The correct order of normal boiling point of \[{{O}_{2}},{{N}_{2}},N{{H}_{3}}\] and
\[C{{H}_{4}}\] for whom the values of Vander Waals constant ‘a’ are \[1.360,1.360,4.170\] and \[2.253{{L}^{2}}atm.mo{{l}^{-2}}\]respectively is:
A. \[{{O}_{2}}<{{N}_{2}}<N{{H}_{3}}<C{{H}_{4}}\]
B. \[{{O}_{2}}<{{N}_{2}}<C{{H}_{4}}<N{{H}_{3}}\]
C. \[N{{H}_{3}}<C{{H}_{4}}<{{N}_{2}}<{{O}_{2}}\]
D. \[N{{H}_{3}}<C{{H}_{4}}<{{O}_{2}}<{{N}_{2}}\]
Answer
603k+ views
Hint: The estimation of the Van der Waals constant, a, of a gaseous substance relies upon the strength of attractions between its component molecules. Atoms experiencing the most weakest attractive forces will have the littlest a constant while those with the most strongest attractive forces will have the biggest values.
Complete step by step solution:
The simplicity of liquefaction of a gas relies upon their intermolecular force of attraction thus is measured regarding van der Waals' constant 'a'. Consequently, higher the value of 'a', more noteworthy the intermolecular force of attraction, simpler the liquefaction. In the current case, \[N{{H}_{3}}\] has the highest 'a', and can most effectively be liquefied. Simplicity of liquefaction is directly proportional to 'a'. If the value of 'a' is more than if we apply even less pressure then the gas can be liquefied, so as to liquefy effectively we ought to have more 'a'. So more 'a' more effectively liquefies.
As \[N{{H}_{3}}\] has higher *a value, it very well may be effectively liquefiable. With the end goal that [rate of liquefaction is directly proportional to a value].
The van der Waals steady 'a' speaks to the magnitude of intermolecular forces of attraction and the Van der Waals constant 'b' speaks to the effective size of the molecules. Vander Waals constant ‘a’ depicts the intermolecular forces. Higher the value of ‘a’ stronger the intermolecular forces.
Hence the correct option is B.
Note: The van der Waals equation is an equation of state that adjusts for two properties of real gases: the prohibited volume of gas particles and attractive forces between gas molecules.
Complete step by step solution:
The simplicity of liquefaction of a gas relies upon their intermolecular force of attraction thus is measured regarding van der Waals' constant 'a'. Consequently, higher the value of 'a', more noteworthy the intermolecular force of attraction, simpler the liquefaction. In the current case, \[N{{H}_{3}}\] has the highest 'a', and can most effectively be liquefied. Simplicity of liquefaction is directly proportional to 'a'. If the value of 'a' is more than if we apply even less pressure then the gas can be liquefied, so as to liquefy effectively we ought to have more 'a'. So more 'a' more effectively liquefies.
As \[N{{H}_{3}}\] has higher *a value, it very well may be effectively liquefiable. With the end goal that [rate of liquefaction is directly proportional to a value].
The van der Waals steady 'a' speaks to the magnitude of intermolecular forces of attraction and the Van der Waals constant 'b' speaks to the effective size of the molecules. Vander Waals constant ‘a’ depicts the intermolecular forces. Higher the value of ‘a’ stronger the intermolecular forces.
Hence the correct option is B.
Note: The van der Waals equation is an equation of state that adjusts for two properties of real gases: the prohibited volume of gas particles and attractive forces between gas molecules.
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