Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q12P

Expert-verified
Fundamentals Of Physics
Found in: Page 578

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

When the US submarine Squalus became disabled at a depth of , a cylindrical chamber was lowered from a ship to rescue the crew. The chamber had a radius of 1m and a height of 4m, 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 the chamber 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 that the interior air pressure matched the water pressure at depth h as given by role="math" localid="1662369677002" p0+ρgh, where p0=1.000 atm

is the surface pressure and ρ=1024kgm3 is the density of sea water.

Assume a surface temperature of 20°C and a submerged water temperature of -30°C.

  1. What is the air volume in the chamber at the surface?
  2. If air had not been released from the tanks, what would have been the air volume in the chamber at depth h = 80.0m?
  3. How many moles of air were needed to be released to maintain the original air volume in the chamber?

  1. The air volume in the chamber at the surface is12.6 m3.
  2. The air volume in the chamber at depth h=80.0 m  is   1.16 m3.
  3. The number of moles of air needed to be released to maintain the original air volume in the chamberis. 5.1×103 mol

See the step by step solution

Step by Step Solution

Step 1: Determine the concept

Find the volume of air in the chamber using the formula for volume of cylinder. Air volume at height h can be calculated from pressure and temperature at that height and pressure, volume, and temperature at the surface using gas law. Number of moles can be calculated by using gas law.

Formulae are as follow:

P=P0+ρgh

PV=nRT

Here, P is pressure, V is volume, T is temperature, is density, h is height, g is an acceleration due to gravity,R is universal gas constant and n is number of moles.

Step 3: (a) Determine the air volume in the chamber at the surface

Air volume in the chamber at the surface can be found as,

V=πr2h'V=3.142(1)2(4)

V=12.568 12.6 m3

Hence, the air volume in the chamber at the surface is12.6 m3.

Step 4: (b) Determine the air volume in the chamber at depth

Pressure at depth hcan be found as,

P=P0+ρgh

P=1.01×105+1024(9.8)(80)

P=9.038×105 Pa

According to the gas law,

PV=nRTP1V1P2V2=T1T2

Volume of the air at depth hcan be found as,

V2=P1V1P2T2T1

1.01×105(12.568)(243)9.038×105(293)V2=1.16 m3

Hence, the air volume in the chamber at depth h=80.0 m  is   1.16 m3.

Step 5: (c) Determine the number of moles of air needed to be released to maintain the original air volume in the chamber

The number of moles of air needed to be released to maintain the original air volume in the chamber is given by gas law as,

n=PVRT9.038×105(12.61.16)8.314×243n=5.1×103 mol

Hence, the number of moles of air needed to be released to maintain the original air volume in the chamber is.5.1×103 mol

Therefore, the volume and the number of moles of air at height hcan be found from its pressure, volume, and temperature at the surface using gas law.

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.