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UPSC Combined Geo-Scientist

Chemist

  • 3 descriptive papers
  • 200 marks each
  • 3 h per paper

Mains Paper II covers physical chemistry.

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.

Model answer to Q1(b), 2018 Mains Paper II: solution pages 1 and 2
Solution pages 1 and 2 of 17

CoversFugacity under repulsionReal gases