question_answer

Out of \[{{N}_{2}}\] and \[N{{H}_{3}},\] which one will have greater value for van der Waals constant 'a' and which one will have greater value for van der Waals constant 'b'?

Answer:

                            (i) As \[N{{H}_{3}}\] is more easily liquefiable (due to hydrogen bonding), intermolecular forces of attraction are stronger than in \[{{N}_{2}}\]. Hence, \[N{{H}_{3}}\] will have greater value for 'a'. (ii) As \[N{{H}_{3}}\] molecule is larger in size than \[{{N}_{2}}\], hence \[N{{H}_{3}}\]  will have greater value for 'b'. (For \[N{{H}_{3}},\,\,a=4.17\,{{L}^{2}}\,\]atm \[mo{{l}^{-2}},\]\[b=0.0371\,\,L\,\,mo{{l}^{-1}}\] For \[{{N}_{2}},\,\,a=1.39\,{{L}^{2}}\,\] atm \[mo{{l}^{-2}},\]\[b=0.0319\,L\,mo{{l}^{-1}}\])