Chapter 11, Problem 38GQ

The following data are the equilibrium vapor pressure of limonene, C₁₀H₁₆, at various temperatures. (Limonene is used as a scent in commercial products.)

1. (a) Plot these data as ln P versus 1/T so that you have a graph resembling the one in Figure 11.13.
2. (b) At what temperature does the liquid have an equilibrium vapor pressure of 250 mm Hg? At what temperature is it 650 mm Hg?
3. (c) What is the normal boiling point of limonene?
4. (d) Calculate the molar enthalpy of vaporization for limonene using the Clausius-Clapeyron equation.

Interpretation Introduction

Interpretation:

The graph of $\text{ln}\text{P}\text{versus}\text{1}/\text{T}$ using the given data has to be plotted.

Concept Introduction:

Clausius-Clapeyron equation:

  $\text{ln}\text{P}\text{=}-\left(\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{RT}}\right)\text{+}\text{C}$

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{p}}_{\text{1}}}\text{=}\text{-}\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Explanation of Solution

The graph of $\text{ln}\text{P}$ versus $\text{1}/\text{T}$ is plotted using given data.

Given:

  $\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 287& 1\\ 326.8& 10\\ 357.3& 40\\ 381.3& 100\\ 424.4& 400\end{array}$

The values of $\text{ln}\text{P}$ and $\text{1}/\text{T}$ is determined

  $\begin{array}{cc}\text{ln}\text{P}& \text{1}/\text{T}\\ 0& \text{3}{\text{.48×10}}^{\text{-3}}\\ 2.3025& \text{3}{\text{.059×10}}^{\text{-3}}\\ 3.6888& \text{2}{\text{.798×10}}^{\text{-3}}\\ \begin{array}{l}4.6051\hfill \end{array}& \text{2}{\text{.622×10}}^{\text{-3}}\\ \begin{array}{l}5.9914\hfill \end{array}& \text{2}{\text{.356×10}}^{\text{-3}}\end{array}$

The graph of $\text{ln}\text{P}$ versus $\text{1}/\text{T}$ is plotted

Interpretation Introduction

Interpretation:

The temperature at which the liquid has vapour pressures of $\text{250}\text{mm}\text{Hg}$ and $\text{650}\text{mm}\text{Hg}$ has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

  $\text{ln}\text{P}\text{=}-\left(\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{RT}}\right)\text{+}\text{C}$

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{p}}_{\text{1}}}\text{=}\text{-}\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Answer to Problem 38GQ

The temperature at which liquid has a vapour pressure of $\text{250}\text{mm}\text{Hg}$ is $\text{396}\text{.96}\text{K}$.

The temperature at which liquid has a vapour pressure of $\text{650}\text{mm}\text{Hg}$ is $434.50\text{K}$.

Explanation of Solution

The temperature at which the liquid has a pressure of $\text{250}\text{mm}\text{Hg}\text{and}\text{650}\text{mm}\text{Hg}$ is determined.

Given:

  $\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 287& 1\\ 326.8& 10\\ 357.3& 40\\ 381.3& 100\\ 424.4& 400\end{array}$

The values of $\text{ln}\text{P}$ and $\text{1}/\text{T}$ is determined

  $\begin{array}{cc}\text{ln}\text{P}& \text{1}/\text{T}\\ 0& \text{3}{\text{.48×10}}^{\text{-3}}\\ 2.3025& \text{3}{\text{.059×10}}^{\text{-3}}\\ 3.6888& \text{2}{\text{.798×10}}^{\text{-3}}\\ \begin{array}{l}4.6051\hfill \end{array}& \text{2}{\text{.622×10}}^{\text{-3}}\\ \begin{array}{l}5.9914\hfill \end{array}& \text{2}{\text{.356×10}}^{\text{-3}}\end{array}$

The graph of $\text{ln}\text{P}$ versus $\text{1}/\text{T}$ is plotted

From the slope of the graph we can calculate the molar enthalpy of vaporization.

$\text{Slope}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}$

From the graph drawn,

  $\begin{array}{c}\text{Slope}\text{=}\frac{{\text{y}}_{\text{2}}-{\text{y}}_{\text{1}}}{{\text{x}}_{\text{2}}-{\text{x}}_{\text{1}}}\\ \\ \text{=}\frac{\left(\text{4}\text{.6051}\right)-\left(\text{2}\text{.3025}\right)}{\left(\text{0}\text{.00279}\right)-\left(\text{0}\text{.00305}\right)}\\ \\ \text{=}-\text{8856}\text{.1538}\end{array}$

$\begin{array}{l}-\text{8856}\text{.1538}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\\ \\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}\end{array}$

• The temperature at which liquid has a vapour pressure of $\underset{\_}{\text{250}\text{mm}\text{Hg}}$ is calculated

We know the equation

$\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{p}}_{\text{1}}}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

$\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{100}\text{mm}\text{Hg}\text{,}{\text{P}}_{\text{2}}\text{=}\text{250}\text{mm}\text{Hg}\\ {\text{T}}_{\text{1}}\text{=}381.3\text{K}\\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}\end{array}$

Substituting the values

$\begin{array}{c}\text{ln}\left[\frac{\text{250}}{\text{100}}\right]\text{=}-\frac{\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{\text{381}\text{.3}\text{K}}\right]\\ \\ \text{0}\text{.9612}\text{=}-\frac{\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{\text{381}\text{.3}\text{K}}\right]\\ \\ {\text{T}}_{\text{2}}\text{=}\text{396}\text{.96}\text{K}\end{array}$

Thus the temperature at which liquid has a vapour pressure of $\text{250}\text{mm}\text{Hg}$ is $396.96K$

• The temperature at which liquid has a vapour pressure of $\underset{\_}{\text{650}\text{mm}\text{Hg}}$ is calculated

$\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{400}\text{mm}\text{Hg}\text{,}{\text{P}}_{\text{2}}\text{=}\text{650}\text{mm}\text{Hg}\\ {\text{T}}_{\text{1}}\text{=}\text{424}\text{.4}\text{K}\\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}\end{array}$

Substituting the values

$\begin{array}{c}\text{ln}\left[\frac{\text{650}}{\text{400}}\right]\text{=}-\frac{\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}\text{K}}-\frac{\text{1}}{\text{424}\text{.4}\text{K}}\right]\\ \\ \text{0}\text{.4855}\text{=}-\frac{\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{\text{424}\text{.4}\text{K}}\right]\\ \\ {\text{T}}_{\text{2}}\text{=}\text{434}\text{.51}\text{K}\end{array}$

Thus the temperature at which liquid has a vapour pressure of $\text{650}\text{mm}\text{Hg}$ is $434.51K$

Interpretation Introduction

Interpretation:

The normal boiling point of limonene has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

  $\text{ln}\text{P}\text{=}-\left(\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{RT}}\right)\text{+}\text{C}$

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{p}}_{\text{1}}}\text{=}\text{-}\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is $\text{760}\text{mm}\text{Hg}$ we can call it as normal boiling point.

Answer to Problem 38GQ

The normal boiling point of limonene is $\text{437}\text{.86}\text{K}$.

The molar enthalpy of vaporization of limonene is $\text{45}\text{.104}\text{kJ}/\text{mol}$.

Explanation of Solution

The normal boiling point of limonene is determined.

Normal boiling point is the temperature when the external pressure is $\text{760}\text{mm}\text{Hg}$

$\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{400}\text{mm}\text{Hg}\text{,}{\text{P}}_{\text{2}}\text{=}760\text{mm}\text{Hg}\\ {\text{T}}_{\text{1}}\text{=}\text{424}\text{.4}\text{K}\\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}\end{array}$

Substituting the values

$\begin{array}{c}\text{ln}\left[\frac{\text{760}}{\text{400}}\right]\text{=}-\frac{\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{\text{424}\text{.4}\text{K}}\right]\\ \\ \text{0}\text{.6418}\text{=}-\frac{\text{73}\text{.63}\text{kJ}/\text{mol}\text{.K}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{\text{424}\text{.4}\text{K}}\right]\\ \\ {\text{T}}_{\text{2}}\text{=}\text{437}\text{.86}\text{K}\end{array}$

The normal boiling point of benzene is $\text{437}\text{.86}\text{K}$

Interpretation Introduction

Interpretation:

The molar enthalpy of vaporization has to be determined

Concept Introduction:

Clausius-Clapeyron equation:

  $\text{ln}\text{P}\text{=}-\left(\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{RT}}\right)\text{+}\text{C}$

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{p}}_{\text{1}}}\text{=}\text{-}\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Molar enthalpy of vaporization: The energy required to convert liquid to gas of 1mol of a substance is called molar enthalpy of vaporization

Answer to Problem 38GQ

The molar enthalpy of vaporization of limonene is $\text{45}\text{.11}\text{kJ}/\text{mol}$.

Explanation of Solution

The molar enthalpy of vaporization using Clausius-Clapeyron equation is determined

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{p}}_{\text{1}}}\text{=}\text{-}\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]$

Given:

  $\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 287& 1\\ 326.8& 10\\ 357.3& 40\\ 381.3& 100\\ 424.4& 400\end{array}$

$\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{1}\text{mm}\text{Hg}\text{,}{\text{P}}_{\text{2}}\text{=}\text{10}\text{mm}\text{Hg}\\ {\text{T}}_{\text{1}}\text{=}\text{287}\text{K}\text{,}{\text{T}}_2\text{=}\text{326}\text{.8}\text{K}\\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{?}\end{array}$

Substituting the values

$\begin{array}{c}\text{ln}\left[\frac{\text{10}}{\text{1}}\right]\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{326}\text{.8K}}-\frac{\text{1}}{\text{287}\text{K}}\right]\\ \\ \text{2}\text{.3025}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{326}\text{.8K}}-\frac{\text{1}}{\text{287}\text{K}}\right]\\ \\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{45}\text{.11}\text{kJ}/\text{mol}\end{array}$

The molar enthalpy of vaporization of limonene is $\text{45}\text{.11}\text{kJ}/\text{mol}$.