The compressibility factor `(Z=PV//nRT)` for `N_(2)` at `223 K` and `81.06 MPa` is `1.95`, and at `373 K` and `20.265 MPa`, it is `1.10`. A certain mass of `N_(2)` occupies a volume of `1.0 dm^(3)` at `223 K` and `81.06 MPa`. Calculate the volume occupied by the same quantity of `N_(2)` at `373 K` and `20.265 MPa`.

Explanation

For `T=223 K`, `P=81.06 MPa`, `Z=1.95` and
`V=1.0 dm^(3)=10^(3) cm^(3)`, we have
`n=(PV)/(nRT)=(81.06xx10^(3))/(1.95xx8.314xx223)=22.42 mol`
Now, at `T=373 K`, `P=20.65 MPa`, `Z=1.10`, the volume occupied will be
`V=(ZnRT)/(P)=(1.10xx22.42xx8.314xx373)/(20.265)=3774.0 cm^(3)`
`:. V=3.774 dm^(3)`