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

Options

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

Answer

A

Explanation

We know that,
`Z = (PV)/(RT)`
`0.5 = (100 xx V)/(0.0821 xx 273)`
V = 0.112 litre
According to van der Waals' equation,
`(P + (a)/(V^(2))) (V - b) = RT` for 1 mole
`[100 + (a)/((0.112)^(2))] [0.112 - 0] = 0.0821 xx 273`
On solving, we get `a = 1.253 L^(2) mol^(-2) atm`