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- Thread starter Bird
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Google for bouyancy equations. Sorry I couldn't help more, I have no time.

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russ_watters

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Bird said:

If you assume the baloon is flexible it won't "pop", but it will shrink significantly under the pressure. It would be rather an odd turn of phrase to call a rigid diving bell a "baloon" IMO, so I would tend to assume the baloon would shrink under the pressure.

Assuming constant density, the pressure at a depth d will be rho*g*d, rho being the density of water, g the acceleration of earths' gravity, and d the depth.

At 10km, I get approximately 1000 bars (atmospheres) of pressure. If one assumes the ideal gass law, PV=nrT. PV will be constant if one assumes constant temperature, one might also be interested in what would happen if the compression was adiabatic (i.e. if no heat was transfered to or from the baloon). Because the baloon is being compressed, adiabatic compression would mean that the air inside would get hotter. But offhand I'd guess that the temperature of the air inside the baloon couldn't get much above the boiling point of water (dont' know what that is at that depth, though), so it probably wouldn't be adiabatic. So two reasonable assumptions might be to have the final temperature be the boiling point of water at that depth (for a fairly fast descent), or to hvae the final temperature be the water temperature (for a slow descent).

For adiabatiic compression I think the proper equatons are

[tex]

T2/T1 = (V1/V2)^{\gamma -1} = (P2/P1)^{1-1/\gamma}

[/tex]

(you might want to double check this). [tex]\mbox{\gamma}[/tex] is a thermodynamic constant associated with air.

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DaveC426913

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So, how many atmospheres are there at the bottom of the trench? (Google atmospheres depth PSI)

What is the relationship between atmospheres and volume? (Boyle's Law)

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