The compressibility factor for definite amount of van der Waal's gas at `0^(@)C` and 100 atm is found to be `0.5`. Assuming the volume of gas molecules negligible, the van der Waal's constant a for a gas is

Options

A. 1.256 atm `L^(2) mol^(-2)`
B. 0.256 atm `L^(2) mol^(-2)`
C. 2.256 atm `L^(2) mol^(-2)`
D. 0.0256 atm `L^(2) mol^(-2)`

Answer

A

Explanation

`V_(m) = (zRT)/(P) = 0.11207 , (P + (a)/(V^(2))) V = RT , PV + (a)/(V) = RT`
`z = 1 - (a)/(RTV_(m)) implies a = 1.256 "atm" L^(2) mol^(-1)`