# Question on boyle's law

1. Dec 5, 2008

### aamphys

1. The problem statement, all variables and given/known data
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)

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

3. 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 cant 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 ???

2. Dec 5, 2008

### alphysicist

Hi aamphys,

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. Dec 5, 2008

### aamphys

isnt 2.6 suppose to be the pressure of the water column on the top of the tube ?

4. Dec 5, 2008

### LowlyPion

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?