When the U.S. submarine Squalus became disabled at a depth of 80 m, a cylindrica
ID: 1406087 • Letter: W
Question
When the U.S. submarine Squalus became disabled at a depth of 80 m, a cylindrical chamber was lowered from a ship to rescue the crew. The chamber had a radius of 1.20 m and a height of 4.70 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 p0 + gh, where p0 = 1.000 atm is the surface pressure and = 1024 kg/m3 is the density of seawater. Assume a surface temperature of 20.0°C and a submerged water temperature of 30.0°C.
(a) What is the air volume in the chamber at the surface?
m3
(b) If air had not been released from the tanks, what would have been the air volume in the chamber at depth h = 80.0 m?
m3
(c) How many moles of air were needed to be released to maintain the original air volume in the chamber?
mol
Explanation / Answer
A. At the surface, the air volume is
V1 = A*h = (pi*(1.00 m)^2)*4.70 m = 14.758 m^3
B. The temperature and pressure of the air inside the submarine at the surface are
T1 = 293 K
p1 = p0 = 1 atm
At depth h = 80 m
T2 = 243 K
p2 = p0 + rho*g*h = 1 atm + 1024*9.8*80*(1 atm/(1.01*10^5 Pa)) = 1 + 7.95 = 8.95 atm
Using ideal gas law
PV = nRT
Air volume at this depth will be
p1V1/p2V2 = T1/T2
V2 = (p1/p2)*(T2/T1)*V1 = (1/8.95)*(243/293)*14.758 = 1.36 m^3
C. Decrease in Volume is
Delta V = V1-V2 = 14.758 - 1.36 = 13.398 m^3
n = p*delta V/(R*T) = 8.95*1.01*10^5*13.398/(8.314*243)
n = 5.994*10^3 mol
in order for the submarine to maintain the original air volume in the chamber, 5.994*10^3 mol of air must be released.
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