due to water at the bottom of a cylinder 1 m deep?"

In these questions, I simply input the values for the following:
pressure = density of water x g x height, and don't consider the pressure of the atmosphere.

My attempt at answering my own questions:

1) Because the air pressure is on top of the water, and as a result, is pushing on the water. Therefore, the pressure experienced by an object in water is not only feeling the pressure of the water pushing on it directly, but also the indirect push of the atmosphere as well.

2) Because those questions specifically state "what is the pressure due to the water." In practical situations, the atmospheric pressure would also have to be considered.

Best Answer

  1. This is an important consideration for SCUBA divers and other experiments. Air pressure is incredibly powerful, but we normally don't notice it. Consider my juvenile experiment: Put a cup on your face around your mouth, and start sucking the air out until the cup sticks to your face (stupid I know, but it makes the point). If you try to pull the cup off, you feel a force holding it there. What is that force? Vacuum? No, it's the atmosphere around you pushing it onto your face!! Seriously. The atmosphere has mass (it's made of atoms), and that mass experiences gravity, so the entire atmosphere is weighing on you every day. How much so? It is putting 14.7 pounds per square inch on you, and everything else it touches. That includes the ocean! So when you're 0.1mm below the surface of the water, you're experiencing just over 14.7 pounds per square inch of pressure. As you go lower, the water adds its own weight (and pressure).
  2. Exactly as you say, the question asked for the pressure due to the water, but if you actually measured it, you'd find 14.7 psi + water pressure.
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