Can someone please help me with my physics?

Question

Please show all formulas and steps.

A cylindrical diving bell 3.0 m in diameter and 4.1 m tall with an open bottom is submerged to a depth of 225 m in the ocean. The surface temperature of the water is 25°C, and the temperature 225 m down is 5.0°C. The density of sea water is 1025 kg/m3. How high does the sea water rise in the bell when the bell is submerged?


First to have correct answers receives ten points.

Thank you very much.

Answer 1

First, the air will be compressed until the pressure matches that of the sea at a depth of 225 m. Pressure at sea level is about 1.01 x 10^5 N / m^2. The increase in pressure at that depth is

ρ g h

where ρ is the density of the water ( given ), g the acceleration of gravity ( 9.8 m/s^2 ), and h the depth ( 225 m ). In your problem, that increase is about 2.3 x 10^6 N / m^2, or about 22.4 atmospheres. This will compress the air in the bell to about 4.5 percent of its original volume, and the water will rise to about 18 cm below the top of the bell.

Second, the density of air is inversely proportional to temperature. When the air is cooled from 25 oC ( 298 K ) to 5 oC ( 278 K ) the volume decreases by about five percent. This will be an insignificant effect in this problem.