Height of Mercury in a pipe after removing it from a tub

[Balint]

Homework Statement

We have a pipe open at both ends and having the length of 1 meter. After putting it into a tub of Mercury and thus filling it to the half, we close the upper end of it, and then remove it from the tub. What is the height of the residual column of the Mercury in the pipe?
p_ex=10⁵ Pa

Homework Equations

The Attempt at a Solution

I don't even know where to start, as there is air inside the pipe at the beginning.

 

[Chestermiller]

Hint: This will involve the use of the ideal gas law.

 

[Balint]

I only know that the force that comes from the difference of the inner and outer pressure equals that of the Mercury column inside the pipe. But I don't know how to calculate with this as long as I don't have the density of Mercury, which I won't at the exam. Is that something I should learn or is there a way to get past that?

Here is my progress so far:
pV=nRT, since nRT is a constant, pV also is, so if I decrease pressure (or gravity does so) it will be $\frac{p_0}{x}×V_0x=nRT$
As the diameter of the pipe is constant, I can use l instead of V, naturally only by itself, not in the ideal gas law, but that only helps me at the end.
Somehow $\Delta p\sim F_u$ and $(1-Vx)\sim F_d$, but I don't know what to do next, as I would need a mass type multiplier somehow.

 

[PietKuip]

Do you know the pressure in units of Torr (mmHg) ?

 

[Balint]

I can convert it, I was given $10^5 Pa$, that's 750,062 Torr, but I think he just wanted to make it easier to calculate in SI...

 

[SteamKing]

1 torr = 1/760 atmosphere. Since 1 atmosphere is the pressure required to support a column of mercury to a height of 760 mm, I think you have a means to use the density of mercury without knowing the density of mercury.

 

[Chestermiller]

I only know that the force that comes from the difference of the inner and outer pressure equals that of the Mercury column inside the pipe. But I don't know how to calculate with this as long as I don't have the density of Mercury, which I won't at the exam. Is that something I should learn or is there a way to get past that?

Here is my progress so far:
pV=nRT, since nRT is a constant, pV also is, so if I decrease pressure (or gravity does so) it will be $\frac{p_0}{x}×V_0x=nRT$
As the diameter of the pipe is constant, I can use l instead of V, naturally only by itself, not in the ideal gas law, but that only helps me at the end.
Somehow $\Delta p\sim F_u$ and $(1-Vx)\sim F_d$, but I don't know what to do next, as I would need a mass type multiplier somehow.

You're on the right track. The initial height of the mercury in the column is 500 mm and 500 mm air. What is the initial pressure of air in the head space? Let 500+x be the final mm of air and 500-x be the final mm of mercury. In terms of x, what is the final mercury pressure at the top of the mercury column (in mm mercury)? In terms of x, what is the final air pressure in the head space (in mm mercury)? How do you think the air pressure in the head space is related to the mercury pressure at the top of the mercury column (i.e., at the exact same location)?