Discussion
In the van der Waals equation of state, the volume correction factor (b) arises from the repulsive interactions between the gas molecules. Considering the molecules to behave as hard spheres of radius r, b = Y × Vmolecule × NA. The value of Y is _____.
[Given: NA is the Avogadro's Number, Vmolecule is the volume of a gas molecule]
Answer: D
The value of Y is 4.
Two hard spheres of radius \(r\) cannot approach closer than \(2r\) centre to centre, so each pair of molecules excludes a sphere of radius \(2r\) from the other's centre:
\[ \begin{aligned} V_{pair} &= \frac{4}{3}\pi(2r)^{3} \\ &= 8 \times \frac{4}{3}\pi r^{3} = 8\,V_{molecule} \end{aligned} \]
That excluded volume belongs to a pair, so each molecule is charged half of it, \(4\,V_{molecule}\). For a mole of gas, \(b = 4 \times V_{molecule} \times N_{A}\).
The answer 8 is the excluded volume of the pair without halving it. 2 and 16 come from halving or doubling at the wrong step; the correct factor is four times the actual volume of the molecules.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development