Question

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

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

Answer 1

The correct option is C $1.52$

$\frac{r_x}{r_{O_2}}=\sqrt{\frac{M_{O_2}}{M_x}}={\left(\frac{4}{5}\right)}^2=\frac{32}{M_x}=M_x=50$
$d_x=0.80kg/m^3$,
$V_m=\frac{1000}{800}\times 50=62.5L$.
$Z=\frac{P{V}_m}{RT}=\frac{1\times 62.5}{0.0821\times 500}=1.52$.

Hence, the correct answer is option $\text{C}$.