The pressure exerted by `1 mol` of `CO_(2)` at `273 K` is `34.98 atm`. Assuming that volume occupied by `CO_(2)` molecules is negligible, the value of van der Waals constant for attraction of `CO_(2)` gas is

Options

A. `3.59 dm^(6) atm mol^(-2)`
B. `2.59 dm^(6) atm mol^(-2)`
C. `1.25 dm^(6) atm mol^(-2)`
D. `1.59 dm^(6) atm mol^(-2)`

Explanation

`[P+(a)/(V^(2))][V-b]=RT`
`:. [P+(a)/(V^(2))]V=RT`
or `V^(2)P-RTV+a=0`
`V=(+-RT+-sqrt(R^(2)T^(2)-4Pa))/(2P)`
Since, `V` is constant at given `P` and `T`, `V` can have only one value or discriminant `=0`
`:. R^(2)T^(2)=4Pa` or `a=(R^(2)T^(2))/(4P)`
`=((0.821)^(2)xx(273)^(2))/(4xx34.98)`
`=3.59 dm^(6) atm mol^(-2)`