Using van der Waals equation, calculate the constant a when two moles of a gas confined in a 4L flask exert a pressure of 11.0 atm at a temperature of 300 K. The value of b is 0.05 L `mol^(-1)`

Options

A. 7.52 `L^(2)` atm `mol^(-2)`
B. 6.46 `L^(2)` atm `mol^(-2)`
C. 12.241.atm `mol^(-2)`
D. 18.63 `L^(2) atm `mol^(-2)`

Answer

B

Explanation

Substituting the given data in the van der Waals equation of state `(p + (n^(2) a)/(V^(2))) ( V - nb) = nRT ` we get `[ 11 "Atm" + ((2 mol)^(2))/((4L)^(2)) ] [4L - 2(mol) (0.05 L mol^(-1))]`
`= (2 mol) (0.082 L "atm" K^(-1) mol^(-1) ) (300 K) or (11 "atm" + (a)/(4) "mol"^(2) L^(-2)) (4L - 0.1 L) = 49.2` L atm
or `a = 4 ((49.2)/(3.9) "atm" - 11 "atm") L^(2) mol^(-2) = 4 (12.615 - 11) L^(2)` atm `mol^(-2) = 6.46 L^(2)` atm `mol^(-2)`