• Support PF! Buy your school textbooks, materials and every day products Here!

Air bubble in water at 50m

  • Thread starter soopo
  • Start date
  • #1
225
0

Homework Statement


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


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).
 
Last edited by a moderator:

Answers and Replies

  • #2
Doc Al
Mentor
44,882
1,129
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.)
 
  • #3
225
0
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 = [tex] \frac { T1 / P1 } { T2 / P2} [/tex]


V1 / V2 = [tex] \frac {290(p1 + 5atm)} {p1 * 300}[/tex]
= 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.
 
  • #4
Doc Al
Mentor
44,882
1,129
I have the following

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


V1 / V2 = [tex] \frac {290(p1 + 5atm)} {p1 * 300}[/tex]
= 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?
 
  • #5
225
0
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 celcius and 300 kelvins is 27 celcius.

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.
 
  • #6
LowlyPion
Homework Helper
3,090
4
Are you sure it's not 6*300/290 ?
 
  • #7
Doc Al
Mentor
44,882
1,129
It is one atm :)
Good. :wink:

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.
 
  • #8
225
0
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!
 

Related Threads for: Air bubble in water at 50m

  • Last Post
Replies
21
Views
938
Replies
8
Views
4K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
963
  • Last Post
Replies
2
Views
3K
Top