Chapter 1, Problem 1.3PR

(a)

Interpretation Introduction

Interpretation:

The van der Waals equation of state has to be expressed as the power series of $\frac{\text{1}}{{\text{V}}_{\text{m}}}$ using the exponential form of ${\text{(1-x)}}^{\text{-1}}$.

Concept Introduction:

van der Waals equation:

van der Waals equation represents the real gas equation.  Real gas molecules have their own volume and there is force of attraction and repulsion constantly working between the real gas molecules unlike the ideal gas molecules.  Hence ideal gas equation is subjected to modify with pressure and volume correction and thus van der Waals equation has been formed for real gases.

  $\begin{array}{l}\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}\\ \\ \text{P}\text{=}\text{pressure}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{V}\text{=}\text{volume}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{n}\text{=}\text{no}\text{.}\text{of}\text{moles}\text{of}\text{real}\text{gas}\text{molecules}\\ \\ \text{R}\text{=}\text{universal}\text{gas}\text{constant}\\ \\ \text{T}\text{=}\text{temperature}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{a}\text{=}\text{pressure}\text{correction}\text{term}\\ \\ \text{b}\text{=}\text{volume}\text{occupied}\text{by}\text{each}\text{molecule}\text{(volume}\text{correction}\text{term}\text{)}\end{array}$

Explanation of Solution

A mathematical function of the form ${\text{(1-x)}}^{\text{-1}}$ may be expressed as a series of expansion

${\text{(1-x)}}^{\text{-1}}\text{=}\text{1+}\text{x}\text{+}{\text{x}}^{\text{2}}\text{+}{\text{x}}^{\text{3}}\text{+}\mathrm{...}\text{(x<1)}$

Now the van der Waals equation of state has to be expressed as an expansion series of $\frac{\text{1}}{{\text{V}}_{\text{m}}}$.

${\text{V}}_{\text{m}}$ is the molar volume i.e. volume per mole.

Hence van der Waals equation can be represented as,

  $\left(\text{P}\text{+}\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\right)\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)\text{=}\text{RT}$

Now the form needed for expansion is ${\text{(1-x)}}^{\text{-1}}$ and hence van der Waals equation has to be expressed in this form.

$\begin{array}{l}\left(\text{P}\text{+}\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\right)\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)\text{=}\text{RT}\\ \\ \text{P}\text{+}\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\text{=}\frac{\text{RT}}{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}\\ \\ \text{P}=\frac{\text{RT}}{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}-\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\\ \\ \text{P}=\frac{\text{RT}}{{\text{V}}_{\text{m}}\times \left(1\text{-}\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}-\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\\ \\ \text{P}=\frac{\text{RT}}{{\text{V}}_{\text{m}}}{\left(1\text{-}\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{-1}-\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\end{array}$

Now according to the series expansion,

  ${\text{(1-x)}}^{\text{-1}}\text{=}\text{1+}\text{x}\text{+}{\text{x}}^{\text{2}}\text{+}{\text{x}}^{\text{3}}\text{+}\mathrm{...}\text{(x<1)}$

Hence here $\text{x}=\frac{\text{b}}{{\text{V}}_{\text{m}}}$

So the expansion form can be written as,

  $\begin{array}{l}\text{P}=\frac{\text{RT}}{{\text{V}}_{\text{m}}}{\left(1\text{-}\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{-1}-\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\\ \\ \text{P}=\frac{\text{RT}}{{\text{V}}_{\text{m}}}\left(1+\frac{\text{b}}{{\text{V}}_{\text{m}}}+{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^2+{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^3+\mathrm{...}\right)-\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\end{array}$

The equation can be rearranged as,

  $\begin{array}{l}\text{P=}\frac{\text{RT}}{{\text{V}}_{\text{m}}}\left(\text{1}\text{+}\frac{\text{b}}{{\text{V}}_{\text{m}}}\text{-}\frac{\text{a}}{{\text{RTV}}_{\text{m}}}\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{3}}\text{+}\mathrm{...}\right)\\ \\ \text{P=}\frac{\text{RT}}{{\text{V}}_{\text{m}}}\left(\text{1}\text{+}\frac{\text{1}}{{\text{V}}_{\text{m}}}\left(\text{b-}\frac{\text{a}}{\text{RT}}\right)\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{3}}\text{+}\mathrm{...}\right)\end{array}$

Thus the van der Waals equation can be represented as the power series of $\frac{\text{1}}{{\text{V}}_{\text{m}}}$ using the exponential form of ${\text{(1-x)}}^{\text{-1}}$ as $\text{P=}\frac{\text{RT}}{{\text{V}}_{\text{m}}}\left(\text{1}\text{+}\frac{\text{1}}{{\text{V}}_{\text{m}}}\left(\text{b-}\frac{\text{a}}{\text{RT}}\right)\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{3}}\text{+}\mathrm{...}\right)$.

(b)

Interpretation Introduction

Interpretation:

An expression of the Boyle temperature has to be derived in the terms of van der Waals constants $\text{a}$ and $\text{b}$ for a van der Waals gas.

Concept Introduction:

van der Waals equation:

van der Waals equation represents the real gas equation.  Real gas molecules have their own volume and there is force of attraction and repulsion constantly working between the real gas molecules unlike the ideal gas molecules.  Hence ideal gas equation is subjected to modify with pressure and volume correction and thus van der Waals equation has been formed for real gases.

  $\begin{array}{l}\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}\\ \\ \text{P}\text{=}\text{pressure}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{V}\text{=}\text{volume}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{n}\text{=}\text{no}\text{.}\text{of}\text{moles}\text{of}\text{real}\text{gas}\text{molecules}\\ \\ \text{R}\text{=}\text{universal}\text{gas}\text{constant}\\ \\ \text{T}\text{=}\text{temperature}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{a}\text{=}\text{pressure}\text{correction}\text{term}\\ \\ \text{b}\text{=}\text{volume}\text{occupied}\text{by}\text{each}\text{molecule}\text{(volume}\text{correction}\text{term}\text{)}\end{array}$

Virial equation:

General equation of states for real gases is virial equation which is proposed by Kammerlingh-Onnes. He proposed the equation as,

  ${\text{PV}}_{\text{m}}\text{=RT}\left(\text{1}\text{+}\frac{{\text{B}}_2\text{(T)}}{{\text{V}}_{\text{m}}}\text{+}\frac{{\text{B}}_3\text{(T)}}{{\text{V}}_{\text{m}}^{\text{2}}}\text{+}\frac{{\text{B}}_4\text{(T)}}{{\text{V}}_{\text{m}}^{\text{3}}}\text{+}\mathrm{...}\right)$

Here, ${\text{B}}_{\text{2}}\text{(T)}$, ${\text{B}}_{\text{3}}\text{(T)}$, ${\text{B}}_4\text{(T)}$ etc. are the 2^nd, 3^rd.,4^th .virial coefficients respectively.

The 1^st virial coefficient is $\text{RT}$ which is always positive and increases with the increase in temperature.  The 2^nd virial coefficient is negative at low temperature, zero at a particular temperature and increases positively if the temperature is increased after that temperature.

Boyle temperature:

The temperature at which real gas starts behaving ideally is called the Boyle temperature.

At this temperature the 2^nd virial coefficient becomes zero and 3^rd, 4^th and higher virial coefficients become insignificant.  The virial equation for real gas becomes ideal gas equation ${\text{PV}}_{\text{m}}\text{=RT}$.

Explanation of Solution

At Boyle temperature the 2^nd virial coefficient becomes zero.

Now, from part (a) it has been obtained that,

  $\text{P=}\frac{\text{RT}}{{\text{V}}_{\text{m}}}\left(\text{1}\text{+}\frac{\text{1}}{{\text{V}}_{\text{m}}}\left(\text{b-}\frac{\text{a}}{\text{RT}}\right)\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{b}}{{\text{V}}_{\text{m}}}\right)}^{\text{3}}\text{+}\mathrm{...}\right)$

Virial equation is,

  ${\text{PV}}_{\text{m}}\text{=RT}\left(\text{1}\text{+}\frac{{\text{B}}_2\text{(T)}}{{\text{V}}_{\text{m}}}\text{+}\frac{{\text{B}}_3\text{(T)}}{{\text{V}}_{\text{m}}^{\text{2}}}\text{+}\frac{{\text{B}}_4\text{(T)}}{{\text{V}}_{\text{m}}^{\text{3}}}\text{+}\mathrm{...}\right)$

Thus comparing the above two equations the 2^nd virial coefficient is,

  ${\text{B}}_2\text{(T)}\text{=}\left(\text{b-}\frac{\text{a}}{\text{RT}}\right)$

At Boyle temperature ${\text{B}}_2\text{(T)}$ is zero.

Thus,

  $\begin{array}{l}\left(\text{b-}\frac{\text{a}}{{\text{RT}}_{\text{B}}}\right)\text{=}\text{0}\text{(}\because \text{at}{\text{T=T}}_{\text{B}}{\text{B}}_{\text{2}}\text{(T)=0)}\\ \\ \text{b}\text{=}\frac{\text{a}}{{\text{RT}}_{\text{B}}}\\ \\ {\text{T}}_{\text{B}}\text{=}\frac{\text{a}}{\text{Rb}}\end{array}$

Thus the expression for Boyle temperature is ${\text{T}}_{\text{B}}\text{=}\frac{\text{a}}{\text{Rb}}$.

(c)

Interpretation Introduction

Interpretation:

Boyle temperature for carbon dioxide has to be calculated for given van der Waals constant values.

Concept Introduction:

van der Waals equation:

van der Waals equation represents the real gas equation.  Real gas molecules have their own volume and there is force of attraction and repulsion constantly working between the real gas molecules unlike the ideal gas molecules.  Hence ideal gas equation is subjected to modify with pressure and volume correction and thus van der Waals equation has been formed for real gases.

  $\begin{array}{l}\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}\\ \\ \text{P}\text{=}\text{pressure}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{V}\text{=}\text{volume}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{n}\text{=}\text{no}\text{.}\text{of}\text{moles}\text{of}\text{real}\text{gas}\text{molecules}\\ \\ \text{R}\text{=}\text{universal}\text{gas}\text{constant}\\ \\ \text{T}\text{=}\text{temperature}\text{of}\text{the}\text{real}\text{gas}\\ \\ \text{a}\text{=}\text{pressure}\text{correction}\text{term}\\ \\ \text{b}\text{=}\text{volume}\text{occupied}\text{by}\text{each}\text{molecule}\text{(volume}\text{correction}\text{term}\text{)}\end{array}$

Virial equation:

General equation of states for real gases is virial equation which is proposed by Kammerlingh-Onnes. He proposed the equation as,

  ${\text{PV}}_{\text{m}}\text{=RT}\left(\text{1}\text{+}\frac{{\text{B}}_2\text{(T)}}{{\text{V}}_{\text{m}}}\text{+}\frac{{\text{B}}_3\text{(T)}}{{\text{V}}_{\text{m}}^{\text{2}}}\text{+}\frac{{\text{B}}_4\text{(T)}}{{\text{V}}_{\text{m}}^{\text{3}}}\text{+}\mathrm{...}\right)$

Here, ${\text{B}}_{\text{2}}\text{(T)}$, ${\text{B}}_{\text{3}}\text{(T)}$, ${\text{B}}_4\text{(T)}$ etc. are the 2^nd, 3^rd.,4^th .virial coefficients respectively.

The 1^st virial coefficient is $\text{RT}$ which is always positive and increases with the increase in temperature.  The 2^nd virial coefficient is negative at low temperature, zero at a particular temperature and increases positively if the temperature is increased after that temperature.

Boyle temperature:

The temperature at which real gas starts behaving ideally is called the Boyle temperature.

At this temperature the 2^nd virial coefficient becomes zero and 3^rd, 4^th and higher virial coefficients become insignificant.  The virial equation for real gas becomes ideal gas equation ${\text{PV}}_{\text{m}}\text{=RT}$.

Answer to Problem 1.3PR

The Boyle temperature for carbon dioxide is ${\text{T}}_{\text{B}}\text{=}\text{751}{\text{.96}}^{\text{ο}}\text{C}$.

Explanation of Solution

From the above part (b) the expression for Boyle temperature has been obtained as,

  ${\text{T}}_{\text{B}}\text{=}\frac{\text{a}}{\text{Rb}}$

Given that the values of van der Waals constants are,

  $\begin{array}{l}\text{a}\text{=}\text{3}\text{.610}\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{\text{-2}}\\ \\ \text{b}\text{=}\left(\text{4}{\text{.29×10}}^{\text{-2}}\right){\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\\ \\ \text{R}=\text{0}{\text{.0821 atm L mol}}^{-1}{\text{ K}}^{-1}\end{array}$

Thus the value of Boyle temperature for carbon dioxide is,

  $\begin{array}{l}{\text{T}}_{\text{B}}\text{=}\frac{\text{a}}{\text{Rb}}\\ \\ {\text{T}}_{\text{B}}\text{=}\frac{\text{3}\text{.610}\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{\text{-2}}}{\text{0}{\text{.0821 atm L mol}}^{\text{-1}}{\text{ K}}^{\text{-1}}\text{×}\left(\text{4}{\text{.29×10}}^{\text{-2}}\right){\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}}\\ \\ {\text{T}}_{\text{B}}\text{=}\text{1024}\text{.96}\text{K}\\ \\ {\text{T}}_{\text{B}}\text{=}\left(\text{1024}\text{.96}\text{-}\text{273}\right)\text{K}\\ \\ {\text{T}}_{\text{B}}\text{=}\text{751}{\text{.96}}^{\text{ο}}\text{C}\end{array}$

Hence the Boyle temperature for carbon dioxide is ${\text{T}}_{\text{B}}\text{=}\text{751}{\text{.96}}^{\text{ο}}\text{C}$.