Chapter 11, Problem 21PS

Equilibrium vapor pressures of benzene, C₆H₆, at various temperatures are given in the table.

1. (a) What is the normal boiling point of benzene?
2. (b) Plot these data so that you have a graph resembling the one in Figure 11.12. At what temperature does the liquid have an equilibrium vapor pressure of 250 mm Hg? At what temperature is the vapor pressure 650 mm Hg?
3. (c) Calculate the molar enthalpy of vaporization for benzene using the Clausius–Clapeyron equation.

Interpretation Introduction

Interpretation:

The normal boiling point of benzene has to be determined.

Concept Introduction:

Boiling point: It is the temperature at which liquid converts to vapor. At boiling point the vapor pressure of liquid and the pressure of the surroundings are equal.

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 21PS

The normal boiling point of benzene is $\text{80}\text{.1°C}$.

Explanation of Solution

The normal boiling point of benzene is calculated

Given:

$\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 280.6& 40\\ 299.1& 100\\ 333.6& 400\\ \begin{array}{l}353.1\hfill \end{array}& 760\end{array}$

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

From the given data it is clear that the temperature at which the pressure is $\text{760}\text{mm}\text{Hg}$ is $\text{80}\text{.1°C}$

Thus the normal boiling point of benzene is $\text{80}\text{.1°C}$.

Interpretation Introduction

Interpretation:

The temperature versus vapor pressure graph should be plotted. The temperatures at which the liquid has vapour pressures of $\text{250}\text{mm}\text{Hg}\text{and}\text{650}\text{mm}\text{Hg}$ has to be determined

Concept Introduction:

Vapor pressure is nothing but the pressure of a vapor in contact with its liquid or solid form.

When a liquid and vapor are in equilibrium the pressure exerted by the vapor is called the equilibrium vapor pressure

Answer to Problem 21PS

The temperatures at which liquid have a vapour pressures of $\text{250}\text{mm}\text{Hg}\text{and}\text{650}\text{mm}\text{Hg}$ are $\text{47°C}\text{and}\text{75°C}$ respectively.

Explanation of Solution

Given,

$\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 280.6& 40\\ 299.1& 100\\ 333.6& 400\\ \begin{array}{l}353.1\hfill \end{array}& 760\end{array}$

The temperatures at which liquid have a vapour pressures of $\text{250}\text{mm}\text{Hg}\text{and}\text{650}\text{mm}\text{Hg}$ is determined

Using the given data we can plot the graph of $\text{P}\text{versus}\text{T}$.

From the graph we can find the approximate temperatures at which the pressures are $\text{250}\text{mm}\text{Hg}\text{and}\text{650}\text{mm}\text{Hg}$.

Therefore,

The temperature at which the pressure is $\text{250}\text{mm}\text{Hg}\text{and}\text{650}\text{mm}\text{Hg}$ are approximately $\text{47°C}\text{and}\text{75°C}$

Interpretation Introduction

Interpretation:

The molar enthalpy of vaporization using Clausius-Clapeyron 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 21PS

The molar enthalpy of vaporization of is $34.5\text{kJ}/\text{mol}$.

Explanation of Solution

The molar enthalpy of vaporization is calculated using the given data,

Given:

  $\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 280.6& 40\\ 299.1& 100\\ 333.6& 400\\ \begin{array}{l}353.1\hfill \end{array}& 760\end{array}$

  $\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{40}\text{mm}\text{Hg}\text{,}{\text{P}}_{\text{2}}\text{=}\text{100}\text{mm}\text{Hg}\\ {\text{T}}_{\text{1}}\text{=}\text{280}\text{.6}\text{K}\text{,}{\text{T}}_{\text{2}}\text{=}\text{299}\text{.1}\text{K}\\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{?}\end{array}$

Clausius-Clapeyron equation is,

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

Substituting the values

  $\begin{array}{c}\text{ln}\left[\frac{\text{100}}{\text{40}}\right]\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{299}\text{.1K}}-\frac{\text{1}}{\text{280}\text{.6}\text{K}}\right]\\ \\ \text{0}\text{.9162}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{299}\text{.1K}}-\frac{\text{1}}{\text{280}\text{.6}\text{K}}\right]\\ \\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{34}\text{.5}\text{kJ}/\text{mol}\end{array}$

The molar enthalpy of vaporization of is $34.5\text{kJ}/\text{mol}$.