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

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SUMMARY

The discussion focuses on calculating the depth of the top of a barometer tube submerged in water, where the atmospheric pressure is equivalent to a 10.4m head of water. The problem involves applying Boyle's Law, represented as p1v1 = p2v2, to determine the pressure and height relationships. The correct depth of the tube's top is established as 1.8m below the water surface by subtracting the height of the water column inside the tube (0.8m) from the total pressure head (2.6m). This calculation clarifies the relationship between pressure, volume, and height in fluid mechanics.

PREREQUISITES
  • Understanding of Boyle's Law and its application in fluid mechanics.
  • Basic knowledge of pressure calculations in fluids, specifically using the formula P = ρgh.
  • Familiarity with the concept of fluid head and its implications in barometric measurements.
  • Ability to manipulate algebraic equations to solve for unknown variables.
NEXT STEPS
  • Study the principles of fluid statics and dynamics to deepen understanding of pressure relationships.
  • Learn about the applications of Boyle's Law in real-world scenarios, such as barometers and gas laws.
  • Explore the concept of hydrostatic pressure and its calculations in various fluid systems.
  • Investigate the effects of temperature on gas behavior and pressure in closed systems.
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Students in physics or engineering, educators teaching fluid mechanics, and anyone interested in understanding the principles of pressure and buoyancy in fluids.

aamphys
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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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