If volume occupied by `CO_(2)` molecules is negligible, then calculate pressure `((P)/(5.277))` exerted by one mole of `CO_(2)` gas at `300 K`. `(a=3.592 atm L^(2)mol^(-2))`

Explanation

`(P+(an^(2))/(V^(2)))(V-nb)=nRT`
`n=1`
`(P+(a)/(V^(2)))(V-nb)=RT`
If `b` is negligible
`P=(RT)/(V)-(a)/(V^(2))`
The equation is quadratic in `V` thus
`V=(+RT+-sqrt(R^(2)T^(2)-4aP))/(2P)`
Since `V` has one value at given `P` and `T`, thus numerical value of discriminant `=0`
`R^(2)T^(2)=4aP`
`P=(R^(2)T^(2))/(4a)=((0.821)^(2)(300)^(2))/(4xx3.592)`
`:. (P)/(5.277)=8`