Chapter 4, Problem 4.2PR

(a)

Interpretation Introduction

Interpretation:

The expression for the change in molar Gibbs energy has to be derived.

Concept Introduction:

The van der Waals equation of state can be expressed as

$\left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$

Where,

$V_m=\frac{V}{n}=$ Molar volume

$a=$ A constant for attractive effect

$b=$ A constant for repulsive effect

$P=$ Pressure

$T=$ Absolute temperature

$R=$ Universal gas constant

Ideal gas equation can also be represented as

$\boxed{PV=nRT}$

Molar Gibbs energy can be mathematically represented in terms of T and P,

$d{G}_m=V_{m}dP-SdT$

At constant T,

$\boxed{d{G}_m=V_{m}dP}$

The change in molar Gibbs energy for an ideal gas is,

$\Delta G_m^{Ideal}=RT\ln \left(\frac{P_f}{P_i}\right)$

Where,

$P_f=$ Final pressure

$P_i=$ Initial pressure

Explanation of Solution

The van der Waals equation of state can be expressed as

$\left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$

According to the give question, we can neglect attractive effects.  Then the equation becomes,

$P\left(V_m-b\right)=RT$

Solving for $V_m$,

$\begin{array}{l}P\left(V_m-b\right)=RT\\ \\ P{V}_m-Pb=RT\\ \\ V_m=\left(\frac{RT}{P}\right)+b.\end{array}$

Where,

$\frac{RT}{P}=$ Molar volume of a perfect gas

By taking integration from limit initial value to final value,

$\begin{array}{l}{\displaystyle \underset{G_i}{\overset{G_f}{\int }}d{G}_m}={\displaystyle \underset{P_i}{\overset{P_f}{\int }}{V}_m}dP\\ \\ G_f-G_i={\displaystyle \underset{P_i}{\overset{P_f}{\int }}\left\{\left(\frac{RT}{P}\right)+b\right\}}dP\\ \\ \boxed{\Delta G_m=RT\ln \left(\frac{P_f}{P_i}\right)+b\left(P_f-P_i\right)}\end{array}$

The expression in the above box represents the change in molar Gibbs energy for a real gas.

(b)

Interpretation Introduction

Interpretation:

Whether the change in molar Gibbs energy is greater or smaller than for a perfect gas has to be decided.

Concept Introduction:

The change in molar Gibbs energy for an ideal gas is,

$\boxed{\Delta G_m^{Ideal}=RT\ln \left(\frac{P_f}{P_i}\right)}$

The change in molar Gibbs energy for a perfect gas is,

$\boxed{\Delta G_m=RT\ln \left(\frac{P_f}{P_i}\right)+b\left(P_f-P_i\right)}$

Answer to Problem 4.2PR

The change in molar Gibbs energy for a perfect gas is greater than the change in molar Gibbs energy for an ideal gas by a factor $b\left(P_f-P_i\right)$.

Explanation of Solution

The change in molar Gibbs energy for an ideal gas is,

$\boxed{\Delta G_m^{Ideal}=RT\ln \left(\frac{P_f}{P_i}\right)}$

The change in molar Gibbs energy for a perfect gas is,

$\boxed{\Delta G_m=RT\ln \left(\frac{P_f}{P_i}\right)+b\left(P_f-P_i\right)}$

The change,

$\begin{array}{l}\Delta G_m-\Delta G_m^{Ideal}=\left[RT\ln \left(\frac{P_f}{P_i}\right)+b\left(P_f-P_i\right)\right]-\left[RT\ln \left(\frac{P_f}{P_i}\right)\right]\\ \\ =b\left(P_f-P_i\right)\end{array}$

Therefore, the change in molar Gibbs energy for a perfect gas is greater than the change in molar Gibbs energy for an ideal gas by a factor $b\left(P_f-P_i\right)$.

(c)

Interpretation Introduction

Interpretation:

The percentage difference between van der Walls and perfect gas for carbon dioxide has to be calculated.

Concept Introduction:

The change in molar Gibbs energy for a perfect gas is greater than the change in molar Gibbs energy for an ideal gas by a factor $b\left(P_f-P_i\right)$.

Percentage increase $=\frac{Newnoumber-Originalnoumber}{Originalnoumber}\times 100$

Answer to Problem 4.2PR

The percentage difference has been calculated to be $0.6861\%.$

Explanation of Solution

Given data:

$\begin{array}{l}{P}_i=1.0atm\\ \\ P_f=10.0atm\end{array}$

Calculation of percentage difference:

$\begin{array}{l}\Delta G_m^{Ideal}=RT\ln \left(\frac{P_f}{P_i}\right)\\ \\ =\left(0.0821L.atm.K^{-1}.mo{l}^{-1}\right)\left(298K\right)\ln \left(\frac{10.0}{1.0}\right)\\ \\ =56.27L.atm.mo{l}^{-1}.\end{array}$

$\begin{array}{l}\Delta G_m-\Delta G_m^{Ideal}=\left[RT\ln \left(\frac{P_f}{P_i}\right)+b\left(P_f-P_i\right)\right]-\left[RT\ln \left(\frac{P_f}{P_i}\right)\right]\\ \\ =b\left(P_f-P_i\right)=\left(4.29\right)\left(10.0-1.0\right)=38.61\times {10}^{-2}L.mo{l}^{-1}.\end{array}$

Percentage difference:

$\begin{array}{c}\frac{\Delta G_m-\Delta G_m^{Ideal}}{\Delta G_m^{Ideal}}\times 100\\ \\ =\frac{38.61\times {10}^{-2}L.mo{l}^{-1}}{56.27L.atm.mo{l}^{-1}}\times 100\\ \\ =0.6861\%.\end{array}$

Therefore, the calculated percentage difference is $\text{0}\text{.6861\%}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The critical constants for the given gas in terms of the parameters ‘a’ and ‘b’ has to be derived.

Explanation of Solution

A gas described by the equation of state

$P=\frac{nRT}{V}-\frac{a{n}^2}{V^2}+\frac{b{n}^3}{V^3}\mathrm{........}(1)$

If we plot a graph between P and V which is called an isotherm, we will get a flat inflexion where $\frac{dP}{dV}=0and\frac{d^{2}P}{d{V}^2}=0$.  It implies that the gas described by the above equation shows critical behavior.

Calculation of critical constants:

Critical Volume$\left(V_c\right)$:

Taking the first and second derivative and equalizing it to zero, we can get two equations,

$\begin{array}{l}\frac{2a{n}^2}{V^3}-\frac{nRT}{V^2}-\frac{3b{n}^3}{V^4}=0\mathrm{.........}(2)\\ \\ \frac{-6a{n}^2}{V^4}+\frac{2nRT}{V^3}+\frac{12b{n}^3}{V^5}=0\mathrm{.........}(3)\end{array}$

Consider equation (1). We can express T in terms of V as below,

$T=-\left[\frac{3b{n}^3}{V^4}-\frac{2a{n}^2}{V^3}\right]/nR=-\left[\frac{3b{n}^3-2a{n}^{2}V}{nR{V}^2}\right]\mathrm{........}(4)$

Now substituting the value of T in equation (2) and solving for $V_c$,

$\boxed{V_c=\frac{3bn}{a}}$

Critical Temperature$\left(T_c\right)$:

By substituting the value of $V_c$ in equation (3) and solving for $T_c$,

$\boxed{T_c=\frac{a^2}{3bR}}$

By substituting the value of $V_c$ and $T_c$ in the equation (1) and solving for $P_c$,

$\boxed{P_c=\frac{a^3}{27b^2}}$