Chapter 5, Problem 5A.3P

(a)

Interpretation Introduction

Interpretation:

The entropy of mixing of a gaseous mixture with mass percentage composition $\text{75}\text{.5}\left({\text{N}}_{\text{2}}\right)$, $\text{23}\text{.2}\left({\text{O}}_{\text{2}}\right)$, and $\text{1}\text{.3}\left(\text{Ar}\right)$, has to be calculated.

Concept Introduction:

In the mixing of two or more perfect gases the entropy of mixing is calculated by the formula below.

    $\Delta S_{\text{mix}}=-nR{\displaystyle \sum x_i\ln x_i}$

The value of $\Delta V_{\text{mix}}$ and $\Delta H_{\text{mix}}$ is always zero for mixing of perfect gases.  As the number of gases being mixed increases, the entropy also increases.

Answer to Problem 5A.3P

The entropy of mixing of a gaseous mixture with mass percentage composition $\text{75}\text{.5}\left({\text{N}}_{\text{2}}\right)$, $\text{23}\text{.2}\left({\text{O}}_{\text{2}}\right)$, and, $\text{1}\text{.3}\left(\text{Ar}\right)$ is $\underset{\_}{5.052J{K}^{-1}}$.

Explanation of Solution

The entropy of mixing of perfect gases ($\Delta S_{\text{mix}}$) is calculated by the formula given below.

    $\Delta S_{\text{mix}}=-nR\left(x_{{\text{N}}_{\text{2}}}\ln x_{{\text{N}}_{\text{2}}}+x_{{\text{O}}_{\text{2}}}\ln x_{{\text{O}}_{\text{2}}}+x_{\text{Ar}}\ln x_{\text{Ar}}\right)\left(1\right)$

Where,

$n$ is the number of moles.

$R$ is the gas constant. $\left(8.31\text{4}{\text{JK}}^{-1}{\text{mol}}^{-1}\right)$.

$x_{{\text{N}}_{\text{2}}}$, $x_{{\text{O}}_{\text{2}}}$, $x_{\text{Ar}}$ are the mole fractions of ${\text{N}}_{\text{2}}$, ${\text{O}}_{\text{2}}$, and $\text{Ar}$.

The mass percentage of ${\text{N}}_{\text{2}}$ is $\text{75}\text{.5}$.

Therefore, the mole fraction of ${\text{N}}_{\text{2}}$ ($x_{{\text{N}}_{\text{2}}}$) is calculated below.

  $\begin{array}{c}\text{Mole}\text{fraction}=\frac{\text{Mass}\text{percentage}}{100}\\ =\frac{\text{75}\text{.5}}{100}\\ =0.755\end{array}$

The mass percentage of ${\text{O}}_{\text{2}}$ is $\text{23}\text{.2}$.

Therefore, the mole fraction of ${\text{O}}_{\text{2}}$ ($x_{{\text{O}}_2}$) is calculated below.

  $\begin{array}{c}\text{Mole}\text{fraction}=\frac{\text{Mass}\text{percentage}}{100}\\ =\frac{\text{23}\text{.2}}{100}\\ =0.232\end{array}$

The mass percentage of $\text{Ar}$ is $\text{1}\text{.3}$.

Therefore, the mole fraction of $\text{Ar}$ ($x_{\text{Ar}}$) is calculated below.

  $\begin{array}{c}\text{Mole}\text{fraction}=\frac{\text{Mass}\text{percentage}}{100}\\ =\frac{\text{1}\text{.3}}{100}\\ =0.013\end{array}$

Therefore, the entropy of mixing $\text{1}\text{mol}$ of gases is calculated below.

    $\begin{array}{c}\Delta S_{\text{mix}}=-1\text{mol}\times 8.31\text{4}\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times \left(\begin{array}{l}\left(\text{0}\text{.755}\times \ln \text{0}\text{.755}\right)\\ +\left(0.232\times \ln 0.232\right)+\left(0.013\times \ln 0.013\right)\end{array}\right)\\ =-1\text{mol}\times 8.31\text{4}\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times \left(\left(-0.2122\right)+\left(-0.3389\right)+\left(-0.05645\right)\right)\\ =-1\text{mol}\times 8.31\text{4}\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times \left(-0.6075\right)\\ =\underset{\_}{5.052J{K}^{-1}}\end{array}$

Therefore, the entropy of mixing of a gaseous mixture with mass percentage composition $\text{75}\text{.5}\left({\text{N}}_{\text{2}}\right)$, $\text{23}\text{.2}\left({\text{O}}_{\text{2}}\right)$, and $\text{1}\text{.3}\left(\text{Ar}\right)$, is $\underset{\_}{5.052J{K}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The change in entropy from part (a) when air is taken as a mixture with mass percentage composition $\text{75}\text{.52}\left({\text{N}}_{\text{2}}\right)$, $\text{23}\text{.15}\left({\text{O}}_{\text{2}}\right)$, $\text{1}\text{.28}\left(\text{Ar}\right)$, and $\text{0}\text{.046}\left({\text{CO}}_{\text{2}}\right)$, has to be calculated.

Concept Introduction:

Refer to (a)

Answer to Problem 5A.3P

The change in entropy from part (a) when air is taken as a mixture with mass percentage composition $\text{75}\text{.52}\left({\text{N}}_{\text{2}}\right)$, $\text{23}\text{.15}\left({\text{O}}_{\text{2}}\right)$, $\text{1}\text{.28}\left(\text{Ar}\right)$, and $\text{0}\text{.046}\left({\text{CO}}_{\text{2}}\right)$ is $\underset{\_}{0.02J{K}^{-1}}$.

Explanation of Solution

The entropy of mixing of perfect gases ($\Delta S_{\text{mix}}$) is calculated by the formula given below.

    $\Delta S_{\text{mix}}=-nR\left(x_{{\text{N}}_{\text{2}}}\ln x_{{\text{N}}_{\text{2}}}+x_{{\text{O}}_{\text{2}}}\ln x_{{\text{O}}_{\text{2}}}+x_{\text{Ar}}\ln x_{\text{Ar}}+x_{{\text{CO}}_{\text{2}}}\ln x_{{\text{CO}}_{\text{2}}}\right)\left(2\right)$

Where,

$n$ is the number of moles.

$R$ is the gas constant. $\left(8.31\text{4}{\text{JK}}^{-1}{\text{mol}}^{-1}\right)$.

$x_{{\text{N}}_{\text{2}}}$, $x_{{\text{O}}_{\text{2}}}$, $x_{\text{Ar}}$, $x_{{\text{CO}}_{\text{2}}}$ are the mole fractions of ${\text{N}}_{\text{2}}$, ${\text{O}}_{\text{2}}$, $\text{Ar}$, and ${\text{CO}}_{\text{2}}$ .

The mass percentage of ${\text{N}}_{\text{2}}$ is $\text{75}\text{.52}$.

Therefore, the mole fraction of ${\text{N}}_{\text{2}}$ ($x_{{\text{N}}_{\text{2}}}$) is calculated below.

  $\begin{array}{c}\text{Mole}\text{fraction}=\frac{\text{Mass}\text{percentage}}{100}\\ =\frac{\text{75}\text{.52}}{100}\\ =0.7552\end{array}$

The mass percentage of ${\text{O}}_{\text{2}}$ is $\text{23}\text{.15}$.

Therefore, the mole fraction of ${\text{O}}_{\text{2}}$ ($x_{{\text{O}}_2}$) is calculated below.

  $\begin{array}{c}\text{Mole}\text{fraction}=\frac{\text{Mass}\text{percentage}}{100}\\ =\frac{\text{23}\text{.15}}{100}\\ =0.2315\end{array}$

The mass percentage of $\text{Ar}$ is $\text{1}\text{.28}$.

Therefore, the mole fraction of $\text{Ar}$ ($x_{\text{Ar}}$) is calculated below.

  $\begin{array}{c}\text{Mole}\text{fraction}=\frac{\text{Mass}\text{percentage}}{100}\\ =\frac{\text{1}\text{.28}}{100}\\ =0.0128\end{array}$

The mass percentage of ${\text{CO}}_{\text{2}}$ is $\text{0}\text{.046}$.

Therefore, the mole fraction of ${\text{CO}}_{\text{2}}$ ($x_{{\text{CO}}_{\text{2}}}$) is calculated below.

  $\begin{array}{c}\text{Mole}\text{fraction}=\frac{\text{Mass}\text{percentage}}{100}\\ =\frac{\text{0}\text{.046}}{100}\\ =0.000\text{46}\end{array}$

Therefore, the entropy of mixing $\text{1}\text{mol}$ of gases is calculated below.

    $\begin{array}{c}\Delta S_{\text{mix}}=-1\text{mol}\times 8.31\text{4}{\text{JK}}^{-1}{\text{mol}}^{-1}\times \left(\begin{array}{l}\left(\text{0}\text{.7552}\times \ln \text{0}\text{.7552}\right)+\\ \left(0.2315\times \ln 0.2315\right)+\\ \left(0.0128\times \ln 0.0128\right)+\\ \left(0.00046\times \ln 0.00046\right)\end{array}\right)\\ =-1\text{mol}\times 8.31\text{4}{\text{JK}}^{-1}{\text{mol}}^{-1}\times \left(\begin{array}{l}\left(-0.2120\right)+\left(-0.3387\right)+\\ \left(-0.05579\right)+\left(-0.000353\right)\end{array}\right)\\ =-1\text{mol}\times 8.31\text{4}{\text{JK}}^{-1}{\text{mol}}^{-1}\times \left(-0.61\right)\\ =\text{5}\text{.072}\text{J}{\text{K}}^{-\text{1}}\end{array}$

Therefore, the entropy of mixing of a gaseous mixture with mass percentage composition $\text{75}\text{.5}\left({\text{N}}_{\text{2}}\right)$, $\text{23}\text{.2}\left({\text{O}}_{\text{2}}\right)$, $\text{1}\text{.3}\left(\text{Ar}\right)$, and $\text{0}\text{.046}\left({\text{CO}}_{\text{2}}\right)$ is $\text{5}\text{.072}{\text{JK}}^{-\text{1}}$.

Therefore, the change in entropy from part (a) is calculated below.

    $\begin{array}{c}\text{Change}\text{in}\text{entropy}=\text{5}\text{.072}{\text{JK}}^{-\text{1}}-\text{5}\text{.052}{\text{JK}}^{-\text{1}}\\ =\underset{\_}{0.02J{K}^{-1}}\end{array}$