What is the force needed to displace 1,000,000 cu. ft. of water?

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To determine the force needed to displace a 1,000,000 cu ft balloon filled with air at the Mariana Trench, buoyancy principles must be applied, considering the weight density of saltwater at approximately 64 pounds per cubic foot. The pressure at a depth of 10 km reaches about 1000 bars, which would significantly compress the balloon, potentially causing it to shrink rather than pop if it is flexible. The ideal gas law (PV=nRT) can be used to analyze the changes in pressure and volume, while adiabatic compression may also affect the temperature of the air inside the balloon. Understanding the relationship between pressure, volume, and temperature is crucial for accurate calculations. Overall, the discussion emphasizes the importance of buoyancy and gas laws in understanding the behavior of objects submerged in deep water.
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I was recently asked what would happen to a 1,000,000 cu ft balloon that was filled with are if it was taken down to the Mariana trench. So, I am wondering what the formula would be for finding out how much force it would take to pull the balloon down that far. Also, how strong would the balloon have to be to not pop under such pressure. Thank you in advance. - Josiah
 
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What you are looking for is bouyancy. A ship can displace water without any force other than gravity...

Google for bouyancy equations. Sorry I couldn't help more, I have no time.
 
The weight density of salt water is about 64 pounds per cubic foot. Multiply that by the volume displaced and that's all there is to it.
 
Bird said:
I was recently asked what would happen to a 1,000,000 cu ft balloon that was filled with are if it was taken down to the Mariana trench. So, I am wondering what the formula would be for finding out how much force it would take to pull the balloon down that far. Also, how strong would the balloon have to be to not pop under such pressure. Thank you in advance. - Josiah

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 Earth's' 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 transferred 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

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

(you might want to double check this). \mbox{\gamma} is a thermodynamic constant associated with air.
 
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I thought they would have been asking how much the balloon would compress under pressure.

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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