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

  • Thread starter Thread starter Frostfire
  • Start date Start date
  • Tags Tags
    Length
Click For Summary

Homework Help Overview

The problem involves a U-shaped tube filled with air and mercury, where the air is compressed by the mercury poured into the tube. The goal is to determine the length L of the mercury column when the system reaches equilibrium, considering the initial conditions of temperature and pressure.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the lengths of the mercury and air columns and the total length of the tube. There is uncertainty about how to relate these lengths to pressure and other variables, with some questioning the relevance of certain equations.

Discussion Status

Participants are exploring different interpretations of the problem and discussing the need for additional equations to relate the lengths of the air and mercury columns. Some have suggested potential relationships but express confusion about how to apply them to find L.

Contextual Notes

There is a lack of specific information regarding the density of mercury and the volume of the air column, which may be impacting the discussion. Participants are also considering the implications of pressure measurements in the context of the problem.

Frostfire
Messages
50
Reaction score
0

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?
 
Physics news on Phys.org
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?
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
10
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
24
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
3
Views
3K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K