Chapter 11, Problem 22PS

Interpretation Introduction

Interpretation:

The molar enthalpy of vaporization and boiling point of octane 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 22PS

The molar enthalpy of vaporization is $\text{38}\text{.51}\text{kJ}/\text{mol}$

The boiling point of octane is calculated is$\text{402}\text{.17}\text{K}$.

Explanation of Solution

Given:

  $\begin{array}{cc}\begin{array}{l}Temperature\\ \left({}^{o}C\right)\end{array}& \begin{array}{l}VaporPressure\\ \left(\text{mm}\text{Hg}\right)\end{array}\\ 25& 13.6\\ 50& 45.3\\ 75& 127.2\\ 100& \begin{array}{l}310.8\hfill \end{array}\end{array}$

Temperatures and corresponding pressures are given. We can calculate the molar enthalpy of vaporization by using the Clausius-Clapeyron equation

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

• The molar enthalpy of vaporization is calculated using given data.

  $\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{13}\text{.6}\text{mm}\text{Hg}\text{,}{\text{P}}_{\text{2}}\text{=}\text{45}\text{.3}\text{mm}\text{Hg}\\ {\text{T}}_{\text{1}}\text{=}\text{25°C}\text{=}\text{298}\text{K}\text{,}{\text{T}}_{\text{2}}\text{=}\text{50°C}\text{=}\text{323}\text{K}\end{array}$

Substituting the values

  $\begin{array}{c}\text{ln}\left[\frac{\text{45}\text{.3}}{\text{13}\text{.6}}\right]\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{323K}}-\frac{\text{1}}{\text{298}\text{K}}\right]\\ \\ \text{ln}\text{(1}\text{.2032)}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{323K}}-\frac{\text{1}}{\text{298}\text{K}}\right]\\ \\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{38}\text{.51}\text{kJ}/\text{mol}\end{array}$

The molar enthalpy of vaporization is.

• The boiling point is calculated by using given data

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

$\begin{array}{c}{\text{P}}_{\text{1}}\text{=}\text{13}\text{.6}\text{mm}\text{Hg}\text{,}{\text{P}}_{\text{2}}\text{=}\text{760}\text{mm}\text{Hg}\\ {\text{T}}_{\text{1}}\text{=}\text{25°C}\text{=}\text{298}\text{K}\text{,}{\text{T}}_{\text{2}}\text{=}\text{?}\\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}\text{38}\text{.51}\text{kJ}/\text{mol}\end{array}$

Substituting the values

  $\begin{array}{c}\text{ln}\left[\frac{\text{760}}{\text{13}\text{.6}}\right]\text{=}-\frac{\text{38}\text{.51}\text{kJ}/\text{mol}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{\text{298}\text{K}}\right]\\ \\ \text{4}\text{.023}\text{=}-\text{4631}\text{.946}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}-\frac{\text{1}}{\text{298}\text{K}}\right]\\ \\ \frac{\text{1}}{{\text{T}}_{\text{2}}}\text{=}\text{2}{\text{.4865×10}}^{-\text{3}}\\ \\ {\text{T}}_{\text{2}}\text{=}\text{402}\text{.07}\text{K}\end{array}$

The boiling point of octane is calculated is$\text{402}\text{.07}\text{K}$.

Conclusion

The molar enthalpy of vaporization and boiling point of octane was determined.