Question

The van der Waal's constants for a gas are a=3.6atmL 2 mol 2 ,b=0.6Lmol −1 . If R=0.08LatmK −1 mol −1 and the Boyle's temperature (K) is

Answer 1

a= 3.6 atm L2 mol2

TB =a/Rb

3.6/6.08×0.6

atm L2 mol-2 /Lmol-1 L atmK-1 mol-1

TB= 75K

hope it helps u