`2 mol` of chlorine gas occupies a volume of `800 mL` at `300 K` and `5xx10^(6) Pa` pressure. Calculate the compressibility factor of the gas. `(R=0.083 L "bar" K^(-1)mol^(-1))`. Comment, whether the gas is more compressible or less compressible under these conditions.

Explanation

Calculation of ideal volume `(V_(ideal))`
`V_(ideal)=(nRT)/(P)=((2.0 mol)(0.083 "bar" K^(-1) mol^(-1))(300 K))/((5xx10^(6)//10^(5))"bar")`
`=1.004 L`
Now,
`Z=(V_(real))/(V_(ideal))=(800 mL)/(1.004xx10^(3)mL)=0.796`
As `Z` is less than `1`, it means that the gas is more compressible under these conditions.