Question

The 3.0-m-long pipe in the figure (Figure 1) 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. Assume that the process is isothermal.

What is the length L of the trapped volume of air?

The 3.0-m-long pipe in the figure (Figure 1) 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. Assume that the process is isothermal. What is the length L of the trapped volume of air?

Answer 1

Concepts and reason

The concepts used to solve this problem are the isothermal expression for ideal gas and the pressure.

Use the mathematical expression for Boyle’s law for isothermal process. Calculate the initial pressure, the initial volume, the final volume and the final pressure. Calculate the volume by using the expression in terms of area and length.

The quadratic equation comes after solving the expression. Calculate the roots of the equation by using the discriminant method to get the value of the length.

Fundamentals

Write the mathematical expression for Boyle’s law.

${P_1}{V_1} = {P_2}{V_2}$

Here, ${P_1}$ is the initial pressure, is the final pressure, ${V_1}$ is the initial volume and ${V_2}$ is the final volume.

Write the expression for the net pressure in terms of atmospheric pressure and pressure due to the liquid column.

$P = {P_0} + \rho gh$

Here, ${P_0}$ is the atmospheric pressure, $\rho$ is the density, $g$ is the acceleration due to gravity and $h$ is the distance.

Write the expression of volume.

$V = aL$

Here, $a$ is the area of the cross-section and $L$ is the length.

Consider the provided diagram.

Calculate the initial pressure.

${P_1} = {P_0}$

Calculate the initial volume.

${V_1} = aL$

Substitute $3.0{\rm{ m}}$ for $a$ .

${V_1} = a\left( {3.0{\rm{ m}}} \right)$

Calculate the final pressure.

${P_2} = {P_0} + \rho gL$

Calculate the final volume.

${V_2} = aL$

Use the expression for mathematical expression for Boyle’s law

${P_1}{V_1} = {P_2}{V_2}$

Substitute ${P_0}$ for ${P_1}$ , $\left( {{P_0} + \rho gL} \right)$ for ${P_2}$ , $a\left( {3.0{\rm{ m}}} \right)$ for ${V_1}$ and $aL$ for ${V_2}$ .

$\begin{array}{c}\\{P_0}\left( {a\left( {3.0{\rm{ m}}} \right)} \right) = \left( {{P_0} + \rho gL} \right)\left( {aL} \right)\\\\{P_0}\left( {3.0{\rm{ m}}} \right) = \left( {{P_0} + \rho gL} \right)\left( L \right)\\\\{P_0}\left( {3.0{\rm{ m}}} \right) = {P_0}L + \rho g{L^2}\\\end{array}$

Rearrange the expression.

$\rho g{L^2} + {P_0}L - {P_0}\left( {3.0{\rm{ m}}} \right) = 0$

Calculate the roots by using the discriminant method.

$\alpha ,\beta = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Substitute ${P_0}$ for $b$ , $\rho g$ for $a$ and $\left( { - {P_0}\left( {3.0{\rm{ m}}} \right)} \right)$ for $c$ .

$\alpha ,\beta = \frac{{ - {P_0} \pm \sqrt {{P_0}^2 - 4\left( {\rho g} \right)\left( { - {P_0}\left( {3.0{\rm{ m}}} \right)} \right)} }}{{2\rho g}}$

Substitute $101325$ for ${P_0}$ , $9.8$ for $g$ and $1000{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}$ for $\rho$ .

$\begin{array}{c}\\L = \frac{{ - 101325{\rm{ Pa}} \pm \sqrt {{{\left( {101325{\rm{ Pa}}} \right)}^2} + 12\left( {1000{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}} \right)\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)\left( {101325{\rm{ Pa}}} \right)} }}{{2\left( {1000{\rm{ kg}} \cdot {{\rm{m}}^{ - 3}}} \right)\left( {9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right)}}\\\\ = - 12.77{\rm{ m}},2.43{\rm{ m}}\\\end{array}$

Ignore the negative value of $L$ . Consider only positive value.

$L = 2.43{\rm{ m}}$

Ans:

The length of the trapped volume of air is $2.43{\rm{ m}}$ .