Question

The density of vapour of a substance $\left(X\right)$ at $1atm$ pressure and $500K$ is $0.8kg/m^3.$ The vapour effuse 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

Answer: (C)

$1.52$

$S=0.8kg/m^3=0.8g/L$

From Graham's law of diffusion,

$\frac{r_x}{r_{o_2}}=\sqrt{\frac{M_{o_2}}{M_x}}\therefore \frac{4}{5}=\sqrt{\frac{32}{M_x}}\therefore M_x=50g/mol$

Now, $\frac{0.8}{50}$ mole occupies $1L$ volume

$\therefore 1$ mole occupies $\frac{50}{0.8}L=62.5L$

Assuming ideal behaviour,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}\Rightarrow 22.4\times 500/273=41.025L$

$\therefore Z=\frac{{\left(PV\right)}_{obs}}{{\left(PV\right)}_{ideal}}=\frac{1\times 62.5}{1\times 41.025}=1.52$