Chapter 16, Problem 16A.7BE

(i)

Interpretation Introduction

Interpretation:

The viscosity of benzene vapour 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.7BE

The viscosity of benzene vapour at $273\text{K}$ is $\underset{\_}{9.44\mu P}$

Explanation of Solution

The viscosity of benzene vapour 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 of benzene is $78\text{g}{\text{mol}}^{-1}=78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$.

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

    $\begin{array}{c}m=\frac{78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ =78\times 1.6605\times {10}^{-27}\text{kg}\end{array}$

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 $0.88{\text{nm}}^{\text{2}}$.

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 $0.88{\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

    $0.88{\text{nm}}^{\text{2}}=0.88\times {10}^{-18}{\text{m}}^2$

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 benzene vapour.

    $\begin{array}{c}\eta =\left(\frac{78\times 1.6605\times {10}^{-27}\text{kg}}{3\times 0.88\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{4\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 273\text{K}}{3.14\times 78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =\left(\frac{1.295\times {10}^{-25}\text{kg}}{2.64\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{9078.888\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}/\text{mol}}{0.245\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =4.905\times {10}^{-8}\times 192.50\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ =9.44\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\end{array}$

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 $9.44\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}9.44\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}=9.44\times {10}^{-6}\times {10}^6\text{μP}\\ =\underset{\_}{9.44\mu P}\end{array}$

Therefore, the viscosity of benzene vapour at $273\text{K}$ is $\underset{\_}{9.44\mu P}$.

(ii)

Interpretation Introduction

Interpretation:

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

Concept introduction:

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

Answer to Problem 16A.7BE

The viscosity of benzene vapour at $298\text{K}$ is $\underset{\_}{9.865\mu P}$.

Explanation of Solution

The viscosity of benzene vapour 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 of benzene is $78\text{g}{\text{mol}}^{-1}=78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$.

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

    $\begin{array}{c}m=\frac{78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ =78\times 1.6605\times {10}^{-27}\text{kg}\end{array}$

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 $0.88{\text{nm}}^{\text{2}}$.

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 $0.88{\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

    $0.88{\text{nm}}^{\text{2}}=0.88\times {10}^{-18}{\text{m}}^2$

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 benzene vapour.

    $\begin{array}{c}\eta =\left(\frac{78\times 1.6605\times {10}^{-27}\text{kg}}{3\times 0.88\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{4\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 298\text{K}}{3.14\times 78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =\left(\frac{1.295\times {10}^{-25}\text{kg}}{2.64\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{9910.288\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}/\text{mol}}{0.245\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =4.905\times {10}^{-8}\times 201.12\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ =9.865\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\end{array}$

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 $9.865\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}9.865\times {10}^{-6}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}=9.865\times {10}^{-6}\times {10}^6\text{μP}\\ =\underset{\_}{9.865\mu P}\end{array}$

Therefore, the viscosity of benzene vapour at $298\text{K}$ is $\underset{\_}{9.865\mu P}$.

(iii)

Interpretation Introduction

Interpretation:

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

Concept introduction:

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

Answer to Problem 16A.7BE

The viscosity of benzene vapour at $1000\text{K}$ is $\underset{\_}{18\mu P}$.

Explanation of Solution

The viscosity of benzene vapour 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 of benzene is $78\text{g}{\text{mol}}^{-1}=78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}$.

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

    $\begin{array}{c}m=\frac{78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{6.022\times {10}^{23}{\text{mol}}^{-1}}\\ =78\times 1.6605\times {10}^{-27}\text{kg}\end{array}$

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 $0.88{\text{nm}}^{\text{2}}$.

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 $0.88{\text{nm}}^{\text{2}}$ into ${\text{m}}^{\text{2}}$ is done as shown below.

    $0.88{\text{nm}}^{\text{2}}=0.88\times {10}^{-18}{\text{m}}^2$

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 benzene vapour.

    $\begin{array}{c}\eta =\left(\frac{78\times 1.6605\times {10}^{-27}\text{kg}}{3\times 0.88\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{4\times 8.314\text{J}/\text{mol}\cdot \text{K}\times 1000\text{K}}{3.14\times 78\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =\left(\frac{1.295\times {10}^{-25}\text{kg}}{2.64\times {10}^{-18}{\text{m}}^2}\right){\left(\frac{33256\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}/\text{mol}}{0.245\text{kg}{\text{mol}}^{-1}}\right)}^{1/2}\\ =4.905\times {10}^{-8}\times 368.427\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\\ =1.80\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\end{array}$

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 $1.80\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$ into $\text{μP}$ is done as shown below.

    $\begin{array}{c}1.80\times {10}^{-5}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}=1.80\times {10}^{-5}\times {10}^6\text{μP}\\ =\underset{\_}{18\mu P}\end{array}$

Therefore, the viscosity of benzene vapour at $1000\text{K}$ is $\underset{\_}{18\mu P}$.