Chemistry - Ideal Gas Question with Solution | TestHub

$\mathrm{d}=\frac{\mathrm{PM}}{\mathrm{ZRT}}$ or $\mathrm{Z}=\frac{\mathrm{PM}}{\mathrm{dRT}}$ $\Rightarrow {\mathrm{Z}}_{{\mathrm{O}}_2}=\frac{1\times 32}{0.8\times \frac{1}{12}\times 400}=1.2$

(positive deviation, so less compressible than ideal gas) ${\mathrm{Z}}_{{\mathrm{SO}}_2}=\frac{1\times 64}{\frac{32}{15}\times \frac{1}{12}\times 400}=0.9$ (negative deviation)

${\mathrm{Z}}_{{\mathrm{H}}_2}=\frac{1\times 2}{\frac{1}{25}\times \frac{1}{12}\times 400}=1.5$ (positive deviation from ideal gas).

1 atm is a low pressure, and at $400\mathrm{K},{\mathrm{Z}}_{{\mathrm{O}}_2}$ is greater than 1 . Hence ${\mathrm{Z}}_{{\mathrm{O}}_2}$ will be equal to 1 at temperature lower than 400 K . Hence Boyle's temperature of ${\mathrm{O}}_2$ will be less than 400 K .