Calculating Change in Volume at 8.80 km Deep

[starfish794]

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?

 

[OlderDan]

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

 

[kyle8921]

I would like to clarify that the change in volume of 1.00 m3 of water carried from the surface to a depth of 8.80 km in the ocean would not be zero. This is because as the water is carried deeper into the ocean, the pressure increases significantly. This increase in pressure causes the water to compress and decrease in volume.

To calculate the change in volume, we can use the formula pv = pv, where p is the pressure and v is the volume. However, we need to make sure that the units are consistent. As the pressure is given in N/m2, we need to convert the volume from m3 to cm3. This gives us a pressure of 9.08 x 107 N/m2 and a volume of 1.00 x 106 cm3.

Using the formula, we can calculate the final volume at a depth of 8.80 km:

pv = pv
(9.08 x 107 N/m2) (1.00 x 106 cm3) = (p2) (v2)
v2 = (9.08 x 107 N/m2) (1.00 x 106 cm3) / p2
v2 = (9.08 x 107 N/m2) (1.00 x 106 cm3) / (1052.88 kg/m3) (9.8 m/s2) (8.80 x 103 m)
v2 = 9.08 x 103 cm3 = 0.00908 m3

Therefore, the change in volume of 1.00 m3 of water carried from the surface to a depth of 8.80 km in the ocean would be approximately 0.00908 m3 or 9.08 liters. This is a significant decrease in volume, highlighting the immense pressure at such depths in the ocean.