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)

Explanation

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