Chapter 4, Problem 4B.13AE

(i)

Interpretation Introduction

Interpretation: The enthalpy of vaporization of naphthalene has to be calculated.

Concept introduction: Clausius-Clapeyron’s expression correlates the pressure and temperature for the processes that includes the phase transitions.  Clausius-Clapeyron’s equation is shown below.

    $\ln \left(\frac{p_2}{p_1}\right)=-\frac{\Delta H_{\text{vaporization}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Answer to Problem 4B.13AE

The enthalpy of vaporization of naphthalene is $\underset{\_}{58322.71Jmo{l}^{-1}}$.

Explanation of Solution

The enthalpy of vaporization is calculated by the Clausius-Clapeyron’s formula.

    $\ln \left(\frac{p_2}{p_1}\right)=-\frac{\Delta H_{\text{vaporization}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$                                                                 (1)

Where,

• $p_1$ and $p_2$ are the pressures.
• $T_1$ and $T_2$ are the temperatures.
• $\Delta H_{\text{vaporization}}$ is the enthalpy of vaporization of naphthalene.
• $R$ is the gas constant.

It is given that the temperature $\left(T_1\right)$ is $85.8{}^{°}\text{C}$.

The conversion of ${}^{°}\text{C}$ to $\text{K}$ is done as,

    $1{}^{°}\text{C}=273\text{}\text{K}$

Therefore, the conversion of $85.8{}^{°}\text{C}$ to $\text{K}$ is done as,

    $\begin{array}{c}85.8{}^{°}\text{C}=85.8+273\text{K}\\ =358.8\text{K}\end{array}$

Therefore, the temperature $\left(T_1\right)$ is $358.8\text{K}$.

It is given that the temperature $\left(T_2\right)$ is $119.3{}^{°}\text{C}$.

The conversion of ${}^{°}\text{C}$ to $\text{K}$ is done as,

    $1{}^{°}\text{C}=273\text{}\text{K}$

Therefore, the conversion of $119.3{}^{°}\text{C}$ to $\text{K}$ is done as,

    $\begin{array}{c}119.3{}^{°}\text{C}=119.3+273\text{K}\\ =392.3\text{K}\end{array}$

Therefore, the temperature $\left(T_2\right)$ is $392.3\text{K}$.

It is given that the pressure $\left(p_1\right)$ is $1.3\text{kPa}$.

The conversion of $\text{kPa}$ to $\text{Pa}$ is done as,

    $\text{1}\text{kPa}=1\times {10}^3\text{Pa}$

Therefore, the conversion of $1.3\text{kPa}$ to $\text{Pa}$ is done as,

    $\begin{array}{c}1.3\text{kPa}=1.3\times {10}^3\text{Pa}\\ =1300\text{Pa}\end{array}$

Therefore, the pressure $\left(p_1\right)$ is $1300\text{Pa}$.

It is given that the pressure $\left(p_2\right)$ is $5.3\text{kPa}$.

The conversion of $\text{kPa}$ to $\text{Pa}$ is done as,

    $\text{1}\text{kPa}=1\times {10}^3\text{Pa}$

Therefore, the conversion of $5.3\text{kPa}$ to $\text{Pa}$ is done as,

    $\begin{array}{c}5.3\text{kPa}=5.3\times {10}^3\text{Pa}\\ =5300\text{Pa}\end{array}$

Therefore, the pressure $\left(p_2\right)$ is $5300\text{Pa}$.

The gas constant is $8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the values of $p_1,p_2,T_1,T_2$ and $R$ in the equation (1).

    $\begin{array}{c}\ln \left(\frac{5300\text{Pa}}{1300\text{Pa}}\right)=-\frac{\Delta H_{\text{vaporization}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\left(\frac{1}{392.3\text{K}}-\frac{1}{358.8\text{K}}\right)\\ \ln \left(4.07\right)=-\frac{\Delta H_{\text{vaporization}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\left(0.0025{\text{K}}^{-1}-0.0027{\text{K}}^{-1}\right)\\ 1.403=-\frac{\Delta H_{\text{vaporization}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\times \left(-0.0002{\text{K}}^{-1}\right)\end{array}$

Rearrange the above equation as,

    $\begin{array}{c}\Delta H_{\text{vaporization}}=\frac{-1.403\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}{-0.0002{\text{K}}^{-1}}\\ =\underset{\_}{58322.71Jmo{l}^{-1}}\end{array}$

Hence, the enthalpy of vaporization of naphthalene is $\underset{\_}{58322.71Jmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation: The normal boiling point of naphthalene has to be calculated.

Concept introduction: Clausius-Clapeyron’s expression correlates the pressure and temperature for the processes that includes the phase transitions.  Clausius-Clapeyron’s equation is shown below.

    $\ln \left(\frac{p_2}{p_1}\right)=-\frac{\Delta H_{\text{vaporization}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Answer to Problem 4B.13AE

The normal boiling point of naphthalene is $\underset{\_}{480.77K}$.

Explanation of Solution

For the normal boiling point of naphthalene pressure $\left(p_2\right)$ is $1.013\times {10}^5\text{Pa}$.

The temperature $\left(T_1\right)$ is $358.8\text{K}$.

The pressure $\left(p_1\right)$ is $1300\text{Pa}$.

The enthalpy of vaporization of naphthalene is $58322.71\text{J}{\text{mol}}^{\text{-1}}$.

Substitute the values of $p_1,p_2,T_1,\Delta H_{\text{vaporization}}$ and $R$ in the equation (1).

    $\begin{array}{c}\ln \left(\frac{1.013\times {10}^5\text{Pa}}{1300\text{Pa}}\right)=-\frac{58322.71\text{J}{\text{mol}}^{\text{-1}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\left(\frac{1}{T_2}-\frac{1}{358.8\text{K}}\right)\\ \ln \left(77.92\right)=-7015\text{K}\times \left(\frac{1}{T_2}-0.0027{\text{K}}^{-1}\right)\\ 4.355=-7015\text{K}\times \left(\frac{1}{T_2}-0.0027{\text{K}}^{-1}\right)\end{array}$

Rearrange the above equation as,

    $\begin{array}{c}\left(\frac{1}{T_2}-0.0027{\text{K}}^{-1}\right)=\frac{4.355}{-7015\text{K}}\\ \left(\frac{1}{T_2}-0.0027{\text{K}}^{-1}\right)=-0.00062{\text{K}}^{-1}\\ \frac{1}{T_2}=-0.00062{\text{K}}^{-1}+0.0027{\text{K}}^{-1}\\ =2.08\times {10}^{-3}{\text{K}}^{-1}\end{array}$

Reverse the value of $\frac{1}{T_2}$.

    $\begin{array}{c}{T}_2=\frac{1}{2.08\times {10}^{-3}{\text{K}}^{-1}}\\ =\underset{\_}{480.77K}\end{array}$

Hence, the normal boiling point of naphthalene is $\underset{\_}{480.77K}$.

(iii)

Interpretation Introduction

Interpretation: The entropy of vaporization at the boiling point of naphthalene has to be calculated.

Concept introduction: Clausius-Clapeyron’s expression correlates the pressure and temperature for the processes that includes the phase transitions.  Clausius-Clapeyron’s equation is shown below.

    $\ln \left(\frac{p_2}{p_1}\right)=-\frac{\Delta H_{\text{vaporization}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Answer to Problem 4B.13AE

The entropy of vaporization at the boiling point of naphthalene is $\underset{\_}{121.31J{K}^{-1}mo{l}^{-1}}$.

Explanation of Solution

The entropy of vaporization at the boiling point of naphthalene is calculated by the following formula.

    $\Delta S_{\text{vaporization}}=\frac{\Delta H_{\text{vaporization}}}{T_{\text{b}}}$                                                                               (2)

Where,

• $\Delta S_{\text{vaporization}}$ is the entropy of vaporization at the boiling point.
• $\Delta H_{\text{vaporization}}$ is the enthalpy of vaporization.
• $T_{\text{b}}$ is the boiling point of naphthalene.

The enthalpy of vaporization of naphthalene is $58322.71\text{J}{\text{mol}}^{\text{-1}}$.

The boiling point of naphthalene is $480.77\text{K}$.

Substitute the value of enthalpy of vaporization and boiling point of naphthalene in equation (4).

    $\begin{array}{c}\Delta S_{\text{vaporization}}=\frac{58322.71\text{J}{\text{mol}}^{\text{-1}}}{480.77\text{K}}\\ =\underset{\_}{121.31J{K}^{-1}mo{l}^{-1}}\end{array}$

Hence, the value of the entropy of vaporization at the boiling point of naphthalene is $\underset{\_}{121.31J{K}^{-1}mo{l}^{-1}}$.