Chapter 6.6, Problem 23P

(A)

Interpretation Introduction

Interpretation:

The heat added to the gas.

Concept Introduction:

The analogous time-independent form of energy balance for the gas is written as:

$\Delta \left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{m}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-m_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +W_{\text{S}}+W_{\text{EC}}+Q\end{array}\right]$

Here, time is $t$, total mass is $M$, specific internal energy is $\widehat{U}$, velocity is $V$, acceleration due to gravity is $g$, height is $h$, initial mass flow is $m_{in}$, initial specific enthalpy is ${\widehat{H}}_{in}$, initial velocity is $V_{in}$, initial height of the gas is $h_{in}$, final mass flow is $m_{out}$, final height of the gas is $h_{out}$, shaft work is added to the system is $W_{\text{S}}$, work is added to the system through expansion or contraction of the system is $W_{\text{EC}}$, and total heat added throughout the process is Q.

Explanation of Solution

The analogous time-independent form of energy balance for the gas is written as:

$\Delta \left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{m}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-m_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +W_{\text{S}}+W_{\text{EC}}+Q\end{array}\right]$        (1)

Neglecting the potential and kinetic energy for a closed system with no shaft work and no expansion work, equation (1) can be rewritten as:

$\begin{array}{l}\Delta \left\{N\left(\underset{\_}{U}\right)\right\}=Q\\ \Delta U=Q\end{array}$        (2)

Here, number of moles is $N$, and molar internal energy is $\underset{\_}{U}$.

The equation for the change in internal energy $\left(d\underset{\_}{U}\right)$ is,

$\begin{array}{l}d\underset{\_}{U}=\left({\underset{\_}{U}}_2-{\underset{\_}{U}}_2^{ig}\right)+\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)-\left({\underset{\_}{U}}_1-{\underset{\_}{U}}_1^{ig}\right)\\ {\underset{\_}{U}}_2-{\underset{\_}{U}}_1=\left({\underset{\_}{U}}_2-{\underset{\_}{U}}_2^{ig}\right)+\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)-\left({\underset{\_}{U}}_1-{\underset{\_}{U}}_1^{ig}\right)\end{array}$        (3)

Here, molar internal energy at state 2 is ${\underset{\_}{U}}_2$, molar internal energy at state 1 is ${\underset{\_}{U}}_1$, and molar internal energy for an ideal gas state at state 2 and 1 is ${\underset{\_}{U}}_2^{ig},\text{and}{\underset{\_}{U}}_1^{ig}$ respectively.

To calculate the residual molar internal energy $\left({\underset{\_}{U}}^R\right)$,

$\begin{array}{c}{\underset{\_}{U}}^R=\underset{\_}{U}-{\underset{\_}{U}}^{ig}\\ ={\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{T{\left(\delta P/\delta T\right)}_{\underset{\_}{V}}-P\right\}}d\underset{\_}{V}\end{array}$        (4)

Here, molar volume is $\underset{\_}{V}$, temperature is $T$, pressure is $P$, change in pressure and temperature at constant molar volume is ${\left(\delta P\right)}_{\underset{\_}{V}},\text{and}{\left(\delta T\right)}_{\underset{\_}{V}}$ respectively.

Partially differentiate the Van der Waals equation to obtain ${\left(\delta P/\delta T\right)}_{\underset{\_}{V}}$.

$\begin{array}{l}P=\frac{RT}{\underset{\_}{V}-b}-\frac{a}{{\underset{\_}{V}}^2}\\ {\left(\delta P/\delta T\right)}_{\underset{\_}{V}}=\frac{R}{\underset{\_}{V}-b}\end{array}$

Here, gas constant is $R$, Van der Waals parameter is $a,\text{and}b$.

Substitute $\frac{R}{\underset{\_}{V}-b}$ for ${\left(\delta P/\delta T\right)}_{\underset{\_}{V}}$, and $\frac{RT}{\underset{\_}{V}-b}-\frac{a}{{\underset{\_}{V}}^2}$ for $P$ in Equation (4).

$\begin{array}{c}\underset{\_}{U}-{\underset{\_}{U}}^{ig}={\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{T{\left(\delta P/\delta T\right)}_{\underset{\_}{V}}-P\right\}}d\underset{\_}{V}\\ ={\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{T\left(\frac{R}{\underset{\_}{V}-b}\right)-\left(\frac{RT}{\underset{\_}{V}-b}-\frac{a}{{\underset{\_}{V}}^2}\right)\right\}}d\underset{\_}{V}\\ ={\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{\frac{a}{{\underset{\_}{V}}^2}\right\}}d\underset{\_}{V}\\ =-\frac{a}{\underset{\_}{V}}\end{array}$

The expression for the change in internal energy $\left(d\underset{\_}{U}\right)$ of an ideal gas is:

$\begin{array}{c}d\underset{\_}{U}=C_V^{\ast }dT\\ =C_V^{\ast }\left(T_2-T_1\right)\end{array}$

Here, constant volume heat capacity for ideal gas is $C_V^{\ast }$, change in temperature is $dT$, final temperature is $T_2$, and initial temperature is $T_1$.

Substitute $2.5R$ for $C_V^{\ast }$, $\text{600}\text{K}$ for $T_2$, and $\text{300}\text{K}$ for $T_1$, $\text{8}\text{.314}\text{J}/\text{K}\cdot \text{mol}$ for $R$.

$\begin{array}{l}d\underset{\_}{U}=C_V^{\ast }dT\\ d\underset{\_}{U}=2.5R\left(\text{600}\text{K}-\text{300}\text{K}\right)\\ d\underset{\_}{U}=2.5\left(\text{8}\text{.314}\text{J}/\text{K}\cdot \text{mol}\right)\left(\text{600}\text{K}-\text{300}\text{K}\right)\\ {\underset{\_}{U}}_2-{\underset{\_}{U}}_1=6,235.5\text{J}/\text{mol}\left(1\text{kJ}/\text{mol}/1,000\text{J}/\text{mol}\right)\\ =6.235\text{kJ}/\text{mol}\end{array}$

Substitute $6.235\text{kJ}/\text{mol}$ for ${\underset{\_}{U}}_2-{\underset{\_}{U}}_1$, $-\frac{a}{\underset{\_}{V}}$ for $\left({\underset{\_}{U}}_2-{\underset{\_}{U}}_2^{ig}\right),\text{and}\left({\underset{\_}{U}}_1-{\underset{\_}{U}}_1^{ig}\right)$ in Equation (3).

$\begin{array}{l}{\underset{\_}{U}}_2-{\underset{\_}{U}}_1=\left({\underset{\_}{U}}_2-{\underset{\_}{U}}_2^{ig}\right)+\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)-\left({\underset{\_}{U}}_1-{\underset{\_}{U}}_1^{ig}\right)\\ 6.235\text{kJ}/\text{mol}=\left[-\frac{a}{\underset{\_}{V}}\right]+\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)-\left[-\frac{a}{\underset{\_}{V}}\right]\\ 6.235\text{kJ}/\text{mol}=\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)\end{array}$

Calculate the change in molar internal energy $\left(\Delta U\right)$.

$\Delta U=N\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)$        (5)

Substitute $\text{10}\text{moles}$ for $N$, and $6.235\text{kJ}/\text{mol}$ for $\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)$ in Equation (5).

$\begin{array}{c}\Delta U=\text{10}\text{moles}\left(6.235\text{kJ}/\text{mol}\right)\\ =62.35\text{kJ}\end{array}$

Substitute $\text{62}\text{.35}\text{kJ}$ for $\Delta U$ in Equation (2).

$\begin{array}{l}\Delta U=Q\\ \text{62}\text{.35}\text{kJ}=Q\end{array}$

The heat added to the gas is $\text{62}\text{.35}\text{kJ}$.

(B)

Interpretation Introduction

Interpretation:

The heat added to the container.

Explanation of Solution

Rewriting Equation (2):

$\begin{array}{c}\Delta U=Q\\ =M{\widehat{C}}_{V}dT\\ =M{\widehat{C}}_V\left(T_2-T_1\right)\end{array}$        (6)

Here, mass is $M$, and constant volume heat capacity on mass basis is ${\widehat{C}}_V$.

Substitute $\text{10}\text{kg}$ for $M$, $\text{1}\text{.5}\text{J}/\text{g}\cdot \text{K}$ for ${\widehat{C}}_V$, $\text{600}\text{K}$ for $T_2$, and $\text{300}\text{K}$ for $T_1$ in Equation (6).

$\begin{array}{c}\Delta U=Q=\left(\text{10}\text{kg}\right)\left(\text{1}\text{.5}\text{J}/\text{g}\cdot \text{K}\right)\left(\text{600}\text{K}-\text{300}\text{K}\right)\\ Q=\left(\text{10}\text{kg}\right)\left[\left(\text{1}\text{.5}\text{J}/\text{g}\cdot \text{K}\right)\left(\frac{\text{1,000}\text{g}}{\text{kg}}\right)\right]\left(\text{300}\text{K}\right)\\ Q=4,500,000\text{J}\left(1\text{kJ}/1,000\text{J}\right)\\ Q=4,500\text{kJ}\end{array}$

The heat added to the container is $\text{4,500}\text{kJ}$.

(C)

Interpretation Introduction

Interpretation:

The change in entropy of the gas.

Concept Introduction:

The residual entropy $\left({\underset{\_}{S}}^R\right)$ is given by:

$\begin{array}{c}{\underset{\_}{S}}^R=\underset{\_}{S}-{\underset{\_}{S}}^{ig}\\ =R\ln Z+{\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{{\left(\delta P/\delta T\right)}_{\underset{\_}{V}}-\left(R/\underset{\_}{V}\right)\right\}}d\underset{\_}{V}\end{array}$

Here, molar volume is $\underset{\_}{V}$, temperature is $T$, change in pressure and temperature at constant molar volume is ${\left(\delta P\right)}_{\underset{\_}{V}},\text{and}{\left(\delta T\right)}_{\underset{\_}{V}}$ respectively, and gas constant is $R$.

Explanation of Solution

The equation for the change in entropy $\left(d\underset{\_}{S}\right)$ is:

$\begin{array}{l}d\underset{\_}{S}=\left({\underset{\_}{S}}_2-{\underset{\_}{S}}_2^{ig}\right)+\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)\\ {\underset{\_}{S}}_2-{\underset{\_}{S}}_1=\left({\underset{\_}{S}}_2-{\underset{\_}{S}}_2^{ig}\right)+\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)\end{array}$

Here, molar entropy at state 2 is ${\underset{\_}{S}}_2$, molar entropy at state 1 is ${\underset{\_}{S}}_1$, and molar entropy for an ideal gas state at state 2 and 1 is ${\underset{\_}{S}}_2^{ig},\text{and}{\underset{\_}{S}}_1^{ig}$ respectively.

The residual entropy $\left({\underset{\_}{S}}^R\right)$ is given by:

${\underset{\_}{S}}^R=R\ln Z+{\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{{\left(\delta P/\delta T\right)}_{\underset{\_}{V}}-\left(R/\underset{\_}{V}\right)\right\}}d\underset{\_}{V}$        (7)

Partially differentiate Van der Waals equation to find ${\left(\delta P/\delta T\right)}_{\underset{\_}{V}}$.

$\begin{array}{l}P=\frac{RT}{\underset{\_}{V}-b}-\frac{a}{{\underset{\_}{V}}^2}\\ {\left(\delta P/\delta T\right)}_{\underset{\_}{V}}=\frac{R}{\underset{\_}{V}-b}\end{array}$

Here, gas constant is $R$, Van der Waals parameter is $a,\text{and}b$.

Substitute $\frac{R}{\underset{\_}{V}-b}$ for ${\left(\delta P/\delta T\right)}_{\underset{\_}{V}}$ in Equation (7).

$\begin{array}{c}\underset{\_}{S}-{\underset{\_}{S}}^{ig}=R\ln Z+{\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{{\left(\delta P/\delta T\right)}_{\underset{\_}{V}}-\left(R/\underset{\_}{V}\right)\right\}}d\underset{\_}{V}\\ =R\ln Z+{\displaystyle \underset{\underset{\_}{V}=\infty ,T=T}{\overset{\underset{\_}{V}=\underset{\_}{V},T=T}{\int }}\left\{\frac{R}{\underset{\_}{V}-b}-\frac{R}{\underset{\_}{V}}\right\}}d\underset{\_}{V}\\ =R\ln Z+{\left[R\ln \left(\underset{\_}{V}-b\right)-R\ln \underset{\_}{V}\right]}_{V=\infty }^{V=V}\\ =R\ln Z+R\ln \left[\frac{\underset{\_}{V}-b}{\underset{\_}{V}}\right]\end{array}$

  $\begin{array}{c}=R\ln \left[Z+\frac{\underset{\_}{V}-b}{\underset{\_}{V}}\right]\\ =R\ln \left[\frac{P\left(\underset{\_}{V}-b\right)}{RT}\right]\end{array}$

Using Van der Waals expression, calculate the initial volume $\left({\underset{\_}{V}}_1\right)$.

$P_1=\frac{R{T}_1}{{\underset{\_}{V}}_1-b}-\frac{a}{{\underset{\_}{V}}_1^2}$

Here, initial pressure is $P_1$.

Substitute $\text{5}\text{bar}$ for $P_1$, $\text{83}\text{.14}\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{K}\cdot \text{mol}$ for $R$, $\text{300}\text{K}$ for $T_1$, $8\times {10}^5{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}$ for $a$, and $\text{100}{\text{cm}}^{\text{3}}/\text{mol}$ for $b$.

$\begin{array}{l}\text{5}\text{bar}=\frac{\left(\text{83}\text{.14}\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{K}\cdot \text{mol}\right)\left(\text{300}\text{K}\right)}{{\underset{\_}{V}}_1-\text{100}{\text{cm}}^{\text{3}}/\text{mol}}-\frac{8\times {10}^5{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}}{{\underset{\_}{V}}_1^2}\\ {\underset{\_}{V}}_1=5,057{\text{cm}}^{\text{3}}/\text{mol}\end{array}$

The number of moles and volume remains constant so this value will also apply at final state. Therefore, the final volume is

$\begin{array}{c}{\underset{\_}{V}}_1={\underset{\_}{V}}_2\\ =5,057{\text{cm}}^{\text{3}}/\text{mol}\end{array}$

The final pressure $\left(P_2\right)$ can be calculated using the expression:

$P_2=\frac{R{T}_2}{{\underset{\_}{V}}_2-b}-\frac{a}{{\underset{\_}{V}}_2^2}$

Substitute $5,057{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}_2$, $\text{83}\text{.14}\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{K}\cdot \text{mol}$ for $R$, $\text{600}\text{K}$ for $T_1$, $8\times {10}^5{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}$ for $a$, and $\text{100}{\text{cm}}^{\text{3}}/\text{mol}$ for $b$.

$\begin{array}{c}{P}_2=\frac{\left(\text{83}\text{.14}\text{bar}\cdot {\text{cm}}^{\text{3}}/\text{K}\cdot \text{mol}\right)\left(\text{600}\text{K}\right)}{5,057{\text{cm}}^{\text{3}}/\text{mol}-\text{100}{\text{cm}}^{\text{3}}/\text{mol}}-\frac{8\times {10}^5{\text{cm}}^{\text{6}}\cdot \text{bar}/{\text{mol}}^{\text{2}}}{{\left(5,057{\text{cm}}^{\text{3}}/\text{mol}\right)}^2}\\ =10.03\text{bar}\end{array}$

Write the expression of the parameter ${\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}$.

${\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}=C_V^{\ast }\ln \left(T_2/T_1\right)+R\ln \left({\underset{\_}{V}}_2/{\underset{\_}{V}}_1\right)$

Here, constant volume heat capacity for ideal gas is $C_V^{\ast }$

Substitute $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$, $\text{600}\text{K}$ for $T_2$, $\text{300}\text{K}$ for $T_1$, $5,057{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}_1,\text{and}{\underset{\_}{V}}_2$, $2.5R$ for $C_V^{\ast }$.

$\begin{array}{c}{\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}=\left(2.5R\right)\ln \left(T_2/T_1\right)+R\ln \left({\underset{\_}{V}}_2/{\underset{\_}{V}}_1\right)\\ {\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}=\left\{\begin{array}{l}\left[\left(2.5\right)\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\right]\left[\ln \left(\text{600}\text{K}/\text{300}\text{K}\right)\right]+\\ \left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left[\ln \left(5,057{\text{cm}}^{\text{3}}/\text{mol}/5,057{\text{cm}}^{\text{3}}/\text{mol}\right)\right]\end{array}\right\}\\ =\left[\left(2.5\right)\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\right]\left[\ln \left(\text{600}\text{K}/\text{300}\text{K}\right)\right]+0\\ =14.40\text{J}/\text{mol}\cdot \text{K}\left(\text{1}\text{kJ}/\text{mol}\cdot \text{K}/1,000\text{J}/\text{mol}\cdot \text{K}\right)\\ =0.0144\text{kJ}/\text{mol}\cdot \text{K}\end{array}$

The change in molar entropy $\left(d{\underset{\_}{S}}_{gas}\right)$ of gas is:

$\begin{array}{c}d{\underset{\_}{S}}_{gas}={\underset{\_}{S}}_2-{\underset{\_}{S}}_1\\ =R\ln \left[\frac{P_2\left({\underset{\_}{V}}_2-b\right)}{R{T}_2}\right]+{\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}-R\ln \left[\frac{P_1\left({\underset{\_}{V}}_1-b\right)}{R{T}_1}\right]\end{array}$        (8)

Substitute $\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}$ for $R$, $\text{600}\text{K}$ for $T_2$, $\text{300}\text{K}$ for $T_1$, $5,057{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}_1,\text{and}{\underset{\_}{V}}_2$ respectively, $\text{5}\text{bar}$ for $P_1$, $\text{100}{\text{cm}}^{\text{3}}/\text{mol}$ for $b$, $\text{10}\text{.03}\text{bar}$ for $P_2$, and $0.0144\text{kJ}/\text{mol}\cdot \text{K}$ for ${\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}$ in Equation (8).

$\begin{array}{c}d{\underset{\_}{S}}_{gas}=\left\{\begin{array}{l}\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\ln \left[\frac{\text{10}\text{.03}\text{bar}\left(5,057{\text{cm}}^{\text{3}}/\text{mol}-\text{100}{\text{cm}}^{\text{3}}/\text{mol}\right)}{\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(\text{600}\text{K}\right)}\right]\\ +\left(0.0144\text{kJ}/\text{mol}\cdot \text{K}\right)-\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\\ \ln \left[\frac{\text{5}\text{bar}\left(5,057{\text{cm}}^{\text{3}}/\text{mol}-\text{100}{\text{cm}}^{\text{3}}/\text{mol}\right)}{\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\left(\text{300}\text{K}\right)}\right]\end{array}\right\}\\ =\left\{\begin{array}{l}\left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\ln \left[\frac{\text{10}\text{.03}\text{bar}\left(4,957{\text{cm}}^{\text{3}}/\text{mol}\right)}{\left(\text{4,988}\text{.4}\text{J}/\text{mol}\right)}\right]+\left(0.0144\text{kJ}/\text{mol}\cdot \text{K}\right)-\\ \left(\text{8}\text{.314}\text{J}/\text{mol}\cdot \text{K}\right)\ln \left[\frac{\text{5}\text{bar}\left(4,957{\text{cm}}^{\text{3}}/\text{mol}\right)}{\left(\text{2,494}\text{.2}\text{J}/\text{mol}\right)}\right]\end{array}\right\}\\ =0.0144\text{kJ}/\text{K}\end{array}$

The change in entropy $\left(\Delta S\right)$ of gas is calculated by using the expression,

$\Delta S_{gas}=N\left(d{\underset{\_}{S}}_{gas}\right)$

Substitute $\text{10}\text{moles}$ for $N$, and $0.0144\text{kJ}/\text{K}$ for $d{\underset{\_}{S}}_{gas}$.

$\begin{array}{c}\Delta S_{gas}=\text{10}\text{moles}\left(0.0144\text{kJ}/\text{K}\right)\\ =0.144\text{kJ}/\text{K}\end{array}$

Hence, the change in entropy $\left(\Delta S_{gas}\right)$ is $0.144\text{kJ}/\text{K}$.

(D)

Interpretation Introduction

Interpretation:

The change in entropy of the universe.

Concept Introduction:

The formula to calculate the change in entropy of the universe $\left(\Delta S_{universe}\right)$ is given by:

$\Delta S_{universe}=\Delta S_{gas}+\Delta S_{container}+\Delta S_{furnace}$

Here, change in entropy of the gas is $\Delta S_{gas}$, change in entropy of the container is $\Delta S_{container}$, and change in entropy of the furnace is $\Delta S_{furnace}$.

Write the expression of the change in entropy of the container $\left(\Delta S_{container}\right)$.

$\Delta S_{container}=\frac{d{Q}_{rev}}{T}$

Here, change in reversible heat is $d{Q}_{rev}$.

Explanation of Solution

The formula to calculate the change in entropy of the universe $\left(\Delta S_{universe}\right)$ is given by:

$\Delta S_{universe}=\Delta S_{gas}+\Delta S_{container}+\Delta S_{furnace}$        (9)

Write the expression of the change in entropy of the container $\left(\Delta S_{container}\right)$.

$\Delta S_{container}=\frac{d{Q}_{rev}}{T}$        (10)

Substitute $M{\widehat{C}}_{V}dT$ for $d{Q}_{rev}$ in Equation (10).

$\Delta S_{container}=\frac{M{\widehat{C}}_{V}dT}{T}$        (11)

Integrating Equation (11), we get,

$\begin{array}{c}\Delta S_{container}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{M{\widehat{C}}_V}{T}}dT\\ =M{\widehat{C}}_V\ln \left(T_2/T_1\right)\end{array}$        (12)

Here, mass is $M$.

Substitute $\text{10}\text{kg}$ for $M$, $\text{1}\text{.5}\text{J}/\text{g}\cdot \text{K}$ for ${\widehat{C}}_V$, $\text{600}\text{K}$ for $T_2$, and $\text{300}\text{K}$ for $T_1$ in Equation (13).

$\begin{array}{c}\Delta S_{container}=\left(\text{10}\text{kg}\right)\left(\text{1}\text{.5}\text{J}/\text{g}\cdot \text{K}\right)\left[\ln \left(\text{600}\text{K}/\text{300}\text{K}\right)\right]\\ =\left(\text{10}\text{kg}\right)\left\{\left(\text{1}\text{.5}\text{J}/\text{g}\cdot \text{K}\right)\left(\text{1,000}\text{g}/\text{1}\text{kg}\right)\left(\text{1}\text{kJ}/\text{1,000}\text{J}\right)\right\}\left[\ln \left(\text{600}\text{K}/\text{300}\text{K}\right)\right]\\ =10.39\text{kJ}/\text{K}\end{array}$

Assuming the furnace as a heat reservoir, calculate the change in entropy of the furnace $\left(\Delta S_{furnace}\right)$

$\Delta S_{furnace}=\frac{Q_{furnace}}{T}$        (13)

Here, heat added or removed from furnace is $Q_{furnace}$.

The heat added or removed from furnace $\left(Q_{furnace}\right)$ is given by the expression:

$Q_{furnace}=-\left(Q_{gas}+Q_{container}\right)$        (14)

Here, heat added or removed from gas and container is $Q_{gas},\text{and}{Q}_{container}$ respectively.

Substitute $\text{62}\text{.35}\text{kJ}$ for $Q_{gas}$, and $\text{4,500}\text{kJ}$ for $Q_{container}$ in Equation (14).

$\begin{array}{c}{Q}_{furnace}=-\left(\text{62}\text{.35}\text{kJ}+\text{4,500}\text{kJ}\right)\\ =-\text{4,562}\text{.35}\text{kJ}\end{array}$

Substitute $-\text{4,562}\text{.35}\text{kJ}$ for $Q_{furnace}$, and $\text{600}\text{K}$ for $T$ in Equation (13).

$\begin{array}{c}\Delta S_{furnace}=\frac{Q_{furnace}}{T}\\ =\frac{-\text{4,562}\text{.35}\text{kJ}}{\text{600}\text{K}}\\ =-7.6039\text{kJ}/\text{K}\end{array}$

Substitute $0.144\text{kJ}/\text{K}$ for $\Delta {\underset{\_}{S}}_{gas}$, $10.39\text{kJ}/\text{K}$ for $\Delta S_{container}$, and $-7.6039\text{kJ}/\text{K}$ for $\Delta S_{furnace}$ in Equation (9).

$\begin{array}{c}\Delta S_{universe}=\Delta S_{gas}+\Delta S_{container}+\Delta S_{furnace}\\ =0.144\text{kJ}/\text{K}+10.39\text{kJ}/\text{K}+\left(-7.6039\text{kJ}/\text{K}\right)\\ =2.930\text{kJ}/\text{K}\end{array}$

The change in entropy of the universe $\left(\Delta S_{universe}\right)$ is $2.930\text{kJ}/\text{K}$.

(E)

Interpretation Introduction

Interpretation:

What aspect of this process is irreversible?

Explanation of Solution

The large variation between the initial and final temperature is the irreversible aspect of this process.