2 moles of a gas confined in a 4 L flask exert a pressure of 9.0 atm of 300 K temperature. The value of bis 0.05 L mol, the value of a is

Answer

B

Explanation

Using van der Waals equation , we have `(P + (an^(2))/(V^(2))) (V - nb) = nRT , (an^(2))/(v^(2)) = (nRT)/(V-nb) - P`
`(a xx 4)/(16) = (2 xx 0.0821 xx 300)/(4 - 2 xx 0.5) - 11 implies (a xx 4)/(16) = 1.631 or a = (1.631 xx 16)/(4) = 6.523 L^(2)` atm `mol^(-2)`