Question

The density of vapour of a substance (X) at 1 atm pressure and 500 K is 0.8 $\frac{kg}{m^3}$. The vapour effuses through a small hole at a rate of $\frac{4}{5}$ times slower than oxygen under the same condition. What is the compressibility factor (Z) of the vapour?

• A. 0.974
• B. 1.35
• C. 1.52
• D. 1.22

Answer 1

The correct option is C

1.52

$\frac{r_x}{r_{O_2}}=\sqrt{\frac{M_{O_2}}{M_x}}=⟮\frac{4}{5}{⟯}^2=\frac{32}{M_x}\Rightarrow M_x=50$

$d_x=0.80kg{m}^{-3}$

$V_m=\frac{1000}{800}\times 50=62.5L$

$Z=\frac{P{V}_m}{RT}=\frac{1\times 62.5}{0.0821\times 500}=1.52$