Question

A diving bell is a $3.0-m$-tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is $20°C$. The bell is lowered into the ocean until its lower end is $100m$deep. The temperature at that depth is $10°C$.

a. How high does the water rise in the bell after enough time has passed for the air inside to reach thermal equilibrium?

b. A compressed-air hose from the surface is used to expel all the water from the bell. What minimum air pressure is needed to do this?

Short answer

a. The water rises$2.7m$ in the bell.

b. Minimum air pressure needed to do this is$11atm$

Step-by-step solution

Defining Ideal Gas Law and Process

Given : Height of cylinder : $3.0-m$

Temperature of the air : $20°C$.

Depth of the bell in the ocean : $100m$

Temperature at that depth : $10°C$

Theory used :
The ideal gas law is given by $pV=nRT$
The ideal gas law for the process with constant number of moles$n$for the closed system is
$\frac{p_2{V}_2}{T_2}=\frac{p_1{V}_1}{T_1}$

Calculating how high the water rises in the bell after enough time has passed for the air inside to reach thermal equilibrium 

(a) At the beginning the air has the length $L₁=3.0m$. The temperature is $T₁=20°+273=293K.$The pressure is equal to the atmospheric pressure $p₁=1.013x{10}^{5}Pa$. The volume of the air is $V₁=L₁A=3A$
Where$A$ is the area of the cylinder.
Finally, the length of the air is $L_2$. The temperature is $T₂=10°+273=283K$. The pressure after a depth of $h=100m\mathrm{is}{p}_2=p_{atm}+\rho gh$

$$\begin{gathered} =1.013\times {10}^{5}Pa+(1000kg/m³)(9.8m/s²)(100m) \\ =10.813\times {10}^{5}Pa \end{gathered}$$

The final volume of air is $V₂=L₂A$. Now we calculate$L₂$.
$$\begin{gathered} \frac{p_2{V}_2}{T_2}=\frac{p_1{V}_1}{T_1} \\ \Rightarrow \frac{(10.813x{10}^{5}Pa)(L₂A)}{283K}=\frac{(1.013x109Pa)\left(34\right)}{293K} \\ \Rightarrow 0.038L_2=0.0103 \\ \Rightarrow L_2=0.27m \end{gathered}$$
This is the length of the air inside the cylinder, so the height of the water inside the cylinder is equal to the difference between the initial length and the final length of the air
$$\begin{gathered} ∆h=L₁-L₂=3m-0.027m \\ =\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{}\mathit{m}\mathbf{} \end{gathered}$$

Calculating the minimum air pressure needed 

(b) To repel all the water, the air must maintain its ultimate pressure, which is given by$p=p₁+\rho gh$
$$\begin{gathered} =1.013\times {10}^{5}Pa+(1000kg/m³)(9.8m/s²)(100m) \\ =10.813x{10}^{5}Pa \end{gathered}$$
Convert the pressure from pascals to atm so that it has the formrole="math" localid="1648372473152" $$\begin{gathered} p=(10.813x{10}^{5}Pa)\frac{1atm}{1.013x{10}^{5}Pa} \\ =\mathbf{11}\mathbf{}\mathit{a}\mathit{t}\mathit{m} \end{gathered}$$