Barometer Tube Displacement: Solving for Isothermal Process

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Tidus
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Homework Statement


A thin tube, sealed at both ends, is 1.00 m long. It initially lies horizontally, with the middle 10.0 cm containing mercury, and the two ends containing air at standard atmospheric pressure:

See picture in attachment

If the tube is now turned to a vertical position, by how much will the mercury be displaced, if the process is isothermal? You may assume that no gas passes from one side of the mercury column to the other.


Homework Equations



P1f = hpg + P2f p=density of mercury , h=length of mercury column
P1V1 = P2V2
V1 + V2 = 0.9

The Attempt at a Solution



Tried solving simultaneous equations but got bogus answers
 
Attachments
  • Diagram of a horizontal sealed tube 1.00 m long with a shaded 10.0 cm section in the middle marking the mercury column, and arrows labelling the 10.0 cm and 1.00 m lengths.
    mercury.jpg
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Say the mercury is displaced X cm. How much does the pressure change on each side? How much of a pressure difference is needed to support the mercury?
 


Tidus said:
(P2 + hpg)(0.45-x) = (0.45 +x)P2 ?
Not sure how you got that expression. Do it step by step. Both sides start with the same volume (proportional to their length, which is initially 0.45m) and the same pressure Pa.

So what's the new pressure of the air on the side that compresses by distance X? And the side that expands?