# Homework Help: Boyle's Law

1. Aug 15, 2010

### ghostanime2001

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

2. Relevant equations
I DONT'T UNDERSTAND WHY 10 metres! T_T

3. 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

2. Aug 15, 2010

### 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)

3. Aug 15, 2010

### ghostanime2001

but isn't the water "Underwater" ? so im 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 similiar application of that experiment?

4. Aug 15, 2010

### 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?

5. Aug 15, 2010

### ghostanime2001

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

6. Aug 15, 2010

### ghostanime2001

200 kPa which i calculated. 100 kPa at the surface
Thats a wierd 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 lets 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? Im typing this up really fast. I hope im 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
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?

7. Aug 15, 2010

### 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