Chapter 6, Problem 6.47P

Interpretation Introduction

Interpretation:

Determine the molar volume of the propane in the final state and enthalpy and entropy changes if propane gas is compressed from initial state to final.

Concept Introduction:

From Lee and Kesler generalized correlations, the molar volume of the propane in the final state, enthalpy and entropy changes are calculated as:

  $V=\frac{ZRT}{P}$

Where, $Z=Z^0+\omega Z^1$

                      ....(1)

And enthalpy and entropy changes are calculated as:

  $\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT+}{H}_2{}^R-H_1{}^R$

Where, $H^R={\left(H^R\right)}^0+\omega {\left(H^R\right)}^1$

                      ....(2)

And

  $\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R$

Where, $S^R={\left(S^R\right)}^0+\omega {\left(S^R\right)}^1$

                      ....(3)

Answer to Problem 6.47P

The molar volume is:

  $V=184.2\frac{{\text{cm}}^3}{\text{mol}}$

And

  $\Delta H=6723.42\frac{\text{J}}{\text{mol}}$

  $\Delta S=-15.905\frac{\text{J}}{\text{mol K}}$

Explanation of Solution

Given information:

It is given that in the initial state, propane is assumed as an ideal gas. And at initial state

And final state pressure and temperature are given as, $P_1=1\text{bar, }{T}_1=308.15\text{K, }{P}_2=135\text{bar, }{T}_2=468.15\text{K}$

For pure species propane, the properties can be written down using Appendix B, Table B.1

$\omega =0.152$, $T_c=369.8\text{K}$, $P_c=42.48\text{bar}$, $M=44.1\frac{\text{gm}}{\text{mol}}$ .

Now For final state

  $\begin{array}{l}{T}_r=\frac{T_2}{T_c}\\ T_r=\frac{468.15\text{K}}{369.8\text{K}}\text{=1}\text{.266}\end{array}$

And

  $\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{135\text{bar}}{42.48\text{bar}}\text{=3}\text{.178}\end{array}$

So, at above values of $T_r$ and $P_r$, The values of $Z^0$ and $Z^1$ can be written from Appendix D

$T_r=\text{1}\text{.266}$ lies between reduced temperatures $T_r=\text{1}\text{.2}$ and $T_r=\text{1}\text{.3}$ and $P_r\text{=3}\text{.178}$ lies in between reduced pressures $P_r=\text{3}\text{.00}$ and $P_r=\text{5}\text{.00}$ .

At $T_r=\text{1}\text{.2}$ and $P_r=\text{3}\text{.00}$

$Z^0=0.5425$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-2.801$, ${\frac{\left(S^R\right)}{R}}^0=-1.727$

At $T_r=\text{1}\text{.3}$ and $P_r=\text{3}\text{.00}$

$Z^0=0.6344$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-2.274$, ${\frac{\left(S^R\right)}{R}}^0=-1.299$

At $T_r=\text{1}\text{.2}$ and $P_r=5.\text{00}$

$Z^0=0.7069$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-3.166$, ${\frac{\left(S^R\right)}{R}}^0=-1.827$

At $T_r=\text{1}\text{.3}$ and $P_r=5.\text{00}$

$Z^0=0.7358$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-2.825$, ${\frac{\left(S^R\right)}{R}}^0=-1.554$

And

At $T_r=\text{1}\text{.2}$ and $P_r=\text{3}\text{.00}$

$Z^1=0.1095$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.934$, ${\frac{\left(S^R\right)}{R}}^1=-0.991$

At $T_r=\text{1}\text{.3}$ and $P_r=\text{3}\text{.00}$

$Z^1=0.2079$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.300$, ${\frac{\left(S^R\right)}{R}}^1=-0.481$

At $T_r=\text{1}\text{.2}$ and $P_r=5.\text{00}$

$Z^1=-0.0141$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-1.840$, ${\frac{\left(S^R\right)}{R}}^1=-1.767$

At $T_r=\text{1}\text{.3}$ and $P_r=5.\text{00}$

$Z^1=0.0875$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-1.066$, ${\frac{\left(S^R\right)}{R}}^1=-1.147$

Applying linear interpolation of two independent variables, From linear double interpolation, if $M$ is the function of two independent variable $X$ and $Y$ then the value of quantity $M$ at two independent variables $X$ and $Y$ adjacent to the given values are represented as follows:

  $\begin{array}{l}{X}_{1}X{X}_2\\ Y_1{M}_{1,1}{M}_{1,2}\\ YM=?\\ Y_2{M}_{2,1}{M}_{2,2}\end{array}$

So,

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ Z^0=\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times 0.5425+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times 0.7069\right]\times \frac{\text{1}\text{.3}-1.266}{\text{1}\text{.3}-1.2}\\ +\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times 0.6344+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times 0.7358\right]\times \frac{1.266\text{-1}\text{.2}}{\text{1}\text{.3}-1.2}\\ \\ Z^0=0.614\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -2.801+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -3.166\right]\times \frac{\text{1}\text{.3}-1.266}{\text{1}\text{.3}-1.2}\\ +\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -2.274+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -2.825\right]\times \frac{1.266\text{-1}\text{.2}}{\text{1}\text{.3}-1.2}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=-2.497\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^0=\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -1.727+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -1.827\right]\times \frac{\text{1}\text{.3}-1.266}{\text{1}\text{.3}-1.2}\\ +\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -1.299+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -1.554\right]\times \frac{1.266\text{-1}\text{.2}}{\text{1}\text{.3}-1.2}\\ \\ {\frac{\left(S^R\right)}{R}}^0=-1.462\end{array}$

And

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ Z^1=\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times 0.1095+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -0.0141\right]\times \frac{\text{1}\text{.3}-1.266}{\text{1}\text{.3}-1.2}\\ +\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times 0.2079+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times 0.0875\right]\times \frac{1.266\text{-1}\text{.2}}{\text{1}\text{.3}-1.2}\\ \\ Z^1=0.164\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -0.934+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -1.84\right]\times \frac{\text{1}\text{.3}-1.266}{\text{1}\text{.3}-1.2}\\ +\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -0.300+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -1.066\right]\times \frac{1.266\text{-1}\text{.2}}{\text{1}\text{.3}-1.2}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=-0.588\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^1=\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -0.991+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -1.767\right]\times \frac{\text{1}\text{.3}-1.266}{\text{1}\text{.3}-1.2}\\ +\left[\left(\frac{\text{5}-3.178}{\text{5}-\text{3}}\right)\times -0.481+\left(\frac{3.178-\text{3}}{\text{5}-\text{3}}\right)\times -1.147\right]\times \frac{1.266\text{-1}\text{.2}}{\text{1}\text{.3}-1.2}\\ \\ {\frac{\left(S^R\right)}{R}}^1=-0.717\end{array}$

Now from equation (1) at final state

  $\begin{array}{l}Z=Z^0+\omega Z^1\\ Z=0.614+0.152\times 0.164\\ Z=0.639\end{array}$

And

  $\begin{array}{l}V=\frac{ZRT}{P}\\ V=\frac{0.639\times 83.14\frac{{\text{cm}}^3\text{bar}}{\text{mol K}}\times 468.15\text{K}}{135\text{bar}}\\ V=184.2\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

From equation (2) at final state,

  $H^R={\left(H^R\right)}^0+\omega {\left(H^R\right)}^1$

  $\begin{array}{l}\frac{{\left(H^R\right)}^0}{R{T}_C}=-2.497\\ \Rightarrow {\left(H^R\right)}^0=-2.497\times 8.314\frac{\text{J}}{\text{mol K}}\times 369.8\text{K}\\ {\left(H^R\right)}^0=-7677.069\frac{\text{J}}{\text{mol}}\end{array}$

And

  $\begin{array}{l}{\frac{\left(H^R\right)}{R{T}_C}}^1=-0.588\\ \Rightarrow {\left(H^R\right)}^1=-0.588\times 8.314\frac{\text{J}}{\text{mol K}}\times 369.8\text{K}\\ {\left(H^R\right)}^1=-1807.816\frac{\text{J}}{\text{mol}}\end{array}$

So,

  $\begin{array}{l}{H}^R={\left(H^R\right)}^0+\omega {\left(H^R\right)}^1\\ H^R=-7677.069\frac{\text{J}}{\text{mol}}+0.152\times -1807.816\frac{\text{J}}{\text{mol}}\\ H^R=-7951.86\frac{\text{J}}{\text{mol}}\end{array}$

  $\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT+}{H}_2{}^R-H_1{}^R$

At final state $H_1{}^R=0$

  $\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT+}{H}_2{}^R$

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Where $\tau =\frac{T}{T_0}$

  $\tau =\frac{T}{T_0}=\frac{468.15}{308.15}=1.519$

Values of above constants for propane in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ propane1.21328.785-8.8240\end{array}$

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.213\times 308.15\times (1.519-1)+\frac{28.785\times {10}^{-3}}{2}{308.15}^2({1.519}^2-1)\\ +\frac{-8.824\times {10}^{-6}}{3}\times {308.15}^3({1.519}^3-1)+\frac{0\times {10}^5}{308.15}(\frac{1.519-1}{1.519})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=1765}\text{.1286K}\end{array}$

  $\begin{array}{l}\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT+}{H}_2{}^R\\ \Delta H=8.314\frac{\text{J}}{\text{K}\text{mol}}\times \text{1765}\text{.1286K}-7951.86\frac{\text{J}}{\text{mol}}\\ \Delta H=6723.42\frac{\text{J}}{\text{mol}}\end{array}$

  $\begin{array}{l}\frac{{\left(S^R\right)}^0}{R}=-1.462\\ \Rightarrow {\left(S^R\right)}^0=-1.462\times 8.314\frac{\text{J}}{\text{mol K}}\\ {\left(S^R\right)}^0=-12.15\frac{\text{J}}{\text{mol K}}\end{array}$

And

  $\begin{array}{l}{\frac{\left(S^R\right)}{R}}^1=-0.717\\ {\left(S^R\right)}^1=8.314\frac{\text{J}}{\text{mol K}}\times -0.717\\ {\left(S^R\right)}^1=-5.96\frac{\text{J}}{\text{mol K}}\end{array}$

So,

  $\begin{array}{l}{S}^R={\left(S^R\right)}^0+\omega {\left(S^R\right)}^1\\ S^R=-12.15\frac{\text{J}}{\text{mol K}}+0.152\times -5.96\frac{\text{J}}{\text{mol K}}\\ S^R=-13.06\frac{\text{J}}{\text{mol K}}\end{array}$

  $\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R$

At final state $S_1{}^R=0$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R$

  ${\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau$

  $\tau =\frac{T}{T_0}=\frac{468.15}{308.15}=1.519$

Values of above constants for propane in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ propane1.21328.785-8.8240\end{array}$

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\\ \left[1.213+\left\{28.785\times {10}^{-3}\times 308.15+\left(-8.824\times {10}^{-6}\times {308.15}^2+\frac{0\times {10}^5}{{1.519}^2\times {308.15}^2}\right)\left(\frac{1.519+1}{2}\right)\right\}\left(\frac{1.519-1}{\ln 1.519}\right)\right]\\ \times \ln 1.519\\ \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=4.563\end{array}$

Hence,

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 4.563-8.314\frac{\text{J}}{\text{mol K}}\times \ln \frac{135\text{bar}}{1\text{bar}}+\left(-13.06\frac{\text{J}}{\text{mol K}}\right)\\ \\ \Delta S=-15.905\frac{\text{J}}{\text{mol K}}\end{array}$

Conclusion

The molar volume is:

  $V=184.2\frac{{\text{cm}}^3}{\text{mol}}$

And

  $\Delta H=6723.42\frac{\text{J}}{\text{mol}}$

  $\Delta S=-15.905\frac{\text{J}}{\text{mol K}}$