What is the volume of an air bubble in water at a depth of 50m?

[soopo]

Homework Statement

The problem is at
http://dl.getdropbox.com/u/175564/volumePressureTemperature.JPG

Homework Equations

I know the equation for ideal gases
pV = nRT
and
the definition of pressure
p = F / A

The Attempt at a Solution

The change in T is 10.
V_i = 1 cm^3

We need to first consider how the change in Temperature affects
the depth at which the bubble is.
It will goes up because the volume of the bubble increases.

The increase in Volume causes the boyonce force to increase.
Thus, the net force in f direction is zero at both depths:
F_boyonce - F_gravity = 0
g( m_Water - m_airBubble ) = 0

There is a linear relation between p and T if we assume the
situation for ideal gases.

I have apparently ignored some critical part.
It seems that the ideal Gas equation is not valid here.
The right answer is 6.2cm^3 that is (f).

 

[Doc Al]

It seems that the ideal Gas equation is not valid here.

Sure it is--it's all you need for this problem. How does the pressure change? The temperature? Use the ideal gas law to determine how the volume changes. (Set up ratios.)

 

[soopo]

Sure it is--it's all you need for this problem. How does the pressure change? The temperature? Use the ideal gas law to determine how the volume changes. (Set up ratios.)

I have the following

V1 / V2 = \frac { T1 / P1 } { T2 / P2}V1 / V2 = \frac {290(p1 + 5atm)} {p1 * 300}
= 290/300 + 1450/(P1 * 300)

The problem is now P1 which is the initial pressure.
It seems that we need to approximate it, since we do not know it.

 

[Doc Al]

I have the following

V1 / V2 = \frac { T1 / P1 } { T2 / P2}


V1 / V2 = \frac {290(p1 + 5atm)} {p1 * 300}
= 290/300 + 1450/(P1 * 300)

You have the ratios a bit backwards. Start with this:
P₁V₁/T₁ = P₂V₂/T₂

The problem is now P1 which is the initial pressure.
It seems that we need to approximate it, since we do not know it.

You should know it. What's the pressure at the surface?

 

[soopo]

You have the ratios a bit backwards. Start with this:
P₁V₁/T₁ = P₂V₂/T₂


You should know it. What's the pressure at the surface?

It is one atm :)

I get the following result
V1 / V2 = 6 * 290/300
= 5.8

whrere 290 kelvins is 17 celsius and 300 kelvins is 27 celsius.

Taking inverse

V2 / V1 = 0.17
where V2 is the volume at the surface.

Both answers seem to be wrong: I cannot get 6.2.
I have likely ignored some small volume.

 

[LowlyPion]

Are you sure it's not 6*300/290 ?

 

[Doc Al]

It is one atm :)

Good.

I get the following result
V1 / V2 = 6 * 290/300
= 5.8

Once again, you're flipping things around somehow.

Start with this:
(P₁V₁)/T₁ = (P₂V₂)/T₂

Then solve for V₁.

 

[soopo]

Are you sure it's not 6*300/290 ?

I finally got the right answer.

V1 = P2 V2 T1 / T2 P1

= 6 * 30 / 29
= 6.2

Thank you for your answers!