Chapter 13, Problem 13.160P

(a)

Interpretation Introduction

Interpretation:

The molarity of ${\text{N}}_2$ in blood is to be calculated.

Concept introduction:

Henry’s law gives the quantitative relationship between the pressure of the gas and its solubility. It states that the amount of gas dissolved in a liquid is proportional to the partial pressure of the gas. Higher the partial pressure of the gas, more will be its solubility and vice-versa.

The formula to calculate the solubility of gases according to Henry’s law is as follows:

$S_{\text{gas}}=k_{\text{H}}\times P_{\text{gas}}$ (1)

Here, $S_{\text{gas}}$ is the solubility of the gas.

$k_{\text{H}}$ is Henry’s constant.

$P_{\text{gas}}$ is the partial pressure of the gas.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $M$ and its unit is $\text{mol/L}$.

Answer to Problem 13.160P

The molarity of ${\text{N}}_2$ in blood is $4.8\times {10}^{-4}M$.

Explanation of Solution

The pressure of the nitrogen gas is calculated as follows:

$\begin{array}{c}{P}_{{\text{N}}_2}=\left(\frac{78\%}{100\%}\right)\left(1\text{atm}\right)\\ =0.78\text{atm}\end{array}$

Substitute $0.78\text{atm}$ for $P_{\text{gas}}$ and $6.2\times {10}^{-4}\text{mol/L}\cdot \text{atm}$ for $k_{\text{H}}$ in equation (1).

$\begin{array}{c}{S}_{\text{gas}}=\left(6.2\times {10}^{-4}\text{mol/L}\cdot \text{atm}\right)\left(0.78\text{atm}\right)\\ =4.836\times {10}^{-4}\text{mol/L}\\ =4.8\times {10}^{-4}\text{mol/L}\end{array}$

The solubility is equal to the molarity so the molarity of nitrogen gas in the blood is $4.8\times {10}^{-4}M$.

Conclusion

The molarity of ${\text{N}}_2$ in blood is $4.8\times {10}^{-4}M$.

(b)

Interpretation Introduction

Interpretation:

The molarity of ${\text{N}}_2$ in blood is to be calculated.

Concept introduction:

The formula to calculate the solubility of gases according to Henry’s law is as follows:

$S_{\text{gas}}=k_{\text{H}}\times P_{\text{gas}}$ (1)

The formula to calculate the pressure at any height is as follows:

$P=h\rho g$ (2)

Here $h$ is the height of the substance.

$\rho$ is the density of the solution.

$g$ is the acceleration due to gravity.

The conversion factor to convert $\text{ft}$ to inches is as follows:

$\text{1ft}=\text{12 inch}$

The conversion factor to convert inches to $\text{cm}$ is as follows:

$\text{1 inch}=\text{2}\text{.54 }\text{}\text{cm}$

The conversion factor to convert $\text{atm}$ to $\text{Pa}$ is as follows:

$1\text{atm}=101325\text{}\text{}\text{Pa}$

Answer to Problem 13.160P

The molarity of ${\text{N}}_2$ in blood is $1.2\times {10}^{-3}M$.

Explanation of Solution

Substitute $50\text{ft}$ for $h$, $\text{1}\text{g/mL}$ for $\rho$, $\text{9}\text{.80665}{\text{m/s}}^2$ for $g$ in equation (2).

$\begin{array}{c}P=\left(50\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)\left(\frac{2.54\text{cm}}{1\text{in}}\right)\left(\frac{1\text{g}}{1\text{mL}}\right)\left(\frac{1\text{mL}}{1{\text{cm}}^3}\right){\left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right)}^2\\ \left(\frac{\text{9}\text{.80665}\text{m}}{1{\text{s}}^2}\right)\left(\frac{1\text{Pa}}{1\text{kg/m}\cdot {\text{s}}^2}\right)\left(\frac{1\text{atm}}{1.01325\times {10}^5\text{Pa}}\right)\\ =1.47535\text{atm}\end{array}$

The total pressure is calculated as follows:

$P={\text{pressure of N}}_2+\text{pressure of water}$ (3)

Substitute $\text{1}\text{atm}$ for the pressure of ${\text{N}}_2$, $1.47535\text{atm}$ for the pressure of water in equation (3).

$\begin{array}{c}P={\text{pressure of N}}_2+\text{pressure of water}\\ =\text{1}\text{atm}+1.47535\text{atm}\\ =2.47535\text{atm}\end{array}$

The pressure of the nitrogen gas is calculated as follows:

$\begin{array}{c}{P}_{{\text{N}}_2}=\left(\frac{78\%}{100\%}\right)\left(2.47535\text{atm}\right)\\ =1.930773\text{atm}\end{array}$

Substitute $1.930773\text{atm}$ for $P_{\text{gas}}$ and $6.2\times {10}^{-4}\text{mol/L}\cdot \text{atm}$ for $k_{\text{H}}$ in equation (1).

$\begin{array}{c}{S}_{\text{gas}}=\left(6.2\times {10}^{-4}\text{mol/L}\cdot \text{atm}\right)\left(1.930773\text{atm}\right)\\ =1.18034\times {10}^{-3}\text{mol/L}\\ =1.2\times {10}^{-3}\text{mol/L}\end{array}$

The solubility is equal to the molarity so the molarity of nitrogen is $1.2\times {10}^{-3}M$.

Conclusion

The molarity of ${\text{N}}_2$ in blood is $1.2\times {10}^{-3}M$.

(c)

Interpretation Introduction

Interpretation:

The volume of ${\text{N}}_2$ in blood is to be calculated.

Concept introduction:

The formula of an ideal gas equation is as follows:

$PV=nRT$ (4)

Here, $P$ is the pressure of the gas.

$V$ is the volume of gas.

$n$ is the number of moles of gas.

$R$ is the universal gas constant.

$T$ is the absolute temperature of the gas.

Answer to Problem 13.160P

The volume of ${\text{N}}_2$ in blood is $17\text{mL}$.

Explanation of Solution

Rearrange equation (4) to calculate the volume of gas as follows:

$V=\frac{nRT}{P}$ (5)

The formula to calculate the number of moles is as follows:

${\text{Moles of N}}_2={\text{initial moles of N}}_2-{\text{final moles of N}}_2$ (6)

Substitute $1.18034\times {10}^{-3}\text{mol}$ for the initial moles and $4.8\times {10}^{-4}\text{mol}$ for the final moles in equation (6).

$\begin{array}{c}{\text{Moles of N}}_2={\text{initial moles of N}}_2-{\text{final moles of N}}_2\\ =1.18034\times {10}^{-3}\text{mol}-4.8\times {10}^{-4}\text{mol}\\ =6.9674\times {10}^{-4}\text{mol}\end{array}$

Substitute $6.9674\times {10}^{-4}\text{mol}$ for $n$, $0.0821\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, ${\text{25}}^{°}\text{C}$ for $T$ and $\text{1}\text{atm}$ for $P$ in equation (3).

$\begin{array}{c}{V}_{{\text{N}}_2}=\frac{\left(6.9674\times {10}^{-4}\text{mol}\right)\left(\text{0}\text{.0821}\text{}\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(\left({\text{25}}^{°}\text{C}+\text{273}\text{.15}\right)\text{K}\right)}{\left(\text{1}\text{atm}\right)}\left(\frac{1\text{}\text{mL}}{{10}^{-3}\text{L}}\right)\\ =17.046\text{L}\\ =\text{17}\text{}\text{L}\end{array}$

Conclusion

The volume of ${\text{N}}_2$ in blood is $17\text{mL}$.