Calculate Pressure Underwater for a 1cm3 Air Bubble at 291m

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Homework Help Overview

The problem involves calculating the new volume of a 1-cm³ air bubble as it rises from a depth of 291 meters in a lake, considering changes in pressure and temperature. The discussion centers around the effects of water depth on pressure and the application of gas laws.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore whether the depth of water affects pressure and how to calculate it. There are attempts to apply principles like Boyle's law and the relationship between pressure, volume, and temperature. Questions arise about the necessity of cross-sectional area in pressure calculations.

Discussion Status

The discussion is active, with participants providing insights into pressure calculations at depth and the relevance of atmospheric pressure. Some guidance has been offered regarding the relationship between pressure and depth, but no consensus has been reached on the method to calculate the new volume of the air bubble.

Contextual Notes

Participants note the absence of a calculator and the need for specific values, such as the density of water and the conversion of temperature to Kelvin. There is also mention of the requirement to express the final volume to the nearest tenth of a cm³.

vworange
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A 1-cm3 air bubble at a depth of 291 meters and at a temperature of 4 oC rises to the surface of the lake where the temperature is 10.4 oC, to the nearest tenth of a cm3, what is its new volume?

I guess my real question is: does the depth of the water make any difference or do i assume it's constant pressure? If i can't make that assumption, how do i calculate the pressure the water has on this?

The obvious answer is that if it's to the nearest tenth of a cm3, it's still coming out to 1.0 cm3 which is wrong.
 
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The depth of the water makes a huge difference. The pressure at depth is equal to the weight of a column of water of cross-sectional area A divided by that area A. The A divides out, leaving the pressure at any depth the same. You need to figure out how much a column of water 291 meters high with a given cross-sectional area would weigh, and divide by the area you used. That will give you the pressure.
 
I don't have cross-sectional area. How else can I go about getting the pressure... between surface and the depth given? I assume the initial pressure goes something along the lines of 9.81x291 (g*h).
 
Boyles law --> P.V/T = constant (pressure * volume / Temperature)
1 meter water = 0.0967841105 atmospheres

Just remember T is in Kelvin i.e 4 degrees = 277 degrees K

Surface pressure is 1, hence at 291 meters = 1+291*0.0967...

You have all the data, I have not got a calculator.
 
The area can be anything you want it to be. It divides out of the calculation. The weight of the column of water is the density of water (1gm/cc) times the volume of the water. The volume of a column of water is the area of the base times the height. The pressure is the weight divided by the area, so the area divides out. The pressure from the water is just the density times the height.

The toatal pressure at depth is the pressure from the water plus atmospheric pressure. The previous reply quotes the pressure due to a depth of one meter of water in terms of atmospheres and then states that the pressure at depth is the sum of one atmosphere plus the pressure due to the depth of water. That approach is good now that you know the pressure from one meter of water. The calculation I suggested would give you that number also. If you can use that number to satisfy anyone who might be grading your work, use it. If you have to justify the number, I have told you how to find it.
 

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