Chapter 10, Problem 68QP

Dry air near sea level has the following composition by volume: ${\text{N}}_{\text{2}}$, 78.08 percent; ${\text{O}}_{\text{2}}$, 20.94 percent; Ar, 0.93 percent; ${\text{CO}}_{\text{3}}$, 0.05 percent. The atmospheric pressure is 1.00 atm. Calculate (a) the partial pressure of each gas in atmospheres and (b) the concentration of each gas in mol/L at $\text{0°C}$. (Hint: Because volume is proportional to the number of moles present, mole fractions of gases can be expressed as ratios of volumes at the same temperature and pressure.)

Interpretation Introduction

Interpretation:

The partial pressure of each gas in the atmosphere and concentration of each gas in moles per liter are to be calculated.

Concept introduction:

The mole fraction of an individual gas for the combination of gases is the ratio of the moles of the individual gas to the total number of moles of the gaseous mixture.

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

Here, ${\chi }_i$ is the mole fraction, $n_i$ is the mole fraction of individual gas and $n_{\text{total}}$ is the total number of moles.

Also, the mole fraction of an individual gas for the combination of gases can be calculated from the ratio of the partial pressure of the individual gases to the total pressure of the combination.

${\chi }_i=\frac{P_i}{P_{\text{total}}}$

Here, ${\chi }_i$ is the mole fraction, $P_i$ is the partial pressure of individual gas and $P_{\text{total}}$ is the total pressure.

Ideal Gas Equation is given as:

$\text{PV = nRT}$

where, $\text{P}$, pressure of the gas;  $\text{V}$, volume of the gas;  $\text{n}$, Number of Moles;  $\text{T}$, absolute temperature; $\text{R}$, Ideal Gas constant also known as Boltzmann Constant, $\text{0}{\text{.082057 L atm K}}^{\text{-1}}{\text{ mol}}^{\text{-1}}\text{.}$

Answer to Problem 68QP

Solution:

(a)

The partial pressure of ${\text{O}}_2$ in air is $0.209\text{ atm}$, and those of ${\text{N}}_2$, $\text{Ar}$ and ${\text{CO}}_2$ in air are $0.781\text{ atm}$, $9.3\times {10}^{-3}\text{ atm}$ and $5\times {10}^{-4}\text{ atm}$

respectively.

(b)

The concentration of ${\text{O}}_2$ in air is $9.3\times {10}^{-3}\text{ mol}/\text{L}$, and those of ${\text{N}}_2$, $\text{Ar}$ and ${\text{CO}}_2$

in air are $3.4\times {10}^{-2}\text{ mol}/\text{L}$, $4.1\times {10}^{-4}\text{ mol}/\text{L}$

and $2\times {10}^{-5}\text{ mol}/\text{L}$

respectively.

Explanation of Solution

a) The partial pressure of each gas in atmospheres

For ${\text{O}}_2$,

Calculate the mole fraction of oxygen as follows:

${\chi }_{{\text{O}}_2}=\frac{n_{{\text{O}}_2}}{n_{\text{total}}}$

Substitute $20.94\%$ for $n_{{\text{O}}_2}$ and $100\%$ for $n_{\text{total}}$ in the above equation.

$\begin{array}{c}{\chi }_{{\text{O}}_2}=\frac{20.94\%}{100\%}\\ =0.2094\end{array}$

Calculate the partial pressure of oxygen as follows:

$P_{{\text{O}}_2}={\chi }_{{\text{O}}_2}\times P_{\text{total}}$

Substitute $0.2094$ for ${\chi }_{{\text{O}}_2}$ and $1\text{ atm}$ for $P_{\text{total}}$ in the above equation.

$\begin{array}{c}{P}_{{\text{O}}_2}=0.209\times 1\text{ atm}\\ =0.209\text{ atm}\end{array}$

So, the partial pressure of oxygen is $\text{0}\text{.209 atm}$

For ${\text{N}}_2$,

Calculate the mole fraction of nitrogen as follows:

${\chi }_{{\text{N}}_2}=\frac{n_{{\text{N}}_2}}{n_{\text{total}}}$

Substitute $78.08\%$ for $n_{{\text{N}}_2}$ and $100\%$ for $n_{\text{total}}$ in the above equation.

$\begin{array}{c}{\chi }_{{\text{N}}_2}=\frac{78.08\%}{100\%}\\ =0.7808\end{array}$

Calculate the partial pressure of nitrogen as follows:

$P_{{\text{N}}_2}={\chi }_{{\text{N}}_2}\times P_{\text{total}}$

Substitute $0.7808$ for ${\chi }_{{\text{N}}_2}$ and $1\text{ atm}$ for $P_{\text{total}}$ in the above equation.

$\begin{array}{c}{P}_{{\text{N}}_2}=0.7808\times 1\text{ atm}\\ =0.7808\text{ atm}\\ \approx \text{0}\text{.781 atm}\end{array}$

So, the partial pressure of nitrogen is $\text{0}\text{.781 atm}$.

For $\text{Ar}$,

Calculate the mole fraction of argon as follows:

${\chi }_{\text{Ar}}=\frac{n_{\text{Ar}}}{n_{\text{total}}}$

Substitute $0.93\%$ for $n_{\text{Ar}}$ and $100\%$ for $n_{\text{total}}$ in the above equation.

$\begin{array}{c}{\chi }_{\text{Ar}}=\frac{0.93\%}{100\%}\\ =0.0093\\ =9.3\times {10}^{-3}\end{array}$

Calculate the partial pressure of argon as follows:

$P_{\text{Ar}}={\chi }_{\text{Ar}}\times P_{\text{total}}$

Substitute $9.3\times {10}^{-3}$ for ${\chi }_{\text{Ar}}$ and $1\text{ atm}$ for $P_{\text{total}}$ in the above equation.

$\begin{array}{c}{P}_{\text{Ar}}=9.3\times {10}^{-3}\times 1\text{ atm}\\ =9.3\times {10}^{-3}\text{ atm}\end{array}$

So, the partial pressure of argon is $9.3\times {10}^{-3}\text{ atm}$.

For ${\text{CO}}_2$,

Calculate the mole fraction of carbon dioxide as follows:

${\chi }_{{\text{CO}}_2}=\frac{n_{{\text{CO}}_2}}{n_{\text{total}}}$

Substitute $0.05\%$ for $n_{{\text{CO}}_2}$ and $100\%$ for $n_{\text{total}}$ in the above equation.

$\begin{array}{c}{\chi }_{{\text{CO}}_2}=\frac{0.05\%}{100\%}\\ =0.0005\\ =5\times {10}^{-4}\end{array}$

Calculate the partial pressure of carbon dioxide as follows:

$P_{{\text{CO}}_2}={\chi }_{{\text{CO}}_2}\times P_{\text{total}}$

Substitute $5\times {10}^{-4}$ for ${\chi }_{{\text{CO}}_2}$ and $1\text{ atm}$ for $P_{\text{total}}$ in the above equation.

$\begin{array}{c}{P}_{{\text{CO}}_2}=5\times {10}^{-4}\times 1\text{ atm}\\ =5\times {10}^{-4}\text{ atm}\end{array}$

So, the partial pressure of carbon dioxide is $5\times {10}^{-4}\text{ atm}$.

Thus, the partial pressure of ${\text{O}}_2$ in air is $0.209\text{ atm}$, and those of ${\text{N}}_2$, $\text{Ar}$

and ${\text{CO}}_2$ in air are $0.781\text{ atm}$, $9.3\times {10}^{-3}\text{ atm}$

and $5\times {10}^{-4}\text{ atm}$

respectively.

b) The concentration of each gas in $\text{mol}/\text{L}$

at $0°\text{C}$

The value of the gas constant is $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$.

The temperature is $0°\text{C}$.

The conversion of temperature from degree Celsius to kelvin can be done by using the formula given below:

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =\left(0+273\right)\\ =273\text{ K}\end{array}$

The concentration of gas in $\text{mol}/\text{L}$

can be calculated by using the ideal gas equation as follows:

$c=\frac{n}{V}$ …… (1)

From the ideal gas equation, the ratio of number of moles to the volume is:

$\frac{n}{V}=\frac{P_i}{RT}$ …… (2)

From equation (1) and (2) ,

$c=\frac{P_i}{RT}$

The partial pressure of oxygen is $0.209\text{ atm}$.

Calculate the concentration of oxygen as follows:

$c_{{\text{O}}_2}=\frac{P_{{\text{O}}_2}}{RT}$

Substitute $0.209\text{ atm}$

for $P_{{\text{O}}_2}$, $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$

and $273\text{ K}$ for $T$ in the above equation.

$\begin{array}{c}{c}_{{\text{O}}_2}=\frac{0.209\cancel{\text{atm}}}{\left(0.08206\text{ L}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(273\cancel{\text{K}}\right)}\\ =\frac{0.209}{22.4}\text{ mol}/\text{L}\\ =0.0093\text{ mol}/\text{L}\\ =9.3\times {10}^{-3}\text{ mol}/\text{L}\end{array}$

So, the concentration of oxygen is $9.3\times {10}^{-3}\text{ mol}/\text{L}$.

The partial pressure of nitrogen is $\text{0}\text{.781 atm}$.

Calculate the concentration of nitrogen as follows:

$c_{{\text{N}}_2}=\frac{P_{{\text{N}}_2}}{RT}$

Substitute $\text{0}\text{.781 atm}$

for $P_{{\text{N}}_2}$, $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$

and $273\text{ K}$ for $T$ in the above equation.

$\begin{array}{c}{c}_{{\text{N}}_2}=\frac{0.781\cancel{\text{atm}}}{\left(0.08206\text{ L}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(273\cancel{\text{K}}\right)}\\ =\frac{0.781}{22.4}\text{ mol}/\text{L}\\ =0.034\text{ mol}/\text{L}\\ =3.4\times {10}^{-2}\text{ mol}/\text{L}\end{array}$

So, the concentration of nitrogen is $3.4\times {10}^{-2}\text{ mol}/\text{L}$.

The partial pressure of argon is $9.3\times {10}^{-3}\text{ atm}$

Calculate the concentration of argon as follows:

$c_{\text{Ar}}=\frac{P_{\text{Ar}}}{RT}$

Substitute $9.3\times {10}^{-3}\text{ atm}$

for $P_{\text{Ar}}$, $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$

and $273\text{ K}$ for $T$ in the above equation.

$\begin{array}{c}{c}_{\text{Ar}}=\frac{9.3\times {10}^{-3}\cancel{\text{atm}}}{\left(0.08206\text{ L}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(273\cancel{\text{K}}\right)}\\ =\frac{9.3\times {10}^{-3}}{22.4}\text{ mol}/\text{L}\\ =0.415\times {10}^{-3}\text{ mol}/\text{L}\\ =4.1\times {10}^{-4}\text{ mol}/\text{L}\end{array}$

So, the concentration of argon is $4.1\times {10}^{-4}\text{ mol}/\text{L}$.

The partial pressure of carbon dioxide is $5\times {10}^{-4}\text{ atm}$.

Calculate the concentration of carbon dioxide as follows:

$c_{{\text{CO}}_2}=\frac{P_{{\text{CO}}_2}}{RT}$

Substitute $5\times {10}^{-4}\text{ atm}$

for $P_{{\text{CO}}_2}$, $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$

and $273\text{ K}$ for $T$ in the above equation.

$\begin{array}{c}{c}_{{\text{CO}}_2}=\frac{5\times {10}^{-4}\cancel{\text{atm}}}{\left(0.08206\text{ L}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(273\cancel{\text{K}}\right)}\\ =\frac{5\times {10}^{-4}}{22.4}\text{ mol}/\text{L}\\ =0.223\times {10}^{-4}\text{ mol}/\text{L}\\ \approx 2\times {10}^{-5}\text{ mol}/\text{L}\end{array}$

So, the concentration of carbon dioxide is $2\times {10}^{-5}\text{ mol}/\text{L}$.

Thus, the concentration of ${\text{O}}_2$ in air is $9.3\times {10}^{-3}\text{ mol}/\text{L}$, and those of ${\text{N}}_2$, $\text{Ar}$ and ${\text{CO}}_2$ in air are $3.4\times {10}^{-2}\text{ mol}/\text{L}$, $4.1\times {10}^{-4}\text{ mol}/\text{L}$

and $2\times {10}^{-5}\text{ mol}/\text{L}$

respectively.