Isothermal pressure change in a U-shaped tube

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Homework Help Overview

The discussion revolves around an isothermal pressure change in a U-shaped tube involving mercury and air. Participants are exploring the implications of a vacuum on the behavior of mercury in the tube and the resulting pressure changes.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how the pressure changes affect the mercury levels in the tube, questioning the assumption that all mercury would be sucked into one side. Some participants suggest considering the incompressibility of mercury and its weight, while others reference relevant equations.

Discussion Status

Participants are actively engaging with the problem, offering hints and questioning assumptions about the behavior of mercury in a vacuum. There is no explicit consensus yet, but the discussion is generating useful insights into the physical principles involved.

Contextual Notes

Participants are considering the effects of atmospheric pressure and the nature of a vacuum, with some noting that the concept of a vacuum does not exert force in the way one might intuitively think.

obstinatus
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Homework Statement
The U-tube in figure 2-18, of uniform cross section 1 cm^2, contains mercury to the depth shown. The barometric pressure is 750 Torr. The left side of the tube is now closed at the top and the right side is connected to a good vacuum pump. A) How far does the mercury level fall in the left side and b) what is the final pressure of the trapped air? The temperature remains constant.
Relevant Equations
Pv = RT
IMG_20200212_153819730.jpg


Hi, just reviewing some thermodynamics from the textbook by Sears and Salinger, having a hard time conceptualizing this one. It's an isothermal change in pressure, so the volumes of the mercury and the air both change to reach equilibrium, but if it's a "good vacuum pump", then won't the right side be completely evacuated, and the mercury sucked completely into the right side? Half of the air molecules will be evacuated, so all else being equal, the pressure of the air should be reduced by half, but because of this arrangement, we have one end of the mercury subject to zero pressure and the other subject to normal pressure. I just don't see how any end state can be specified besides all of the mercury sucked into the right side of the tube. Thanks in advance for any hints!
 
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Are you familiar with the equation ##\Delta p=\rho g h##? Do you think that it might apply to this problem?
 
Two hints: Mercury is effectively incompressible, and Mercury is very heavy.
 
The barometric pressure is 750 Torr
Which is how many mm that the atmospheric pressure can push a column of mercury to rise into a vacuum?
obstinatus said:
... mercury sucked completely into the right side...
Perhaps that statement needs to be thought about some more.
 
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obstinatus said:
I just don't see how any end state can be specified besides all of the mercury sucked into the right side of the tube.

Having heard nothing, another hint: Vacuum does not “suck”. The existence of a vacuum above one surface of the Mercury imparts no force on the Mercury.
 

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