Calculate the pressure exerted by one mole of `CO_(2)` gas at `273 K` van der Waals constant `a=3.592 dm^(6) atm mol^(-2)`. Assume that the volume occupied by `CO_(2)` molecules is negligible.

Explanation

For 1 mole `[P+(a)/(V^(2))][V-b] =RT`
if b is negligible `P =(RT)/(V) - (a)/(V^(2))`
`:. PV^(2) - RTV + a =0`
The equation is quadratic in `V` thus
`V =+RT+-sqrt(((RT)^(2)-4Pa)/(2P))`
`V` has only one value at given `P` and `T` and thus numerical value of discriminant =0
or `(RT)^(2) - 4 Pa = 0`
or `P = (R^(2)T^(2))/(4a) = ((0.0821)^(2)xx(273)^(2))/(4 xx 3.592) = 34.96` atm .