Chapter 6, Problem 6.50P

Interpretation Introduction

Interpretation:

Determine the molar volume, enthalpy and entropyfor1,3-butadiene as a saturated vapor and as a saturated liquid at $380\text{K}$ .

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:

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

Where, $H^R={\left(H^R\right)}^0+\omega {\left(H^R\right)}^1$....(2)

And

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

Where, $S^R={\left(S^R\right)}^0+\omega {\left(S^R\right)}^1$....(3)

Answer to Problem 6.50P

The molar volume is

$V^{vap}=872.38\frac{{\text{cm}}^3}{\text{mol}}$, $V^{liq}=109.95\frac{{\text{cm}}^3}{\text{mol}}$

And

$H^{vap}=1680.026\frac{\text{J}}{\text{mol}}$

, $H^{liq}=-12308.09\frac{\text{J}}{\text{mol}}$

$S^{vap}=-14.056\frac{\text{J}}{\text{mol K}}$, $S^{liq}=-50.87\frac{\text{J}}{\text{mol K}}$

Explanation of Solution

Given information:

It is given that enthalpy and entropy are set equal to zero for the ideal gas state at $101.33\text{kPa}$ and $273.15\text{K}$ .

The vapor pressure of 1,3-butadiene at $380\text{K}$ is $1919.4\text{kPa}$

For pure species 1,3-butadiene, the properties can be written down using Appendix B, Table B.1

$\omega =0.19$, $T_c=425.2\text{K}$, $P_c=42.77\text{bar}$, $Z_C=0.267$, $T_n=268.7\text{K}$, $V_C=220.4\frac{{\text{cm}}^3}{\text{mol}}$

The molar volume for1,3-butadiene saturated vapor is calculated directly from equation,

  $V=\frac{ZRT}{P}$

For $Z=Z^0+\omega Z^1$

  $\begin{array}{l}{T}_r=\frac{T_2}{T_c}\\ T_r=\frac{380\text{K}}{425.2\text{K}}\text{=0}\text{.894}\end{array}$

Since vapor pressure is given as $1919.4\text{kPa}$ and is final state pressure.

So,

  $\begin{array}{l}{P}_r=\frac{P}{P_c}\\ P_r=\frac{1919.4\text{kPa}\times \frac{1\text{bar}}{100\text{kPa}}}{42.77\text{bar}}\text{=0}\text{.449}\end{array}$

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

$T_r=\text{0}\text{.894}$ lies between reduced temperatures $T_r=\text{0}\text{.85}$ and $T_r=\text{0}\text{.90}$ and $P_r\text{=0}\text{.449}$ lies in between reduced pressures $P_r=\text{0}\text{.4}$ and $P_r=\text{0}\text{.6}$ .

At $T_r=\text{0}\text{.85}$ and $P_r=\text{0}\text{.4}$

$Z^0=0.0661$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-4.309$, ${\frac{\left(S^R\right)}{R}}^0=-4.785$

At $T_r=\text{0}\text{.85}$ and $P_r=\text{0}\text{.6}$

$Z^0=0.0983$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-4.313$, ${\frac{\left(S^R\right)}{R}}^0=-4.418$

At $T_r=\text{0}\text{.90}$ and $P_r=\text{0}\text{.4}$

$Z^0=0.78$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.596$, ${\frac{\left(S^R\right)}{R}}^0=-0.463$

At $T_r=\text{0}\text{.90}$ and $P_r=\text{0}\text{.6}$

$Z^0=0.1006$, ${\frac{\left(H^R\right)}{R{T}_C}}^0=-4.074$, ${\frac{\left(S^R\right)}{R}}^0=-4.145$

And

At $T_r=\text{0}\text{.85}$ and $P_r=\text{0}\text{.4}$

$Z^1=-0.0268$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-4.753$, ${\frac{\left(S^R\right)}{R}}^1=-4.853$

At $T_r=\text{0}\text{.85}$ and $P_r=\text{0}\text{.6}$

$Z^1=-0.0391$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-4.754$, ${\frac{\left(S^R\right)}{R}}^1=-4.841$

At $T_r=\text{0}\text{.90}$ and $P_r=\text{0}\text{.4}$

$Z^1=-0.1118$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.751$, ${\frac{\left(S^R\right)}{R}}^1=-0.744$

At $T_r=\text{0}\text{.90}$ and $P_r=\text{0}\text{.6}$

$Z^1=-0.0396$, ${\frac{\left(H^R\right)}{R{T}_C}}^1=-4.254$, ${\frac{\left(S^R\right)}{R}}^1=-4.269$

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,

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ Z^0=\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times 0.0661+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times 0.0983\right]\times \frac{\text{0}\text{.9}-\text{0}\text{.894}}{0.9-0.85}\\ +\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times 0.78+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times 0.1006\right]\times \frac{\text{0}\text{.894-0}\text{.85}}{0.9-0.85}\\ \\ Z^0=0.549\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.309+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.313\right]\times \frac{\text{0}\text{.9}-\text{0}\text{.894}}{0.9-0.85}\\ +\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.596+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.074\right]\times \frac{\text{0}\text{.894-0}\text{.85}}{0.9-0.85}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=-1.791\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^0=\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.785+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.418\right]\times \frac{\text{0}\text{.9}-\text{0}\text{.894}}{0.9-0.85}\\ +\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.463+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.145\right]\times \frac{\text{0}\text{.894-0}\text{.85}}{0.9-0.85}\\ \\ {\frac{\left(S^R\right)}{R}}^0=-1.823\end{array}$

And

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ Z^1=\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.0268+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.0391\right]\times \frac{\text{0}\text{.9}-\text{0}\text{.894}}{0.9-0.85}\\ +\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.1118+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.0396\right]\times \frac{\text{0}\text{.894-0}\text{.85}}{0.9-0.85}\\ \\ Z^1=-0.086\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.753+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.754\right]\times \frac{\text{0}\text{.9}-\text{0}\text{.894}}{0.9-0.85}\\ +\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.751+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.254\right]\times \frac{\text{0}\text{.894-0}\text{.85}}{0.9-0.85}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=-1.987\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^1=\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.853+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.841\right]\times \frac{\text{0}\text{.9}-\text{0}\text{.894}}{0.9-0.85}\\ +\left[\left(\frac{\text{0}\text{.6}-\text{0}\text{.449}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -0.744+\left(\frac{\text{0}\text{.449}-\text{0}\text{.4}}{\text{0}\text{.6}-\text{0}\text{.4}}\right)\times -4.269\right]\times \frac{\text{0}\text{.894-0}\text{.85}}{0.9-0.85}\\ \\ {\frac{\left(S^R\right)}{R}}^1=-1.997\end{array}$

Now from equation (1) at final state

  $\begin{array}{l}Z=Z^0+\omega Z^1\\ Z=0.549+0.19\times -0.086\\ Z=0.53\end{array}$

And

  $\begin{array}{l}V=\frac{ZRT}{P}\\ V=\frac{0.53\times 8.314\frac{\text{l kPa}}{\text{mol K}}\times \frac{1000{\text{cm}}^3}{1\text{l}}\times 380\text{K}}{1919.4\text{kPa}}\\ V^{vap}=872.38\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

From equation (2) at final state,

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

  $\begin{array}{l}{\frac{\left(H^R\right)}{R{T}_C}}^0=-1.791\\ \Rightarrow {\left(H^R\right)}^0=-1.791\times 8.314\frac{\text{J}}{\text{mol K}}\times 425.2\text{K}\\ {\left(H^R\right)}^0=-6331.39\frac{\text{J}}{\text{mol}}\end{array}$

And

  $\begin{array}{l}{\frac{\left(H^R\right)}{R{T}_C}}^1=-1.987\\ \Rightarrow {\left(H^R\right)}^1=-1.987\times 8.314\frac{\text{J}}{\text{mol K}}\times 425.2\text{K}\\ {\left(H^R\right)}^1=-7024.27\frac{\text{J}}{\text{mol}}\end{array}$

So,

  $\begin{array}{l}{H}^R={\left(H^R\right)}^0+\omega {\left(H^R\right)}^1\\ H^R=-6331.39\frac{\text{J}}{\text{mol}}+0.19\times -7024.27\frac{\text{J}}{\text{mol}}\\ H^R=-7666.0013\frac{\text{J}}{\text{mol}}\end{array}$

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

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

  $\tau =\frac{T}{T_0}=\frac{380}{273.15}=1.391$

Values of above constants for 1,3-Butadiene in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ 1,3-butadiene2.73426.786-8.8820\end{array}$

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=2.734\times 273.15\times (1.391-1)+\frac{26.786\times {10}^{-3}}{2}{273.15}^2({1.391}^2-1)\\ +\frac{-8.882\times {10}^{-6}}{3}\times {273.15}^3({1.391}^3-1)+\frac{0\times {10}^5}{273.15}(\frac{1.391-1}{1.391})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=1124}\text{.13K}\end{array}$

  $\begin{array}{l}\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT+}{H}_2{}^R\\ \Delta H=8.314\frac{\text{J}}{\text{K}\text{mol}}\times \text{1124}\text{.13K}-7666.0013\frac{\text{J}}{\text{mol}}\\ \Delta H=1680.026\frac{\text{J}}{\text{mol}}=H^{vap}\end{array}$

  $\begin{array}{l}{\frac{\left(S^R\right)}{R}}^0=-1.823\\ \Rightarrow {\left(S^R\right)}^0=-1.823\times 8.314\frac{\text{J}}{\text{mol K}}\\ {\left(S^R\right)}^0=-15.16\frac{\text{J}}{\text{mol K}}\end{array}$

And

  $\begin{array}{l}{\frac{\left(S^R\right)}{R}}^1=-1.997\\ {\left(S^R\right)}^1=8.314\frac{\text{J}}{\text{mol K}}\times -1.997\\ {\left(S^R\right)}^1=-16.6\frac{\text{J}}{\text{mol K}}\end{array}$

So,

  $\begin{array}{l}{S}^R={\left(S^R\right)}^0+\omega {\left(S^R\right)}^1\\ S^R=-15.16\frac{\text{J}}{\text{mol K}}+0.19\times -16.6\frac{\text{J}}{\text{mol K}}\\ S^R=-18.31\frac{\text{J}}{\text{mol K}}\end{array}$

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

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

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

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

  $\tau =\frac{T}{T_0}=\frac{380}{273.15}=1.391$

Values of above constants for 1,3-Butadiene in above equation are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ 1,3-butadiene2.73426.786-8.8820\end{array}$

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\\ \left[2.734+\left\{26.786\times {10}^{-3}\times 273.15+\left(-8.882\times {10}^{-6}\times {273.15}^2+\frac{0\times {10}^5}{{1.391}^2\times {273.15}^2}\right)\left(\frac{1.391+1}{2}\right)\right\}\left(\frac{1.391-1}{\ln 1.391}\right)\right]\\ \times \ln 1.391\\ \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=3.453\end{array}$

Hence,

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 3.453-8.314\frac{\text{J}}{\text{mol K}}\times \ln \frac{1919.4\text{kPa}}{101.33\text{kPa}}+\left(-18.31\frac{\text{J}}{\text{mol K}}\right)\\ \\ \Delta S=S^{vap}=-14.056\frac{\text{J}}{\text{mol K}}\end{array}$

Now, for saturated liquid

Molar volume of saturated liquid can be found out using following correlation proposed by Rackett

  $\begin{array}{l}{V}^{sat}=V_C{Z}_C{}^{{(1-T_r)}^{0.2857}}\\ V^{sat}=220.4\frac{{\text{cm}}^3}{\text{mol}}\times {0.267}^{{(1-\text{0}\text{.894})}^{0.2857}}\\ \\ V^{liq}=109.95\frac{{\text{cm}}^3}{\text{mol}}\end{array}$

Now, for enthalpy and entropy of saturated liquid

  $\begin{array}{l}\frac{\Delta H_n}{R{T}_n}=\frac{1.092\left(\ln P_C-1.013\right)}{0.93-\frac{T_n}{T_c}}\\ \frac{\Delta H_n}{R{T}_n}=\frac{1.092\left(\ln 42.77-1.013\right)}{0.93-\frac{268.7\text{K}}{425.2\text{K}}}\\ \\ \frac{\Delta H_n}{R{T}_n}=10.05\\ \Delta H_n=10.05\times 8.314\frac{\text{J}}{\text{mol K}}\times 268.7\text{K}\\ \Delta H_n=22448.8\frac{\text{J}}{\text{mol}}\end{array}$

  $\begin{array}{l}\frac{\Delta H_2}{\Delta H_1}={\left(\frac{1-T_{r2}}{1-T_{r1}}\right)}^{0.38}\\ \frac{\Delta H_2}{22448.8\frac{\text{J}}{\text{mol}}}={\left(\frac{1-\text{0}\text{.894}}{1-\frac{268.7\text{K}}{425.2\text{K}}}\right)}^{0.38}\\ \\ \Delta H_2=13988.12\frac{\text{J}}{\text{mol}}\end{array}$

  $\begin{array}{l}\Delta H_2=H^{vap}-H^{liq}\\ H^{liq}=H^{vap}-\Delta H_2\\ H^{liq}=1680.026\frac{\text{J}}{\text{mol}}-13988.12\frac{\text{J}}{\text{mol}}\\ H^{liq}=-12308.09\frac{\text{J}}{\text{mol}}\end{array}$

  $\begin{array}{l}\frac{\Delta H_2}{T}=S^{vap}-S^{liq}\\ S^{liq}=S^{vap}-\frac{\Delta H_2}{T}\\ S^{liq}=-14.056\frac{\text{J}}{\text{mol K}}-\frac{13988.12\frac{\text{J}}{\text{mol}}}{380\text{K}}\\ S^{liq}=-50.87\frac{\text{J}}{\text{mol K}}\end{array}$