Question

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$\mathbf{1}\mathbf{m}$and a height of$\mathbf{4}\mathbf{m}$, 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 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 byrole="math" localid="1662369677002" ${\mathbf{\text{p}}}_{\mathbf{\text{0}}}\mathbf{\text{+ρgh}}$, where${\text{p}}_{\text{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{000}\mathbf{}\mathbf{atm}$

is the surface pressure and$\mathbf{\text{ρ=}}$$\mathbf{1024}\frac{\mathrm{kg}}{{\mathbf{m}}^3}$is the density of sea water.

Assume a surface temperature of$\mathbf{20}\mathbf{°}\mathbf{C}$and a submerged water temperature of$\mathbf{-}\mathbf{30}\mathbf{°}\mathbf{C}$.

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 =$\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{m}$?
3. How many moles of air were needed to be released to maintain the original air volume in the chamber?

Short answer

1. The air volume in the chamber at the surface is$12.6{\text{\hspace{0.17em}m}}^3.$
2. The air volume in the chamber at depth$h=80.0\text{\hspace{0.17em}m\hspace{0.17em}}\text{is\hspace{0.17em}\hspace{0.17em}}1.16{\text{m}}^3.$
3. The number of moles of air needed to be released to maintain the original air volume in the chamberis.$5.1\times {10}^3\text{mol}$

Step-by-step solution

Determine the concept

Find thevolume 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:

$P=P_0+\rho gh$

$PV=nRT$

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.

(a) Determine the air volume in the chamber at the surface

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

$$\begin{gathered} V=\pi r^{2}h\text{'} \\ V=3.142{\left(1\right)}^2\left(4\right) \end{gathered}$$

$V=12.568\approx 12.6{\text{\hspace{0.17em}m}}^3$

Hence, the air volume in the chamber at the surface is$12.6{\text{\hspace{0.17em}m}}^3.$

(b) Determine the air volume in the chamber at depth

Pressure at depth hcan be found as,

$P=P_0+\rho gh$

$P=1.01\times {10}^5+1024\left(9.8\right)\left(80\right)$

$P=9.038\times {10}^5\text{Pa}$

According to the gas law,

$$\begin{gathered} PV=nRT \\ \frac{P_1{V}_1}{P_2{V}_2}=\frac{T_1}{T_2} \end{gathered}$$

Volume of the air at depth hcan be found as,

$\therefore V_2=\frac{P_1{V}_1}{P_2}\frac{T_2}{T_1}$

$$\begin{gathered} \frac{1.01\times {10}^5\left(12.568\right)\left(243\right)}{9.038\times {10}^5\left(293\right)} \\ \therefore V_2=1.16{\text{\hspace{0.17em}m}}^3 \end{gathered}$$

Hence, the air volume in the chamber at depth$h=80.0\text{\hspace{0.17em}m\hspace{0.17em}}\text{is\hspace{0.17em}\hspace{0.17em}}1.16{\text{m}}^3.$

(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,

$$\begin{gathered} n=\frac{P△V}{RT} \\ \frac{9.038\times {10}^5(12.6-1.16)}{8.314\times 243} \\ n=5.1\times {10}^3\text{mol} \end{gathered}$$

Hence,the number of moles of air needed to be released to maintain the original air volume in the chamber is.$5.1\times {10}^3\text{mol}$

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.