Question

When the U.S. submarine Squalus became disabled at a depth of 56.5 m, a cylindrical chamber was lowered from a ship to rescue the crew. The chamber had a radius of 2.18 m and a height of 3.88 m, was open at the bottom, and held two rescuers. It slid along a guide cable that a diver had attached to a hatch on the submarine. Once it reached the hatch and clamped to the hull, the crew could escape into the chamber. During the descent, air was released from tanks to prevent water from flooding the chamber. Assume the interior air pressure matched the water pressure at depth h as given by p=p0+ρ⁢g⁢h, where p0 = 1.00 atm is the surface pressure and ρ= 1040 kg/m³ is the density of seawater. Assume a surface temperature of 19.6 ˚C and a submerged water temperature of -34.4 ˚C. (a) What is the air volume in the chamber at the surface? (b) If air had not been released from the tanks, what would have been the air volume in the chamber at depth h = 56.5 m? Assume that h is a depth of water surface inside the chamber below the sea level. (c) How many moles of air were needed to be released to maintain the original air volume in the chamber?

Answer 1

for answer (a),

the air volume in the chamber at the surface is, v=(3.14)r^2h

for answer (b),

the air volume in the chamber at depth h=56.5 m, is

V2 = (P1*T2*V1)/(P1+pgh)*T1

For answer (c) number of moles,

n=P*(delta V)/R*T