Barometer Tube Displacement: Solving for Isothermal Process

AI Thread Summary
The discussion revolves around calculating the displacement of mercury in a sealed tube when it is turned vertically, assuming an isothermal process. The initial setup includes a 1.00 m tube with mercury occupying the middle 10.0 cm and air at standard atmospheric pressure on both ends. Participants are attempting to solve for the pressure changes on either side of the mercury column as it displaces, using equations related to pressure and volume. There is confusion regarding the application of simultaneous equations and how to express the pressure difference needed to support the mercury. The conversation emphasizes the need for a step-by-step approach to clarify the calculations involved in determining the new pressures and displacement.
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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
 

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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?
 


(P2 + hpg)(0.45-x) = (0.45 +x)P2 ?
 


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?
 

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