Chapter 6, Problem 6.57P

Interpretation Introduction

Interpretation:

The value of $\Delta S$for the given process is to be calculated.

Concept Introduction:

The formula to calculate reduced enthalpy $\left(H^R\right)$ is:

  $H^R=R{T}_c{P}_r\left(B^0-T_r\frac{d{B}^0}{d{T}_r}+\omega \left(B^1+T_r\frac{d{B}^1}{d{T}_r}\right)\right)$    ......(1)

The formula to calculate reduced entropy $\left(S^R\right)$is:

  $S^R=-R{P}_r\left(\frac{d{B}^0}{d{T}_r}+\omega \frac{d{B}^1}{d{T}_r}\right)$    ......(2)

Here, $P_r$is reduced pressure, $T_r$is reduced temperature, and $T_c$is critical temperature of the given system. Values of $B^0,B^1,\frac{d{B}^0}{d{T}_r},\text{ and }\frac{d{B}^1}{d{T}_r}$are calculated by:

  $\begin{array}{c}{B}^0=0.083-\frac{0.422}{T_r^{1.6}}\\ B^1=0.139-\frac{0.172}{T_r^{4.2}}\\ \frac{d{B}^0}{d{T}_r}=\frac{0.675}{T_r^{2.6}}\\ \frac{d{B}^1}{d{T}_r}=\frac{0.722}{T_r^{5.2}}\end{array}$    ......(3)

The formula to calculate reduced enthalpy $\left(H^R\right)$in terms of specific heat is:

  $\begin{array}{c}\underset{H^R}{\underbrace{H^{ig}-H_0^{ig}}}={\displaystyle {\int }_{T_0}^T{C}_p^{ig}dT}\\ H^R=R\left(A{T}_0\left(\tau -1\right)+\frac{B}{2}{T}_0^2\left({\tau }^2-1\right)+\frac{C}{3}{T}_0^3\left({\tau }^3-1\right)+\frac{D}{T^0}\left(\frac{\tau -1}{\tau }\right)\right)\mathrm{......}\text{ (4)}\end{array}$

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

The formula to calculate entropy change of vapor $\left(\Delta S^{ig}\right)$for an ideal gas from temperature $T_1\left(\text{K}\right)\text{ to }{T}_2\left(\text{K}\right)$is:

  $\Delta S^{ig}=R\left[{\displaystyle {\int }_{T_0}^T{C}_p^{ig}\frac{dT}{T}}-\ln \frac{P_0}{P}\right]$    ......(5)

Value of $C_p^{ig}$is taken from table C.1.

The formula to calculate entropy change of vapor $\left(\Delta S\right)$from temperature $T_1\left(\text{K}\right)\text{ to }{T}_2\left(\text{K}\right)$is:

  $\Delta S=\Delta S^{ig}-S^R$    ......(6)