The van der Waals equation adjusts for properties of real gases with correction
ID: 826492 • Letter: T
Question
The van der Waals equation adjusts for properties of real gases with correction factors 'a' and 'b'. Between NO and N2, which would have larger correction factor 'b' and why?
A.
NO will have a larger 'b' value because it is larger
B.
N2 will have a larger 'b' value because it is larger
C.
NO will have a larger 'b' value because it has stronger intermolecular forces
D.
N2 will have a larger 'b' value because it has stronger intermolecular forces
A.
NO will have a larger 'b' value because it is larger
B.
N2 will have a larger 'b' value because it is larger
C.
NO will have a larger 'b' value because it has stronger intermolecular forces
D.
N2 will have a larger 'b' value because it has stronger intermolecular forces
Explanation / Answer
b value denotes intermolecular forces
NO has greater intermolecular forces due to interation of N of one molecule with O of other
Hence:
NO will have a larger 'b' value because it has stronger intermolecular forces
NO will have a larger 'b' value because it has stronger intermolecular forces
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.