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March 31, 2015

March 31, 2015

Posted by **Tufan** on Tuesday, March 25, 2008 at 12:50pm.

- Science, Physical Chemistry -
**DrBob222**, Tuesday, March 25, 2008 at 1:40pmTry 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.

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