Chapter 13, Problem 116P

(a)

To determine

The time taken to last air inside the tank at a depth of $2.0\text{m}$.

Answer to Problem 116P

The time taken to last air inside the tank at a depth of $2.0\text{m}$ is $28\min$.

Explanation of Solution

Write the expression to calculate the final volume of the tank.

  $V^{\prime }=\left(\frac{\Delta V}{\Delta t}\right)\Delta t$

Here, $V^{\prime }$ is the final volume, $\frac{\Delta V}{\Delta t}$ is the rate of change of volume or rate of breaths and $\Delta t$ is the time taken.

Write the expression for the Boyle’s law.

  $\frac{V^{\prime }}{V}=\frac{P}{P^{\prime }}$

Here, V is the initial volume of the tank, $P^{\prime }$ is the final pressure or the pressure at the given depth and P is the initial pressure of the tank.

Rewrite the above expression using the expression for $V^{\prime }$.

  $\frac{\left(\frac{\Delta V}{\Delta t}\right)\Delta t}{V}=\frac{P}{P^{\prime }}$                                                                                                        (I)

Write the expression for the pressure at the given depth.

  $P^{\prime }=P_0+\rho gd$

Here, $P_0$ is the atmospheric pressure, $\rho$ is the density of water, g is the acceleration due to gravity and d is the depth.

Substitute the above expression in (I) to rewrite in terms of $\Delta t$.

  $\begin{array}{c}\frac{\left(\frac{\Delta V}{\Delta t}\right)\Delta t}{V}=\frac{P}{P_0+\rho gd}\\ \left(\frac{\Delta V}{\Delta t}\right)\Delta t=\frac{PV}{P_0+\rho gd}\\ \Delta t=\frac{1}{\left(\frac{\Delta V}{\Delta t}\right)}\left(\frac{PV}{P_0+\rho gd}\right)\end{array}$

Substitute $0.0100{\text{m}}^3$ for V, $1.0\times {10}^7\text{Pa}$, P, $0.500\text{L}/\text{s}$, for $\frac{\Delta V}{\Delta t}$, $1.013\times {10}^5\text{Pa}$ for $P_0$, $1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for g and $2.0\text{m}$ for d in the above equation to calculate $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{1}{\left(0.500\text{L}/\text{s}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\right)}\left(\frac{0.0100{\text{m}}^3\left(1.0\times {10}^7\text{Pa}\right)}{1.013\times {10}^5\text{Pa}+\left(1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)2.0\text{m}}\right)\\ =\frac{1}{\left(0.500\text{L}/\text{s}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\right)}\left(\frac{1.0\times {10}^5\text{Pa}\cdot {\text{m}}^3}{1.21\times {10}^5\text{Pa}}\right)\\ =1.653\times {10}^3\text{s}\left(\frac{1\min }{60\text{s}}\right)\\ =27.55\min \\ \sim 28\min \end{array}$

Conclusion:

Therefore, the time taken to last air inside the tank at a depth of $2.0\text{m}$ is $28\min$.

(b)

To determine

The time taken to last air inside the tank at a depth of $20.0\text{m}$.

Answer to Problem 116P

The time taken to last air inside the tank at a depth of $20.0\text{m}$ is $11\min$.

Explanation of Solution

Write the expression to calculate $\Delta t$.

  $\Delta t=\frac{1}{\left(\frac{\Delta V}{\Delta t}\right)}\left(\frac{PV}{P_0+\rho gd}\right)$

Substitute $0.0100{\text{m}}^3$ for V, $1.0\times {10}^7\text{Pa}$, P, $0.500\text{L}/\text{s}$, for $\frac{\Delta V}{\Delta t}$, $1.013\times {10}^5\text{Pa}$ for $P_0$, $1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for g and $20.0\text{m}$ for d in the above equation to calculate $\Delta t$.

  $\begin{array}{c}\Delta t=\frac{1}{\left(0.500\text{L}/\text{s}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\right)}\left(\frac{0.0100{\text{m}}^3\left(1.0\times {10}^7\text{Pa}\right)}{1.013\times {10}^5\text{Pa}+\left(1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)20.0\text{m}}\right)\\ =\frac{1}{\left(0.500\text{L}/\text{s}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\right)}\left(\frac{1.0\times {10}^5\text{Pa}\cdot {\text{m}}^3}{2.97\times {10}^5\text{Pa}}\right)\\ =0.673\times {10}^3\text{s}\left(\frac{1\min }{60\text{s}}\right)\\ =11.22\min \\ \sim 11\min \end{array}$

Conclusion:

Therefore, the time taken to last air inside the tank at a depth of $20.0\text{m}$ is $11\min$.