Chapter 10, Problem 10.34P

Interpretation Introduction

Interpretation:

Estimate V, H^R, G^R and S^R for a binary vapor mixture of acetone and 1,3 butadiene.

Concept introduction:

Volume can be calculated with the help of compressibility factor (Z) which is defined as the ratio of real gas volume to ideal gas volume as described in equation 3.36 in book:

  $\begin{array}{l}Z=\frac{PV}{RT}=1+\frac{BP}{RT}\\ V=\frac{RTZ}{P}\end{array}$

Residual properties in thermodynamics is the difference between real gas property and ideal gas property at the same temperature, pressure and composition.

Equation 6.54, 6.55, 6.56 and 10.62 mentioned in question are:

  $\begin{array}{l}\frac{H^R}{RT}=\frac{P}{R}\left(\frac{B}{T}-\frac{dB}{dT}\right)\\ \frac{S^R}{R}=-\frac{P}{R}\frac{dB}{dT}\\ G^R=BP\\ B={\displaystyle \sum _i{\displaystyle \sum _j{y}_1{y}_j{B}_{i,j}}}\end{array}$

Where,

  $\begin{array}{l}{B}_{i,j}=\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[B_{i,j}^0+{\omega }_{i,j}{B}_{i,j}^1\right]\\ B_{i,j}^0=0.083-\frac{0.422}{{\left(T_{r_{i,j}}\right)}^{1.6}}\\ B_{i,j}^1=0.139-\frac{0.172}{{\left(T_{r_{i,j}}\right)}^{4.2}}\\ {\omega }_{i,j}=\frac{{\omega }_i+{\omega }_j}{2}\\ T_{c_{i,j}}=\sqrt{T_{c_i}.T_{c_j}}\\ Z_{c_{i,j}}=\frac{Z_{c_i}+Z_{c_j}}{2}\\ V_{C_{i,j}}={\left[\frac{{\left(V_{c_i}\right)}^{1/3}+{\left(V_{c_j}\right)}^{1/3}}{2}\right]}^3\\ P_{c_{i,j}}=\frac{Z_{c_{i,j}}R{T}_{c_{i,j}}}{V_{C_{i,j}}}\end{array}$

$P_{c_{i,j}}$ = Critical pressure of gases

$V_{c_{i,j}}$ = Volume of gases

T = temperature

$B^0$ and $B^1$ are the correlation coefficient given in the book.

  $P_r$ = Reduced pressure

  $P_r=\frac{P_i^{sat}}{P_c}$

Where,

$P_c$ = Critical pressure

Similarly, $T_r$ = reduced temperature = $T_{r_{i,j}}=\frac{T}{T_{C_{i,j}}}$

Answer to Problem 10.34P

The volume of vapor mixture of acetone and 1,3 butadiene is 15694 cm³/mol.

Reduced enthalpy, Gibbs free energy and entropy of acetone and 1,3 butadiene mixture is -344. J/mol, -101.7 J/mol and -0.0727 J/mol.K respectively.

Explanation of Solution

Composition for binary mixture of gases (Given):

Y₁ = 0.28 and Y₂ = 0.72

Temperature (given) = 60⁰C or 333.15K (60+273.15)

Pressure (given) = 170 kpa

Refer APPENDIX-B and Table-B.1 to determine critical properties and acentric factor of acetone(1)/1,3 butadiene (2) as:

Component	Tc (K)	Vc (cm3/mol)	$\omega$	Zc
Acetone (1)	508.2	209	0.307	0.233
1,3 Butadiene (2)	425.2	220.4	0.190	0.267

Where,

P_c = critical pressure

T_c = critical temperature

V_c = critical volume

Z_c = critical compressibility factor

  $\omega$ = eccentric factor

Reduced temperature for both gases can be calculated as:

  $\begin{array}{l}{T}_{c_{i,j}}=\sqrt{T_{c_i}.T_{c_j}}\\ T_{c_{1,1}}=\sqrt{T_{c_1}.T_{c_1}}=T=508.2K\\ T_{c_{1,2}}=\sqrt{T_{c_1}.T_{c_2}}\\ T_{c_{1,2}}=\sqrt{508.2\times 425.2}\\ T_{c_{1,2}}=464.85K=T_{c_{2,1}}\\ T_{c_{2,2}}=\sqrt{T_{c_2}.T_{c_2}}=T_{c_2}=425.2\\ T_{c_{i,j}}=\left(\begin{array}{cc}508.2& 464.85\\ 464.85& 425.2\end{array}\right)\end{array}$

Similar to the above calculated matrix we will calculate the reduced temperature as follows:

  $\begin{array}{l}{T}_{r_{i,j}}=\left(\begin{array}{cc}\frac{333.15}{508.2}& \frac{333.15}{464.85}\\ \frac{333.15}{464.85}& \frac{333.15}{425.2}\end{array}\right)\\ T_{r_{i,j}}=\left(\begin{array}{cc}0.656& 0.717\\ 0.717& 0.784\end{array}\right)\end{array}$

Critical volume of both the gases in the mixture can be calculated as given below:

With the help of critical volume and critical temperature, critical pressure of gases in the mixture can be calculated as follows:

  $P_{c_{i,j}}=\frac{Z_{c_{i,j}}R{T}_{c_{i,j}}}{V_{C_{i,j}}}$

The reduced compressibility factor can be calculated as:

  $\begin{array}{l}{Z}_{c_{i,j}}=\frac{Z_{c_i}+Z_{c_j}}{2}\\ Z_{c_{i,j}}=\left(\begin{array}{cc}\frac{Z_{c_1}+Z_{c_1}}{2}& \frac{Z_{c_1}+Z_{c_2}}{2}\\ \frac{Z_{c_2}+Z_{c_1}}{2}& \frac{Z_{c_2}+Z_{c_2}}{2}\end{array}\right)\\ Z_{c_{i,j}}=\left(\begin{array}{cc}\frac{0.233+0.233}{2}& \frac{0.233+0.267}{2}\\ \frac{0.267+0.233}{2}& \frac{0.267+0.267}{2}\end{array}\right)\\ Z_{c_{i,j}}=\left(\begin{array}{cc}0.233& 0.25\\ 0.267& 0.267\end{array}\right)\end{array}$

After putting all values in the critical pressure formula, we will get:

  $\begin{array}{l}{P}_{c_{i,j}}=\frac{Z_{c_{i,j}}R{T}_{c_{i,j}}}{V_{C_{i,j}}}\\ P_{c_{i,j}}=\left(\begin{array}{cc}47.104& 45.013\\ 45.013& 42.826\end{array}\right)\end{array}$

At this calculated value of reduced temperature, we can calculate correlation constant as:

  $\begin{array}{l}{B}_{i,j}^0=0.083-\frac{0.422}{{\left(T_{r_{i,j}}\right)}^{1.6}}\\ B_{i,j}^0=\left(\begin{array}{cc}-0.7463& -0.6361\\ -0.6361& -0.5405\end{array}\right)\\ B_{i,j}^1=0.139-\frac{0.172}{{\left(T_{r_{i,j}}\right)}^{4.2}}\\ B_{i,j}^1=\left(\begin{array}{cc}-0.874& -0.558\\ -0.558& -0.34\end{array}\right)\end{array}$

Overall correlation constant value in gas mixture can be calculated by equation:

  $B_{i,j}=\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[B_{i,j}^0+{\omega }_{i,j}{B}_{i,j}^1\right]$

  $\begin{array}{l}{\omega }_{i,j}=\frac{{\omega }_i+{\omega }_j}{2}\\ {\omega }_{i,j}=\left(\begin{array}{cc}0.307& 0.2485\\ 0.2485& 0.19\end{array}\right)\end{array}$

  $\begin{array}{l}{B}_{i,j}=\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[B_{i,j}^0+{\omega }_{i,j}{B}_{i,j}^1\right]\\ B_{i,j}=\left(\begin{array}{cc}-910.278& -665.188\\ -665.188& -499.527\end{array}\right)\frac{c{m}^3}{mol}\end{array}$

  $\begin{array}{l}B={\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j{B}_{i,j}}}\\ B=y_1^2{B}_{1,1}+2y_1{y}_2{B}_{1,2}+y_2^2{B}_{2,2}\\ B=\left({0.28}^2\times \left(-910.278\right)\right)+\left(2\times 0.28\times 0.72\times \left(-665.188\right)\right)+\left({0.72}^2\times \left(-499.527\right)\right)\\ B=-598.524{\text{cm}}^{\text{3}}\text{/mol}\end{array}$

Using above calculated value compressibility factor and molar volume of gas can be calculated as given below:

  $\begin{array}{l}Z=\frac{PV}{RT}=1+\frac{BP}{RT}\\ Z=1+\frac{\left(-598.524\times 170\right)}{8314\times 333.15}\\ Z=0.9632\\ V=\frac{RTZ}{P}\\ V=\frac{8314\times 333.15\times 0.9632}{170}\\ V=15694.47{\text{cm}}^{\text{3}}\text{/mol}\end{array}$

Reduced property can be calculated by finding the value of dB/dT and putting in equations given in book as follows:

  $\begin{array}{l}\frac{d{B}_{i,j}}{dT}={\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j}}\left[\frac{R}{P_{c_{i,j}}}\left[\frac{d{B}_{i,j}^0}{dT}+{\omega }_{i,j}\frac{d{B}_{i,j}^1}{dT}\right]\right]\\ B_{i,j}^0=0.083-\frac{0.422}{{\left(T_{r_{i,j}}\right)}^{1.6}}\\ \frac{d{B}_{i,j}^0}{dT}=\frac{0.675}{{\left(T_{r_{i,j}}\right)}^{2.6}}\\ \frac{d{B}_{i,j}^0}{dT}=\left(\begin{array}{cc}2.02& 1.6\\ 1.6& 1.271\end{array}\right)\\ B_{i,j}^1=0.139-\frac{0.172}{{\left(T_{r_{i,j}}\right)}^{4.2}}\\ \frac{d{B}_{i,j}^1}{dT}=\frac{0.722}{{\left(T_{r_{i,j}}\right)}^{5.2}}\\ \frac{d{B}_{i,j}^1}{dT}=\left(\begin{array}{cc}6.466& 2.072\\ 2.072& 2.559\end{array}\right)\end{array}$

  $\begin{array}{l}\frac{d{B}_{i,j}}{dT}={\displaystyle \sum _i{\displaystyle \sum _j{y}_i{y}_j}}\left[\frac{R{T}_{c_{i,j}}}{P_{c_{i,j}}}\left[\frac{d{B}_{i,j}^0}{dT}+{\omega }_{i,j}\frac{d{B}_{i,j}^1}{dT}\right]\right]\\ \frac{d{B}_{i,j}}{dT}=y_1^2\left[\frac{R}{P_{c_{1,1}}}\left[\frac{d{B}_{1,1}^0}{dT}+{\omega }_{1,1}\frac{d{B}_{1,1}^1}{dT}\right]\right]+2y_1{y}_2\left[\frac{R}{P_{c_{1,2}}}\left[\frac{d{B}_{1,2}^0}{dT}+{\omega }_{1,2}\frac{d{B}_{1,2}^1}{dT}\right]\right]+y_2^2\left[\frac{R}{P_{c_{2,2}}}\left[\frac{d{B}_{2,2}^0}{dT}+{\omega }_{2,2}\frac{d{B}_{2,2}^1}{dT}\right]\right]\\ \frac{d{B}_{i,j}}{dT}=4.2764\end{array}$ Residual properties of acetone/1,3 butadiene binary mixture from equations mentioned in question can be estimated as given below:

  $\begin{array}{l}\frac{H^R}{RT}=\frac{P}{R}\left(\frac{B}{T}-\frac{dB}{dT}\right)\\ H^R=PT\left(\frac{B}{T}-\frac{dB}{dT}\right)\\ H^R=0.170\times 333.15\times \left(\frac{-598.524}{333.15}-4.2764\right)\\ H^R=-322.49\text{J/mol}\\ \frac{S^R}{R}=-\frac{P}{R}\frac{dB}{dT}\\ S^R=-P\frac{dB}{dT}\\ S^R=-\text{0}\text{.170 bar×4}{\text{.2764 cm}}^{\text{3}}\text{/molK×}\frac{\text{1 J}}{\text{10 bar}{\text{.cm}}^{\text{3}}}\\ S^R=-0.0727\text{J/mol}\text{.K}\\ G^R=BP\\ G^R=-598.524\times 0.17\\ G^R=-101.75\text{J/mol}\end{array}$

Conclusion

The volume of vapor mixture of acetone and 1,3 butadiene is 15694 cm³/mol.

Reduced enthalpy, Gibbs free energy and entropy of acetone and 1,3 butadiene mixture is -344. J/mol, -101.7 J/mol and -0.0727 J/mol.K respectively.