Chapter 4, Problem 4C.6E

(a)

Interpretation Introduction

Interpretation:

The additional change (from previous exercise) in molar Gibbs energy and entropy of mixing when the nitrogen, oxygen and argon are mixed to form air at $\text{298}\text{K}$ has to be given.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $\Delta G$.

Molal Gibbs free energy can be determined using the following formulas as show below,

$\Delta G=RT\ln \frac{V_1}{V_2}\boxed{\begin{array}{l}Where,\\ \Delta G:ChangeinGibbsfreeenergy\\ R:Gascons\tan t\\ T:Temperature\\ V_1:Initialvolume\\ V_2:Finalvolume\end{array}}$

The equation for finding ${\text{ΔG}}_{\text{mix}}$ is given below:

  $\boxed{\begin{array}{c}{\text{ΔG}}_{\text{mix}}\text{=}\text{nRT}{\text{x}}_{\text{a}}\text{ln}{\text{x}}_{\text{a}}\text{+}{\text{x}}_{\text{b}}\text{ln}{\text{x}}_{\text{b}}\\ \\ \text{n}\text{-}\text{Number}\text{of}\text{moles}\\ {\text{x}}_{\text{a}}\text{-}\text{mole}\text{fraction}\text{of}\text{a}\text{component}\\ {\text{x}}_{\text{b}}\text{-}\text{mole}\text{fraction}\text{of}\text{b}\text{component}\\ \text{R-}\text{Universal}\text{gas}\text{constant}\\ \text{T}\text{-}\text{Temperature}\end{array}}$

The equation for finding ${\text{ΔS}}_{\text{mix}}$ is given below:

$\begin{array}{c}{\text{ΔS}}_{\text{mix}}\text{=}\text{-nR}\left({\text{x}}_{\text{a}}\text{ln}{\text{x}}_{\text{a}}\text{+}{\text{x}}_{\text{b}}\text{ln}{\text{x}}_{\text{b}}\right)\\ \\ \text{n}\text{-}\text{Number}\text{of}\text{moles}\\ {\text{x}}_{\text{a}}\text{-}\text{mole}\text{fraction}\text{of}\text{a}\text{component}\\ {\text{x}}_{\text{b}}\text{-}\text{mole}\text{fraction}\text{of}\text{b}\text{component}\\ \text{R-}\text{Universal}\text{gas}\text{constant}\end{array}$

Explanation of Solution

Given data:

$\begin{array}{l}{\text{χ}}_{{\text{N}}_{\text{2}}}\text{=}\text{0}\text{.78}\\ \\ {\text{χ}}_{{\text{O}}_{\text{2}}}\text{=}\text{0}\text{.210}\\ \\ {\text{χ}}_{Ar}\text{=}\text{0}\text{.0096}\\ \\ \text{T}\text{=}\text{298}\text{K}\end{array}$

The equation for finding ${\text{ΔG}}_{\text{mix}}$ is given below:

${\text{ΔG}}_{\text{mix}}\text{=}\text{RT}\left({\text{x}}_{\text{a}}\text{ln}{\text{x}}_{\text{a}}\text{+}{\text{x}}_{\text{b}}\text{ln}{\text{x}}_{\text{b}}+{\text{x}}_c\text{ln}{\text{x}}_c\right)$

$\begin{array}{c}{\text{ΔG}}_{\text{mix}}\text{=}\text{RT}\left({\text{x}}_{\text{a}}\text{ln}{\text{x}}_{\text{a}}\text{+}{\text{x}}_{\text{b}}\text{ln}{\text{x}}_{\text{b}}+{\text{x}}_c\text{ln}{\text{x}}_c\right)\\ \\ \text{=}\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{298}\text{K}\left(\text{0}\text{.78}\text{ln}\text{0}\text{.78}\text{+}\text{0}\text{.210}\text{ln}\text{0}\text{.210}\text{+}\text{0}\text{.0096}\text{ln}\text{0}\text{.0096}\right)\\ \\ \text{=}\text{-1402}\text{.65}\text{J}/\text{mol}\end{array}$

In the previous problem, the value of ${\text{ΔG}}_{\text{mix}}$ is found to be $\text{-1303}\text{.2}\text{J}/\text{mol}$, thus the additional change in molar Gibbs energy is $-99.45J/mol$ $\because \left(-\text{1402}\text{.65}\text{J/mol-}\left(\text{-1303}\text{.2J/mol}\right)\right)$

Molar entropy of mixing can be determined as follows,

Given data:

$\begin{array}{l}{\text{χ}}_{{\text{N}}_{\text{2}}}\text{=}\text{0}\text{.78}\\ \\ {\text{χ}}_{{\text{O}}_{\text{2}}}\text{=}\text{0}\text{.210}\\ \\ {\text{χ}}_{Ar}\text{=}\text{0}\text{.0096}\\ \\ \text{T}\text{=}\text{298}\text{K}\end{array}$

The equation for finding ${\text{ΔS}}_{\text{mix}}$ is given below:

${\text{ΔS}}_{\text{mix}}\text{=}\text{-R}\left({\text{x}}_{\text{a}}\text{ln}{\text{x}}_{\text{a}}\text{+}{\text{x}}_{\text{b}}\text{ln}{\text{x}}_{\text{b}}+\text{+}{\text{x}}_c\text{ln}{\text{x}}_c\right)$

$\begin{array}{c}{\text{ΔS}}_{\text{mix}}\text{=}\text{-R}\left({\text{x}}_{\text{a}}\text{ln}{\text{x}}_{\text{a}}\text{+}{\text{x}}_{\text{b}}\text{ln}{\text{x}}_{\text{b}}+\text{+}{\text{x}}_c\text{ln}{\text{x}}_c\right)\left(\frac{\pi }{2}-\theta \right)\\ \\ \text{=}\text{-8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\left(\text{0}\text{.78}\text{ln}\text{0}\text{.78}\text{+}\text{0}\text{.210}\text{ln}\text{0}\text{.210}\text{+}\text{0}\text{.0096}\text{ln}\text{0}\text{.0096}\right)\\ \\ \text{=}\text{4}\text{.707}\text{J}/\text{K}\end{array}$

In the previous problem, the value of ${\text{ΔS}}_{\text{mix}}$ is found to be $\text{4}\text{.373}\text{J}/\text{K}$, thus the additional change in molar entropy of mixing is $0.334J/K$ $\because \left(\text{4}\text{.707}\text{J/K}\text{-4}\text{.373}\text{J/K}\text{=}\text{0}\text{.334}\right)$

(c)

Interpretation Introduction

Interpretation:

The mixing is whether spontaneous or not has to be given.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by $\Delta G$.

Molal Gibbs free energy can be determined using the following formulas as show below,

$\Delta G=RT\ln \frac{V_1}{V_2}\boxed{\begin{array}{l}Where,\\ \Delta G:ChangeinGibbsfreeenergy\\ R:Gascons\tan t\\ T:Temperature\\ V_1:Initialvolume\\ V_2:Finalvolume\end{array}}$

$\Delta G=\Delta H-T\Delta S\boxed{\begin{array}{l}where,\\ \Delta H:Enthalpychange\\ T:Temperature\\ \Delta S:Changeinentropy\end{array}}$

For an isomermal reaction enthalpy change is zero, thus this equation can be written as,

$\begin{array}{c}\Delta G=-T\Delta S\\ =-T\left(C_P\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}\right)\\ =-T\left(0-R\ln \frac{P_2}{P_1}\right)\end{array}$

Explanation of Solution

The molar Gibbs energy of mixing when the nitrogen, oxygen and nitrogen are mixed to form air at $\text{298}\text{K}$ is calculated to be negative and the molar entropy of mixing when the nitrogen, oxygen and nitrogen are mixed to form air at $\text{298}\text{K}$ is calculated to be positive.

$\begin{array}{l}{\text{ΔG}}_{\text{mix}}\text{=}\text{-ve}\\ \\ {\text{ΔS}}_{\text{mix}}\text{=}\text{+ve}\end{array}$

The above conditions satisfies the criteria for spontaneous.

Therefore, the mixing is spontaneous.