If the volume occupied by `CO_(2)` molecules is negligible, then calculate the pressure exerted by one mole of `CO_(2)` gas at 273 K (a=3.592 atm `"litre"^(-2)mol^(-2)`)

Explanation

For 1 mole, `(P+(a)/(V_(2)))(V-b)=RT`
If b is negligible, `(P+(a)/(V_(2)))V=RT" or " P=(RT)/(V)-(a)/(V_(2))" or "PV^(2)-RTV+a=0`
This equation is quadratic in V. Hence, `V=(+RT+-sqrt(R^(2)T^(2)-4aP))/(2P)`
As V can have only one value at P and T, hence we must have
`R^(2)T^(2)-4aP=0 " or " R^(2)T^(2)=4 a P`
or `" " P=(R^(2)T^(2))/(4a)=((0.0821" L atm "K^(-1))^(2)(273 K)^(2))/(4xx3.592" atm "L6(-2)mol^(-2))=34.96" atm"`