Chapter 1, Problem 1A.8AE

(i)

Interpretation Introduction

Interpretation: The mole fraction and partial pressure of nitrogen and oxygen assuming that air consists of only these two gases has to be calculated.

Concept introduction: The partial pressure of any gas in a mixture of gases is calculated using the total pressure of the mixture and the mole fraction of the gas in the mixture.  This is represented by the formula given below as,

  $p_{\text{total}}=\frac{p_{\text{a}}}{x_{\text{a}}}$

The mole fraction is calculated using the formula given below as,

    $x_{\text{a}}=\frac{n_{\text{a}}}{n_{\text{Total}}}$

Answer to Problem 1A.8AE

The mole fraction and partial pressure of nitrogen has been calculated as $\underset{\_}{0.756}$ and $\underset{\_}{0.746bar}$ respectively and for oxygen as $\underset{\_}{0.243}$ and $\underset{\_}{0.241bar}$ respectively assuming that air consists of only these two gases.

Explanation of Solution

The mass density of air at $0.987\text{bar}$ and $27°\text{C}$ is given as $1.146\text{kg}{\text{m}}^{-3}$. The mass of air is calculated using the formula given below as,

    $\text{Mass}=\text{Density}\times \text{Volume}$

Substitute the corresponding values in the above equation as given below.

    $\begin{array}{c}\text{Mass}=\text{Density}\times \text{Volume}\\ =1.146\text{kg}{\text{m}}^{-3}\times 1{\text{m}}^3\\ =1.146\text{kg}\end{array}$

The perfect gas equation is given by the formula as,

    $pV=nRT$

Where,

• Ø  $p$ is the total pressure.
• Ø  $V$ is the volume.
• Ø  $n$ is the number of moles.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.

Rearrange the above equation for number of moles $\left(n\right)$ as follows.

    $\begin{array}{c}pV=nRT\\ n=\frac{pV}{RT}\end{array}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}n=\frac{pV}{RT}\\ =\frac{\left(0.987\text{bar}\right)\left(1{\text{m}}^3\right)}{\left(8.314\times {10}^{-5}\text{bar}{\text{m}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(27+273.15\right)\text{K}}\\ =39.55\text{mol}\end{array}$

The total number of moles is represented as,

    $n_{\text{total}}=n_{{\text{N}}_{\text{2}}}+n_{{\text{O}}_{\text{2}}}$

The number of moles of nitrogen is calculated as,

    $\begin{array}{c}{n}_{\text{total}}=n_{{\text{N}}_{\text{2}}}+n_{{\text{O}}_{\text{2}}}\\ n_{{\text{N}}_{\text{2}}}=n_{\text{total}}-n_{{\text{O}}_{\text{2}}}\end{array}$                                                                                            (1)

The mass of nitrogen is calculated as,

    $m_{{\text{N}}_{\text{2}}}=n_{{\text{N}}_{\text{2}}}\times {\text{M}}_{{\text{N}}_{\text{2}}}$                                                                                            (2)

Where,

• Ø  ${\text{M}}_{{\text{N}}_{\text{2}}}$ is the molar mass of nitrogen molecule.
• Ø  $n_{{\text{N}}_{\text{2}}}$ is the number of moles of nitrogen molecule.

Similarly, the mass of oxygen is calculated as,

    $m_{{\text{O}}_{\text{2}}}=n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}$                                                                                             (3)

The total mass of air can be written as,

    $m_{\text{air}}=m_{{\text{N}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}$

Substitute the equation (2) and (3) in the above equation as follows.

    $\begin{array}{l}{m}_{\text{air}}=m_{{\text{N}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}\\ m_{\text{air}}=n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}+n_{{\text{N}}_{\text{2}}}\times {\text{M}}_{{\text{N}}_{\text{2}}}\end{array}$

Rearrange the above equation as follows.

    $\begin{array}{l}{m}_{\text{air}}=n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}+n_{{\text{N}}_{\text{2}}}\times {\text{M}}_{{\text{N}}_{\text{2}}}\\ n_{{\text{N}}_{\text{2}}}=\frac{m_{\text{air}}-\left(n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\end{array}$

The above equation can be written in terms of $n_{{\text{O}}_{\text{2}}}$ as given below.

    $\begin{array}{c}{n}_{{\text{N}}_{\text{2}}}=\frac{m_{\text{air}}-\left(n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\\ \frac{\left(n_{\text{total}}-n_{{\text{O}}_{\text{2}}}\right)}{n_{{\text{O}}_{\text{2}}}}=\frac{1}{n_{{\text{O}}_{\text{2}}}}\left(\frac{m_{\text{air}}-\left(n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)\\ n_{{\text{O}}_{\text{2}}}=\frac{n_{\text{total}}-\left(\frac{m_{\text{air}}}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}{1-\left(\frac{{\text{M}}_{{\text{O}}_{\text{2}}}}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}\end{array}$

The molar mass of nitrogen and oxygen are $0.028\text{kg}/\text{mol}$ and $0.032\text{kg}/\text{mol}$ respectively.

Substitute the corresponding values in the above equation as follows.

    $\begin{array}{l}{n}_{{\text{O}}_{\text{2}}}=\frac{n_{\text{total}}-\left(\frac{m_{\text{air}}}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}{1-\left(\frac{{\text{M}}_{{\text{O}}_{\text{2}}}}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}\\ =\frac{39.55\text{mol}-\left(\frac{1.146\text{kg}}{0.028\text{kg}/\text{mol}}\right)}{1-\left(\frac{0.032\text{kg}/\text{mol}}{0.028\text{kg}/\text{mol}}\right)}\\ =9.65\text{mol}\end{array}$

Substitute the corresponding values in equation (1) as given below.

    $\begin{array}{c}{n}_{{\text{N}}_{\text{2}}}=n_{\text{total}}-n_{{\text{O}}_{\text{2}}}\\ =39.55\text{mol}-9.65\text{mol}\\ =29.9\text{mol}\end{array}$

The mole fraction of nitrogen is calculated using the formula given below as,

    $x_{{\text{N}}_{\text{2}}}=\frac{n_{{\text{N}}_{\text{2}}}}{n_{\text{Total}}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{x}_{{\text{N}}_{\text{2}}}=\frac{n_{{\text{N}}_{\text{2}}}}{n_{\text{Total}}}\\ =\frac{29.9\text{mol}}{39.55\text{mol}}\\ =\underset{\_}{0.756}\end{array}$

The mole fraction of oxygen is calculated using the formula given below as,

    $x_{{\text{O}}_{\text{2}}}=\frac{n_{{\text{O}}_{\text{2}}}}{n_{\text{Total}}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{x}_{{\text{O}}_{\text{2}}}=\frac{n_{{\text{O}}_{\text{2}}}}{n_{\text{Total}}}\\ =\frac{9.65\text{mol}}{39.55\text{mol}}\\ =\underset{\_}{0.243}\end{array}$

The partial pressure is calculated using the perfect gas law given below as,

    $\begin{array}{c}pV=nRT\\ p=\frac{nRT}{V}\end{array}$                                                                                                    (4)

Substitute the values in equation (4) for nitrogen as follows.

    $\begin{array}{c}{p}_{{\text{N}}_{\text{2}}}=\frac{n_{{\text{N}}_{\text{2}}}RT}{V}\\ =\frac{\left(29.9\text{mol}\right)\left(8.314\times {10}^{-5}\text{bar}{\text{m}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(300.15\text{K}\right)}{\text{1}{\text{m}}^{\text{3}}}\\ =\underset{\_}{0.746bar}\end{array}$

Substitute the values in equation (4) for oxygen as follows.

    $\begin{array}{c}{p}_{{\text{O}}_{\text{2}}}=\frac{n_{{\text{O}}_{\text{2}}}RT}{V}\\ =\frac{\left(9.65\text{mol}\right)\left(8.314\times {10}^{-5}\text{bar}{\text{m}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(300.15\text{K}\right)}{\text{1}{\text{m}}^{\text{3}}}\\ =\underset{\_}{0.241bar}\end{array}$

Thus, the mole fraction and partial pressure of nitrogen is $\underset{\_}{0.756}$ and $\underset{\_}{0.746bar}$ respectively and for oxygen is $\underset{\_}{0.243}$ and $\underset{\_}{0.241bar}$ respectively.

(ii)

Interpretation Introduction

Interpretation: The mole fraction and partial pressure of nitrogen and oxygen when $1.0\text{mole}\text{percent}$ of $\text{Ar}$ is also present in air has to be calculated.

Concept introduction: The partial pressure of any gas in a mixture of gases is calculated using the total pressure of the mixture and the mole fraction of the gas in the mixture.  This is represented by the formula given below as,

  $p_{\text{total}}=\frac{p_{\text{a}}}{x_{\text{a}}}$

The mole fraction is calculated using the formula given below as,

    $x_{\text{a}}=\frac{n_{\text{a}}}{n_{\text{Total}}}$

Answer to Problem 1A.8AE

The mole fraction and partial pressure of nitrogen has been calculated as $\underset{\_}{0.77}$ and $\underset{\_}{0.76bar}$ respectively and for oxygen as $\underset{\_}{0.22}$ and $\underset{\_}{0.21bar}$ respectively when $1.0\text{mole}\text{percent}$ of $\text{Ar}$ is also present in air.

Explanation of Solution

The total number of moles is represented as,

    $\begin{array}{l}{n}_{\text{total}}=n_{{\text{N}}_{\text{2}}}+n_{{\text{O}}_{\text{2}}}+n_{\text{Ar}}\\ n_{\text{total}}=n_{{\text{N}}_{\text{2}}}+n_{{\text{O}}_{\text{2}}}+0.01n\end{array}$

The number of moles of nitrogen is calculated as,

    $\begin{array}{c}{n}_{\text{total}}=n_{{\text{N}}_{\text{2}}}+n_{{\text{O}}_{\text{2}}}+0.01n\\ n_{{\text{N}}_{\text{2}}}=n_{\text{total}}-n_{{\text{O}}_{\text{2}}}-0.01n\end{array}$                                                                                (5)

The total mass of air can be written as,

    $m_{\text{air}}=m_{{\text{N}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}+m_{\text{Ar}}$

Substitute the corresponding values in the above equation as follows.

    $\begin{array}{l}{m}_{\text{air}}=m_{{\text{N}}_{\text{2}}}+m_{{\text{O}}_{\text{2}}}+m_{\text{Ar}}\\ m_{\text{air}}=n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}+n_{{\text{N}}_{\text{2}}}\times {\text{M}}_{{\text{N}}_{\text{2}}}+n_{\text{Ar}}\times {\text{M}}_{\text{Ar}}\end{array}$

Rearrange the above equation as follows.

    $\begin{array}{l}{m}_{\text{air}}=n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}+n_{{\text{N}}_{\text{2}}}\times {\text{M}}_{{\text{N}}_{\text{2}}}+n_{\text{Ar}}\times {\text{M}}_{\text{Ar}}\\ n_{{\text{N}}_{\text{2}}}=\frac{m_{\text{air}}-\left(n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}\right)-\left(n_{\text{Ar}}\times {\text{M}}_{\text{Ar}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\end{array}$

The above equation can be written in terms of $n_{{\text{O}}_{\text{2}}}$ as given below.

    $\begin{array}{c}{n}_{{\text{N}}_{\text{2}}}=\frac{m_{\text{air}}-\left(n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}\right)-\left(n_{\text{Ar}}\times {\text{M}}_{\text{Ar}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\\ \frac{\left(n_{\text{total}}-n_{{\text{O}}_{\text{2}}}-0.01n\right)}{n_{{\text{O}}_{\text{2}}}}=\frac{1}{n_{{\text{O}}_{\text{2}}}}\left(\frac{m_{\text{air}}-\left(n_{{\text{O}}_{\text{2}}}\times {\text{M}}_{{\text{O}}_{\text{2}}}\right)-\left(n_{\text{Ar}}\times {\text{M}}_{\text{Ar}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)\\ n_{{\text{O}}_{\text{2}}}=\frac{0.9900n-\left(\frac{m_{\text{air}}-\left(0.01n\times {\text{M}}_{\text{Ar}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}{1-\left(\frac{{\text{M}}_{{\text{O}}_{\text{2}}}}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}\end{array}$

Substitute the corresponding values in the above equation as follows.

    $\begin{array}{c}{n}_{{\text{O}}_{\text{2}}}=\frac{0.9900n-\left(\frac{m_{\text{air}}-\left(0.01n\times {\text{M}}_{\text{Ar}}\right)}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}{1-\left(\frac{{\text{M}}_{{\text{O}}_{\text{2}}}}{{\text{M}}_{{\text{N}}_{\text{2}}}}\right)}\\ =\frac{0.9900\left(39.55\text{mol}\right)-\left(\frac{1.146\text{kg}-\left(0.01\left(39.55\text{mol}\right)\times \left(\text{0}\text{.039}\text{kg}/\text{mol}\right)\right)}{0.028\text{kg}/\text{mol}}\right)}{1-\left(\frac{0.032\text{kg}/\text{mol}}{0.028\text{kg}/\text{mol}}\right)}\\ =8.56\text{mol}\end{array}$

Substitute the corresponding values in equation (5) to calculate the number of moles of nitrogen as follows.

    $\begin{array}{c}{n}_{{\text{N}}_{\text{2}}}=n_{\text{total}}-n_{{\text{O}}_{\text{2}}}-0.01n\\ =39.55\text{mol}-8.56\text{mol}-\left(0.01\times 39.55\text{mol}\right)\\ =30.59\text{mol}\end{array}$

The mole fraction of nitrogen is calculated using the formula given below as,

    $x_{{\text{N}}_{\text{2}}}=\frac{n_{{\text{N}}_{\text{2}}}}{n_{\text{Total}}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{x}_{{\text{N}}_{\text{2}}}=\frac{n_{{\text{N}}_{\text{2}}}}{n_{\text{Total}}}\\ =\frac{30.59\text{mol}}{39.55\text{mol}}\\ =\underset{\_}{0.77}\end{array}$

The mole fraction of oxygen is calculated using the formula given below as,

    $x_{{\text{O}}_{\text{2}}}=\frac{n_{{\text{O}}_{\text{2}}}}{n_{\text{Total}}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{x}_{{\text{O}}_{\text{2}}}=\frac{n_{{\text{O}}_{\text{2}}}}{n_{\text{Total}}}\\ =\frac{8.56\text{mol}}{39.55\text{mol}}\\ =\underset{\_}{0.22}\end{array}$

The partial pressure is calculated using the perfect gas law given below as,

    $\begin{array}{c}pV=nRT\\ p=\frac{nRT}{V}\end{array}$                                                                                                    (4)

Substitute the values in equation (4) for nitrogen as follows.

    $\begin{array}{c}{p}_{{\text{N}}_{\text{2}}}=\frac{n_{{\text{N}}_{\text{2}}}RT}{V}\\ =\frac{\left(30.59\text{mol}\right)\left(8.314\times {10}^{-5}\text{bar}{\text{m}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(300.15\text{K}\right)}{\text{1}{\text{m}}^{\text{3}}}\\ =\underset{\_}{0.76bar}\end{array}$

Substitute the values in equation (4) for oxygen as follows.

    $\begin{array}{c}{p}_{{\text{O}}_{\text{2}}}=\frac{n_{{\text{O}}_{\text{2}}}RT}{V}\\ =\frac{\left(8.56\text{mol}\right)\left(8.314\times {10}^{-5}\text{bar}{\text{m}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(300.15\text{K}\right)}{\text{1}{\text{m}}^{\text{3}}}\\ =\underset{\_}{0.21bar}\end{array}$

Thus, the mole fraction and partial pressure of nitrogen is $\underset{\_}{0.77}$ and $\underset{\_}{0.76bar}$ respectively and for oxygen is $\underset{\_}{0.22}$ and $\underset{\_}{0.21bar}$ respectively.