Depth of a jam tin submerged open-end down using Boyle's Law

[ghostanime2001]

Homework Statement

Atmospheric pressure is 100 kPa and a depth of 1m in water is equivalent in pressure to 10 kPa. A jam tin, open end down, is forced under water until the water half fills the can. At this stage, how far beneath the surface is the can? (ans is 10 m)

Homework Equations

I DONT'T UNDERSTAND WHY 10 metres! T_T

The Attempt at a Solution

So from what I know, I must use Boyle's Law.
P1 = 100 kpa
P2 = ?
V1= ?
V2 = 1/2 V1

P1V1=P2V2
100V1=P2(1/2 V1)
100V1=1/2 P2V1 <---Don't the V1's cancel out?
200=P2

using ratio's:
$$\frac{1 m}{10 kPa}$$ = $$\frac{x}{200 kPa}$$

x=20 m

Why is my answer different from my worksheet (10m) ??

Any help is greatly appreciated!
Thanks

 

[mgb_phys]

when the water half fills the can there must be twice as much pressure as at the surface
so the question is asking what depth of water s equal to atmopsheric pressure.

Atmopsheric pressure is given as 100KPa - what does this mean?
(what column of water would have the same weight)

 

[ghostanime2001]

but isn't the water "Underwater" ? so I am getting the picture has not floating on top but under.. and.. isn't there an experiment where u have a bottle or something and u flip it and force it down the inside of the bottle has air in it and not water. Isn't this a similar application of that experiment?

 

[ghostanime2001]

I understand that outside the water the pressure of the atmosphere is 100 kPa and doesn't "half fills the can" mean the volume is reduced by half of the original volume of the can? so 1/2 V1 = V2 does make sense right?

 

[ghostanime2001]

How would I show algebraically that the pressure is 100 kPa?

 

[ghostanime2001]

when the water half fills the can there must be twice as much pressure as at the surface

200 kPa which i calculated. 100 kPa at the surface

so the question is asking what depth of water s equal to atmopsheric pressure.

Thats a weird question.. the pressure underneath the water PLUS the pressure from the atmosphere totals 200 kPa!

I think this is where I went wrong. I assumed that 200 kPa was the pressure the gas inside the can was experiencing and I had wrongly assumed that the 200 kPa was from the water only!

From what i can understand from this question. At first when the can is opened there is already pressure on it right? from the atmosphere and then let's say i forced the can down to 1 m underneath water that means the total pressure is 100 kPa + 10 kPa (from 1m in water) = 110 kPa total pressure. since 100 kpa is already forced upon the water and the can is INSIDE the water all i had left was 100 kpa more pressure from water therefore totalling 200 kpa. Am I right in this line of thinking? I am typing this up really fast. I hope I am not wrong :( If I am incorrect in my line of thinking.. please help me correct it.. Thanks guys! that was puzzling :S but good!

ORRR i can use boyle's law where it says that PV = constant!

lets say:
P1=100
P2=200
V1=?
V2=1/2 V1

plug and play into boyle's expression:
P1V1=P2V2
100V1=200(1/2 V1)
100V1=100V1
which both of them are equal!

does the 1/2 factor simplify the pressure down to 100? and have nothing to do with the volume? just like u said before

so the question is asking what depth of water s equal to atmopsheric pressure.

If I didn't know all this would I use 200 kPa or simplify it down like that and use 100 kPa using 1/2 factor from volume?

I know its a lot of questions.. but please help me clarify this.. thanks.

 

[mgb_phys]

P1V1 = P2V2
The pressure * volume is the same before and after

So if the volume after is half then the pressure is twice as much.
V2=0.5V1
P2 = P1/0.5 =2*P1

Pressure at surface is 100KPa so pressure underwater must be 200KPa
You are told that each m of water gives you 10KPa so to get an extar 100Kpa you need 100/10 = 10m of water