Finding the volume of air in a box when it's lowered into water?

In summary, the problem involves a box being lowered into the sea, trapping air at a pressure of 101 kPa and a volume of 2.5 m^3. The box and air have a combined volume of 3 m^3. With constant temperature and a mass of 4000 kg, the volume of air at a depth of 19 m below the sea surface can be calculated using the ideal gas law. The initial and final pressures and volume of air are used to apply Boyle's Law to solve for the final volume of air.
  • #1
PhyIsOhSoHard
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Homework Statement


A box that is open at the bottom is lowered into the sea (density like water). The outer volume of the box and the air inside it is [itex]V_{out}=3 m^3[/itex].
The moment the box touches the sea surface the air inside it gets trapped and has a volume at [itex]V_0=2.5 m^3[/itex] and a pressure at [itex]p_0=101 kPa[/itex]. As the box is lowered the temperature is constant and the mass of the box and air is [itex]m=4000 kg[/itex].
Calculate the volume of the air in the box as it is lowered to [itex]h=19 m[/itex] below the sea surface.

Homework Equations


[itex]p=p_0+ρgh[/itex]

[itex]pV=nRT[/itex]

The Attempt at a Solution


I was thinking about using the first equation to calculate the pressure at 19 m below the sea surface since I know the pressure at the sea surface and I know the density of water:
[itex]p=101000 Pa+9.8\frac{m}{s^2}\cdot 19 m=287000 Pa[/itex]

I'm supposed to use the ideal-gas equation to calcuate the volume of the air trapped in the box at 19 m below the sea surface... So I just calculated the pressure at this level but I'm completely lost on what else to do now.
 
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  • #2
The thing unclear to me is the volume of air in the box is stated as 3.0 and then re-stated again as 2.5. But perhaps it doesn't change the answer.

At any rate, you can apply the ideal gas law since you know the final pressure P1 and the initial pressure p0 and volume V0.
 
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  • #3
paisiello2 said:
The thing unclear to me is the volume of air in the box is stated as 3.0 and then re-stated again as 2.5. But perhaps it doesn't change the answer.

At any rate, you can apply the ideal gas law since you know the final pressure P1 and the initial pressure p0 and volume V0.


The 3.0 is the volume for the air and the box. The 2.5 is only for the air.

In the ideal gas law what do I use for n and T?
 
  • #4
Yes, of course, that would make the most sense. It's confusing since Vout is then irrelevant.

n and T are constant, therefore nRT is also constant. So you can assume any value you like.
 
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  • #5
paisiello2 said:
Yes, of course, that would make the most sense. It's confusing since Vout is then irrelevant.

n and T are constant, therefore nRT is also constant. So you can assume any value you like.

But what about [itex]V_0=2.5[/itex]? Am I not supposed to use that for anything?
 
  • #6
PhyIsOhSoHard said:
But what about [itex]V_0=2.5[/itex]? Am I not supposed to use that for anything?
You are using Boyle's Law, ## PV=constant##
where you are given ##V_0## and required to find ##V_1##
 

1. How do you find the volume of air in a box when it's lowered into water?

To find the volume of air in a box when it's lowered into water, you will need to use the displacement method. This involves measuring the change in water level when the box is submerged in the water. The difference in water level represents the volume of air in the box.

2. What is the purpose of finding the volume of air in a box when it's lowered into water?

The purpose of finding the volume of air in a box when it's lowered into water is to determine the buoyant force acting on the box. This is important in understanding the stability of objects in water and for calculating the weight of objects.

3. Does the size or shape of the box affect the volume of air in water?

Yes, the size and shape of the box can affect the volume of air in water. A larger box will displace more water and therefore have a larger volume of air, while the shape of the box can also impact the amount of water displaced.

4. What units are used to measure the volume of air in a box submerged in water?

The volume of air in a box submerged in water is typically measured in cubic meters (m3) or cubic centimeters (cm3). These units represent the amount of space the air occupies in the box.

5. Can the volume of air in a box submerged in water change?

Yes, the volume of air in a box submerged in water can change if the water level changes or if the box is moved. However, the amount of air in the box will remain the same as long as the box is not opened and no air is added or removed.

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