Converting pressure to atomspheres

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SUMMARY

The pressure at a depth of 10,900 meters in seawater, with a mass density of 1030 kg/m³, is calculated using the formula pressure = mass density x gravity x height, resulting in 1.10x10⁸ N/m². This pressure converts to approximately 1086 atmospheres using the conversion factor of 1 atmosphere = 1.013x10⁵ N/m². The discussion highlights a common misconception regarding atmospheric pressure at depth, emphasizing the importance of accounting for the atmospheric pressure above the ocean surface. Participants reaffirm the accuracy of the calculated pressure and encourage a deeper understanding of physics principles.

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  • Understanding of fluid mechanics principles
  • Knowledge of pressure calculations in physics
  • Familiarity with units of measurement (N/m² and atmospheres)
  • Basic understanding of gravity's role in pressure
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  • Study the derivation of pressure formulas in fluid mechanics
  • Learn about the effects of atmospheric pressure on underwater measurements
  • Explore the concept of hydrostatic pressure in different fluids
  • Investigate the relationship between depth and pressure in various environments
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Students in introductory physics courses, educators teaching fluid mechanics, and anyone interested in understanding pressure dynamics in aquatic environments.

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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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