Chapter 11, Problem 39IL

(a)

Interpretation Introduction

Interpretation:

The normal boiling point of dichlorodimethylsilane has to be determined

Concept Introduction:

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 39IL

The normal boiling point of dichlorodimethylsilane is $\text{343}\text{.3}\text{K}$.

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 $\text{302}\text{.33}\text{K}$ and $\text{327}\text{.51}\text{K}$ respectively.

The molar enthalpy of vaporization of is $33.6\text{kJ}/\text{mol}$.

Explanation of Solution

The normal boiling point of dichlorodimethylsilane is calculated

Given:

$\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 272.6& 40\\ 290.5& 100\\ 324.9& 400\\ 343.3& 760\end{array}$

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 $\text{343}\text{.3}\text{K}$

Thus the normal boiling point of dichlorodimethylsilane is $\text{343}\text{.3}\text{K}$.

(b)

Interpretation Introduction

Interpretation:

The graph of $\text{ln}\text{P}$ versus $\text{1}/\text{T}$ has to be plotted. The temperatures at which liquid have a 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 39IL

Using the given data we can plot the graph of $\text{ln}\text{P}$ versus $\text{1}/\text{T}$

The temperature at which liquid has a vapour pressure of $\text{250}\text{mm}\text{Hg}$ are $\text{302}\text{.33}\text{K}$.

The temperature at which liquid has a vapour pressure $\text{650}\text{mm}\text{Hg}$ are $\text{327}\text{.51}\text{K}$.

Explanation of Solution

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

Given:

  $\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 272.6& 40\\ 290.5& 100\\ 324.9& 400\\ 343.3& 760\end{array}$

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

  $\begin{array}{cc}\text{ln}\text{P}& \text{1}/\text{T}\\ 3.68& \text{3}{\text{.668×10}}^{\text{-3}}\\ 4.60& \text{3}{\text{.423×10}}^{\text{-3}}\\ 5.99& \text{3}{\text{.077×10}}^{\text{-3}}\\ 6.63& \text{2}{\text{.9112×10}}^{\text{-3}}\end{array}$

Using the given data we can plot the graph of $\text{ln}\text{P}$ versus $\text{1}/\text{T}$

From the slope of the graph we can find the value of $-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}$

  $\begin{array}{l}\text{Slope}\text{=}\frac{{\text{y}}_{\text{2}}-{\text{y}}_{\text{1}}}{{\text{x}}_{\text{2}}-{\text{x}}_{\text{1}}}\\ \\ \text{=}\frac{\left(\text{5}\text{.9}\right)-\left(\text{4}\text{.6}\right)}{\left(\text{0}\text{.003077}\right)-\left(\text{0}\text{.003423}\right)}\\ \\ \text{=}-\text{3757}\text{.22}\end{array}$

Using the equation for the straight line in the plot

$\text{ln}\text{P}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\text{×}\frac{\text{1}}{\text{T}}\text{+}\text{C}$

C, the constant value can be calculated by substituting any one of the value of pressure and temperature from the table given in the equation.

Substituting the values

  $\text{ln}\text{P}\text{=}\text{-3757}\text{.22}\text{×}\frac{\text{1}}{\text{T}}\text{+17}\text{.949}$

From this equation we can calculate the temperature at which the pressures are $\text{250}\text{mm}\text{Hg}\text{and}\text{650}\text{mm}\text{Hg}$.

When the pressure is $\text{250}\text{mm}\text{Hg}$

  $\begin{array}{l}\text{ln}\text{250}\text{=}-\text{3757}\text{.22}\text{×}\frac{\text{1}}{\text{T}}\text{+17}\text{.949}\\ \\ -\text{12}\text{.4275}\text{=}-\text{3757}\text{.22}\text{×}\frac{\text{1}}{\text{T}}\\ \\ \text{3}{\text{.3076×10}}^{\text{-3}}\text{=}\frac{\text{1}}{\text{T}}\\ \\ \text{T}\text{=}\text{302}\text{.33}\text{K}\end{array}$

The temperature at which the pressures is $\text{250}\text{mm}\text{Hg}$ is $\text{302}\text{.33}\text{K}$

When the pressure is $\text{650}\text{mm}\text{Hg}$

  $\begin{array}{l}\text{ln}\text{650}\text{=}-\text{3757}\text{.22}\text{×}\frac{\text{1}}{\text{T}}\text{+17}\text{.949}\\ \\ -\text{11}\text{.4720}\text{=}-\text{3757}\text{.22}\text{×}\frac{\text{1}}{\text{T}}\\ \\ \text{3}{\text{.0533×10}}^{\text{-3}}\text{=}\frac{\text{1}}{\text{T}}\\ \\ \text{T}\text{=}\text{327}\text{.51}\text{K}\end{array}$

The temperature at which the pressures is $\text{650}\text{mm}\text{Hg}$ is $\text{327}\text{.51}\text{K}$

(c)

Interpretation Introduction

Interpretation:

The molar enthalpy of vaporization has to be explained.

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.

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 39IL

The molar enthalpy of vaporization of is $33.6\text{kJ}/\text{mol}$.

Explanation of Solution

Given:

  $\begin{array}{cc}\begin{array}{l}Temperature\\ \left(K\right)\end{array}& \begin{array}{l}VapourPressure\\ \left(mmHg\right)\end{array}\\ 272.6& 40\\ 290.5& 100\\ 324.9& 400\\ 343.3& 760\end{array}$

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

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

Substituting the values

$\begin{array}{c}\text{ln}\left[\frac{\text{100}}{40}\right]\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{290}\text{.5K}}-\frac{\text{1}}{\text{272}\text{.6}\text{K}}\right]\\ 0.9162\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{0}\text{.008314}\text{kJ}/\text{K}\text{.mol}}\left[\frac{\text{1}}{\text{290}\text{.5K}}-\frac{\text{1}}{\text{272}\text{.6}\text{K}}\right]\\ {\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}\text{=}33.6\text{kJ}/\text{mol}\end{array}$

The molar enthalpy of vaporization using the given data is