Calculating Change in Volume at 8.80 km Deep

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At a depth of 8.80 km in the ocean, the pressure is approximately 9.08 x 10^7 N/m², significantly affecting water density, which is calculated to be 1052.88 kg/m³. The discussion centers on calculating the change in volume of 1.00 m³ of water under this pressure, with some participants questioning why the expected change isn't zero. The use of the formula pv = pv is suggested to determine the volume change. Participants also inquire about the methods used to derive the density at this depth. Understanding these calculations is crucial for accurately assessing the effects of deep-sea pressure on water volume.
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Consider a point in the ocean about 8.80 km deep. The pressure at that point is huge, about 9.08 107 N/m2.
(a) Calculate the change in volume of 1.00 m3 of water carried from the surface to this point in the ocean.

I don't get why the answer isn't zero. But anyway, I found the density at that point to be 1052.88 kg/m^3. Should I be using pv=pv to find the change in volume?
 
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starfish794 said:
Consider a point in the ocean about 8.80 km deep. The pressure at that point is huge, about 9.08 107 N/m2.
(a) Calculate the change in volume of 1.00 m3 of water carried from the surface to this point in the ocean.

I don't get why the answer isn't zero. But anyway, I found the density at that point to be 1052.88 kg/m^3. Should I be using pv=pv to find the change in volume?
Someone else is looking at this same problem but with a somewhat different depth number. What information did you use to get that density?

https://www.physicsforums.com/showthread.php?t=143100
 

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