How Do You Determine the Length L in a Compressed Air and Mercury Experiment?

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To determine the length L of the air column in a tube filled with mercury, the relationship between the pressures and volumes of the air and mercury must be established. The ideal gas law (PV=NRT) is relevant, but additional equations are needed to relate the lengths of the air and mercury columns. The total length of the tube is fixed at 1 m, leading to the equation La + Lm = Ltot, where La is the length of the air column and Lm is the length of the mercury column. The final pressure exerted by the mercury must also be considered, as it affects the compression of the air. Understanding the pressure of mercury in relation to atmospheric pressure is crucial for solving the problem.
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Homework Statement



Au shaped tube has a total length of 1 m, it is initially filled with air at 20* C and 1 atm, mercury is poured in without letting air escape, compressing the air. This continues until the mercury is filled to a level L, how long is L

Homework Equations



Im guessing PV=NRT,
It seems like it should involve the density of HG but its not given, nor is the volume


The Attempt at a Solution



I don't even know where to start, It seems like the final pressure is needed, Any suggestions?
 
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As more mercury is piled on, the length of the air column goes down. This continues until mercury and the compressed air fills the entire tube.
 
ideasrule said:
As more mercury is piled on, the length of the air column goes down. This continues until mercury and the compressed air fills the entire tube.

I'm still missing something, what does that lead to? Is there a formula I am just not seeing, W/E or ...
 
Well, it leads to the formula La+Lm=Ltot, where La and Lm are the lengths of mercury and air. You need one more equation to solve the problem; this equation has to relate La with Lm.
 
ideasrule said:
Well, it leads to the formula La+Lm=Ltot, where La and Lm are the lengths of mercury and air. You need one more equation to solve the problem; this equation has to relate La with Lm.

I don't want to appear dense( no pun intended) but I still don't get it. None of the equations I see are in terms of L, only pressure, velocity or area. I can't see how the youngs modulus would have anything to do with it
 
Frostfire said:
I don't want to appear dense( no pun intended) but I still don't get it. None of the equations I see are in terms of L, only pressure, velocity or area. I can't see how the youngs modulus would have anything to do with it

Does it involve the pressure of mercury as 1 atm is 760 mm HG?
 
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