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Submarine rescue . When the U.S. submarine Squalus became disabled at a depth of

ID: 1666228 • Letter: S

Question

Submarine rescue. When the U.S. submarineSqualus became disabled at a depth of 80 m, a cylindricalchamber was lowered from a ship to rescue the crew. The chamber hada radius of 1.20 m and a height of4.60 m, was open at the bottom, and heldtwo rescuers. It slid along a guide cable that a diver had attachedto a hatch on the submarine. Once it reached the hatch and clampedto the hull, the crew could escape into the chamber. During thedescent, air was released from tanks to prevent water from floodingthe chamber. Assume the interior air pressure matched the waterpressure at depth h as given by p0 +gh, where p0 =1.000 atm is the surface pressure and = 1024kg/m3 is the density of seawater. Assume a surfacetemperature of 20.0°C and a submerged water temperature of-30.0°C. (a) What is the air volume in the chamber atthe surface?
1 m3

(b) If air had not been released from the tanks, what would havebeen the air volume in the chamber at depth h =89.0 m?
2 m3

(c) How many moles of air were needed to be released to maintainthe original air volume in the chamber?
3 mol
(a) What is the air volume in the chamber atthe surface?
1 m3

(b) If air had not been released from the tanks, what would havebeen the air volume in the chamber at depth h =89.0 m?
2 m3

(c) How many moles of air were needed to be released to maintainthe original air volume in the chamber?
3 mol

Explanation / Answer

a, The volume of the chambe at the surface is Vo =S l = r^2 l = x 1.2^2 x 4.6 = 20.8 m^3 the initial mol of air in the chamber is no =PVo / RTo = 854.7 mol b, The pressure at the depth of 80m is P = Po+ gh = 919 200 Pa If air was not released, the volume at that depth should be V= noRT / P = 1.88 m^3 c, If air was released to fill the chamber, the number of molshoud be n / no = Vo / V => n = noVo / V = 9468.2 mol So the number of mol of air needed to be released is n =n - no = 8613.5 mol