Chapter 9.8, Problem 28P

Interpretation Introduction

Interpretation:

To calculate the change in enthalpy and entropy of the given system assuming ideal gas model.

Concept introduction:

The enthalpy during mixing remains constant and considering the gas as an ideal gas, the pressure change also has no effect on the enthalpy. It may seem like the change in enthalpy is zero, but since the temperature changes and enthalpy being a function of temperature also changes.

Similarly entropy during the mixing process remains constant and is also a function of temperature as well as pressure. Hence to know the entropy change, both the final pressure and temperature after mixing is required.

Explanation of Solution

The energy balance equation for the gas system on both sides of the partition is:

$U_{final}-U_{initial}=0$

Here, initial and final internal energy is $U_{initial}$ and $U_{final}$ respectively.

Being a closed adiabatic system with rigid boundaries, the internal energy of blending is 0, so:

${\underset{\_}{U}}_{final}=y_{Ar}{\underset{\_}{U}}_{Ar,final}+y_{N_2}{\underset{\_}{U}}_{N_2,final}$        (1)

Here, vapor phase mole fraction of argon and nitrogen is $y_{Ar}\text{and}{y}_{N_2}$, molar internal energy of argon and nitrogen is ${\underset{\_}{U}}_{Ar,final}$ and ${\underset{\_}{U}}_{N_2,final}$ respectively.

The internal energy of argon and nitrogen which is not known at the moment must be calculated at the last temperature. Adding this into the energy balance:

$N{\underset{\_}{U}}_{final}-N_{Ar}{\underset{\_}{U}}_{Ar,initial}+N_{N_2}{\underset{\_}{U}}_{N_2,initial}=0$

Here, total number of moles is N.

Substitute the value of ${\underset{\_}{U}}_{final}$ from Equation (1).

$\begin{array}{l}N\left(y_{Ar}{\underset{\_}{U}}_{Ar,final}+y_{N_2}{\underset{\_}{U}}_{N_2,final}\right)-N_{Ar}{\underset{\_}{U}}_{Ar,initial}+N_{N_2}{\underset{\_}{U}}_{N_2,initial}=0\\ y_{Ar}{\underset{\_}{U}}_{Ar,final}+y_{N_2}{\underset{\_}{U}}_{N_2,final}-y_{Ar}{\underset{\_}{U}}_{Ar,initial}-y_{N_2}{\underset{\_}{U}}_{N_2,initial}=0\\ y_{Ar}\left({\underset{\_}{U}}_{Ar,final}-{\underset{\_}{U}}_{Ar,initial}\right)+y_{N_2}\left({\underset{\_}{U}}_{N_2,final}-{\underset{\_}{U}}_{N_2,initial}\right)=0\end{array}$        (2)

There are 2.5 moles of argon and 112 g of nitrogen.

$\left(112g\right)\left(\frac{\text{1}\text{mol}}{\text{28}\text{g}}\right)=4\text{mol}{\text{N}}_{\text{2}}$

The value of $y_{N_2}$ can be calculated as:

$y_{N_2}=\frac{n_{N_2}}{n_{N_2}+n_{Ar}}$

Here, number of moles in nitrogen and argon is $n_{N_2}\text{and}{n}_{Ar}$ respectively.

Substitute 4 mol for $n_{N_2}$ and 2.5 mol for $n_{Ar}$.

$\begin{array}{c}{y}_{N_2}=\frac{4\text{mol}}{4+2.5\text{mol}}\\ =0.615\end{array}$

The value of $y_{Ar}$ is calculated as:

$y_{Ar}=1-y_{N_2}$

Substitute 0.615 for $y_{N_2}$.

$\begin{array}{c}{y}_{Ar}=1-0.615\\ =0.385\end{array}$

On modelling argon a monoatomic gas into an ideal gas, it has $C_V=1.5R$.

${\underset{\_}{U}}_{Ar,final}-{\underset{\_}{U}}_{Ar,initial}=C_V\left(T_{final}-T_{initial}\right)$

Here, heat capacity at constant volume is $C_V$, initial and final temperature is $T_{initial}\text{and}{T}_{final}$ respectively.

Substitute $130°\text{C}$ for $T_{initial}$ and 1.5 R for $C_V$.

${\underset{\_}{U}}_{Ar,final}-{\underset{\_}{U}}_{Ar,initial}=1.5R\left(T_{final}-130°\text{C}\right)$

Referring to Appendix D, “ideal gas heat capacity”, the value of ideal gas heat capacity at constant pressure is obtained as:

$\begin{array}{l}\frac{C_P^{*}}{R}=A+BT+C{T}^2+D{T}^3+E{T}^4\\ C_P^{*}=R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]\end{array}$

Here, constants are A, B, C, D, and E, and gas constant is R.

The ideal gas heat capacity at constant volume is:

$C_V^{*}=C_P^{*}-R$        (3)

Substitute $R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]$ for $C_P^{*}$ in Equation (3).

$\begin{array}{c}{C}_V^{*}=R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]-R\\ =R\left[A+BT+C{T}^2+D{T}^3+E{T}^4-1\right]\end{array}$

The difference of molar internal energy of both gas is written as:

${\underset{\_}{U}}_{N_2,final}-{\underset{\_}{U}}_{N_2,initial}={\displaystyle \underset{T=T_{initial}}{\overset{T=T_{final}}{\int }}{C}_V^{*}}dT$

Substitute $R\left[A+BT+C{T}^2+D{T}^3+E{T}^4-1\right]$ for $C_V^{*}$.

$\begin{array}{c}{\underset{\_}{U}}_{N_2,final}-{\underset{\_}{U}}_{N_2,initial}={\displaystyle \underset{T=T_{initial}}{\overset{T=T_{final}}{\int }}R\left[A+BT+C{T}^2+D{T}^3+E{T}^4-1\right]}dT\\ =R{\left[\left(A-1\right)T+B\frac{T^2}{2}+C\frac{T^3}{3}+D\frac{T^4}{4}+E\frac{T^5}{5}\right]}_{T=75°\text{C}}^{T=T_{final}}\end{array}$

From Equation (2),

$y_{Ar}\left({\underset{\_}{U}}_{Ar,final}-{\underset{\_}{U}}_{Ar,initial}\right)+y_{N_2}\left({\underset{\_}{U}}_{N_2,final}-{\underset{\_}{U}}_{N_2,initial}\right)=0$

Substitute 0.385 for $y_{Ar}$, 0.615 for $y_{N_2}$, $R{\left[\left(A-1\right)T+B\frac{T^2}{2}+C\frac{T^3}{3}+D\frac{T^4}{4}+E\frac{T^5}{5}\right]}_{T=75°\text{C}}^{T=T_{final}}$ for ${\underset{\_}{U}}_{N_2,final}-{\underset{\_}{U}}_{N_2,initial}$ and $1.5R\left(T_{final}-130°\text{C}\right)$ for ${\underset{\_}{U}}_{Ar,final}-{\underset{\_}{U}}_{Ar,initial}$.

$0.385\left(1.5R\left(T_{final}-130°\text{C}\right)\right)+0.615\left(R{\left[\begin{array}{l}\left(A-1\right)T+B\frac{T^2}{2}+\\ C\frac{T^3}{3}+D\frac{T^4}{4}+E\frac{T^5}{5}\end{array}\right]}_{T=75°\text{C}}^{T=T_{final}}\right)=0$        (4)

Referring to Appendix D.1, “Ideal gas heat capacity”, the parameters A, B, C, D, and E for nitrogen are obtained and are given in Table (1).

Compound	A	$B\times {10}^3$	$C\times {10}^4$	$D\times {10}^8$	$E\times {10}^{11}$
Nitrogen	3.539	–0.261	0.007	0.157	–0.099

Substituting $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R and Table (1) values in Equation (4), we get

$T_{final}$ as $90°\text{C}$.

At this stage, we are able to find the enthalpy change in argon and nitrogen. Since there is no enthalpy change during mixing, the total of these two changes in enthalpy gives the enthalpy of the system.

$\begin{array}{c}\Delta H_{Ar}=N_{Ar}{\displaystyle \underset{T=T_{initial}}{\overset{T=T_{final}}{\int }}{C}_P^{*}dT}\\ =N_{Ar}{C}_P^{*}\left[T_{final}-T_{initial}\right]\end{array}$        (5)

Here, change in enthalpy of argon is $\Delta H_{Ar}$.

Substitute 2.5 mol for $N_{Ar}$, $\frac{5}{2}R$ for $C_P^{*}$, $90°\text{C}$ for $T_{final}$, and $130°\text{C}$ for $T_{initial}$ in Equation (5).

$\Delta H_{Ar}=\left(\text{2}\text{.5}\text{mol}\right)\left(\frac{5}{2}R\right)\left[90°\text{C}-130°\text{C}\right]$

Substitute $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R.

$\begin{array}{c}\Delta H_{Ar}=\left(\text{2}\text{.5}\text{mol}\right)\left(\frac{5}{2}\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)\right)\left[90°\text{C}-130°\text{C}\right]\\ =\left(\text{2}\text{.5}\text{mol}\right)\left(\frac{5}{2}\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)\right)\left[\left(90+273\right)\text{K}-\left(130+273\right)\text{K}\right]\\ =-2079\text{J}\end{array}$

The change in enthalpy of nitrogen is calculated as:

$\Delta H_{N_2}=N_{N_2}{C}_P^{*}\left[T_{final}-T_{initial}\right]$

Substitute 4 mol for $N_{N_2}$, $R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]$ for $C_P^{*}$, $90°\text{C}$ for $T_{final}$, and $75°\text{C}$ for $T_{initial}$.

$\Delta H_{N_2}=\left(4\text{mol}\right)\left(R\left[A+BT+C{T}^2+D{T}^3+E{T}^4\right]\right)\left[\left(90+273\right)\text{K}-\left(75+273\right)\text{K}\right]$

Substituting $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R and the values from Table (1), we get,

$\Delta H_{N_2}=1747\text{J}$

The total change in enthalpy of system is:

$\Delta H_{sys}=\Delta H_{N_2}+\Delta H_{Ar}$

Substitute 1747 J for $\Delta H_{N_2}$ and ­2079 J for $\Delta H_{Ar}$.

$\begin{array}{c}\Delta H_{sys}=1747\text{J}-2079\text{J}\\ \Delta H_{sys}=-332\text{J}\end{array}$

The change in enthalpy of the system is $-332\text{J}$.

This is a minimal change in enthalpy but it is not zero. From this it can be seen that even if $\Delta U=0$, it does not necessarily mean that $\Delta H=0$.

Use ideal gas equation to find the final pressure.

$P=\frac{NRT}{V}$        (6)

Here, final temperature and volume is T and V respectively.

On removing the partition, the volume of the system does not change and the total volume is the sum of the individual volume of argon and nitrogen.

$V=\frac{N_{1}R{T}_1}{P_1}+\frac{N_{2}R{T}_2}{P_2}$        (7)

Here, initial pressure and temperature of argon and nitrogen is $P_1$, $P_2$, $T_1,\text{and}{T}_2$ respectively.

Substitute 2.5 mol for $N_1$, $83.14\frac{\text{bar}\cdot {\text{cm}}^{\text{3}}}{\text{mol}\cdot \text{K}}$ for R, 403.15 K for $T_1$, 20 bar for $P_1$, 4 mol for $N_2$, 348.15 for $T_2$, and 30 bar for $P_2$ in Equation (7).

$\begin{array}{c}V=\frac{\text{2}\text{.5}\text{mol}\left(83.14\frac{\text{bar}\cdot {\text{cm}}^{\text{3}}}{\text{mol}\cdot \text{K}}\right)403.15\text{K}}{20\text{bar}}+\frac{4\text{mol}\left(83.14\frac{\text{bar}\cdot {\text{cm}}^{\text{3}}}{\text{mol}\cdot \text{K}}\right)348.15\text{K}}{30\text{bar}}\\ =8049{\text{cm}}^{\text{3}}\end{array}$

Substitute $8049{\text{cm}}^{\text{3}}$ for V, 6.5 mol for N, 363.15 K for T, and $83.14\frac{\text{bar}\cdot {\text{cm}}^{\text{3}}}{\text{mol}\cdot \text{K}}$ for R in equation (6).

$\begin{array}{c}P=\frac{\left(\text{6}\text{.5}\text{mol}\right)\left(83.14\frac{\text{bar}\cdot {\text{cm}}^{\text{3}}}{\text{mol}\cdot \text{K}}\right)363.15\text{K}}{8049{\text{cm}}^{\text{3}}}\\ =24.38\text{bar}\end{array}$

The change in entropy is calculated as:

$\begin{array}{c}\Delta S=S_{final}-S_{initial}\\ =N{\underset{\_}{S}}_{final}-N_{Ar}{\underset{\_}{S}}_{Ar,initial}-N_{N_2}{\underset{\_}{S}}_{N_2,initial}\\ =N\left({\underset{\_}{S}}_{final}-y_{Ar}{\underset{\_}{S}}_{Ar,initial}-y_{N_2}{\underset{\_}{S}}_{N_2,initial}\right)\end{array}$        (8)

Here, molar entropy of argon and nitrogen is ${\underset{\_}{S}}_{Ar,initial}$ and ${\underset{\_}{S}}_{N_2,initial}$ respectively.

Substitute 6.5 moles for N in Equation (8).

$\Delta S=6.5\text{moles}\left({\underset{\_}{S}}_{final}-y_{Ar}{\underset{\_}{S}}_{Ar,initial}-y_{N_2}{\underset{\_}{S}}_{N_2,initial}\right)$        (9)

The final molar entropy for an ideal gas mixture is calculated as:

${\underset{\_}{S}}_{final}=y_{Ar}{\underset{\_}{S}}_{Ar,final}+y_{N_2}{\underset{\_}{S}}_{N_2,final}-R{y}_1\ln y_1-R{y}_2\ln y_2$

Substitute the value of ${\underset{\_}{S}}_{final}$ in Equation (9).

$\begin{array}{c}\Delta S=6.5\text{moles}\left(\begin{array}{l}{y}_{Ar}{\underset{\_}{S}}_{Ar,final}+y_{N_2}{\underset{\_}{S}}_{N_2,final}-R{y}_1\ln y_1-R{y}_2\ln y_2-y_{Ar}{\underset{\_}{S}}_{Ar,initial}\\ -y_{N_2}{\underset{\_}{S}}_{N_2,initial}\end{array}\right)\\ =6.5\text{moles}\left[\begin{array}{l}{y}_{Ar}\left({\underset{\_}{S}}_{Ar,final}-{\underset{\_}{S}}_{Ar,initial}\right)+y_{N_2}\left({\underset{\_}{S}}_{N_2,final}-{\underset{\_}{S}}_{N_2,initial}\right)-\\ R{y}_1\ln y_1-R{y}_2\ln y_2\end{array}\right]\end{array}$        (10)

This accounts for two thing namely the change in entropy of argon and nitrogen due to pressure and temperature and the entropy of mixing the aspect of mixing in equation (10) is:

$-R{y}_1\ln y_1-R{y}_2\ln y_2$

Substitute $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R, 0.615 for $y_1$, and 0.385 for $y_2$.

$\begin{array}{c}-R{y}_1\ln y_1-R{y}_2\ln y_2=\left[\begin{array}{l}-\left(8.314\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K}\cdot \text{mol}$}\right.\right)0.615\ln 0.615-\\ \left(8.314\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{K}\cdot \text{mol}$}\right.\right)0.385\ln 0.385\end{array}\right]\\ =5.54\frac{\text{J}}{\text{K}\cdot \text{mol}}\end{array}$

The change in entropy for a pure ideal gas is expressed as:

$d\underset{\_}{S}=\frac{C_V^{*}}{T}dT+\frac{R}{\underset{\_}{V}}d\underset{\_}{V}$        (11)

Here, change in molar volume is $d\underset{\_}{V}$, change in temperature is dT, temperature is T, and ideal gas heat capacity at constant volume is $C_V^{*}$.

The $C_V^{*}$ being constant for argon, integrate the difference between the molar entropy ${\underset{\_}{S}}_{Ar,final}-{\underset{\_}{S}}_{Ar,initial}=C_P^{*}\ln \left(\frac{T_{final}}{T_{initial}}\right)+R\ln \left(\frac{P_{initial}}{P_{final}}\right)$

Substitute $\frac{5}{2}R$ for $C_P^{*}$, 363.15 K for $T_{final}$, 403.15 K for $T_{initial}$, 20 bar for $P_{initial}$, and 24.38 bar for $P_{final}$.

${\underset{\_}{S}}_{Ar,final}-{\underset{\_}{S}}_{Ar,initial}=\frac{5}{2}R\ln \left(\frac{363.15\text{K}}{403.15\text{K}}\right)+R\ln \left(\frac{20\text{bar}}{24.38\text{bar}}\right)$        (12)

Substitute $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R in Equation (12),

$\begin{array}{c}{\underset{\_}{S}}_{Ar,final}-{\underset{\_}{S}}_{Ar,initial}=\frac{5}{2}\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)\ln \left(\frac{363.15\text{K}}{403.15\text{K}}\right)+\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)\ln \left(\frac{20\text{bar}}{24.38\text{bar}}\right)\\ =-3.82\frac{\text{J}}{\text{K}\cdot \text{mol}}\end{array}$

For nitrogen $C_V^{*}$ is not constant, so integrate the difference between the molar entropy for nitrogen from Equation (11).

Substitute $R\left[A+BT+C{T}^2+D{T}^3+E{T}^4-1\right]$ for $C_V^{*}$ and integrate the difference between the molar entropy for nitrogen.

$\begin{array}{c}{\underset{\_}{S}}_{N_2,final}-{\underset{\_}{S}}_{N_2,initial}=R\left[A+BT+C{T}^2+D{T}^3+E{T}^4-1\right]\ln \left(\frac{T_{final}}{T_{initial}}\right)+R\ln \left(\frac{P_{initial}}{P_{final}}\right)\\ ={\displaystyle \underset{T_{initial}}{\overset{T_{final}}{\int }}\frac{R\left[A+BT+C{T}^2+D{T}^3+E{T}^4-1\right]dT}{T}}+{\displaystyle \underset{V_{initial}}{\overset{V_{final}}{\int }}\frac{R}{\underset{\_}{V}}}d\underset{\_}{V}\\ =\left\{\begin{array}{c}R{\left[\left(A-1\right)\ln T+BT+C\frac{T^2}{2}+D\frac{T^3}{3}+E\frac{T^4}{4}\right]}_{T=348.15\text{K}}^{T=363.15\text{K}}+\\ R\ln \left(\frac{\frac{363.15\text{K}}{24.38\text{bar}}}{\frac{348.15\text{K}}{30\text{bar}}}\right)\end{array}\right\}\\ =2.95\frac{\text{J}}{\text{mol}\cdot \text{K}}\end{array}$

From Equation (10),

$\Delta S=6.5\text{moles}\left[y_{Ar}\left({\underset{\_}{S}}_{Ar,final}-{\underset{\_}{S}}_{Ar,initial}\right)+y_{N_2}\left({\underset{\_}{S}}_{N_2,final}-{\underset{\_}{S}}_{N_2,initial}\right)-R{y}_1\ln y_1-R{y}_2\ln y_2\right]$

Substitute $2.95\frac{\text{J}}{\text{mol}\cdot \text{K}}$ for ${\underset{\_}{S}}_{N_2,final}-{\underset{\_}{S}}_{N_2,initial}$, 0.385 for $y_{Ar}$, 0.615 for $y_{N_2}$, $-3.82\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for ${\underset{\_}{S}}_{Ar,final}-{\underset{\_}{S}}_{Ar,initial}$, $5.54\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for $-R{y}_1\ln y_1-R{y}_2\ln y_2$.

$\begin{array}{c}\Delta S=6.5\text{moles}\left[0.385\left(-3.82\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)+0.615\left(2.95\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)+5.54\frac{\text{J}}{\text{K}\cdot \text{mol}}\right]\\ =38.2\frac{\text{J}}{\text{K}}\end{array}$

Hence, the change in entropy is $38.2\frac{\text{J}}{\text{K}}$.