Chapter 13, Problem 36QP

Interpretation Introduction

Interpretation: The amount of nitrogen gas released in litres is to be calculated.

Concept introduction:

The solubility of gas depends upon the pressure of a gas.

The relation between solubility and pressure of a gas is expressed as follows:

$c=kP$

Here, $P$ is the pressure of the gas, $k$ is the henry’s constant, and $c$ is concentration of gas.

Answer to Problem 36QP

Solution: $\text{0}\text{.28 L}$

Explanation of Solution

Given: The solubility of ${\text{N}}_{\text{2}}$

in blood is $\text{5}\text{.6}\times {\text{10}}^{-4}\text{ mol}/\text{L}$.

The total volume of blood in body is $\text{5}\text{.0 L}$.

In liquid, the solubility of a gas is directly proportional to the pressure of the gas, which is expressed as follows:

$c=kP$ …… (1)

Here, $k$ is the Henry’s law constant, $P$ is the pressure, and $c$ is the molar concentration of gas.

Substitute $\text{5}\text{.6}\times {\text{10}}^{-4}\text{ mol}/\text{L}$

for $c$ and $\text{0}\text{.80 atm}$

for $P$ in equation (1) as follows:

$\begin{array}{c}\text{5}\text{.6}\times {\text{10}}^{-4}\text{ mol}/\text{L}=k\left(\text{0}\text{.80 atm}\right)\\ k=\frac{\left(\text{5}\text{.6}\times {\text{10}}^{-4}\text{ mol}/\text{L}\right)}{\left(\text{0}\text{.80 atm}\right)}\\ =7.0\times {\text{10}}^{-4}\text{ mol}/\text{L}\cdot \text{atm}\end{array}$

So, the Henry’s law constant is $7.0\times {\text{10}}^{-4}\text{ mol}/\text{L}\cdot \text{atm}$.

At $\text{4}\text{.0 atm}$, the concentration of nitrogen in blood is calculated as follows:

Substitute $7.0\times {\text{10}}^{-4}\text{ mol}/\text{L}\cdot \text{atm}$

for $k$ and $\text{4}\text{.0 atm}$

for $P$ in equation (1) as follows:

$\begin{array}{c}c=\left(7.0\times {\text{10}}^{-4}\text{ mol}/\text{L}\cdot \cancel{\text{atm}}\right)\left(\text{4}\text{.0 }\cancel{\text{atm}}\right)\\ =2.8\times {10}^{-3}\text{ mol}/\text{L}\end{array}$

So, the concentration of nitrogen in blood is $2.8\times {10}^{-3}\text{ mol}/\text{L}$.

At $\text{0}\text{.80 atm}$,

The number of moles of nitrogen in blood is given as follows:

${\text{Moles N}}_2=\text{molarity}\times \text{liter of solution}$ …… (2)

Substitute $\text{5}\text{.6}\times {\text{10}}^{-4}\text{ mol}/\text{L}$

for $\text{molarity}$

and $5.0\text{ L}$

for $\text{liter of solution}$

in equation (2) as follows:

$\begin{array}{c}{\text{Moles N}}_2=\left(\text{5}\text{.6}\times {\text{10}}^{-4}\text{ mol}/\cancel{\text{L}}\right)\times \left(5.0\cancel{\text{L}}\right)\\ =2.8\times {\text{10}}^{-3}\text{ mol}\end{array}$

At $\text{4}\text{.0 atm}$,

The number of moles of nitrogen in blood is given as follows:

${\text{Moles N}}_2=\text{molarity}\times \text{liter of solution}$ …… (3)

Substitute $2.8\times {10}^{-3}\text{ mol}/\text{L}$

for $\text{molarity}$

and $5.0\text{ L}$

for $\text{liter of solution}$

in equation (3) as follows:

$\begin{array}{c}{\text{Moles N}}_2=\left(2.8\times {10}^{-3}\text{ mol}/\cancel{\text{L}}\right)\times \left(5.0\cancel{\text{L}}\right)\\ =1.4\times {\text{10}}^{-2}\text{ mol}\end{array}$

The amount of nitrogen released in number of moles is calculated as follows:

$\begin{array}{c}{\text{Moles N}}_2=\left(1.4\times {\text{10}}^{-2}\text{ mol}\right)-\left(\text{2}\text{.8}\times {\text{10}}^{-3}\text{ mol}\right)\\ =1.1\times {\text{10}}^{-2}\text{ mol}\end{array}$

The ideal gas equation is represented as follows:

$PV=n\text{R}T$ …… (4)

Here, $P$

is the pressure, $V$

is the volume, $n$

is the number of moles of nitrogen, $\text{R}$

is the gas constant, and $T$

is the temperature.

$\begin{array}{c}T\left(\text{K}\right)=T\left({}^{\text{o}}\text{C}\right)+273\\ =37+273\\ =310\text{ K}\end{array}$

Substitute $1.1\times {\text{10}}^{-2}\text{ mol}$

for $n$, $310\text{ K}$

for $T$, $0.0821\text{ L}\text{.atm/mol}\text{.K}$

for $\text{R}$, and $1.0\text{ atm}$

for $P$

in equation (4) as follows:

$\begin{array}{c}\left(1.0\text{ atm}\right)V=\left(1.1\times {\text{10}}^{-2}\text{ mol}\right)\left(0.0821\text{ L}\text{.atm}/\text{mol}\text{.K}\right)\left(310\text{ K}\right)\\ V=\frac{\left(1.1\times {\text{10}}^{-2}\cancel{\text{mol}}\right)\left(0.0821\text{ L}\text{.}\cancel{\text{atm}}/\cancel{\text{mol}}\text{.}\cancel{\text{K}}\right)\left(310\cancel{\text{K}}\right)}{\left(1.0\cancel{\text{atm}}\right)}\\ =0.28\text{ L}\end{array}$

Conclusion

The amount of nitrogen gas released in litres is $0.28\text{ L}$.