
If \[{\text{N}}{{\text{H}}_3}\] gas is liquefied more easily than \[{{\text{N}}_2}\]. Hence:
(A). Van der Waals' constant \[a\] and \[b\] of \[{\text{N}}{{\text{H}}_3}\] > that of \[{{\text{N}}_2}\]
(B). Van der Waals' constant \[a\] and \[b\] of \[{\text{N}}{{\text{H}}_3}\] < that of \[{{\text{N}}_2}\]
(C). \[a{\text{ (N}}{{\text{H}}_3})\] > \[a{\text{ (}}{{\text{N}}_2})\] but \[b{\text{ (N}}{{\text{H}}_3})\] < \[b{\text{ (}}{{\text{N}}_2})\]
(D). \[a{\text{ (N}}{{\text{H}}_3})\] < \[a{\text{ (}}{{\text{N}}_2})\] but \[b{\text{ (N}}{{\text{H}}_3})\] > \[b{\text{ (}}{{\text{N}}_2})\]
Answer
476.7k+ views
Hint: In the liquid phase, the molecules are closer than in the gas phase. The van der Waals' constant \[a\] is an indirect measure of magnitude of attractive forces between molecules. The van der Waals' constant \[b\] gives an idea of the effective size of molecules.
Complete step by step answer:
For \[n\] moles of gas, the van der Waals equation is:
\[\left( {p + \dfrac{{a{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT\].
In this equation, \[p\] is pressure, \[V\] is volume, \[R\] is universal gas constant and \[T\] is absolute temperature. The Values of \[a\] and \[b\] are constant for a gas.
The constant \[a\] gives an idea of the force of attraction between molecules. Molecules are closer in the liquid phase than those in the gas phase. Thus, if the attraction force between molecules is more then it will be easier to bring those molecules closer. We can say that gases with large values of \[a\] can be liquefied easier than those with small values of \[a\].
Therefore, \[a{\text{ (N}}{{\text{H}}_3})\] > \[a{\text{ (}}{{\text{N}}_2})\].
The constant \[b\] gives an idea of volume occupied by one mole of molecules themselves. Due to larger size of nitrogen atom than hydrogen atom, volume of \[{{\text{N}}_2}\] molecule is more than that of \[{\text{N}}{{\text{H}}_3}\].
Therefore, \[b{\text{ (N}}{{\text{H}}_3})\] < \[b{\text{ (}}{{\text{N}}_2})\].
Hence, the correct option is (C).
Note: The values of \[a\] for both the gases are very different. This is the reason why \[{\text{N}}{{\text{H}}_3}\] gas can be liquefied more easily than\[{{\text{N}}_2}\]. Whereas, the values of \[b\] for both the gases are very close to each other.
Complete step by step answer:
For \[n\] moles of gas, the van der Waals equation is:
\[\left( {p + \dfrac{{a{n^2}}}{{{V^2}}}} \right)(V - nb) = nRT\].
In this equation, \[p\] is pressure, \[V\] is volume, \[R\] is universal gas constant and \[T\] is absolute temperature. The Values of \[a\] and \[b\] are constant for a gas.
The constant \[a\] gives an idea of the force of attraction between molecules. Molecules are closer in the liquid phase than those in the gas phase. Thus, if the attraction force between molecules is more then it will be easier to bring those molecules closer. We can say that gases with large values of \[a\] can be liquefied easier than those with small values of \[a\].
Therefore, \[a{\text{ (N}}{{\text{H}}_3})\] > \[a{\text{ (}}{{\text{N}}_2})\].
The constant \[b\] gives an idea of volume occupied by one mole of molecules themselves. Due to larger size of nitrogen atom than hydrogen atom, volume of \[{{\text{N}}_2}\] molecule is more than that of \[{\text{N}}{{\text{H}}_3}\].
Therefore, \[b{\text{ (N}}{{\text{H}}_3})\] < \[b{\text{ (}}{{\text{N}}_2})\].
Hence, the correct option is (C).
Note: The values of \[a\] for both the gases are very different. This is the reason why \[{\text{N}}{{\text{H}}_3}\] gas can be liquefied more easily than\[{{\text{N}}_2}\]. Whereas, the values of \[b\] for both the gases are very close to each other.
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