A lake diver releases an air bubble of 1.55mm radius at a depth of 42.4 m under

Question

A lake diver releases an air bubble of 1.55mm radius at a depth of 42.4 m under water. Assuming that the temperature at the 42.4 m depth is 4.58 degrees Celsius and that the surface temperature is 26.3 degree s Celsius, determine the bubble radius when it reaches the (fresh) water surface. Note: the density of fresh water is 103 kg/m3. If you have 3 moles of air at room temperature in a cylinder with a length of 1m and radius of 0.03m, what is the pressure of the gas inside and outside the cylinder?

Explanation / Answer

P1V1 = P2V2 P1= atmospheric pressure = 1 atm P2= Pressure at a depth of 42.4 m = 1+(103*9.8*42.4*1000/760) = 57.3 atm V2 = (4/3)pr23 at a depth of 42.2m....r2= 1.55 mm 1*r13=57.3*r23 r1= 5.98mm

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