Question

The density of vapours of a substance of molar mass 18 g at 1 atm pressure and 500k is 0.36kgm^-3 the value of compressibility factor z for the vapours will be

Answer 1

Molar mass of the substance=18 g

Density of the substance=0.36 kg m⁻³= 0.36 g dm⁻³

Pressure=1 atm=101.3 kPa

temperature=500K

As ,

$Density=\frac{mass}{volume}$

$volume=\frac{mass}{Density}$

$volume=\frac{mass}{Density}$

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

$volume=50.25$

Compressibility factor,

$z=\frac{PV}{RT}$,

R=gas constant=8.314 J K⁻¹ mol⁻¹

Putting all the values in the equation,

$z=\frac{101.3\times50.25}{8.314\times500}$

z=1.22