Find out the value of the van der Waals constant 'a' when two moles of a gas confined in a 4 L flask exerts a pressure of `11.0 atm ` at `300 K. ` `(b= 0.05 L mol^(-1))`.

Explanation

Given V = 4 L , P = 11.0 atm , T = 300 K , b = 0.05 L `mol^(-1) , n = 2`
The van der Waal.s equation for n moles of gas is `(P + (n^(2) a)/(V^(2))) ( V- nb) = n R T`
`(1.1 + (2^2 a)/(4^2)) (4 L - 2 xx 0.5 L mol^(-1)) = 2 mol xx 0.0821 L` atm `K^(-1) mol^(-1) xx 300 K`
a= `6.46` atm `L^(2) mol^(-2)`