Chapter 9.7, Problem 1E

Interpretation Introduction

Interpretation:

To determine the entropy change $\left(\Delta S^{ig}\right)$.

Concept Introduction:

The partial entropy of mixing $\left(\Delta {\underset{\_}{S}}^{ig}\right)$ assumed as an ideal gas.

$\Delta {\underset{\_}{S}}^{ig}=-R{y}_1\ln y_1-R{y}_2\ln y_2$

Here, gas constant is $R$, mole fraction of Propanol is $y_1$, and mole fraction of ethanol is $y_2$.

The mole fraction $\left(y_n\right)$ is expressed as:

$y_n=\frac{N_1}{N_{Total}}$

Here, $n$ refers the respective components and total number of moles is $N_{Total}$.

The entropy change $\left(\Delta S^{ig}\right)$ is expressed as:

$\Delta S^{ig}=N_{Total}\left(\Delta {\underset{\_}{S}}^{ig}\right)$

Explanation of Solution

Given Information

The number of moles in propanol $N_1=1\text{mole}$.

The number of moles in ethanol $N_2=4\text{moles}$.

Write the expression to calculate the total number of moles.

$N_{Total}=N_1+N_2$        (1)

Here, number of moles in Propanol is $N_1$, and number of moles in ethanol is $N_2$.

Substitute $\text{1}\text{mole}$ for $N_1$, and $\text{4}\text{moles}$ of $N_2$ in Equation (2).

$\begin{array}{c}{N}_{Total}=\text{1}\text{mole}+\text{4}\text{mole}\\ \text{= 5}\text{mole}\end{array}$

Write the expression to calculate the mole fraction of Propanol $\left(y_1\right)$.

$y_1=\frac{N_1}{N_{Total}}$        (2)

Substitute $\text{1}\text{mole}$ for $N_1$, and $\text{5}\text{moles}$ for $N_{Total}$ in Equation (2).

$\begin{array}{c}{y}_1=\frac{1\text{mole}}{5\text{mole}}\\ =0.2\end{array}$

Write the expression to calculate the mole fraction of ethanol $\left(y_2\right)$.

$y_2=\frac{N_2}{N_{Total}}$        (3)

Substitute $\text{4}\text{moles}$ of $N_2$ and $\text{5}\text{moles}$ for $N_{Total}$ in Equation (3).

$\begin{array}{c}{y}_2=\frac{\text{4}\text{moles}}{5\text{moles}}\\ =0.8\end{array}$

The value of gas constant as $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$.

Write the expression to calculate the partial entropy of mixing $\left(\Delta {\underset{\_}{S}}^{ig}\right)$ assumed as an ideal gas.

$\Delta {\underset{\_}{S}}^{ig}=-R{y}_1\ln y_1-R{y}_2\ln y_2$        (4)

Substitute $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$, $0.2$ for $y_1$, $0.8$ for $y_2$ in Equation (4).

$\begin{array}{c}\Delta {\underset{\_}{S}}^{ig}=-R{y}_1\ln y_1-R{y}_2\ln y_2\\ =-\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(0.2\right)\left(\ln 0.2\right)-\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(0.8\right)\left(\ln 0.8\right)\\ =2.676\text{J}/\text{mol}\cdot \text{K}+1.484\text{J}/\text{mol}\cdot \text{K}\\ =4.16\text{J}/\text{mol}\cdot \text{K}\end{array}$

Write the expression to calculate the entropy change $\left(\Delta S^{ig}\right)$.

$\Delta S^{ig}=N_{Total}\left(\Delta {\underset{\_}{S}}^{ig}\right)$        (5)

Substitute $\text{5}\text{moles}$ for $N_{Total}$, and $4.16\text{J}/\text{mol}\cdot \text{K}$ for $\Delta {\underset{\_}{S}}^{ig}$ in Equation (5).

$\begin{array}{c}\Delta S^{ig}=\left(\text{5}\text{moles}\right)\left(4.16\text{J}/\text{mol}\cdot \text{K}\right)\\ =20.8\text{J}/\text{K}\end{array}$

Thus, the entropy change $\left(\Delta S^{ig}\right)$ is $20.8\text{J}/\text{K}$.