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 \[mo{{l}^{-2}},\]\[b=0.0319\,L\,mo{{l}^{-1}}\])
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