Calculate the compressibility factor for `CO_(2)` if one mole of it occupies `0.4` litre at `300K` and `40 atm`. Comment on the result:

Options

A. `0.65`
B. 1
C. `1.65`
D. `0.9`

Explanation

Compressibility factor (Z) = `(PV)/(nRT)`
`Z = (40 xx 0.4)/(1 xx 0.0821 xx 300) = 0.65`
Since Z value is lesser than 1 and thus , `n RT gt RV` . In order to have Z = 1 , volume of `CO_(2)` must have been more at same P and T or `CO_2` is more compressible than ideal gas .