UPSC Combined Geo-Scientist, Mains · 2018 · Mains Paper II
Q1(b). Fugacity under repulsion
Question
Justify why the fugacity f of a real gas exceeds its observed pressure p when intermolecular repulsive interactions dominate.
Answer in brief
Repulsion raises the escaping tendency and chemical potential relative to the ideal gas, giving φ=f/p>1 when the net residual contribution is positive.
Key points
- Define μ=μ°+RTln (f/p°) and the ideal reference at the same pressure as μid=μ°+RTln (p/p°). Hence μ-μid=RTln (f/p).
- At fixed temperature, ln φ=∫₀p (Z-1) dp/p, where Z=pVₘ/(RT). Repulsive/excluded-volume effects tend to make Z>1.
- If the positive repulsive contribution dominates this integral, ln φ>0 and f>p.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

CoversFugacity under repulsionReal gases
All parts of Question 1
- 1(a) Equal molecular speeds5 marks
- 1(b) Fugacity under repulsion5 marks
- 1(c) Wetting of glass5 marks
- 1(d) FCC particle count5 marks
- 1(e) Joule–Thomson cooling5 marks
- 1(f) Gibbs and Helmholtz changes5 marks
- 1(g) Equilibrium-constant conventions5 marks
- 1(h) Very dilute hydrochloric acid5 marks
- 1(i) Pressure controlling boiling5 marks
- 1(j) Primary kinetic salt effect5 marks
- 1(k) Heat of spontaneous adsorption5 marks
- 1(l) Conductivity as an intensive property5 marks
- 1(m) Anthracene photodimerization5 marks
- 1(n) Particle-in-a-box eigenvalue5 marks
- 1(o) Second Balmer line5 marks
- 1(p) Normal-mode counts5 marks