- #1
dave2945
- 5
- 0
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.
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.