0 votes

Given that molar residual Gibbs free energy, $g^{R}$, and molar residual volume, $v^{R}$, are related as $\frac{g^{R}}{RT}=\int _{0}^{P}\left ( \frac{v^{R}}{RT} \right )dP$ , find $g^{R}$ at $T=27^{\circ}C$ and $P=0.2\:MPa$. The gas may be assumed to follow the virial equation of state, $z=1+BP/RT$, where $B=-10^{-4}\:m^{3}$/mol at the given conditions $(R=8.314\:Jmol.K)$. The value of $g^{R}$ in $J$/mol is :

- $0.008$
- $-2.4$
- $20$
- $-20$