Chemistry - Real Gas Question with Solution | TestHub

$\mathrm{P}={\mathrm{P}}_{\mathrm{c}}\mathrm{\pi }=\left(\frac{\mathrm{a}}{27{\mathrm{b}}^2}\right)\mathrm{\pi }$

$\mathrm{V}={\mathrm{V}}_{\mathrm{c}}\mathrm{\varphi }=\left(3\mathrm{b}\right)\mathrm{\varphi }$

$\mathrm{T}={\mathrm{T}}_{\mathrm{c}}\mathrm{\theta }=\left(\frac{8\mathrm{a}}{27\mathrm{R}\mathrm{b}}\right)\mathrm{\theta }$

Hence substituting in the van der Waal's equation

$\left(\mathrm{P}+\frac{\mathrm{a}}{{\mathrm{V}}^2}\right)\left(\mathrm{V}-\mathrm{b}\right)=\mathrm{R}\mathrm{T}$

$\Rightarrow \left(\mathrm{\pi }+\frac{3}{{\mathrm{\varphi }}^2}\right)\left(3\mathrm{\varphi }-1\right)=8\mathrm{\theta }$

This is reduced state equation

$\left(\mathrm{\pi }+\frac{3}{{\mathrm{\varphi }}^2}\right)\left(\frac{3\mathrm{\varphi }}{8\mathrm{\theta }}-\frac{1}{8\mathrm{\theta }}\right)=1$

$\left(\frac{\mathrm{\pi }}{3}+\frac{1}{{\mathrm{\varphi }}^2}\right)\left(3\mathrm{\varphi }-1\right)=\frac{8}{3}\mathrm{\theta }$