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))`

Options

A. 7
B. 8
C. 9
D. 3

Answer

B

Explanation

`(P+(an^(2))/(V^(2)))(V-nb)=nRT`
`(P+(a)/(V^(2)))(V-b)=RT " " (because n = 1)`
If b is negligible `P=(RT)/(V)-(a)/(V^(2))`
The equation is quadratic in V, thus
`V= (+ RT pm 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)=((0.0821)^(2)(300)^(2))/(4xx3.592) therefore (P)/(5.277)=8`