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 Waal's 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)`

Answer

A

Explanation

`[P + (a)/(V^(2))] (V-b) = RT` when volume of `CO_(2)` molecules is negligible , `[P + (a)/(V^(2))] V = RT` or `V^(2) P - RTV + a = 0 , V = (pm RT pm 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
`therefore R^(2) T^(2) = 4Pa or a = (R^(2) T^(2))/(4P) = ((0.0821)^(2) xx (273)^(2))/(4 xx 34.98) = 3.59 dm^(6)` atm `mol^(-2)`