Chapter 6, Problem 6.59P

(a)

Interpretation Introduction

Interpretation:

Determine the temperature of the expanded ethane gas, expands isentropically in a turbine and work produced if properties of ethane are calculated by equations for an ideal gas.

Concept Introduction:

The entropy change for an ideal gas is calculated by,

  $\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}}$.     ...(1)

Answer to Problem 6.59P

Temperature is

  $T=367.4\text{K}$

Work done is

  $W=-8746.9\frac{\text{J}}{\text{mol}}$

Explanation of Solution

Given information:

It is given that ethane expands isentropically from $30\text{bar}$ and $493.15\text{K}$to $2.6\text{bar}$ .

Process is isentropically, so $\Delta S=0$

From equation (1)

  $\begin{array}{l}\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}}\\ 0=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}-R\ln \frac{P_2}{P_1}}\\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}=\ln \frac{P_2}{P_1}}\end{array}$

wherE

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

  $\tau =\frac{T}{T_0}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ Ethane1.13119.225-5.5610\end{array}$

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\\ \left[1.131+\left\{19.225\times {10}^{-3}\times 493.15+\left(-5.561\times {10}^{-6}\times {493.15}^2+\frac{0}{{\tau }^2\times {493.15}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ \\ \ln \frac{P_2}{P_1}=\left[1.131+\left\{19.225\times {10}^{-3}\times 493.15+\left(-5.561\times {10}^{-6}\times {493.15}^2+\frac{0}{{\tau }^2\times {493.15}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ \\ \ln \frac{2.6}{30}=\left[1.131+\left\{19.225\times {10}^{-3}\times 493.15+\left(-5.561\times {10}^{-6}\times {493.15}^2\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ \\ \tau =0.745\end{array}$

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ 0.745=\frac{T}{493.15\text{K}}\\ T=367.4\text{K}\end{array}$

For work produced

  $W=\Delta H$ for ideal gas

And for ideal gas

  $\begin{array}{l}\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}dT}\\ \Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\end{array}$

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

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

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\\ =1.131\times 493.15\times (0.745-1)+\frac{19.225\times {10}^{-3}}{2}\times {493.15}^2\times ({0.745}^2-1)+\frac{-5.561\times {10}^{-6}}{3}\times {493.15}^3({0.745}^3-1)\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=-1052.07\text{K}\end{array}$

Hence,

  $\begin{array}{l}W=\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\\ W=8.314\frac{\text{J}}{\text{mol K}}\times -1052.07\text{K}\\ W=-8746.9\frac{\text{J}}{\text{mol}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Determine the temperature of the expanded ethane gas, expands isentropically in a turbine and work produced if properties of ethylene are calculated by appropriate generalized correlations.

Concept Introduction:

The entropy change by appropriate generalized correlations is

  $\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R$.     ...(2)

Answer to Problem 6.59P

Temperature is

  $T=362.56\text{K}$

Work done is

  $W=-8453.15\frac{\text{J}}{\text{mol}}$

Explanation of Solution

Given information:

It is given that ethane expands isentropically from $30\text{bar}$ and $493.15\text{K}$to $2.6\text{bar}$ .

Process is isentropically, so $\Delta S=0$

From equation (2),

  $\begin{array}{l}\Delta S={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R\\ R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}-R\ln \frac{P_2}{P_1}+}{S}_2{}^R-S_1{}^R=0\\ R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}=R\ln \frac{P_2}{P_1}-}{S}_2{}^R+S_1{}^R\end{array}$

Initial statE

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

$\omega =0.1$, $T_c=305.3\text{K}$, $P_c=48.72\text{bar}$, $Z_C=0.279$

  $\begin{array}{l}{T}_r=\frac{T_1}{T_c}\\ T_r=\frac{493.15\text{K}}{305.3\text{K}}\text{=1}\text{.615}\end{array}$

  $\begin{array}{l}{P}_r=\frac{P_1}{P_c}\\ P_r=\frac{30\text{bar}}{48.72\text{bar}}\text{=0}\text{.61576}\end{array}$

So, at above values of $T_r$ and $P_r$, The values of ${\frac{\left(H^R\right)}{R{T}_C}}^0$ and ${\frac{\left(H^R\right)}{R{T}_C}}^1$ can be written from Appendix D

$T_r=\text{1}\text{.615}$ lies between reduced temperatures $T_r=\text{1}\text{.6}$and $T_r=\text{1}\text{.7}$ and $P_r=\text{0}\text{.61576}$ lies in between reduced pressures $P_r=\text{0}\text{.6}$ and $P_r=\text{0}\text{.8}$ .

At $T_r=\text{1}\text{.6}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.261$, ${\frac{\left(S^R\right)}{R}}^0=-0.12$

At $T_r=\text{1}\text{.6}$ and $P_r=\text{0}\text{.8}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.35$, ${\frac{\left(S^R\right)}{R}}^0=-0.162$

At $T_r=\text{1}\text{.7}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.231$, ${\frac{\left(S^R\right)}{R}}^0=-0.102$

At $T_r=\text{1}\text{.7}$ and $P_r=\text{0}\text{.8}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.309$, ${\frac{\left(S^R\right)}{R}}^0=-0.137$

And

At $T_r=\text{1}\text{.6}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.013$, ${\frac{\left(S^R\right)}{R}}^1=-0.057$

At $T_r=\text{1}\text{.6}$ and $P_r=\text{0}\text{.8}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.011$, ${\frac{\left(S^R\right)}{R}}^1=-0.073$

At $T_r=\text{1}\text{.7}$ and $P_r=\text{0}\text{.6}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=0.009$, ${\frac{\left(S^R\right)}{R}}^1=-0.044$

At $T_r=\text{1}\text{.7}$ and $P_r=\text{0}\text{.8}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=0.017$, ${\frac{\left(S^R\right)}{R}}^1=-0.056$

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:

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

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times -0.261+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times -0.35\right]\times \frac{\text{1}\text{.7}-1.615}{1.7-1.6}\\ +\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times -0.231+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times -0.309\right]\times \frac{1.615-1.6}{1.7-1.6}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=-0.263\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^0=\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times -0.12+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times -0.162\right]\times \frac{\text{1}\text{.7}-1.615}{1.7-1.6}\\ +\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times -0.102+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times -0.137\right]\times \frac{1.615-1.6}{1.7-1.6}\\ \\ {\frac{\left(S^R\right)}{R}}^0=-0.1205\end{array}$

And

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times -0.013+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times -0.011\right]\times \frac{\text{1}\text{.7}-1.615}{1.7-1.6}\\ +\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times 0.009+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times 0.017\right]\times \frac{1.615-1.6}{1.7-1.6}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=-0.0095\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^1=\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times -0.057+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times -0.073\right]\times \frac{\text{1}\text{.7}-1.615}{1.7-1.6}\\ +\left[\left(\frac{0.8-\text{0}\text{.61576}}{0.8-\text{0}\text{.6}}\right)\times -0.044+\left(\frac{\text{0}\text{.61576}-\text{0}\text{.6}}{0.8-\text{0}\text{.6}}\right)\times -0.056\right]\times \frac{1.615-1.6}{1.7-1.6}\\ \\ {\frac{\left(S^R\right)}{R}}^1=-0.056\end{array}$

Now, from equation,

  $H_1{}^R={\left(H_1{}^R\right)}^0+\omega {\left(H_1{}^R\right)}^1$

Or

  $\begin{array}{l}\frac{H_1{}^R}{R{T}_C}=\frac{{\left(H_1{}^R\right)}^0}{R{T}_C}+\omega \frac{{\left(H_1{}^R\right)}^1}{R{T}_C}\\ \frac{H_1{}^R}{R{T}_C}=-0.263+0.1\times -0.0095\\ \frac{H_1{}^R}{R{T}_C}=-0.26395\\ H_1{}^R=-0.26395\times 8.314\frac{\text{J}}{\text{mol K}}\times 305.3\text{K}\\ H_1{}^R=-669.97\frac{\text{J}}{\text{mol}}\end{array}$

And

  $S_1{}^R={\left(S_1{}^R\right)}^0+\omega {\left(S_1{}^R\right)}^1$

Or

  $\begin{array}{l}\frac{S_1{}^R}{R}=\frac{{\left(S_1{}^R\right)}^0}{R}+\omega \frac{{\left(S_1{}^R\right)}^1}{R}\\ \frac{S_1{}^R}{R}=-0.1205+0.1\times -0.056\\ \frac{S_1{}^R}{R}=-0.1261\\ S_1{}^R=-0.1261\times 8.314\frac{\text{J}}{\text{mol K}}\\ S_1{}^R=-1.048\frac{\text{J}}{\text{mol}\text{K}}\end{array}$

Final state

Final state temperature is calculated in part (a), $T=367.4\text{K}$

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ Ethane1.13119.225-5.5610\end{array}$

  $\begin{array}{l}{T}_r=\frac{T_2}{T_c}\\ T_r=\frac{367.4\text{K}}{305.3\text{K}}\text{=1}\text{.204}\end{array}$

  $\begin{array}{l}{P}_r=\frac{P_2}{P_c}\\ P_r=\frac{2.6\text{bar}}{48.72\text{bar}}\text{=0}\text{.0534}\end{array}$

So, at above values of $T_r$ and $P_r$, The values of ${\frac{\left(H^R\right)}{R{T}_C}}^0$ and ${\frac{\left(H^R\right)}{R{T}_C}}^1$ can be written from Appendix D

$T_r=\text{1}\text{.204}$ lies between reduced temperatures $T_r=\text{1}\text{.2}$and $T_r=\text{1}\text{.3}$ and $P_r=\text{0}\text{.0534}$ lies in between reduced pressures $P_r=\text{0}\text{.05}$ and $P_r=\text{0}\text{.1}$ .

At $T_r=\text{1}\text{.2}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.036$, ${\frac{\left(S^R\right)}{R}}^0=-0.021$

At $T_r=\text{1}\text{.2}$ and $P_r=\text{0}\text{.1}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.073$, ${\frac{\left(S^R\right)}{R}}^0=-0.042$

At $T_r=\text{1}\text{.3}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.031$, ${\frac{\left(S^R\right)}{R}}^0=-0.017$

At $T_r=\text{1}\text{.3}$ and $P_r=\text{0}\text{.1}$

${\frac{\left(H^R\right)}{R{T}_C}}^0=-0.063$, ${\frac{\left(S^R\right)}{R}}^0=-0.033$

And

At $T_r=\text{1}\text{.2}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.02$, ${\frac{\left(S^R\right)}{R}}^1=-0.019$

At $T_r=\text{1}\text{.2}$ and $P_r=\text{0}\text{.1}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.04$, ${\frac{\left(S^R\right)}{R}}^1=-0.037$

At $T_r=\text{1}\text{.3}$ and $P_r=\text{0}\text{.05}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.013$, ${\frac{\left(S^R\right)}{R}}^1=-0.013$

At $T_r=\text{1}\text{.3}$ and $P_r=\text{0}\text{.1}$

${\frac{\left(H^R\right)}{R{T}_C}}^1=-0.026$, ${\frac{\left(S^R\right)}{R}}^1=-0.026$

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:

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

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.036+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.073\right]\times \frac{\text{1}\text{.3}-1.204}{1.3-1.2}\\ +\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.031+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.063\right]\times \frac{1.204-1.2}{1.3-1.2}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^0=-0.0383\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^0=\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.021+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.042\right]\times \frac{\text{1}\text{.3}-1.204}{1.3-1.2}\\ +\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.017+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.033\right]\times \frac{1.204-1.2}{1.3-1.2}\\ \\ {\frac{\left(S^R\right)}{R}}^0=-0.022\end{array}$

And

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.02+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.04\right]\times \frac{\text{1}\text{.3}-1.204}{1.3-1.2}\\ +\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.013+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.026\right]\times \frac{1.204-1.2}{1.3-1.2}\\ \\ {\frac{\left(H^R\right)}{R{T}_C}}^1=-0.02106\end{array}$

  $\begin{array}{l}M=\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{1,2}\right]\frac{Y_2-Y}{Y_2-Y_1}\\ +\left[\left(\frac{X_2-X}{X_2-X_1}\right)M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right)M_{2,2}\right]\frac{Y-Y_1}{Y_2-Y_1}\\ \\ {\frac{\left(S^R\right)}{R}}^1=\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.019+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.037\right]\times \frac{\text{1}\text{.3}-1.204}{1.3-1.2}\\ +\left[\left(\frac{0.1-\text{0}\text{.0534}}{0.1-0.05}\right)\times -0.013+\left(\frac{\text{0}\text{.0534}-\text{0}\text{.05}}{0.1-0.05}\right)\times -0.026\right]\times \frac{1.204-1.2}{1.3-1.2}\\ \\ {\frac{\left(S^R\right)}{R}}^1=-0.02\end{array}$

Now, from equation,

  $H_2{}^R={\left(H_2{}^R\right)}^0+\omega {\left(H_2{}^R\right)}^1$

Or

  $\begin{array}{l}\frac{H_2{}^R}{R{T}_C}=\frac{{\left(H_2{}^R\right)}^0}{R{T}_C}+\omega \frac{{\left(H_2{}^R\right)}^1}{R{T}_C}\\ \frac{H_2{}^R}{R{T}_C}=-0.0383+0.1\times -0.02106\\ \frac{H_2{}^R}{R{T}_C}=-0.0404\\ H_2{}^R=-0.0404\times 8.314\frac{\text{J}}{\text{mol K}}\times 305.3\text{K}\\ H_2{}^R=-102.561\frac{\text{J}}{\text{mol}}\end{array}$

And

  $S_2{}^R={\left(S_2{}^R\right)}^0+\omega {\left(S_2{}^R\right)}^1$

Or

  $\begin{array}{l}\frac{S_2{}^R}{R}=\frac{{\left(S_2{}^R\right)}^0}{R}+\omega \frac{{\left(S_2{}^R\right)}^1}{R}\\ \frac{S_2{}^R}{R}=-0.022+0.1\times -0.02\\ \frac{S_2{}^R}{R}=-0.024\\ S_2{}^R=-0.024\times 8.314\frac{\text{J}}{\text{mol K}}\\ S_2{}^R=-0.1995\frac{\text{J}}{\text{mol}\text{K}}\end{array}$

Now,

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}=\ln \frac{P_2}{P_1}-}\frac{S_2{}^R}{R}+\frac{S_1{}^R}{R}\\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}=\ln \frac{2.6}{30}-}\frac{\left(-0.1995\frac{\text{J}}{\text{mol}\text{K}}\right)}{8.314\frac{\text{J}}{\text{mol K}}}+\frac{\left(-1.048\frac{\text{J}}{\text{mol}\text{K}}\right)}{8.314\frac{\text{J}}{\text{mol K}}}\\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_P{}^{ig}\frac{dT}{T}=}-2.5477\end{array}$

Hence,

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

  $\tau =\frac{T}{T_0}$

  $\begin{array}{l}{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\left[A+\left\{B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}\frac{dT}{T}}=\\ \left[1.131+\left\{19.225\times {10}^{-3}\times 493.15+\left(-5.561\times {10}^{-6}\times {493.15}^2+\frac{0}{{\tau }^2\times {493.15}^2}\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ \\ \\ -2.5477=\left[1.131+\left\{19.225\times {10}^{-3}\times 493.15+\left(-5.561\times {10}^{-6}\times {493.15}^2\right)\left(\frac{\tau +1}{2}\right)\right\}\left(\frac{\tau -1}{\ln \tau }\right)\right]\times \ln \tau \\ \\ \tau =0.7352\end{array}$

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ 0.7352=\frac{T}{493.15\text{K}}\\ T=362.56\text{K}\end{array}$

Now, for work produced

  $W=\Delta H$

And enthalpy change by generalized correlations is given by,

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

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

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

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\\ =1.131\times 493.15\times (0.735-1)+\frac{19.225\times {10}^{-3}}{2}\times {493.15}^2\times ({0.735}^2-1)+\frac{-5.561\times {10}^{-6}}{3}\times {493.15}^3({0.735}^3-1)\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=-1088.593\text{K}\end{array}$

Hence,

  $\begin{array}{l}W=\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT+}{H}_2{}^R-H_1{}^R\\ W=8.314\frac{\text{J}}{\text{mol K}}\times -1088.593\text{K+}\left(-102.561\frac{\text{J}}{\text{mol}}\right)-\left(-669.97\frac{\text{J}}{\text{mol}}\right)\\ W=-8453.15\frac{\text{J}}{\text{mol}}\end{array}$

Hence,

  $\begin{array}{l}W=\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P{}^{ig}}{R}dT}\\ W=8.314\frac{\text{J}}{\text{mol K}}\times -1425.78\text{K}\\ W=11853.93\frac{\text{J}}{\text{mol}}\end{array}$