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=�gh
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 �� 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 \rho


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

Maybe like this:

P_1V_1 = P_2V_2

P_1(0.3\times A) = (0.2(9.81)\rho)(0.2\times A)

The value for \rho is 13.534 \frac{g}{cm^3}
The final answer is p=\rho gh \frac{\(\frac{l-h}{2}-\delta l\)\(\frac{l-h}{2}+\delta l\)}{\delta l(l-h)}=0.49*10^5\frac{N}{m^2}
I hope this helps.Thank you.

p=\rho*g*h\frac{\(frac{l-h}{2}-\delta l \)\(\frac{l-h}{2}+\delta l \)}{\delta l*(l-h)} 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 \rho*g*h and [\delta?] should have been \delta l. 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 = h\rho g = 0.2m \times 13534kg/m^3 \times 9.81m/s^2 = 26553.708 Pa

So, using P_1V_1 = P_2V_2

P_1 = \frac{(26553.708)(A\times 30)}{A\times 40} = 19915.281 Pa

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:


Weight = \rho g h A = A(4/3-4/5)P



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:
P_0\delta l(l-h)=A\(\frac{l-h}{2}+\delta l\)\(\frac{l-h}{2}-\delta l\)\rho gh
where:
P_0 is the pressure and \delta l(l-h) the volume when the tube is placed horizontally
the pressure above is A and the volume \frac{l-h}{2}+\delta l
the pressure below is \rho gh and the volume frac{l-h}{2}-\delta l

but ebaines' method is by far much more simple.