Chapter 10, Problem 129AP

Interpretation Introduction

Interpretation:

The partial pressure of each gas is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is thenumber of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is the volume, and $P$ is the pressure.

The formula for the conversion of temperature from degree Celsius to Kelvin is represented as:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Answer to Problem 129AP

Solution: The partial pressure of ${\text{NO}}_2$ and ${\text{N}}_2{\text{O}}_4$ are $0.53\text{ atm}$ and $0.45\text{ atm}$ respectively.

Explanation of Solution

Given information:

Pressure, $P=0.98\text{ atm}$

Temperature, $T=25°\text{C}$

Density, $d=2.7\text{ g}/\text{L}$

Let the volume be $1\text{ L}$.

The mass of the mixture is calculated as follows:

$m=d\times V$

Substitute $1\text{ L}$ for $V$ and $2.7\text{ g}/\text{L}$ for $d$ in the above equation,

$\begin{array}{c}m=\left(2.7\text{ g}/\text{L}\right)\left(1\text{ L}\right)\\ =2.7\text{ g}\end{array}$

Giventemperature is $25°\text{C}$.

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

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(20+273.15\right)\\ =293.15\text{ K}\end{array}$

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

The equation for an ideal gas is:

$P=\frac{nRT}{V}$

Rearrange the above equation for the number of moles as follows:

$n=\frac{PV}{RT}$

Substitute $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$, $293.15\text{ K}$ for $T$, $0.98\text{ atm}$ for $P$, and $1\text{ L}$ for $V$ in the above equation,

$\begin{array}{c}n=\frac{\left(0.98\text{ atm}\right)\left(1\text{ L}\right)}{\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(298.15\text{ K}\right)}\\ =\left(\frac{0.98}{24.47}\right)\text{ mol}\\ =\text{0}.0401\text{ mol}\end{array}$

The molar mass for the mixture of gas is calculated as follows:

$M_{\text{mixture}}=\frac{m}{n}$

Substitute $2.7\text{ g}$ for $m$ and $\text{0}.0401\text{ mol}$ for $n$ in the above equation,

$\begin{array}{c}{M}_{\text{mixture}}=\frac{2.7\text{ g}}{\text{0}.0401\text{ mol}}\\ =67.33\text{ g}/\text{mol}\\ \approx 67\text{ g}/\text{mol}\end{array}$

The mole fraction and molar mass have the following relationship for the mixture of gas:

${\chi }_{{\text{NO}}_{\text{2}}}{M}_{{\text{NO}}_{\text{2}}}+{\chi }_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}{M}_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=M_{\text{mixture}}$

Substitute $67\text{ g}/\text{mol}$ for $M_{\text{mixture}}$, $\text{92 g}/\text{mol}$ for $M_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$, $46\text{ g}/\text{mol}$ for $M_{{\text{NO}}_{\text{2}}}$, and $1-{\chi }_{{\text{NO}}_{\text{2}}}$ for ${\chi }_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}$ in the above equation,

$\begin{array}{c}{\chi }_{{\text{NO}}_{\text{2}}}\left(46\text{ g}/\text{mol}\right)+\left(1-{\chi }_{{\text{NO}}_{\text{2}}}\right)\left(\text{92 g}/\text{mol}\right)=67\text{ g}/\text{mol}\\ 46{\chi }_{{\text{NO}}_{\text{2}}}+\text{92}-\text{92}{\chi }_{{\text{NO}}_{\text{2}}}=67\\ 46{\chi }_{{\text{NO}}_{\text{2}}}-\text{92}{\chi }_{{\text{NO}}_{\text{2}}}=67-\text{92}\\ {\chi }_{{\text{NO}}_{\text{2}}}=0.54\end{array}$

The mole fraction of ${\text{N}}_2{\text{O}}_4$ is calculated as follows:

${\chi }_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=1-{\chi }_{{\text{NO}}_{\text{2}}}$

Substitute $0.54$ for ${\chi }_{{\text{NO}}_{\text{2}}}$ in the above equation,

$\begin{array}{c}{\chi }_{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}=1-0.54\\ =0.46\end{array}$

The partial pressure of ${\text{NO}}_2$ is calculatedas follows:

$P_{{\text{NO}}_2}={\chi }_{{\text{NO}}_2}\times P_{\text{total}}$

Substitute $0.54$ for ${\chi }_{{\text{NO}}_2}$ and $0.98\text{ atm}$ for $P_{\text{total}}$ in the above equation,

$\begin{array}{c}{P}_{{\text{NO}}_2}=0.54\times 0.98\text{ atm}\\ =0.53\text{ atm}\end{array}$

The partial pressure of ${\text{N}}_2{\text{O}}_4$ is calculated as follows:

$P_{{\text{N}}_2{\text{O}}_4}={\chi }_{{\text{N}}_2{\text{O}}_4}\times P_{\text{total}}$

Substitute $0.46$ for ${\chi }_{{\text{N}}_2{\text{O}}_4}$ and $0.98\text{ atm}$ for $P_{\text{total}}$ in the above equation,

$\begin{array}{c}{P}_{{\text{N}}_2{\text{O}}_4}=0.46\times 0.98\text{ atm}\\ =0.45\text{ atm}\end{array}$

Conclusion

The partial pressure of ${\text{NO}}_2$ is $0.53\text{ atm}$ and ${\text{N}}_2{\text{O}}_4$ is $0.45\text{ atm}$.