Question
A cylindrical diving bell 3.0 m in diameter and 4 m tall with an open top of 25 degrees celsius, and the air temp 220 down is 5 degrees celsius. The density of seawater is 1025 kg/m^3. How high does the sea water rise in the bell when the bell is submerged?
Answers
solve for the new volume of air in the bell..
V2=V1P1T2/P2T1
Then the height will be (V1-V2)/V1 *4m
Temps have to be in Kelvins.
To get pressure, consider a column 1 m^2.
The weight of that column is 1m^2*height*density*g
so pressure at the 220m down is
P2=101.3kpa+220*densityseawater*g Pa and
P2=101.3kPa+.220*density*9.8 in kPa
V2=V1P1T2/P2T1
Then the height will be (V1-V2)/V1 *4m
Temps have to be in Kelvins.
To get pressure, consider a column 1 m^2.
The weight of that column is 1m^2*height*density*g
so pressure at the 220m down is
P2=101.3kpa+220*densityseawater*g Pa and
P2=101.3kPa+.220*density*9.8 in kPa