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lemon14 · Oct 4, 2010, 10:00 AM

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 Î”l=10 cm. What is the pressure of the tube in horizontal position?

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

lemon14 · Oct 4, 2010, 10:02 AM

Î”l means Delta and ÃÂ means Rho

Unknown008 · Oct 4, 2010, 10:27 AM

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:

$P_1V_1 = P_2V_2$

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

lemon14 · Oct 4, 2010, 11:06 AM

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 :)

Unknown008 · Oct 4, 2010, 12:48 PM

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?

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 :(

ebaines · Oct 4, 2010, 02:29 PM

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:

$<br /> Weight = \rho g h A = A(4/3-4/5)P<br />$

Solve for P.

lemon14 · Oct 5, 2010, 06:09 AM

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.