Chapter 4, Problem 4.5PR

(a)

Interpretation Introduction

Interpretation:

The volume of oxygen is given up by $100c{m}^3$ of blood flowing from the lungs in the capillary has to be determined.

Answer to Problem 4.5PR

The volume of oxygen is given up by $100c{m}^3$ of blood flowing from the lungs in the capillary has been calculated to be $14.62275c{m}^3$.

Explanation of Solution

Heamoglobin concentration in normal blood is $150gd{m}^{-3}$ which implies that in $1L$ of blood there is $150g$ of Hb.  So in $100c{m}^3$ of blood, the amount of Hb will be

  $\frac{150g}{1000}\times 100=15g.$

$1g$ of Heamoglobin binds about $1.34c{m}^3$ of oxygen per gram.  So $15g$ of Hb will bind

  $15\times 1.34c{m}^3=20.1c{m}^3.$

Which also implies that $100c{m}^3$ of blood binds $20.1c{m}^3$ of oxygen.

Heamoglobin in the lungs is about $97\%$ saturated with oxygen, while in the capillary is only $75\%$ saturated.

According to the question, $100c{m}^3$ of blood is flowing from lungs to capillary.

  $\boxed{Lung}\to \boxed{Capillary}$

In lung, $100c{m}^3$ of blood will bind $19.497c{m}^3$ of oxygen,

  $20.1c{m}^3\times \frac{97}{100}=19.497c{m}^3.$

In capillary, $100c{m}^3$ of blood from lung will bind $14.62275c{m}^3$ of oxygen,

  $19.497c{m}^3\times \frac{75}{100}=14.62275c{m}^3.$

The amount of oxygen taken up by the body tissues will be $\left(19.497-14.62275\right)c{m}^3=4.87425c{m}^3.$

Therefore, the volume of oxygen is given up by $100c{m}^3$ of blood flowing from the lungs in the capillary has been calculated to be $\text{4}{\text{.87425cm}}^{\text{3}}$.

(b)

Interpretation Introduction

Interpretation:

The mass of nitrogen dissolved in $100g$ of water saturated with air at $4.0atmand{20}^{\circ }C$ has to be calculated.

Concept Introduction:

Henry’s law:

The vapour pressure of a volatile solute A is proportional to its mole fraction in a solution. 

Mathematically, it can be represented as

$p_A=K_H^{\text{'}}{X}_A$

Where,

$K_H^{\text{'}}=$ Henry’s law constant

$X_A=$ Mole fraction of a volatile solute A

$p_A=$ Vapour pressure of a volatile solute A

Another version of Henry’s law can be mathematically represented as

$C=K_{H}p$

Where,

$K_H^{\text{'}}=$ Henry’s law constant

$C=$ Concentration of the dissolved gas

$p=$ partial pressure of the gas

Answer to Problem 4.5PR

The mass of nitrogen dissolved in $100g$ of water saturated with air at $4.0atmand1.0atm$ has been calculated to be $56.2167\mu g$ and $14.0544\mu g$ respectively.

Explanation of Solution

According to the given question, in $1g$ of water the amount of nitrogen dissolves is $0.18\mu g$. So in $100g$ of water the amount of dissolved nitrogen will be $0.18\mu g\times 100=18\mu g.$

Air is $78.08$ mole per cent nitrogen which implies that in $1atm$ pressure of air, the partial pressure of nitrogen is $.7808$.  So in $4atm$ pressure of air, the partial pressure of nitrogen will be $.7808\times 4=3.1232atm.$

Substituting all the values in Henry’s law and solving for C,

  $\begin{array}{l}C=K_{H}p\\ \\ =\frac{18\mu g}{atm}\times 3.1232atm\\ \\ =56.2167\mu g.\end{array}$

Therefore, the mass of nitrogen dissolved in $100g$ of water saturated with air at $4.0atmand{20}^{\circ }C$ has been calculated to be $\text{56}\text{.2167}\text{μg}$.

Similarly, at $1atm$

  $\begin{array}{l}C=K_{H}p\\ \\ =\frac{18\mu g}{atm}\times 0.7808atm\\ \\ =14.0544\mu g.\end{array}$

Therefore, the mass of nitrogen dissolved in $100g$ of water saturated with air at $1.0atmand{20}^{\circ }C$ has been calculated to be $14.0544\text{μg}$.

(c)

Interpretation Introduction

Interpretation:

The increase in nitrogen concentration in fatty tissue in going from $1atm$ to $4atm$ has to be determined.

Concept Introduction:

Henry’s law:

The vapour pressure of a volatile solute A is proportional to its mole fraction in a solution.

Mathematically, it can be represented as

$p_A=K_H^{\text{'}}{X}_A$

Where,

$K_H^{\text{'}}=$ Henry’s law constant

$X_A=$ Mole fraction of a volatile solute A

$p_A=$ Vapour pressure of a volatile solute A

Another version of Henry’s law can be mathematically represented as

$C=K_{H}p$

Where,

$K_H^{\text{'}}=$ Henry’s law constant

$C=$ Concentration of the dissolved gas

$p=$ partial pressure of the gas

Answer to Problem 4.5PR

The increase in nitrogen concentration in fatty tissue in going from $1atm$ to $4atm$ has been calculated to be $172.6492\mu g$.

Explanation of Solution

According to the question, nitrogen is 4 times as soluble in fatty tissues as in water.

So at $4atm$,

  $\begin{array}{l}{\left(So\mathrm{lub}ility\right)}_{Fatty}=4{\left(So\mathrm{lub}ility\right)}_{water}\\ \\ =4\times \text{56}\text{.2167}\text{μg}\text{=}\text{224}\text{.8668}\text{μg}.\end{array}$

So at $1atm$,

  $\begin{array}{l}{\left(So\mathrm{lub}ility\right)}_{Fatty}=4{\left(So\mathrm{lub}ility\right)}_{water}\\ \\ =4\times 14.0544\text{μg}\text{=}56.2167\text{μg}.\end{array}$

The increase in nitrogen concentration in fatty tissue in going from $1atm$ to $4atm$ will be

$\left(\text{224}\text{.8668}-\text{56}\text{.2167}\right)\text{μg}=172.6492\text{μg}.$