Converting pressure to atomspheres

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The discussion revolves around calculating the pressure at a depth of 10,900 meters in the ocean, using the formula pressure = mass density x gravity x height. The calculated pressure is approximately 1.10x10^8 N/m^2, which converts to about 1086 atmospheres when divided by the atmospheric pressure of 1.013x10^5 N/m^2. Participants express confusion over the accuracy of this figure, with one suggesting a much lower estimate of 1.3 to 1.4 atmospheres, while others affirm the correctness of the initial calculation. The conversation emphasizes the importance of understanding physics concepts to avoid guessing in calculations. Mastery of these principles is seen as essential for advancing in the subject.
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Homework Statement


The deepest part of the ocean is about 10,900 meters deep. Calculate the pressure at that depth in units of N/m^2. Mass density for seawater is 1030 kg/m^3. Then convert the pressure to the number of atmospheres given one atmosphere is 1.013x10^5 N/m^2.

Homework Equations


pressure = mass density x gravity x height

The Attempt at a Solution


p= (1030 kg/m^3)(9.8 m/s^2)(10,900 m)
p= 1.10x10^8 N/m^2

1 atmosphere = 1.013x10^5 N/m^2

1.10x10^8 N/m^2 / 1.013x10^5 N/m^2 = 1086 atmospheres.

I know that number is wrong. My guess is there are about 1.3 - 1.4 atmospheres of pressure there, but I can't get the math to find an exact number for that conversion. What is the best way to convert pressure to atmospheres?
 
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Imagine a column of water with a cross section of 1 m2 and 10900 m high. It's easy to calculate how much does it weigh... At 1030 kg/m3, its mass is 10900 m3 * 1030 kg/m3 = 11,2270 * 106 kg. Ant it weighs 11,2270 * 106 * 9,8 = 1,10025 *108 N. The pressure is the same, in N/m2

Using the equivalence 1 atm = 1.01325 * 105, I get the same figure as you: 1086 atm.

It's a reasonable figure. The pressure at a point 15 m deep, in fresh water, is already 1,5 atm...
 
In what way is that number wrong? On what do you base your guess of 1.3 - 1.4 atmospheres?

I note that you didn't account for the fact that the atmosphere itself lies above the ocean surface and presses down with its own pressure...
 
dlp248 said:

Homework Statement


The deepest part of the ocean is about 10,900 meters deep. Calculate the pressure at that depth in units of N/m^2. Mass density for seawater is 1030 kg/m^3. Then convert the pressure to the number of atmospheres given one atmosphere is 1.013x10^5 N/m^2.

Homework Equations


pressure = mass density x gravity x height

The Attempt at a Solution


p= (1030 kg/m^3)(9.8 m/s^2)(10,900 m)
p= 1.10x10^8 N/m^2

1 atmosphere = 1.013x10^5 N/m^2

1.10x10^8 N/m^2 / 1.013x10^5 N/m^2 = 1086 atmospheres.

I know that number is wrong. My guess is there are about 1.3 - 1.4 atmospheres of pressure there, but I can't get the math to find an exact number for that conversion. What is the best way to convert pressure to atmospheres?

This is why one learns physics: so you are not guessing about things so much.
 
Thanks guys. I guess when I saw the number I questioned myself. It'll come with more practice.

SteamKing said:
This is why one learns physics: so you are not guessing about things so much.

That is the exact why reason I am loving physics! I'm in the base introductory course at my college this semester so I can get my calc done and then off to bigger and better things! Planning on getting a BS in physics!
 
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