Chapter 5, Problem 5.123QP

(a)

Interpretation Introduction

Interpretation:

The volume of his lungs increase by the time he reaches the surface and the total pressure exerted on the diver has to be explained.

Concept Introduction:

Boyle’s law states that the volume of a sample of a gas at constant temperature is inversely proportional to pressure

$\begin{array}{l}\text{V=}{\text{k}}_{\text{1}}\frac{\text{1}}{\text{P}}\text{(or)}\\ {\text{P}}_{\text{1}}{\text{V}}_{\text{1}}{\text{= P}}_{\text{2}}{\text{V}}_{\text{2}}\left(\text{at constant temperature}\right)\end{array}$

Answer to Problem 5.123QP

The diver’s lungs would increase in volume 2.1 times by the time he reaches the surface.

Explanation of Solution

This is a Boyle’s law problem, pressure and volume vary. Suppose that the pressure at the water outside is 1 atm. The pressure that the diver experiences 36 ft under water is

$\begin{array}{l}\text{1}\text{atm}\text{+}\left(\text{36}\text{ft}\text{×}\frac{\text{1}\text{atm}}{\text{33}\text{ft}}\right)\text{=}\text{2}\text{.1}\text{atm}\\ {\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}{\text{V}}_{\text{2}}\\ \frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\\ \frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{=}\frac{\text{2}\text{.1}\text{atm}}{\text{1}\text{atm}}\text{=}\text{2}\text{.1}\end{array}$

The diver’s lungs would increase in volume 2.1 times by the time he reaches the surface.

Volume is equal to pressure.

Conclusion

The volume of his lungs increase by the time he reaches the surface was calculated.

(b)

Interpretation Introduction

Concept Introduction:

• The pressure exerted by an individual gas in a mixture is known as its partial pressure.
• Assuming we have a mixture of ideal gases, we can use the ideal gas law to solve problems involving gases in a mixture.
• Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases:
• ${\text{P}}_{\text{Total}}\text{}{\text{=P}}_{\text{gas 1}}\text{}{\text{+P}}_{\text{gas 2}}\text{}{\text{+P}}_{\text{gas 3}}\text{}\mathrm{....}$
•       $\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}\text{=}{\text{χ}}_{{\text{O}}_{\text{2}}}{\text{P}}_{\text{total}}\\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\frac{{\text{n}}_{{\text{O}}_{\text{2}}}}{{\text{n}}_{{\text{O}}_{\text{2}}}\text{+}{\text{n}}_{{\text{N}}_{\text{2}}}}{\text{P}}_{\text{total}}\end{array}$

Answer to Problem 5.123QP

The air that the diver breathes must have an oxygen content of $\text{5}\text{.0\%}$ by volume.

Explanation of Solution

The surface and the total pressure exerted on the diver

 $\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}\text{=}{\text{χ}}_{{\text{O}}_{\text{2}}}{\text{P}}_{\text{total}}\\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\frac{{\text{n}}_{{\text{O}}_{\text{2}}}}{{\text{n}}_{{\text{O}}_{\text{2}}}\text{+}{\text{n}}_{{\text{N}}_{\text{2}}}}{\text{P}}_{\text{total}}\end{array}$

At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas.

${\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\frac{{\text{V}}_{{\text{O}}_{\text{2}}}}{{\text{V}}_{{\text{O}}_{\text{2}}}\text{+}{\text{V}}_{{\text{N}}_{\text{2}}}}{\text{P}}_{\text{total}}$

We know the partial pressure of O2 in air, and we know the total pressure exerted on the diver.

Plugging these values into the above equation gives:

$\begin{array}{l}\text{0}\text{.20}\text{atm}\text{=}\frac{{\text{V}}_{{\text{O}}_{\text{2}}}}{{\text{V}}_{{\text{O}}_{\text{2}}}\text{+}{\text{V}}_{{\text{N}}_{\text{2}}}}\text{(4}\text{.00}\text{atm)}\\ \frac{{\text{V}}_{{\text{O}}_{\text{2}}}}{{\text{V}}_{{\text{O}}_{\text{2}}}\text{+}{\text{V}}_{{\text{N}}_{\text{2}}}}\text{=}\frac{\text{0}\text{.20}\text{atm}}{\text{4}\text{.00}\text{atm}}\text{=0}\text{.050}\end{array}$

Plugging these values into the above equation to give the air that the diver breathes must have an oxygen content of $\text{5}\text{.0\%}$ by volume.

Conclusion

The surface and the total pressure exerted on the diver were explained.