Using van der Waals' equation, find the constant 'a' (in atm `L^(2)mol^(-2)`) when two moles of a gas confined in 4 L flask exerts a pressure of 11.0 atmospheres at a temperature of 300 K. The value of b is 0.05 L `mol^(-1)` .(R = 0.082 atm.L/K mol)

Answer

( 6.46 atm `"L"^(2) "mol"^(-2)` )

Explanation

The van der Waals’ equation is `( p + (n^(2)a)/(V^(2))) (V-nb) = nRT`
`rArr a = (V^(2))/(n^(2)) [(nRT)/(V- nb) - p] = ((4)^(2))/((2)^(2)) [(2 xx 0.082 xx 300)/(4 - 2 (0.05)) -11] = 6.46`atm `L^(2) mol^(-2)`