Chapter 10, Problem 10.59QP

(a)

Interpretation Introduction

Interpretation: The solubility of ${\text{O}}_{\text{2}}$ in the blood of a climber on Mt. Everest and a scuba diver at $100$ feet is to be calculated.

Concept introduction: The Henry’s law states that “the amount of gas dissolved in given liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.”

To determine: The solubility of ${\text{O}}_{\text{2}}$ in the blood of a climber on Mt. Everest.

Answer to Problem 10.59QP

Solution

The solubility of ${\text{O}}_{\text{2}}$ in the blood of a climber on Mt. Everest is $\underset{\_}{0.0027mol/L}$.

Explanation of Solution

Explanation

Given

Refer Problem 10.57

The Henry’s law constant for ${\text{O}}_{\text{2}}$ dissolved in arterial blood is $0.037\text{mol}/\text{L}\cdot \text{atm}$.

Atmospheric pressure is $0.35\text{atm}$.

The concentration of ${\text{O}}_{\text{2}}$ in blood is calculated by the formula,

$C=K_{\text{H}}\times p$ (1)

Where,

• $K_{\text{H}}$ is the Henry’s law constant.
• $p$ partial pressure of ${\text{O}}_{\text{2}}$.

Partial pressure of the ${\text{O}}_2$ gas is calculated by the formula,

$p=m\times P$

Where,

• $m$ is the mole fraction of oxygen.
• $P$ is the atmospheric pressure.

The mole fraction of oxygen is 0.209.

Substitute the value of mole fraction and atmospheric pressure in above formula.

$\begin{array}{c}p=0.209\times 0.35\text{atm}\\ =\text{0}\text{.0731}\text{atm}\end{array}$

Partial pressure of ${\text{O}}_2$ gas is $0.0731\text{atm}$.

Substitute the value of partial pressure and Henry’s law constant in equation (1).

$\begin{array}{c}C=0.037\text{mol}/\text{L}\cdot \text{atm}\times 0.0731\text{atm}\\ =\underset{\_}{0.0027mol/L}\end{array}$

Therefore, the solubility of ${\text{O}}_{\text{2}}$ in the blood of a climber on Mt. Everest is $\underset{\_}{0.0027mol/L}$.

(b)

Interpretation Introduction

To determine: The solubility of ${\text{O}}_{\text{2}}$ in the blood of a scuba diver at $100$ feet.

Answer to Problem 10.59QP

Solution

The solubility of ${\text{O}}_{\text{2}}$ in the blood of a scuba diver at $100$ feet is $\underset{\_}{0.0231mol/L}$.

Explanation of Solution

Explanation

Refer Problem 10.57

The Henry’s law constant for ${\text{O}}_{\text{2}}$ dissolved in arterial blood is $0.037\text{mol}/\text{L}\cdot \text{atm}$.

Total pressure is $0.35\text{atm}$.

The concentration of ${\text{O}}_{\text{2}}$ in blood is calculated by the formula,

$C=K_{\text{H}}\times p$ (1)

Where,

• $K_{\text{H}}$ is the Henry’s law constant.
• $p$ partial pressure of ${\text{O}}_{\text{2}}$.

Partial pressure of the ${\text{O}}_2$ gas is calculated by the formula,

$p=m\times P$

Where,

• $m$ is the mole fraction of oxygen.
• $P$ is the atmospheric pressure.

The mole fraction of oxygen is 0.209.

Substitute the value of mole fraction and total pressure in above formula.

$\begin{array}{c}p=0.209\times 3\text{atm}\\ =\text{0}\text{.627}\text{atm}\end{array}$

Partial pressure of ${\text{O}}_2$ gas is $0.627\text{atm}$.

Substitute the value of partial pressure and Henry’s law constant in equation (1).

$\begin{array}{c}C=0.037\text{mol}/\text{L}\cdot \text{atm}\times 0.627\text{atm}\\ =\underset{\_}{0.0231mol/L}\end{array}$

Therefore, the solubility of ${\text{O}}_{\text{2}}$ in the blood of a scuba diver at $100$ feet is $\underset{\_}{0.0231mol/L}$.

Conclusion

1. a. The solubility of ${\text{O}}_{\text{2}}$ in the blood of a climber on Mt. Everest is $\underset{\_}{0.0027mol/L}$.
2. b. The solubility of ${\text{O}}_{\text{2}}$ in the blood of a scuba diver at $100$ feet is $\underset{\_}{0.0231mol/L}$.