Science, Physical Chemistry
posted by Tufan .
On the bottom of a sea (depth:10m) a bubble of a mathane gas if formed (diameter,d bottom=1 cm). What is the diameter of the bubble, when it reaches at the surface? The sea water temperature is 5 C at the bottom and 25 C on the surface.
Try the following:
The pressure on the bubble at a depth of 10 m is Patm + (density of water x gravity x height).
101325 Pa + (1000 kg/m^3 x 9.8 m/s^2 x 10 m) = ?? (The density of water is very near 1000 km/m^3 at 4 degrees C. 5 degrees C won't change it very much.)
The volume of the sphere of methane is 4/3(pi)(r^3).
Now use (P1V1)/T1 = (P2V2)/T2 to convert the volume of methane at the 10 m depth to V2 at the surface of the water where atmospheric pressure is 101325 pa. Then using V2 = 4/3(pi)(r^3) recalculate the radius of the methane bubble at the surface. Check my thinking. Don't forget to use Kelvin for temperature.