A spherical balloon is submerged in the ocean (p = 1021 kg/m^3) all a depth of 5

Question

A spherical balloon is submerged in the ocean (p = 1021 kg/m^3) all a depth of 50 m. The temperature of the water (and air in the ball) is 8 degree C. The beach ball contains 5.0 mol of air molecules. The volume of a sphere is 4/3 pi r^3. (a) What is the pressure in the water at a depth of 50 m? (b) What is the volume of the balloon at a depth of 50 m? (c) What is the buoyant force on the balloon at a depth of 50 m? (d) If the balloon is brought to the surface (so that it is not submerged at all) what is its radius? (Assume it remains spherical.)

Explanation / Answer

1) pressure at 50 m depth=pgh=1021*9.8*50=500290 pa

2)pv=nRT

500290*v=5*8.314*(273+8)

v=0.02335 m3

3) buoyancy=pvg=1021*0.02335*9.8=233.6234 N

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