91Ó°ÊÓ

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:

$P=P_0+\mathrm{ÒÏ}gh$

$PV=nRT$

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,

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

$V=12.568≈12.6{\text{″¾}}^3$

Hence, the air volume in the chamber at the surface is$12.6{\text{″¾}}^3.$

Pressure at depth hcan be found as,

$P=P_0+\mathrm{ÒÏ}gh$

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

$P=9.038×{10}^5\text{Pa}$

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,

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

$$\begin{gathered} \frac{1.01×{10}^5\left(12.568\right)\left(243\right)}{9.038×{10}^5\left(293\right)} \\ ∴V_2=1.16{\text{″¾}}^3 \end{gathered}$$

Hence, the air volume in the chamber at depth$h=80.0\text{″¾â€‰}\text{¾±²õ â¶Ä‰}1.16{\text{m}}^3.$

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,

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

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

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.