Solving for Depth in Boyle's Law: Barometer Tube Submerged in Water

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Homework Help Overview

The problem involves a barometer tube submerged in water, specifically determining the depth of the top of the tube when water rises inside it. The context is rooted in Boyle's Law and fluid pressure concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Boyle's Law to relate pressures and volumes in the context of the submerged barometer tube. There are attempts to derive the height of the water column and its relation to the atmospheric pressure. Questions arise regarding the interpretation of pressure values and the calculations leading to the depth of the tube's top.

Discussion Status

Some participants have provided insights into the reasoning behind the calculations, while others are questioning the assumptions made about pressure and height. There is an ongoing exploration of how to reconcile the calculated pressure with the expected depth of the tube.

Contextual Notes

Participants note the specific measurements given in the problem, such as the atmospheric pressure equivalent and the height of the water column inside the tube. There is a focus on understanding the relationship between these measurements and the final depth calculation.

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Homework Statement


An empty barometer tube,1m long is lowered vertically, mouth downwards, into a tank of water. what will be the depth of the top of the tube when the water has risen 20cm inside the tube?(atmospheric pressure may be assumed to be equal to 10.4m head of water)


Homework Equations


boyle's states p1v1= p2v2 assuming temperature remains constant.



The Attempt at a Solution



p1=10.4 m in m of water
v1= (1 x A) cubic meter

p2= (10.4 +h) in m of water
v2= (0.8 x A) cubic meter where a is area of cross section of tube.

substitute:
(10.4 +h) (0.8A)= 10.4 x 1 x A

10.4 +h = 10.4 divided by 0.8 = 13

and so the pressure due to the water coloum should be 13.10.4= 2.6

after this i can't get it right becase if the pressure is 2.6 then we should be able to use the formula pressure=ht x densityx g to find height of the water but the answer you get is not the answer in the book which says that the top of the tube is 1.8 m below the surface. It shows it as
2.6-0.8= 1.8

how exactly do you get 1.8 and why do you have to subtract 0.8 from 2.6 ?
 
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Hi aamphys,

aamphys said:

Homework Statement


An empty barometer tube,1m long is lowered vertically, mouth downwards, into a tank of water. what will be the depth of the top of the tube when the water has risen 20cm inside the tube?(atmospheric pressure may be assumed to be equal to 10.4m head of water)


Homework Equations


boyle's states p1v1= p2v2 assuming temperature remains constant.



The Attempt at a Solution



p1=10.4 m in m of water
v1= (1 x A) cubic meter

p2= (10.4 +h) in m of water
v2= (0.8 x A) cubic meter where a is area of cross section of tube.

substitute:
(10.4 +h) (0.8A)= 10.4 x 1 x A

10.4 +h = 10.4 divided by 0.8 = 13

and so the pressure due to the water coloum should be 13.10.4= 2.6

after this i can't get it right becase if the pressure is 2.6 then we should be able to use the formula pressure=ht x densityx g to find height of the water but the answer you get is not the answer in the book which says that the top of the tube is 1.8 m below the surface. It shows it as
2.6-0.8= 1.8

how exactly do you get 1.8 and why do you have to subtract 0.8 from 2.6 ?



Because they want the distance from the water surface to the top of the tube. The distance from the water surface to the top of the water inside the tube is 2.6m, and the top of the tube is 0.8m above that, so 1.8m is the answer they are looking for.
 
isnt 2.6 suppose to be the pressure of the water column on the top of the tube ?
 
The fluid head is height already isn't it?

Hence with ψ = P/ρg

and P = ρgh

Haven't they already supplied you with the measurement that the surface between the air/water at the bottom of the tube is .8m from the top of the tube? And since the air/water interface is 2.6 m (of head) below the surface, that the top of the tube needs to be 2.6m - .8m?
 

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