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

In summary: Yes, they have already supplied the information that the air/water interface is 2.6m below the surface.
  • #1
aamphys
8
0

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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  • #2
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.
 
  • #3
isnt 2.6 suppose to be the pressure of the water column on the top of the tube ?
 
  • #4
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?
 

1. What is Boyle's Law?

Boyle's Law is a gas law that states the relationship between pressure and volume of a gas at constant temperature. It states that as the volume of a gas decreases, the pressure increases proportionally.

2. Who discovered Boyle's Law?

Boyle's Law was discovered by chemist and physicist Robert Boyle in 1662.

3. What is the formula for Boyle's Law?

The formula for Boyle's Law is P1V1 = P2V2, where P1 and V1 are the initial pressure and volume of a gas, and P2 and V2 are the final pressure and volume of the gas, respectively.

4. What units are used to measure pressure and volume in Boyle's Law?

Pressure is typically measured in units of atmospheres (atm) or pascals (Pa), while volume is measured in units of liters (L) or cubic meters (m3).

5. What is an example of Boyle's Law in everyday life?

An example of Boyle's Law in everyday life is the use of a spray bottle. When the trigger is pressed, the volume of the gas inside the bottle decreases, causing an increase in pressure, which then forces the liquid out of the nozzle in a fine mist.

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