Chapter 4, Problem 4.1PR

(a)

Interpretation Introduction

Interpretation:

The vapor pressure of water at $\text{308}\text{}\text{K}$ has to be calculated.

Concept Introduction:

Vapor pressure:

Vapor pressure can be defined as the pressure exerted by the vapor when it is in thermodynamic equilibrium with its liquid or condensed phase at a given temperature in a closed system.  It relates to the tendency of a particle to escape from its liquid phase.  High vapor pressure at normal temperature is referred to as volatile.  With increase in temperature the vapor pressure increases.

Enthalpy of vaporization:

Enthalpy of vaporization is defined as the amount of heat that has to be transferred to a liquid so that it can be transformed to its vapor phase. It is dependent on pressure.

Clausius-Clapeyron equation:

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}-\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)$

This equation is used to calculate the vapor pressure at different temperature if pressure at one temperature and enthalpy of vaporization are known.

Where,

${\text{P}}_1\text{=}$ Vapor pressure at temperature ${\text{T}}_{\text{1}}\text{K}$

${\text{P}}_2\text{=}$ Vapor pressure at temperature ${\text{T}}_{\text{2}}\text{K}$

${\text{ΔH}}_{\text{vap}}\text{=}$ Enthalpy of vaporization

$\text{R}\text{=}$ Universal gas constant

Answer to Problem 4.1PR

Pressure at $\text{308}\text{}\text{K}$ is $0.063\text{atm}\text{.}$

Explanation of Solution

According to Clausius-Clapeyron equation,

$\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}-\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)$

Given that,

${\text{ΔH}}_{\text{vap}}\text{=}\text{40}\text{.656}\text{kJ/mol}$

It is also given that ${\text{ΔH}}_{\text{vap}}$ is not dependent on temperature.

At boiling temperature of water the pressure is $\text{1}\text{atm}$.

  ${\text{T}}_{\text{1}}=373\text{K}\text{(boiling temperature of water)}$

  ${\text{P}}_{\text{1}}\text{=}\text{1}\text{atm}$

  $\text{R}\text{=}\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$

The another temperature at which vapor pressure has to calculated is given as ${\text{T}}_{\text{2}}=308\text{K}$

The pressure to be calculated is ${\text{P}}_{\text{2}}$.

Applying Clausius-Clapeyron equation,

  $\begin{array}{l}\text{ln}\frac{{\text{P}}_{\text{2}}}{\text{1}\text{atm}}\text{=}-\frac{\text{40}{\text{.656×10}}^{\text{3}}\text{J}{\text{mol}}^{\text{-1}}}{\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\left(\frac{\text{1}}{\text{308K}}\text{-}\frac{\text{1}}{\text{373K}}\right)\\ \\ \text{ln}{\text{P}}_{\text{2}}\text{=}-\frac{\text{40}{\text{.656×10}}^{\text{3}}\text{J}{\text{mol}}^{\text{-1}}}{\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\left(\frac{\text{1}}{\text{308K}}\text{-}\frac{\text{1}}{\text{373K}}\right)\\ \\ {\text{P}}_{\text{2}}\text{=}0.063\text{atm}\end{array}$

Hence the pressure at $\text{308}\text{}\text{K}$ is $0.063\text{atm}$.

(b)

Interpretation Introduction

Interpretation:

Derivation for the modified version of Clausius-Clapeyron equation has to be done in the given condition of ${\text{ΔH}}_{\text{vap}}\text{=}\text{a}\text{+}\text{bT}$.

Concept Introduction:

Vapor pressure:

Vapor pressure can be defined as the pressure exerted by the vapor when it is in thermodynamic equilibrium with its liquid or condensed phase at a given temperature in a closed system.  It relates to the tendency of a particle to escape from its liquid phase.  High vapor pressure at normal temperature is referred to as volatile.  With increase in temperature the vapor pressure increases.

Enthalpy of vaporization:

Enthalpy of vaporization is defined as the amount of heat that has to be transferred to a liquid so that it can be transformed to its vapor phase. It is dependent on pressure.

Clausius-Clapeyron equation:

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}-\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)$

This equation is used to calculate the vapor pressure at different temperature if pressure at one temperature and enthalpy of vaporization are known.

Where,

${\text{P}}_1\text{=}$ Vapor pressure at temperature ${\text{T}}_{\text{1}}\text{K}$

${\text{P}}_2\text{=}$ Vapor pressure at temperature ${\text{T}}_{\text{2}}\text{K}$

${\text{ΔH}}_{\text{vap}}\text{=}$ Enthalpy of vaporization

$\text{R}\text{=}$ Universal gas constant

Explanation of Solution

According to Clausius-Clapeyron equation,

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}-\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)$

It is given that ${\text{ΔH}}_{\text{vap}}\text{=}\text{a}\text{+}\text{bT}$

Using the method of calculus on Clausius-Clapeyron equation we get,

  $\begin{array}{l}\text{dlnP}\text{=}\frac{{\text{ΔH}}_{\text{vap}}}{{\text{RT}}^{\text{2}}}\text{dT}\\ \\ \frac{\text{dlnP}}{\text{dT}}\text{=}\frac{\text{a}\text{+}\text{bT}}{{\text{RT}}^{\text{2}}}\\ \\ {\displaystyle \int \frac{\text{dP}}{\text{P}}}\text{=}{\displaystyle \int \frac{\text{a}\text{+}\text{bT}}{{\text{RT}}^{\text{2}}}}\text{dT}\\ \\ {\displaystyle \int \frac{\text{dP}}{\text{P}}}\text{=}{\displaystyle \int \frac{\text{a}}{{\text{RT}}^{\text{2}}}}\text{dT}\text{+}{\displaystyle \int \frac{\text{b}}{\text{RT}}}\text{dT}\\ \\ {\displaystyle \int \frac{\text{dP}}{\text{P}}}\text{=}\frac{\text{a}}{\text{R}}{\displaystyle \int \frac{\text{dT}}{{\text{T}}^{\text{2}}}}\text{+}\frac{\text{b}}{\text{R}}{\displaystyle \int \frac{\text{dT}}{\text{T}}}\\ \\ \text{lnP}\text{=}\text{-}\frac{\text{a}}{\text{RT}}\text{+}\frac{\text{b}}{\text{R}}\text{lnT}\end{array}$

Thus the modified version of Clausius-Clapeyron equation when ${\text{ΔH}}_{\text{vap}}\text{=}\text{a}\text{+}\text{bT}$ is derived as $\text{lnP}\text{=}\text{-}\frac{\text{a}}{\text{RT}}\text{+}\frac{\text{b}}{\text{R}}\text{lnT}$.

(c)

Interpretation Introduction

Interpretation:

With the help of modified version of Clausius-Clapeyron equation, vapor pressure of water at $\text{T}\text{=}\text{308}\text{K}$ has to be calculated.

Concept Introduction:

Vapor pressure:

Vapor pressure can be defined as the pressure exerted by the vapor when it is in thermodynamic equilibrium with its liquid or condensed phase at a given temperature in a closed system.  It relates to the tendency of a particle to escape from its liquid phase.  High vapor pressure at normal temperature is referred to as volatile.  With increase in temperature the vapor pressure increases.

Enthalpy of vaporization:

Enthalpy of vaporization is defined as the amount of heat that has to be transferred to a liquid so that it can be transformed to its vapor phase. It is dependent on pressure.

Clausius-Clapeyron equation:

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}-\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)$

This equation is used to calculate the vapor pressure at different temperature if pressure at one temperature and enthalpy of vaporization are known.

Where,

${\text{P}}_1\text{=}$ Vapor pressure at temperature ${\text{T}}_{\text{1}}\text{K}$

${\text{P}}_2\text{=}$ Vapor pressure at temperature ${\text{T}}_{\text{2}}\text{K}$

${\text{ΔH}}_{\text{vap}}\text{=}$ Enthalpy of vaporization

$\text{R}\text{=}$ Universal gas constant

Answer to Problem 4.1PR

Vapor pressure calculated at $\text{T}\text{=}\text{308}\text{K}$ is $\text{0}\text{.0566}\text{atm}$.

Explanation of Solution

According to Clausius-Clapeyron equation,

  $\text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}-\frac{{\text{ΔH}}_{\text{vap}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)$

According to modified version of Clausius-Clapeyron equation when ${\text{ΔH}}_{\text{vap}}\text{=}\text{a}\text{+}\text{bT}$,

  $\text{lnP}\text{=}\text{-}\frac{\text{a}}{\text{RT}}\text{+}\frac{\text{b}}{\text{R}}\text{lnT}$

At boiling temperature of water the pressure is $\text{1}\text{atm}$.

  ${\text{T}}_{\text{1}}=373\text{K}\text{(boiling temperature of water)}$

${\text{P}}_{\text{1}}\text{=}\text{1}\text{atm}$

$\text{R}\text{=}\text{8}\text{.314}\text{J}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}$

The another temperature at which vapor pressure has to calculated is given as ${\text{T}}_{\text{2}}=308\text{K}$

The pressure to be calculated is ${\text{P}}_{\text{2}}$.

Applying Clausius-Clapeyron equation,

  $\text{lnP}\text{=}\text{-}\frac{\text{a}}{\text{RT}}\text{+}\frac{\text{b}}{\text{R}}\text{lnT}$

  $\begin{array}{l}\\ \text{ln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\text{-}\frac{\text{a}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right)\text{+}\frac{\text{b}}{\text{R}}\text{ln}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\\ \\ \text{ln}\frac{{\text{P}}_{\text{2}}}{\text{1}\text{atm}}\text{=}\text{-}\frac{\text{57}\text{.373}{\text{kJmol}}^{\text{-1}}}{\text{8}\text{.314}{\text{Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\left(\frac{\text{1}}{\text{308}\text{K}}\text{-}\frac{\text{1}}{\text{373}\text{K}}\right)\text{+}\frac{\text{(-44}\text{.801}{\text{kJmol}}^{\text{-1}}\text{)}}{\text{8}\text{.314}{\text{Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\text{ln}\frac{\text{308}\text{K}}{\text{373K}}\\ \\ \text{ln}{\text{P}}_{\text{2}}\text{=}\text{-}\frac{\text{57}\text{.373}{\text{×10}}^{\text{3}}{\text{Jmol}}^{\text{-1}}}{\text{8}\text{.314}{\text{Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\left(\frac{\text{1}}{\text{308}\text{K}}\text{-}\frac{\text{1}}{\text{373}\text{K}}\right)\text{+}\frac{\text{(-44}\text{.801}{\text{×10}}^{\text{3}}{\text{Jmol}}^{\text{-1}}\text{)}}{\text{8}\text{.314}{\text{Jmol}}^{\text{-1}}{\text{K}}^{\text{-1}}}\text{ln}\frac{\text{308}\text{K}}{\text{373K}}\\ \\ {\text{P}}_{\text{2}}\text{=}\text{0}\text{.0566}\text{atm}\end{array}$

With the help of the modified Clausius-Clapeyron equation the vapor pressure calculated at $\text{T}\text{=}\text{308}\text{K}$ is $\text{0}\text{.0566}\text{atm}$.