Problem involving gas laws and hydrostatics

[1729]

Homework Statement

A glass tube filled with air at room temperature is 1.54m long. The tube is closed on one end, and open on the other. When submerging the open end in water, the water in the tube rises by 0.14m.
How much of the tube is above the water's surface?

The correct answer is 0.40m.

Homework Equations

Boyle's law
Run-of-the-mill hydrostatics equations

The Attempt at a Solution

First comes Boyle's law: $p_2=\frac{p_1V_1}{V_2}=\frac{p_1Ah_1}{Ah_2}=\frac{p_1h_1}{h_2}$, where $p_1$ is the air pressure in the tube outside of water; $p_2$ is the pressure of the air in the tube whilst submerged; and $h$ represents the height of the air.

We can fill this in: $p_2=\frac{10^5 \ \mathrm{Pa}\cdot 1.54\ \mathrm{m}}{1.40\ \mathrm{m}}=1.1\cdot 10^5 \ \mathrm{Pa}$.
Since this is the air's pressure on the surface of the water in the tube, the pressure in the water at the highest point is equal to $p_2$.

How to continue from this point? I've tried thinking about a mercury barometer but I'm not sure. My idea was to look at the gauge pressure at the very top of the water ($10^4\ \mathrm{Pa}$), and use some hydrostatics equation. I don't know how I would go about that last part though.

Thanks in advance.

 

[J Hann]

So the difference between the atmosphere and the pressure in the tube is .1 atmosphere.
What height of a column of water produces .1 atmosphere of pressure?

 

[Chestermiller]

Are you sure it doesn't sink 0.14 meters?

 

[1729]

So the difference between the atmosphere and the pressure in the tube is .1 atmosphere.
What height of a column of water produces .1 atmosphere of pressure?

A column of approximately one metre. (using gauge pressure formula)
Where does this metre start however? Does it start at the very end of the submerged part of the tube, up the tube to the very top of the water column?

Are you sure it doesn't sink 0.14 meters?

I'm absolutely certain it doesn't sink. I double-checked when I translated the problem statement.

EDIT: I guess I did translate it incorrectly, because I myself didn't understand the ambiguous wording in the statement.

Here's a sketch of the problem:

 

[Chestermiller]

Have you tried the experiment with a straw?

Let x represent the amount of the tube above the surface. Then the amount of tube below the water surface is 1.54 - x. If the water actually does rise into the tube (which I doubt), the amount of air in the tube is now of length x - 0.14. So, if the initial pressure of the air in the 1.54 meter tube was 1 bar, what is the pressure in the air within the submerged tube? This must match the hydrostatic pressure at the surface of the water in the tube. What is that?

 

[1729]

You are absolutely right. There is no way it can rise above the water surface. Now the problem actually makes sense to me; I can see what the gauge pressure formula is doing. The pressure at the water surface inside the tube is equal to 1000 Pa, therefore you can use the formula on the basin itself. This means that, between the water surface in the basin and the water surface in the tube, there is a one metre difference.

1.54m - 0.14m - 1.00m = 0.40m

Thank you both for taking the time to reply!

 

[Chestermiller]

I guess I was mistaken. I was thinking you were saying that the water rises in the tube 0.14 meters above the water surface in the tank. What it's really saying is that it rises 0.14 m above the base of the tube. So the actual amount of air in the tube is now 1.4 m, and the new pressure in the tube air is now 100000 x 1.54/1.4 Pa = 110000 Pa. If y is the submerged depth of the tube, then the pressure of the air in the tube is 100000+1000g(y-0.14). If I set the two pressure expressions equal and solve for y, I get y = 1.14 m. So the part above the water is 0.4 m.