In the middle of a tube made of glass set horizontally (length: l=1m) there is a column of quicksilver (length: h=20 cm). When the tube is put vertically, the column descends with Δl=10 cm. What is the pressure of the tube in horizontal position?

I tried to use the formula p1v1=p2v2
v1=(l-h)Δl
p2=ρgh
then I wrote the two pressures in vertical position: (l-h)/2-Δl and (l-h)/2+Δl but I don't know how to put them together for v2.

Δl means Delta and ρ means Rho

Do you have any data? I don't know where to start for this one?

I think we'll need the value of


Type in (math)\rho(/math) but replace the () by []

Maybe like this:

The value for is
The final answer is
I hope this helps.Thank you.

this is the correct form

or not... sorry I am in a hurry in the first parenthesis I should have written "-" instead of "+"; [\rhogh] should have been and [\delta?] should have been . I'm sorry again and thank you (I finally did it :)

EDIT: Look at ebaines' answer. I was hesitant about whether to consider the effect of both gases and he confirmed that we needed to take both into consideration.


I don't know what you did exactly :confused:

Am I missing something?

http://p1cture.me/images/65589453008987891793.png

The pressure where the silver is on the gas that I marked on the arrow is where the pressure of the gas and the pressure of the silver are equal.

So, pressure of gas =

So, using



Or am I missing something?

I never had such a problem with pressure like this before though :(

It's hard to follow the math that you're trying to write out, but I would approach it like this:

In the vertical position the quicksilcver is supported by the difference between the air pressure below and the air pressure above. You know that the pressure below is 4/3 times the original pressure P (this comes from the concept that P1V1 = P2V2), and similarly the air pressure above is 4/5 times the original. Hence the difference in air pressure times the area it works on must equal the weight of the quicksilver:




Solve for P.

I succeeded in solving it with both methods (mine and ebaines') and I got the same result, so I think it must be correct. Thank you for help.

My method:

where:
is the pressure and the volume when the tube is placed horizontally
the pressure above is A and the volume
the pressure below is and the volume

but ebaines' method is by far much more simple.