Chemistry - Real Gas Question with Solution | TestHub

$\left(\mathrm{A}\right)$Maximum density of$\mathrm{C}{\mathrm{O}}_2$in gaseous phase will be at point 'A'
Now:$\mathrm{d}=\frac{\mathrm{P}\mathrm{M}}{\mathrm{R}\mathrm{T}}=\frac{60\times 44}{\mathrm{R}\times 300}=107\mathrm{ }\mathrm{g}\mathrm{m}/\mathrm{l}\mathrm{t}=0.107\mathrm{ }\mathrm{g}\mathrm{m}/\mathrm{c}\mathrm{c}$
Or$\mathrm{d}=\frac{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{g}\mathrm{a}\mathrm{s}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{g}\mathrm{a}\mathrm{s}}=\frac{44\mathrm{ }\mathrm{g}\mathrm{m}}{440\mathrm{ }\mathrm{c}\mathrm{c}}=0.1$
$\left(\mathrm{B}\right)$Density of liquid$\mathrm{C}{\mathrm{O}}_2$at 60 atm (at point B)
At point B there will be only liquid$\mathrm{C}{\mathrm{O}}_2$. So we can find out the density of density of liquid$\mathrm{C}{\mathrm{O}}_2$only at point B.
$\text{d}=\frac{\text{M}}{\text{V}}=\frac{44}{0.044}=1\text{ gm/cc}$
$\left(\mathrm{C}\right)$at point C the volume will be$\frac{0.44+0.044}{2}\approx 0.22\text{L}$
And pressure is same so obviously approx. half of$\mathrm{C}{\mathrm{O}}_2$is liquified
$\left(\mathrm{D}\right)$Z can be greater than 1 at$27^{\mathrm{o}}\mathrm{C}$. For example at point 'A' Z is 1.07, So wrong.