Question

The $3.0-m$ -long pipe in FIGURE P18.50 is closed at the top end. It is slowly pushed straight down into the water until the top end of the pipe is level with the water’s surface. What is the length $L$of the trapped volume of air?

Short answer

The length$L$of the trapped volume of air is $\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}$

Step-by-step solution

: Given information and Formula used 

Given : Length of pipe :$3.0-m$

It is slowly pushed straight down into the water until the top end of the pipe is level with the water’s surface

Theory used :

In thermodynamics, an Isothermal process is a type of thermodynamic process in which the temperature $T$of a system remains constant: $\Delta T=0$

Calculating L

The process isisothermal because the glass was slowly pushed into the water and the mass of water would cool any increased temperature. Therefore, we can write:
$p_{a}V=pV₁$,
where $V=AH\mathrm{is}\mathrm{the}\mathrm{total}\mathrm{volume}\mathrm{and}V₁=AL$is the volume of air during immersion. By substitution, we can express$L$ as
$$\begin{gathered} p_{a}AH=pAL \\ \Rightarrow L=\frac{p_a}{p}H \end{gathered}$$
The pressure in the compressed volume is equal to the sum of the atmospheric pressure and the water column above the free surface of the water, i.e., the pressure from the height$L$ of the water. So our pressure is
$p=p_a+\rho gL$
If we substitute the expression for L, we get
$$\begin{gathered} L=\frac{p_a}{p_a+\rho gL}H \\ \Rightarrow p_{a}H=L{p}_a+\rho gL² \\ \Rightarrow \rho gL²+p_{a}L-p_{a}H=0. \end{gathered}$$
Its roots are
$$\begin{gathered} L=\frac{-p_a\pm \sqrt{p^2+4\rho g{p}_{a}H}}{2\rho g} \\ =\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m} \end{gathered}$$
It is clear that the length cannot be negative, so we will take only the positive roots of this equation.