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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2021 · Paper I

Q1(d). Vapour pressure of water at 100 °C from its value at 95 °C

Question

The vapour pressure of water at 95 °C is found to be 634 mm. What would be the vapour pressure at a temperature of 100 °C? The heat of vapourization in this range of temperature may be taken as 40593 J mol⁻¹. [R = 8.314 J K⁻¹ mol⁻¹]

Answer in brief

By the integrated Clausius–Clapeyron equation, ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂) = 0.1777, so P₂ = 634 × 1.1945 = 757 mm Hg.

Key points

  • T₁ = 368.15 K and T₂ = 373.15 K; ΔHvap/R = 40593/8.314 = 4882.5 K.
  • 1/T₁ − 1/T₂ = 3.640 × 10⁻⁵ K⁻¹, so ln(P₂/P₁) = 0.1777 and P₂/P₁ = 1.1945.
  • P₂ = 757.3 mm Hg ≈ 757 mm Hg (0.996 atm).
  • Cross-check: 100 °C is the normal boiling point, so the value is only 3 mm Hg (0.4 %) short of 760 mm Hg.
  • A rise of 5 K raises the vapour pressure by 19 %.

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(d), 2021 Paper I: solution pages 5 and 6
Solution pages 5 and 6 of 59
Model answer to Q1(d), 2021 Paper I: solution pages 7 and 8
Solution pages 7 and 8 of 59
Read the whole 2021 solution2021 question paper

Topic notes: the phase rule, phase diagrams, the Clausius–Clapeyron equation, eutectics and partially miscible liquids, with every Civil Services, IFoS and Geo-Scientist PYQ →

CoversClausius–Clapeyron equationvapour pressureenthalpy of vaporisationboiling point of water

All parts of Question 1