Chapter 7.7, Problem 24P

(A)

Interpretation Introduction

Interpretation:

The change in internal energy for a gas

Concept Introduction:

The change in molar internal energy using residual properties.

${\underset{\_}{U}}_2-{\underset{\_}{U}}_1=\left({\underset{\_}{U}}_2-{\underset{\_}{U}}_2^{ig}\right)+\left({\underset{\_}{U}}_2^{ig}-{\underset{\_}{U}}_1^{ig}\right)-\left({\underset{\_}{U}}_1-{\underset{\_}{U}}_1^{ig}\right)$

Here, molar internal energy at state 1 (vapor) and 2 (liquid) are ${\underset{\_}{H}}_1$ and ${\underset{\_}{H}}_2$, molar internal energy at state 1 and 2 at inert gas are ${\underset{\_}{H}}_1^{ig}$ and ${\underset{\_}{H}}_2^{ig}$ respectively.

The reduced temperature $\left(T_r\right)$.

$T_r=\frac{T}{T_c}$

Here, critical temperature is $T_c$ and temperature is T.

The reduced pressure $\left(P_r\right)$.

$P_r=\frac{P}{P_c}$

Here, pressure is $P$.

The acentric factor in a manner analogous to Soave’s $m$.

$\kappa =0.37464+1.54226\omega -0.26993{\omega }^2$

Write the $\alpha$ as a function of the reduced temperature.

$\alpha ={\left[1+\kappa \left(1-T_r^{0.5}\right)\right]}^2$

The Peng-Robinson parameter a at the critical point.

$a_c=\frac{0.45724R^2{T}_c^2}{P_c}$

The van der Waals parameter $a$.

$a=a_c\alpha$

The van der Waals parameter $b$.

$b=\frac{0.07780R{T}_c}{P_c}$

The Peng-Robinson equation.

$P=\frac{RT}{\underset{\_}{V}-b}-\frac{a}{\underset{\_}{V}\left(\underset{\_}{V}+b\right)+b\left(\underset{\_}{V}-b\right)}$

Here, molar volume is $\underset{\_}{V}$, parameters of Robinson equation are a, b, gas constant is R, temperature and pressure is T and P respectively.

The compressibility factor.

$Z=\frac{P\underset{\_}{V}}{RT}$

The residual properties of A.

$A=\frac{aP}{R^2{T}^2}$

The residual properties of B.

$B=\frac{bP}{RT}$

The expression for residual molar internal energy using Equation (2).

$\begin{array}{l}\frac{{\underset{\_}{U}}^R}{R{T}_2}=\left\{\left(\frac{A}{B\sqrt{8}}\right)\left(1+\frac{\kappa \sqrt{T_r}}{\sqrt{\alpha }}\right)\ln \left[\frac{Z+\left(1+\sqrt{2}\right)B}{Z+\left(1-\sqrt{2}\right)B}\right]\right\}\\ {\underset{\_}{U}}_2-{\underset{\_}{U}}_1=R{T}_2\left\{\left(\frac{A}{B\sqrt{8}}\right)\left(1+\frac{\kappa \sqrt{T_r}}{\sqrt{\alpha }}\right)\ln \left[\frac{Z+\left(1+\sqrt{2}\right)B}{Z+\left(1-\sqrt{2}\right)B}\right]\right\}\end{array}$

Here, compressibility factor is Z, constants of residual properties are A and B.

(B)

Interpretation Introduction

Interpretation:

The change in molar entropy for the gas in this process.

Concept Introduction:

Write the change in molar entropy using residual properties.

${\underset{\_}{S}}_2-{\underset{\_}{S}}_1=\left({\underset{\_}{S}}_2-{\underset{\_}{S}}_2^{ig}\right)+\left({\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}\right)-\left({\underset{\_}{S}}_1-{\underset{\_}{S}}_1^{ig}\right)$

Here, molar entropy at state 1 (vapor) and 2 (liquid) is ${\underset{\_}{S}}_1$ and ${\underset{\_}{S}}_2$, molar entropy at state 1 and 2 at inert gas is ${\underset{\_}{S}}_1^{ig}$ and ${\underset{\_}{S}}_2^{ig}$ respectively.

Write the expression for residual molar entropy.

$\frac{{\underset{\_}{S}}^R}{R}=\ln \left(Z-B\right)-\left\{\left(\frac{A}{B\sqrt{8}}\right)\left(1+\frac{\kappa \sqrt{T_r}}{\sqrt{a}}\right)\ln \left[\frac{Z+\left(1+\sqrt{2}\right)B}{Z+\left(1-\sqrt{2}\right)B}\right]\right\}$

Write the ideal gas component.

$\begin{array}{l}d\underset{\_}{S}=\frac{C_V^{*}}{T}dT+\frac{R}{\underset{\_}{V}}d\underset{\_}{V}\\ {\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}={\displaystyle \underset{T_1=400\text{K}}{\overset{T_2=400\text{K}}{\int }}\frac{C_V^{*}}{T}dT+\frac{R}{\underset{\_}{V}}d\underset{\_}{V}}\\ {\underset{\_}{S}}_2^{ig}-{\underset{\_}{S}}_1^{ig}={\displaystyle \underset{T_1=400\text{K}}{\overset{T_2=400\text{K}}{\int }}\frac{C_V^{*}}{T}dT+R\ln \frac{{\underset{\_}{V}}_2}{{\underset{\_}{V}}_1}}\end{array}$