Chapter 1, Problem 1.4PR

(a)

Interpretation Introduction

Interpretation:

The expression for the 1^st and 2^nd derivative of has to be calculated by differentiating the van der Waals equation of state and the expression for the critical constants has to be calculated also in terms of van der Waals constants.

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.

  $\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}$

Where,

P is pressure of the real gas

V is volume of the real gas

‘n’ is the number of moles of real gas molecules.

R is the gas constant

T is the temperature

‘a’ is the pressure correction term

‘b’ is the volume occupied by each molecules (volume correction term).

Critical constants:

There are some particular values of temperature, pressure and volume that are called as critical constants.

The temperature above which the vapor of a substance cannot be liquefied no matter what amount of pressure is given is called the critical temperature of the substance.

The pressure which is required to liquefy the vapor of a substance at critical temperature is called the critical pressure of the substance.

Explanation of Solution

Van der Waals equation of state can be represented as,

  $\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}$

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,

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

The critical points of van der Waals gas occurs where the isotherm has flat inflexion which is where $\frac{\text{dP}}{{\text{dV}}_{\text{m}}}\text{=}\text{0}$(zero slope) and $\frac{{\text{d}}^{\text{2}}\text{P}}{{\text{dV}}_{\text{m}}^{\text{2}}}\text{=}\text{0}$(zero curvature).

Now the 1^st and 2^nd derivatives have to be calculated by differentiating van der Waals equation.

Hence by applying differentiation on van der Waals equation,

  $\begin{array}{l}\text{P}\text{=}\frac{\text{RT}}{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}\text{-}\frac{\text{a}}{{\text{V}}_{\text{m}}^{\text{2}}}\\ \\ \frac{\text{dP}}{{\text{dV}}_{\text{m}}}\text{=}\text{-}\frac{\text{RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{2}}}\text{+}\frac{\text{2a}}{{\text{V}}_{\text{m}}^{\text{3}}}\mathrm{...}\text{(1)}\\ \\ \frac{{\text{d}}^{\text{2}}\text{P}}{{\text{dV}}_{\text{m}}^{\text{2}}}\text{=}\frac{\text{2RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{3}}}\text{-}\frac{\text{6a}}{{\text{V}}_{\text{m}}^{\text{4}}}\mathrm{...}\text{(2)}\end{array}$

Now according to the conditions of critical points,

  $\frac{\text{dP}}{{\text{dV}}_{\text{m}}}\text{=}\text{0}$

  $\frac{{\text{d}}^{\text{2}}\text{P}}{{\text{dV}}_{\text{m}}^{\text{2}}}\text{=}\text{0}$

Now applying these conditions on 1^st and 2^nd differential forms,

  $\begin{array}{l}\text{-}\frac{\text{RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{2}}}\text{+}\frac{\text{2a}}{{\text{V}}_{\text{m}}^{\text{3}}}=0\mathrm{...}\text{(1)}\\ \\ \frac{\text{2RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{3}}}\text{-}\frac{\text{6a}}{{\text{V}}_{\text{m}}^{\text{4}}}=0\mathrm{...}\text{(2)}\end{array}$

Now by simplification,

  $\begin{array}{l}\frac{\text{RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{2}}}=\frac{\text{2a}}{{\text{V}}_{\text{m}}^{\text{3}}}\mathrm{...}\text{(3)}\\ \\ \frac{\text{2RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{3}}}=\frac{\text{6a}}{{\text{V}}_{\text{m}}^{\text{4}}}\mathrm{...}\text{(4)}\end{array}$

Now by dividing equation (4) by equation (3),

  $\begin{array}{l}\frac{\frac{\text{2RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{3}}}}{\frac{\text{RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{2}}}}\text{=}\frac{\frac{\text{6a}}{{\text{V}}_{\text{m}}^{\text{4}}}}{\frac{\text{2a}}{{\text{V}}_{\text{m}}^{\text{3}}}}\\ \\ \frac{\text{2RT×}{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{2}}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{3}}\text{×RT}}\text{=}\frac{{\text{6a×V}}_{\text{m}}^{\text{3}}}{{\text{V}}_{\text{m}}^{\text{4}}\text{×2a}}\\ \\ \frac{\text{2}}{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}\text{=}\frac{\text{3}}{{\text{V}}_{\text{m}}}\\ \\ {\text{2V}}_{\text{m}}\text{=}{\text{3V}}_{\text{m}}\text{-}\text{3b}\\ \\ {\text{V}}_{\text{m}}\text{=}\text{3b}\end{array}$

Here ${\text{V}}_{\text{m}}$ is the critical volume and that can be represented as ${\text{V}}_{\text{m,c}}\text{=}\text{3b}$.

Now putting the value of critical volume value in equation (3),

  $\begin{array}{l}\frac{\text{RT}}{{\left({\text{V}}_{\text{m}}\text{-}\text{b}\right)}^{\text{2}}}\text{=}\frac{\text{2a}}{{\text{V}}_{\text{m}}^{\text{3}}}\\ \\ \frac{{\text{RT}}_{\text{c}}}{{\left(\text{3b}\text{-}\text{b}\right)}^{\text{2}}}\text{=}\frac{\text{2a}}{{\text{(3b)}}^{\text{3}}}\\ \\ \frac{{\text{RT}}_{\text{c}}}{{\left(\text{2b}\right)}^{\text{2}}}\text{=}\frac{\text{2a}}{{\text{(3b)}}^{\text{3}}}\\ \\ {\text{T}}_{\text{c}}\text{=}\frac{{\text{2a×4b}}^{\text{2}}}{{\text{27Rb}}^{\text{3}}}\\ \\ {\text{T}}_{\text{c}}\text{=}\frac{\text{8a}}{\text{27Rb}}\end{array}$

Hence the critical temperature is ${\text{T}}_{\text{c}}\text{=}\frac{\text{8a}}{\text{27Rb}}$.

Now putting the value of critical temperature and volume in van der Waals equation,

  $\begin{array}{l}\left({\text{P}}_{\text{c}}\text{+}\frac{\text{a}}{{\text{V}}_{\text{m,c}}^{\text{2}}}\right)\left({\text{V}}_{\text{m,c}}\text{-}\text{b}\right)\text{=}{\text{RT}}_{\text{c}}\\ \\ {\text{P}}_{\text{c}}\text{=}\frac{{\text{RT}}_{\text{c}}}{\left({\text{V}}_{\text{m,c}}\text{-}\text{b}\right)}\text{-}\frac{\text{a}}{{\text{V}}_{\text{m,c}}^{\text{2}}}\\ \\ {\text{P}}_{\text{c}}\text{=}\frac{\text{R×}\frac{\text{8a}}{\text{27Rb}}}{\left(\text{3b}\text{-}\text{b}\right)}\text{-}\frac{\text{a}}{{\text{(3b)}}^{\text{2}}}\\ \\ {\text{P}}_{\text{c}}\text{=}\frac{\text{4a}}{{\text{27b}}^{\text{2}}}\text{-}\frac{\text{a}}{{\text{9b}}^{\text{2}}}\\ \\ {\text{P}}_{\text{c}}\text{=}\frac{\text{a}}{{\text{27b}}^{\text{2}}}\end{array}$

Hence the critical pressure value is ${\text{P}}_{\text{c}}\text{=}\frac{\text{a}}{{\text{27b}}^{\text{2}}}$.

Thus the values of critical temperature, pressure and molar volume are respectively $\frac{\text{8a}}{\text{27Rb}}$, $\frac{\text{a}}{{\text{27b}}^{\text{2}}}$ and $\text{3b}$.

(b)

Interpretation Introduction

Interpretation:

The value of the compression factor at critical point is $\frac{3}{8}$ has to be shown.

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.

  $\left(\text{P}\text{+}\frac{{\text{an}}^{\text{2}}}{{\text{V}}^{\text{2}}}\right)\left(\text{V}\text{-}\text{nb}\right)\text{=}\text{nRT}$

Where,

P is pressure of the real gas

V is volume of the real gas

‘n’ is the number of moles of real gas molecules.

R is the gas constant

T is the temperature

‘a’ is the pressure correction term

‘b’ is the volume occupied by each molecules (volume correction term).

Critical constants:

There are some particular values of temperature, pressure and volume that are called as critical constants.

The temperature above which the vapor of a substance cannot be liquefied no matter what amount of pressure is given is called the critical temperature of the substance.

The pressure which is required to liquefy the vapor of a substance at critical temperature is called the critical pressure of the substance.

Compressibility factor:

The compressibility factor of a gas is the correction factor that describes the deviation of real gas from its ideal behavior.

It is defined as the ratio of the molar volume of the gas to molar volume of ideal gas at same temperature and pressure.

Explanation of Solution

At critical point,

${\text{P}}_{\text{c}}\text{=}\frac{\text{a}}{{\text{27b}}^{\text{2}}}$

${\text{T}}_{\text{c}}\text{=}\frac{\text{8a}}{\text{27Rb}}$

${\text{V}}_{\text{m,c}}\text{=}\text{3b}$

According to ideal gas equation, $\text{PV=nRT}$

Thus, $\text{V=}\frac{\text{nRT}}{\text{P}}$

According to definition of compressibility factor,

  $\begin{array}{l}\text{Z=}\frac{\text{voulme}\text{of}\text{real}\text{gas}}{\text{volume}\text{of}\text{ideal}\text{gas}}\\ \\ \text{Z=}\frac{{\text{V}}_{\text{c,m}}}{\frac{{\text{RT}}_{\text{c}}}{{\text{P}}_{\text{c}}}}\text{(at}\text{critical}\text{condition)}\\ \\ \text{Z=}\frac{{\text{V}}_{\text{c,m}}{\text{P}}_{\text{c}}}{{\text{RT}}_{\text{c}}}\\ \\ \text{Z=}\frac{\text{3b×}\frac{\text{a}}{{\text{27b}}^{\text{2}}}}{\text{R×}\frac{\text{8a}}{\text{27Rb}}}\\ \\ \text{Z=}\frac{\text{3}}{\text{8}}\end{array}$

Hence the compressibility factor is $\frac{\text{3}}{\text{8}}$.

(c)

Interpretation Introduction

Interpretation:

The values of critical constants of nitrogen dioxide have to be calculated along with its compressibility factor.

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.

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

Critical constants:

There are some particular values of temperature, pressure and volume that are called as critical constants.

The temperature above which the vapor of a substance cannot be liquefied no matter what amount of pressure is given is called the critical temperature of the substance.

The pressure which is required to liquefy the vapor of a substance at critical temperature is called the critical pressure of the substance.

Compressibility factor:

The compressibility factor of a gas is the correction factor that describes the deviation of real gas from its ideal behavior.

It is defined as the ratio of the molar volume of the gas to molar volume of ideal gas at same temperature and pressure.

Explanation of Solution

At critical point,

${\text{P}}_{\text{c}}\text{=}\frac{\text{a}}{{\text{27b}}^{\text{2}}}$

${\text{T}}_{\text{c}}\text{=}\frac{\text{8a}}{\text{27Rb}}$

${\text{V}}_{\text{m,c}}\text{=}\text{3b}$

Given that,

  $\begin{array}{l}\text{a}\text{=}\text{5}\text{.354}{\text{dm}}^{\text{6}}\text{bar}{\text{mol}}^{\text{-1}}\\ \\ \text{b}\text{=}\text{0}\text{.04424}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\\ \\ \text{R}\text{=}\text{0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\end{array}$

Now putting the values of van der Waals constants $\text{a}$, $\text{b}$ and universal gas constant $\text{R}$ the values of critical constants can be obtained.

  $\begin{array}{l}{\text{P}}_{\text{c}}\text{=}\frac{\text{a}}{{\text{27b}}^{\text{2}}}\\ \\ {\text{P}}_{\text{c}}\text{=}\frac{\text{5}\text{.354}{\text{dm}}^{\text{6}}\text{bar}{\text{mol}}^{\text{-1}}}{\text{27(0}\text{.04424}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}{)}^{\text{2}}}\\ \\ {\text{P}}_{\text{c}}\text{=}\text{101}\text{.317}\text{bar}{\text{mol}}^{\text{-1}}\end{array}$

This is the molar critical pressure and if one mole of gas is considered then the critical pressure is ${\text{P}}_{\text{c}}\text{=}\text{101}\text{.317}\text{bar}$.

$\begin{array}{l}{\text{T}}_{\text{c}}\text{=}\frac{\text{8a}}{\text{27Rb}}\\ \\ {\text{T}}_{\text{c}}\text{=}\frac{\text{8×5}\text{.354}{\text{dm}}^{\text{6}}\text{bar}{\text{mol}}^{\text{-1}}}{\text{27×0}\text{.04424}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\text{×0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×}\frac{\text{1}\text{.01325}\text{bar}}{\text{1}\text{atm}}}\left[\begin{array}{l}\text{1}\text{atm}\text{=}\text{1}\text{.01325}\text{bar}\\ \text{1}\text{L}\text{=}\text{1}{\text{dm}}^{\text{3}}\end{array}\right]\\ \\ {\text{T}}_{\text{c}}\text{=}\text{442}\text{.51}\text{K}\text{(considering 1 mole of gas)}\end{array}$

  $\begin{array}{l}{\text{V}}_{\text{m,c}}\text{=}\text{3b}\\ \\ {\text{V}}_{\text{m,c}}\text{=}\text{3×0}\text{.04424}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\\ \\ {\text{V}}_{\text{m,c}}\text{=}\text{0}\text{.13272}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\\ \\ {\text{V}}_{\text{c}}\text{=}\text{0}\text{.13272}{\text{dm}}^{\text{3}}\text{(considering}\text{1}\text{mole}\text{of}\text{gas}\text{)}\end{array}$

The formula of compressibility factor calculated is,

  $\text{Z=}\frac{{\text{V}}_{\text{c,m}}{\text{P}}_{\text{c}}}{{\text{RT}}_{\text{c}}}$

Now putting the values of the critical constants,

  $\begin{array}{l}\text{Z=}\frac{{\text{V}}_{\text{c,m}}{\text{P}}_{\text{c}}}{{\text{RT}}_{\text{c}}}\\ \\ \text{Z=}\frac{\text{0}\text{.13272}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}\text{×}\text{101}\text{.317}\text{bar}}{\text{0}\text{.0821}\text{L}\text{atm}{\text{mol}}^{\text{-1}}{\text{K}}^{\text{-1}}\text{×}\frac{\text{1}\text{.01325}\text{bar}}{\text{1}\text{atm}}\text{×442}\text{.51}\text{K}}\\ \\ \text{Z=0}\text{.365}\approx \text{0}\text{.37}\end{array}$

Hence the compressibility factor for nitrogen dioxide is nearly equal to $\frac{\text{3}}{\text{8}}$.

Thus the values of critical temperature, critical pressure, molar critical volume are respectively $\text{442}\text{.51}\text{K}$, $\text{101}\text{.317}\text{bar}$ and $\text{0}\text{.13272}{\text{dm}}^{\text{3}}{\text{mol}}^{\text{-1}}.$

Also it has been calculated that the compressibility factor for nitrogen dioxide is nearly equal to $\frac{\text{3}}{\text{8}}$.