The density of steam at `100^(@)C` and `10^(5) Pa` pressure is `0.6 Kgm^(-3)`. Calculate the compresibility factor of steam.

Explanation

We know
`z=(PV)/(nRT)` ….(`i`)
and `d=(PM)/(RT)` ….(`ii`)
Using (`i`) and (`ii`), we get
`z=(PV)/((w//M)RT)=(MPV)/(wRT)=(MP)/(dRT)`
Therefore, on subsituding all the values, we get
`z=((18xx10^(-3)kg mol^(-1))xx10^(5)(Nm^(-2)))/((0.6kg m^(-3))xx(8.31 Nm K^(-1)mol^(-1))xx(373.15 K))`
`=0.967`