VKC ScienceChemistry for written exams Telegram

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 · 2016 · Mains Paper II

Q1(e). Expansion cooling of helium

Question

Calculate the fall in temperature of He initially at 15° C when it is suddenly expanded to eight times its volume. Given γ=5/3.

Answer in brief

Under the intended reversible-adiabatic model, the fall is 216.1 K. Sudden expansion alone does not specify this model.

Key points

  • For reversible adiabatic expansion of an ideal gas, TVγ-1=constant. With T₁=288.15 K and V₂/V₁=8,T₂=288.15(1/8)2/3=72.04 K.
  • Thus T₁-T₂=216.11 K. The same numerical temperature difference is 216.11° C.
  • The word suddenly does not guarantee reversibility.

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(e), 2016 Mains Paper II: solution pages 3 and 4
Solution pages 3 and 4 of 19
Read the whole 2016 solution2016 question paper

CoversExpansion cooling of heliumAdiabatic processes

All parts of Question 1