What is the radius of the spherical air bubble before it reaches the surface?

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SUMMARY

The problem involves calculating the radius of a spherical air bubble rising from a depth of 10 meters in a lake, where the initial radius is 1 mm at a temperature of 4.0 degrees Celsius and the surface temperature is 27 degrees Celsius. The atmospheric pressure at the surface is 1030 mbar. To solve this, the ideal gas law (P1*V1/T1 = P2*V2/T2) is applicable, along with the volume formula for a sphere to find the new radius after accounting for pressure changes due to depth. The pressure at the bottom of the lake must be calculated using the depth and converted to SI units.

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Homework Statement


Hi,i came across this problem in a past exam paper and i got a bit stuck on it,can anyone enlighten me please?im prob just missing something really obvious and basic...

A spherical air bubble at the bottom of a lake,10m below the surface,has a radius of 1mm. The temperature at the bottom of the lake is 4.0degrees celsius. The bubble gets dislodged and rises slowly to the surface,where the temperature is 27degrees celsius.if the atmospheric pressure is 1030 mbar,what is the radius of the bubble just before it reaches the surface?(neglect surface tension)

Homework Equations


i really don't know...


The Attempt at a Solution


my first problem is that i don't know the conversion factor for millibars to an SI pressure unit,or if i even need to convert the pressure value to SI units?

could i possibly use ratios to solve for the volume of the bigger bubble,such as the ideal gas equation: P1*V1/T1= P2*V2/T2?
and then use the volume of a sphere formula to solve for the radius?

although i don't think this approach is right as i would not be making use of the ten metre figure??any ideas or hints would be fantastic,as they don't give solutions at the back of the papers,just answers,which arent really that much use to me... thanks
 
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Your approach is generally correct. You would be making use of the 10 meters in finding the pressure difference between the two locations. 1 bar is 10^5 Pa

http://en.wikipedia.org/wiki/Pressure
 
thanks for that-much appreciated!:biggrin:
 

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