Chapter 4, Problem 4B.4P

(a)

Interpretation Introduction

Interpretation: The value of $\text{d}p/\text{d}T$ using the Clapeyron equation at the normal boiling point has to be estimated.

Concept introduction: The value of $\text{d}p/\text{d}T$ using the Clapeyron equation is calculated by using the Clapeyron’s equation.  The Clapeyron’s equation is shown below.

    $\left(\frac{\text{d}p}{\text{d}T}\right)=\frac{{\Delta }_{\text{vap}}H}{T\Delta V_{\text{m}}}$

Answer to Problem 4B.4P

The value of $\text{d}p/\text{d}T$ using the Clapeyron equation at the normal boiling point is $\underset{\_}{0.0055Pa{K}^{-1}}$.

Explanation of Solution

The Clapeyron equation is shown below.

    $\frac{\text{d}p}{\text{d}T}=\frac{{\Delta }_{\text{vap}}H}{T\Delta V_{\text{m}}}$                                                                                                  (1)

Where,

• $\frac{\text{d}p}{\text{d}T}$ is the change in pressure with respect to change in temperature.
• ${\Delta }_{\text{vap}}H$ is the enthalpy of vaporization.
• $T$ is the normal boiling point of the liquid.
• $\Delta V_{\text{m}}$ is the change in molar volume.

It is given that the enthalpy of vaporization is $14.4\text{}\text{kJ}{\text{mol}}^{\text{-1}}$.

The conversion of $\text{kJ}{\text{mol}}^{\text{-1}}$ to $\text{J}{\text{mol}}^{\text{-1}}$ is done as,

    $\text{1}\text{kJ}{\text{mol}}^{\text{-1}}=1\times {10}^{-3}\text{J}{\text{mol}}^{\text{-1}}$

Therefore, the conversion of $14.4\text{kJ}{\text{mol}}^{\text{-1}}$ to $\text{J}{\text{mol}}^{\text{-1}}$ is done as,

    $14.4\text{kJ}{\text{mol}}^{\text{-1}}=14.4\times {10}^{-3}\text{J}{\text{mol}}^{\text{-1}}$

Therefore, the enthalpy of vaporization of liquid is $14.4\times {10}^{-3}\text{J}{\text{mol}}^{\text{-1}}$.

It is given that the normal boiling point of the liquid is $180\text{K}$.

It is given that the molar volume of the liquid $\left(V_{{\text{m}}_1}\right)$ and the vapour state $\left(V_{{\text{m}}_{\text{2}}}\right)$ is $115\text{}{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}$ and $14.5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}$ respectively.

The conversion of ${\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}$ to ${\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$ is done as,

    $1{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}=1\times {10}^{-6}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$

Therefore, the conversion of $115{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}$ to ${\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$ is done as,

    $\text{115}{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}=115\times {10}^{-6}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$

Therefore, the molar volume in the liquid state $\left(V_{{\text{m}}_1}\right)$ is $115\times {10}^{-6}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$.

The conversion of ${\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}$ to ${\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$ is done as,

    $1{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}=1\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$

Therefore, the conversion of $14.5{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}$ to ${\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$ is done as,

    $\text{14}\text{.5}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}=14.5\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$

Therefore, the molar volume in the liquid state $\left(V_{{\text{m}}_2}\right)$ is $14.5\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$.

The change in molar volume is calculated by the following formula.

    $\Delta V_{\text{m}}=V_{{\text{m}}_{\text{2}}}-V_{{\text{m}}_1}$                                                                                             (2)

Substitute the value of the molar volume in liquid and vapour state.

    $\begin{array}{c}\Delta V_{\text{m}}=14.5\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}-115\times {10}^{-6}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}\\ =14.5\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}-0.115\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}\\ =14.385\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the change in molar volume $\left(\Delta V_{\text{m}}\right)$ is $14.385\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}$.

Substitute the value of ${\Delta }_{\text{vap}}H,\Delta V_{\text{m}}$ and $T$ in equation (1).

    $\begin{array}{c}\frac{\text{d}p}{\text{d}T}=\frac{14.4\times {10}^{-3}\text{J}{\text{mol}}^{\text{-1}}}{180\text{K}\times 14.385\times {10}^{-3}{\text{m}}^{\text{3}}{\text{mol}}^{\text{-1}}}\\ =0.0055\text{J}{\text{m}}^{\text{-3}}{\text{K}}^{\text{-1}}\end{array}$

The value of $\text{d}p/\text{d}T$ using the Clapeyron equation at the normal boiling point is $0.0055\text{J}{\text{m}}^{\text{-3}}{\text{K}}^{\text{-1}}$.

The conversion of $\text{J}{\text{m}}^{\text{-3}}{\text{K}}^{-1}$ to $\text{Pa}{\text{K}}^{-1}$ is done as,

    $1\text{J}{\text{m}}^{\text{-3}}{\text{K}}^{-1}=1\text{}\text{Pa}{\text{K}}^{-1}$

Therefore the conversion of $0.0055\text{J}{\text{m}}^{\text{-3}}{\text{K}}^{\text{-1}}$ to $\text{Pa}{\text{K}}^{-1}$ is done as,

    $0.0055\text{J}{\text{m}}^{\text{-3}}{\text{K}}^{\text{-1}}=\underset{\_}{0.0055Pa{K}^{-1}}$

Hence, the value of $\text{d}p/\text{d}T$ using the Clapeyron equation at the normal boiling point is $\underset{\_}{0.0055Pa{K}^{-1}}$.

(b)

Interpretation Introduction

Interpretation: The percentage error in the resulting value of $\text{d}p/\text{d}T$ has to be estimated.

Concept introduction: The Clausius-Clapeyron’s equation shows the correlation between the temperature and pressure.  The Clausius-Clapeyron’s equation can be used for the phase transition process.  The Clausius-Clapeyron’s equation is shown below.

    $\frac{\text{d}p}{\text{d}T}=\frac{p{\Delta }_{vap}H}{R{T}^2}$

Answer to Problem 4B.4P

The percentage error in the resulting value of $\text{d}p/\text{d}T$ is $\underset{\_}{1.63\%}$.

Explanation of Solution

The value of $\text{d}p/\text{d}T$ is calculated by the Clausius-Clapeyron’s equation.

    $\frac{\text{d}p}{\text{d}T}=\frac{p{\Delta }_{vap}H}{R{T}^2}$                                                                                               (3)

Where,

• ${\Delta }_{vap}H$ is the enthalpy of vaporization.
• $T$ is the normal boiling point.
• $R$ is the gas constant.

Substitute the value of $p,{\Delta }_{vap}H,R$ and $T$ in equation (3).

    $\begin{array}{c}\frac{\text{d}p}{\text{d}T}=\frac{1.013\times {10}^5\text{Pa}\times 14.4\times {10}^{-3}\text{J}{\text{mol}}^{\text{-1}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times {\left(180\text{K}\right)}^2}\\ =\frac{14.58\times {10}^2\text{Pa}\text{J}{\text{mol}}^{\text{-1}}}{269373.6\text{J}\text{K}{\text{mol}}^{\text{-1}}}\\ =0.00541\text{Pa}{\text{K}}^{-1}\end{array}$

The percentage error in the resulting value of $\text{d}p/\text{d}T$ is calculated by the following formula.

    $\%\text{Error}=\left(\frac{{\left(\text{d}p/\text{d}T\right)}_{\text{Clapeyron's}}-{\left(\text{d}p/\text{d}T\right)}_{\text{Clausius-Clapeyron}}}{{\left(\text{d}p/\text{d}T\right)}_{\text{Clapeyron's}}}\right)\times 100\%$                        (4)

The value of $\text{d}p/\text{d}T$ from Clapeyron equation is $0.0055\text{Pa}{\text{K}}^{\text{-1}}$.

The value of $\text{d}p/\text{d}T$ from Clausius-Clapeyron equation is $0.0055\text{Pa}{\text{K}}^{\text{-1}}$.

Substitute the values of $\text{d}p/\text{d}T$ in equation (4).

    $\begin{array}{c}\%\text{Error}=\left(\frac{0.0055\text{Pa}{\text{K}}^{\text{-1}}-0.00541\text{Pa}{\text{K}}^{-1}}{0.0055\text{Pa}{\text{K}}^{\text{-1}}}\right)\times 100\%\\ =\underset{\_}{1.63\%}\end{array}$

Hence, the percentage error in the resulting value of $\text{d}p/\text{d}T$ is $\underset{\_}{1.63\%}$.