Chapter 11, Problem 66SCQ

(a)

Interpretation Introduction

Interpretation:

The equation for the straight line using the given data has to be written.

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 66SCQ

The equation for straight line is $\text{ln}\text{P}\text{=}-\text{4928}\text{.57}\frac{\text{1}}{\text{T}}+20.486$

Explanation of Solution

Given:

$\text{ln}\text{P}$	$\text{1}/\text{T}$
2.30	0.00369
3.69	0.00342
4.61	0.00325
5.99	0.00297
6.63	0.00285

From the given data we can derive the equation for straight line $\text{y}\text{=}\text{mx}\text{+c}$

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

$\begin{array}{l}\text{Slope}\text{=}\frac{{\text{y}}_{\text{2}}\text{-}{\text{y}}_{\text{1}}}{{\text{x}}_{\text{2}}{\text{-x}}_{\text{1}}}\\ \text{=}\frac{\text{5}\text{.99-4}\text{.61}}{\text{0}\text{.00297-0}\text{.00325}}\\ \text{=}-4928.57\end{array}$

In the given graph $\text{ln}\text{P}$ is plotted against $\text{1}/\text{T}$. Thus we can write the equation as

$\begin{array}{l}\text{ln}\text{P}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}\frac{\text{1}}{\text{T}}\text{+}\text{C}\\ \text{2}\text{.3}\text{=}-\text{4928}\text{.57 ×}\text{0}\text{.00369}\text{+}\text{C}\\ \text{C}\text{=}20.486\\ \text{ln}\text{P}\text{=}-\text{4928}\text{.57}\frac{\text{1}}{\text{T}}+20.486\end{array}$

(b)

Interpretation Introduction

Interpretation:

The way by which enthalpy of vaporization can be calculated using the graph has to be given.

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 66SCQ

The explanation for determining enthalpy of vaporization from the graph is given.

Explanation of Solution

Given:

Temperature $\left(\text{K}\right)$	Vapour Pressure $\left(\text{mm}\text{Hg}\right)$
287	1
326.8	10
357.3	40
381.3	100
424.4	400

Using the data given we get a straight line graph and it satisfies the straight line equation $\text{y}\text{=}\text{mx}\text{+}\text{c}$

In the graph the y axis have $\text{ln}\text{P}$ and x axis have $\text{1}/\text{T}$.

The equation corresponding to the straight line in the graph is

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

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,

$\text{Slope}\text{=}\frac{{\text{y}}_{\text{2}}\text{-}{\text{y}}_{\text{1}}}{{\text{x}}_{\text{2}}{\text{-x}}_{\text{1}}}$

Thus from the graph we can calculate the slope and using the equation $\text{Slope}\text{=}-\frac{{\text{Δ}}_{\text{vap}}{\text{H}}_{\text{0}}}{\text{R}}$, enthalpy of vaporization can be calculated.

(c)

Interpretation Introduction

Interpretation:

The vapour pressure of ethanol at $\text{0°C}\text{and}\text{10°C}$ has to be calculated.

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 66SCQ

The vapour pressure of ethanol at $\text{0°C}\text{and}\text{10°C}$ are $\text{11}\text{.3}\text{atm}$ and $\text{1440}\text{.3}\text{atm}$ respectively.

Explanation of Solution

We know the equation of the line. Substituting the given temperatures in this equation, we can calculate the vapour pressures.

$\text{ln}\text{P}\text{=}\text{4928}\text{.57}\frac{\text{1}}{\text{T}}+20.486$

Vapour pressure at $\text{0°C}$

$\begin{array}{l}\text{ln}\text{P}\text{=}-\text{4928}\text{.57}\frac{\text{1}}{\text{273}}\text{+20}\text{.486}\\ \text{P}\text{=}\text{11}\text{.3}\text{atm}\end{array}$

Vapour pressure at $\text{100°C}$

$\begin{array}{l}\text{ln}\text{P}\text{=}-\text{4928}\text{.57}\frac{\text{1}}{\text{373}}\text{+20}\text{.486}\\ \text{P}\text{=}\text{1440}\text{.3}\text{atm}\end{array}$