Question

15
"DUID TUINM; <b for Na
The density of vapours of a substance of molar
mass 18 g at 1 atm pressure and 500 K is
0.36 kg m . The value of compressibility factor,
Z for the vapours will be
(Take R = 0.082 L atm mole-' K-1)​

Answer 1

Answer:

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