Chapter 16, Problem 16A.7AE

(i)

Interpretation Introduction

Interpretation:

The viscosity of air at $273\text{K}$ is to be calculated.

Concept introduction:

Viscosity is called the property of a fluid that is “a measure of its resistance to flow”.  It is the consequence of a molecular attraction and denotes the internal friction of a moving fluid.  A fluid with high viscosity resists motion due to its high molecular interaction.  The substances with low intermolecular interaction have lower viscosity.

Answer to Problem 16A.7AE

The viscosity of air at $273\text{K}$ is $\underset{\_}{12.67\mu P}$.

Explanation of Solution

The viscosity of air is calculated by the formula shown below.

    $\eta =\left(\frac{m}{3\sigma }\right){\left(\frac{4RT}{\pi M}\right)}^{1/2}$        (1)

Where,

• $R$ is the gas constant.
• $T$ is the temperature.
• $\sigma$ is the collision cross section area.
• $M$ is the molar mass.
• $m$ is equal to $\frac{M}{N_{\text{A}}}$.

The molar mass is $29.0\text{g}{\text{mol}}^{-1}=29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$.

The value of $m$ is calculated as shown below.

    $\begin{array}{c}m=\frac{29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ =29.0\times 1.6605\times {10}^{-27}\text{kg}\end{array}$

The Avogadro’s number is $6.022\times {10}^{23}{\text{mol}}^{-1}$.

The value of gas constant is $8.314\text{J}/\text{mol}\cdot \text{K}$.

The value of $\sigma$ is $0.40{\text{nm}}^{\text{2}}$.

The conversion of ${\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

  $\begin{array}{l}1\text{nm}=1\times {10}^{-9}\text{m}\\ 1{\text{nm}}^2={\left(1\times {10}^{-9}\right)}^2{\text{m}}^2\\ 1{\text{nm}}^2=1\times {10}^{-18}{\text{m}}^2\end{array}$

Therefore, the conversion of $0.40{\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

    $0.40{\text{nm}}^{\text{2}}=0.40\times {10}^{-18}{\text{m}}^2$

Also, $1\text{J}=\text{1}\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}$

Substitute the values of $m$, $R$, $T$, $\sigma$ and $M$ in equation (1) to calculate the viscosity of air.

    $\begin{array}{c}\eta =\left(\frac{29.0\times 1.6605\times {10}^{-27}\text{kg}}{3\times 0.40\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{4\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 273\text{K}}{3.14\times 29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =\left(\frac{4.81545\times {10}^{-26}\text{kg}}{1.2\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{9078.888\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}/\text{mol}}{0.09106\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =4.012\times {10}^{-8}\times 315.756\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ =1.2670\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\end{array}$

The conversion of $\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}\text{1}\text{μP}=1\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ \text{1}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}={10}^6\text{μP}\end{array}$

Therefore, the conversion of $1.2670\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}1.2670\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}=1.2670\times {10}^{-5}\times {10}^6\text{μP}\\ =\underset{\_}{12.67\mu P}\end{array}$

Therefore, the viscosity of air at $273\text{K}$ is $\underset{\_}{12.67\mu P}$.

(ii)

Interpretation Introduction

Interpretation:

The viscosity of air at $298\text{K}$ is to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

Answer to Problem 16A.7AE

The viscosity of air at $298\text{K}$ is $\underset{\_}{13.235\mu P}$.

Explanation of Solution

The viscosity of air is calculated by the formula shown below.

    $\eta =\left(\frac{m}{3\sigma }\right){\left(\frac{4RT}{\pi M}\right)}^{1/2}$        (1)

Where,

• $R$ is the gas constant.
• $T$ is the temperature.
• $\sigma$ is the collision cross section area.
• $M$ is the molar mass.
• $m$ is equal to $\frac{M}{N_{\text{A}}}$.

The molar mass is $29.0\text{g}{\text{mol}}^{-1}=29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$.

The value of $m$ is calculated as shown below.

    $\begin{array}{c}m=\frac{29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ =29.0\times 1.6605\times {10}^{-27}\text{kg}\end{array}$

The Avogadro’s number is $6.022\times {10}^{23}{\text{mol}}^{-1}$.

The value of gas constant is $8.314\text{J}/\text{mol}\cdot \text{K}$.

The value of $\sigma$ is $0.40{\text{nm}}^{\text{2}}$.

The conversion of ${\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

  $\begin{array}{l}1\text{nm}=1\times {10}^{-9}\text{m}\\ 1{\text{nm}}^2={\left(1\times {10}^{-9}\right)}^2{\text{m}}^2\\ 1{\text{nm}}^2=1\times {10}^{-18}{\text{m}}^2\end{array}$

Therefore, the conversion of $0.40{\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

    $0.40{\text{nm}}^{\text{2}}=0.40\times {10}^{-18}{\text{m}}^2$

Also, $1\text{J}=\text{1}\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}$

Substitute the values of $m$, $R$, $T$, $\sigma$ and $M$ in equation (1) to calculate the viscosity of air.

    $\begin{array}{c}\eta =\left(\frac{29.0\times 1.6605\times {10}^{-27}\text{kg}}{3\times 0.40\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{4\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 298\text{K}}{3.14\times 29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =\left(\frac{4.81545\times {10}^{-26}\text{kg}}{1.2\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{9910.288\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}/\text{mol}}{0.09106\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =4.012\times {10}^{-8}\times 329.89\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ =1.3235\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\end{array}$

The conversion of $\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}\text{1}\text{μP}=1\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ \text{1}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}={10}^6\text{μP}\end{array}$

Therefore, the conversion of $1.3235\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}1.3235\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}=1.3235\times {10}^{-5}\times {10}^6\text{μP}\\ =\underset{\_}{13.235\mu P}\end{array}$

Therefore, the viscosity of air at $298\text{K}$ is $\underset{\_}{13.235\mu P}$.

(iii)

Interpretation Introduction

Interpretation:

The viscosity of air at $1000\text{K}$ is to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

Answer to Problem 16A.7AE

The viscosity of air at $1000\text{K}$ is $\underset{\_}{24.245\mu P}$.

Explanation of Solution

The viscosity of air is calculated by the formula shown below.

    $\eta =\left(\frac{m}{3\sigma }\right){\left(\frac{4RT}{\pi M}\right)}^{1/2}$        (1)

Where,

• $R$ is the gas constant.
• $T$ is the temperature.
• $\sigma$ is the collision cross section area.
• $M$ is the molar mass.
• $m$ is equal to $\frac{M}{N_{\text{A}}}$.

The molar mass is $29.0\text{g}{\text{mol}}^{-1}=29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$.

The value of $m$ is calculated as shown below.

    $\begin{array}{c}m=\frac{29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ =29.0\times 1.6605\times {10}^{-27}\text{kg}\end{array}$

The Avogadro’s number is $6.022\times {10}^{23}{\text{mol}}^{-1}$.

The value of gas constant is $8.314\text{J}/\text{mol}\cdot \text{K}$.

The value of $\sigma$ is $0.40{\text{nm}}^{\text{2}}$.

The conversion of ${\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

  $\begin{array}{l}1\text{nm}=1\times {10}^{-9}\text{m}\\ 1{\text{nm}}^2={\left(1\times {10}^{-9}\right)}^2{\text{m}}^2\\ 1{\text{nm}}^2=1\times {10}^{-18}{\text{m}}^2\end{array}$

Therefore, the conversion of $0.40{\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

    $0.40{\text{nm}}^{\text{2}}=0.40\times {10}^{-18}{\text{m}}^2$

Also, $1\text{J}=\text{1}\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}$

Substitute the values of $m$, $R$, $T$, $\sigma$ and $M$ in equation (1) to calculate the viscosity of air.

    $\begin{array}{c}\eta =\left(\frac{29.0\times 1.6605\times {10}^{-27}\text{kg}}{3\times 0.40\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{4\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 1000\text{K}}{3.14\times 29.0\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =\left(\frac{4.81545\times {10}^{-26}\text{kg}}{1.2\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{33256\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}/\text{mol}}{0.09106\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =4.012\times {10}^{-8}\times 604.325\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ =2.4245\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\end{array}$

The conversion of $\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}\text{1}\text{μP}=1\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ \text{1}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}={10}^6\text{μP}\end{array}$

Therefore, the conversion of $2.4245\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}2.4245\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}=2.4245\times {10}^{-5}\times {10}^6\text{μP}\\ =\underset{\_}{24.245\mu P}\end{array}$

Therefore, the viscosity of air at $1000\text{K}$ is $\underset{\_}{24.245\mu P}$.