# Calculating Change in Volume at 8.80 km Deep

• starfish794
In summary, the conversation is about calculating the change in volume of water at a point 8.80 km deep in the ocean, with a pressure of 9.08 x 107 N/m2. The question is whether the change in volume should be zero or not, and the density at that point is found to be 1052.88 kg/m^3. There is also a reference to a forum thread where someone else is discussing a similar problem with a different depth.
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?

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?

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.

## 1. How do you calculate the change in volume at a depth of 8.80 km?

To calculate the change in volume at a depth of 8.80 km, you need to know the initial volume and the final volume. Then, you can use the formula: change in volume = final volume - initial volume. Make sure to convert the depth to meters before calculating.

## 2. What factors affect the change in volume at 8.80 km deep?

There are several factors that can affect the change in volume at a depth of 8.80 km, including temperature, pressure, and the composition of the material. These factors can cause expansion or contraction of the material, resulting in a change in volume.

## 3. Can you calculate the change in volume at 8.80 km deep for any material?

Yes, the change in volume at 8.80 km deep can be calculated for any material as long as you have the necessary data, such as the initial and final volumes and the material's properties. However, the result may vary depending on the material's behavior under high pressure and temperature conditions.

## 4. How does the density of the material affect the change in volume at 8.80 km deep?

The density of the material plays a significant role in determining the change in volume at 8.80 km deep. Materials with higher density are less likely to undergo a significant change in volume compared to materials with lower density. This is because denser materials are more resistant to compression and can withstand higher pressure.

## 5. What units should be used for calculating the change in volume at 8.80 km deep?

The units used for calculating the change in volume at 8.80 km deep should be consistent. It is recommended to use metric units, such as meters and cubic meters, for accurate calculations. However, if using imperial units, make sure to convert the values to the corresponding metric units before calculating.

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