Question

If the density of vapours of a substance of molar mass 18 g at 1 atm pressure and 500 K is 0.36 kg m${}^{-3}$, then calculate the value of Z for the vapours.

[Take R=0.082 L atm mol${}^{-1}$K${}^{-1}$]

Answer 1

The molar mass of the substance $=$ 18 g

Density of substance $=0.36$ kg m${}^{-3}=0.36$ g dm${}^{-3}$

Pressure = 1 atm = 101.3k pa

temperature = 500k

As, Density = $\frac{\text{mass}}{\text{volume}}$

$\therefore$ volume $=\frac{18}{0.36}=50.25$

Compreulbility factor $Z=\frac{PV}{RT},$

R = 8.314 J/k mol

$\therefore Z=\frac{101.3\times 50.25}{8.314\times 500}$

$Z=1.22$