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

Givne : `V= 4L, P= 11.0 atm, T= 300K, 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)= nRT`
`(11+(2^(2)a)/(4^2))(4L- 2xx0.05L mol^(-1))= 2mol xx 0.0821 L atm K^(-1) mol^(-1) xx 300K`
`a= 6.46 atm L^(2) mol^(-2)`.