What is the volume of a diver's air bubble at the surface of the sea?

[dave2945]

Hello everyone. Here is my problem:

At 25.1 meters below the surface of the sea (density is 1025 (kg)/m^3),
where the temp is 3.00 C, a diver exhales an air bubble that has a volume of 1.00 cm^3. If the surface temp of the sea is 15.00 C, what is the volume of the bubble just before it breaks the surface.

Here is what I got: (d=density)

pv=nrt under water P+dgh=p
PV=nrT surface so, (P+dgh)v=nrt and PV=nrT

The nr's cancel, so (PV/((P+dgh)(v)))=T/t
Simplifying, I get that V=((P+dgh)vT)/(Pt)

Only thing is, I don't have pressure at surface. I tried using 1.013X10^5 N/m^2 (1 atm) and got .174 m^3. And, I did remember to change to .01 meters from 1.00 cm^3. Answer seems reasonable but wrong. Any help would be appreciated.

 

[Hootenanny]

And, I did remember to change to .01 meters from 1.00 cm^3. Answer seems reasonable but wrong. Any help would be appreciated.

Just a quick check, you know that 1.00 cm³ \neq 0.01 m³ right?

 

[dave2945]

Oh yeah. It is 1.00X10^-6 m^3, right??

 

[Hootenanny]

Oh yeah. It is 1.00X10^-6 m^3, right??

Yup, that's right.

 

[dave2945]

OK, I tried that, subbing 10^-6 m^3, got 1.74X10^-5 cubic meters and still not right. Hmmmm..

 

[dave2945]

Anyone have any ideas?

 

[JMuse]

I have a question about the equation you use for the pressure of the gas underwater. Shouldn't the pressure of the gas underwater be equal to the pressure of the water since they are both in equilibrium at that height; which is just "dgh," not "P + dgh."

 

[Astronuc]

I have a question about the equation you use for the pressure of the gas underwater. Shouldn't the pressure of the gas underwater be equal to the pressure of the water since they are both in equilibrium at that height; which is just "dgh," not "P + dgh."

One must look at the 'absolute' pressure. At the surface of the water, the local or ambient pressure is 1 atm (~14.7 psia). As one descends in the water, the mass of water above contributes to pressure, and the local pressure in the water is given by P_a + \rhogh, where h is the depth of water, and \rho is density.

When the air bubble is released, it is at same pressure as the water. Now there is also a difference in temperature which must also be considered for this problem. As the temperature of the bubble increases, it would also expand due to temperature, but the bubble is always at the same pressure as the water around it.

 

[Astronuc]

Anyone have any ideas?

Please show the work. Remember that the temperature scale in °C (Celsius) is a 'relative' scale, and one must convert to an absolute scale K, if one uses ratios.

T (K) = T (°C) + 273.15

 

[dave2945]

Oh, ok. I just screwed up by not converting to K's. Thank u much for the help Astronuc.