We are going to use the following corresponding$-$state formulation of residual molar Gibbs free

energy for pure species: $G_{{?}^{?}}(R)=G_{-}{G}_{{?}^{?}}(ig),$ where $T_r=\frac{T}{T_c},P_r=\frac{P}{P_c},$ and $$ are reduced temperature,

reduced pressure, and acentric factor, respectively, and $T_c$ and $P_c$ are critical temperature and critical

pressure $\frac{G_{{?}^{?}}(R)}{RT}=\frac{P_r}{T_r}[0\mathrm{.}083-\frac{0\mathrm{.}422}{T_r^{1\mathrm{.}6}}+(0\mathrm{.}139-\frac{0\mathrm{.}172}{T_r^{4\mathrm{.}2}})]$

$($a$)(15$ pts$)$ Considering $T=T_r{T}_c$ and $P=P_r{P}_c,$ convert $d(\frac{G_{{?}^{?}}(R)}{RT})=\frac{V^R}{RT}dP-\frac{H_{{?}^{?}}(R)}{R{T}^2}dT$ to

$d(\frac{G_{{?}^{?}}(R)}{RT})=\frac{P_c}{T_c}\frac{V^R}{R{T}_r}d{P}_r-\frac{1}{T_c}\frac{H_{{?}^{?}}(R)}{R{T}_r^2}d{T}_r$

$($b$)(15$ pts$)$ Use $($a$)$ to get $V_{{?}^{?}}(R)$ and then $Z-1=\frac{P{V}_{{?}^{?}}(R)}{RT}$ to obtain the corresponding compressibility

factor as $Z=1+\frac{P_r}{T_r}[0\mathrm{.}083-\frac{0\mathrm{.}422}{T_r^{1\mathrm{.}6}}+(0\mathrm{.}139-\frac{0\mathrm{.}172}{T_r^{4\mathrm{.}2}})]$