When the U.S. submarine Squalus became disabled at a depth of 88.0 m, a cylindri

Question

When the U.S. submarine Squalus became disabled at a depth of 88.0 m, a cylindrical chamber was lowered from a ship to rescue the crew. The chamber had a radius of 2.35 m and a height of 2.30 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

density of seawater. Assume a surface temperature of 22.8 C and a submerged water temperature of -32.0 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 = 88.0 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?

Explanation / Answer

(a) The air volume in the chamber at the surface which is given as :

Vair = (pi) r2 h = (3.14) (2.35 m)2 (2.3 m)

Vair = 39.8 m3

(b) If air had not been released from the tanks, then the air volume in the chamber at depth h = 88 m which will be given as :

For the water pressure at depth h, we have

P = P0 + (rho) g h

P = (101325 Pa) + (1020 kg/m3) (9.8 m/s2) (88 m)

P = (101325 Pa) + (879648 Pa)

P = 980973 Pa

P = 980.9 kPa

using a combined gas law, we have

P1 V1 / T1 = P2 V2 / T2

where, T1 = initial temperature = (273 + 22.8) K = 295.8 K

T2 = final temperature = (273 - 32) K = 241 K

(101.3 kPa) (39.8 m3) / (295.8 K) = (980.9 kPa) V2 / (241 K)

(971649.3 kPa m3) / (295.8 K) = (980.9 kPa) V2

V2 = (971649.3 kPa m3.K) / (290150.2 kPa. K)

V2 = 3.34 m3

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