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Q12P

Expert-verifiedFound in: Page 578

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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 $\mathbf{1}\text{\hspace{0.17em}}\mathbf{m}$**** and a height of $\mathbf{4}\text{\hspace{0.17em}}\mathbf{m}$****, 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 by role="math" localid="1662369677002" ${\mathbf{\text{p}}}_{\mathbf{\text{0}}}\mathbf{\text{+\rho gh}}$****, where ${{\text{p}}}_{{\text{0}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{000}}{\mathbf{}}{\mathbf{atm}}$**

**is the surface pressure and ${\mathbf{\text{\rho =}}}$$\mathbf{1024}\text{\hspace{0.17em}}\frac{\mathrm{kg}}{{\mathbf{m}}^{3}}$**** is the density of sea water.**

**Assume a surface temperature of $\mathbf{20}\mathbf{\xb0}\mathbf{C}$**** and a submerged water temperature of ${\mathbf{-}}{\mathbf{30}}{\mathbf{\xb0}}{\mathbf{C}}$****.**

**What is the air volume in the chamber at the surface?****If air had not been released from the tanks, what would have been the air volume in the chamber at depth***h*= ${\mathbf{80}}{\mathbf{.}}{\mathbf{0}}{\text{\hspace{0.17em}}}{\mathbf{m}}$**?****How many moles of air were needed to be released to maintain the original air volume in the chamber?**

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

**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:

$P={P}_{0}+\rho gh$

$PV=nRT$

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.

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

$V=\pi {r}^{2}h\text{'}\phantom{\rule{0ex}{0ex}}V=3.142{\left(1\right)}^{2}\left(4\right)$

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

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

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,

$PV=nRT\phantom{\rule{0ex}{0ex}}\frac{{P}_{1}{V}_{1}}{{P}_{2}{V}_{2}}=\frac{{T}_{1}}{{T}_{2}}$

Volume of the air at depth *h*can be found as,

$\therefore {V}_{2}=\frac{{P}_{1}{V}_{1}}{{P}_{2}}\frac{{T}_{2}}{{T}_{1}}$

$\frac{1.01\times {10}^{5}\left(12.568\right)\left(243\right)}{9.038\times {10}^{5}\left(293\right)}\phantom{\rule{0ex}{0ex}}\therefore {V}_{2}=1.16{\text{\hspace{0.17em}m}}^{3}$

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

** **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,

$n=\frac{P\u25b3V}{RT}\phantom{\rule{0ex}{0ex}}\frac{9.038\times {10}^{5}(12.6-1.16)}{8.314\times 243}\phantom{\rule{0ex}{0ex}}n=5.1\times {10}^{3}\text{mol}$

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 *h*can be found from its pressure, volume, and temperature at the surface using gas law.

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