Converting pressure to atomspheres

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Homework Help Overview

The problem involves calculating the pressure at a depth of 10,900 meters in the ocean, using the mass density of seawater and converting that pressure into atmospheres. The subject area includes fluid mechanics and pressure calculations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of pressure using the formula involving mass density, gravity, and height. There are attempts to convert the calculated pressure into atmospheres, with some participants questioning the accuracy of the original poster's conversion and assumptions.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the assumptions made regarding atmospheric pressure at the ocean's surface. Some guidance has been offered regarding the calculation method, but there is no explicit consensus on the correct interpretation of the results.

Contextual Notes

Participants note the importance of accounting for atmospheric pressure above the ocean surface in their calculations. There is also a mention of the original poster's uncertainty regarding their results, which may stem from a lack of experience in physics.

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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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