20 mol of chlorine gas occupies a volume of 800 ml at 300 K and `5 xx 10^(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 the conditions

Explanation

calculation of ideal volume `(V_("ideal"))` ltbr. `V_("ideal")=(nRT)/(P)`
`=(2.0(mol)xx0.83("bar "LK^(-1)mol^(-1))xx300(K))/(5xx10^(6)//10^(5)("bar"))`
`=1.004L`
Now, `Z=(V_("real"))/(V_("ideal"))=(800(mL))/(1.004xx10^(3)(mL))=0.796`
As Z is less that 1, it means that gas is more compressible under these condition.