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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 · 2026 · Mains Paper II

Q2. Critical, Boyle and inversion temperatures of a van der Waals gas

  • 10 marks
  • Topic: Kinetic theory and the gaseous state
  • Answered
Question

A non-ideal gas has three characteristic temperatures: Critical temperature (T_C), Boyle temperature (T_B) and Inversion temperature (Tᵢ). Write expressions of T_C and T_B when the gas is van der Waals. What are the significances of T_C and T_B?

Answer in brief

T_C=(8a)/(27Rb), T_B=a/Rb (and Tᵢ=(2a)/(Rb)), so T_B=27/8T_C=3.375 T_C. Above T_C a gas cannot be liquefied by pressure; at T_B the second virial coefficient is zero and the gas is nearly ideal over a wide range of P.

Key points

  • from (∂ P/∂ V)=0 and (∂²P/∂ V²)=0 for P=(RT)/(V-b)-(a)/(V²) we get V_c=3b, P_c=(a)/(27b²) and T_C=(8a)/(27Rb).
  • expand Z=(PVₘ)/(RT)=1+(B(T))/(Vₘ)+… with B(T)=b-a/RT. T_B is where B=0: b-(a)/(RT_B)=0 ⇒ T_B=a/Rb
  • μ_(JT)=0 at low pressure gives Tᵢ=(2a)/(Rb)=2T_B.
  • the highest temperature at which liquid and vapour exist together. Above it only one fluid phase (supercritical fluid); no pressure liquefies the gas.

The full model answer

Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.

Expressions

  1. T_C: from (∂ P/∂ V)=0 and (∂²P/∂ V²)=0 for P=(RT)/(V-b)-(a)/(V²) we get V_c=3b, P_c=(a)/(27b²) and T_C=(8a)/(27Rb).
  2. T_B: expand Z=(PVₘ)/(RT)=1+(B(T))/(Vₘ)+… with B(T)=b-a/RT. T_B is where B=0: b-(a)/(RT_B)=0 ⇒ T_B=a/Rb
  3. Tᵢ (maximum inversion temperature): μ_(JT)=0 at low pressure gives Tᵢ=(2a)/(Rb)=2T_B.

Significance

  • T_C: the highest temperature at which liquid and vapour exist together. Above it only one fluid phase (supercritical fluid); no pressure liquefies the gas.
  • T_B: attraction and repulsion exactly cancel at low pressure, Z=1 and (∂ Z/∂ P)_T=0 as P→0: the gas obeys Boyle's law over a wide range of pressure. Z<1 below T_B (attraction), Z>1 above T_B.
  • Tᵢ: the Joule–Thomson effect cools the gas only below Tᵢ (used in Linde liquefaction).

experimental values for N₂: T_C=126 K, T_B=327 K, Tᵢ=621 K; for H₂: T_C=33 K, T_B=110 K, Tᵢ=205 K (the van der Waals formulas give only approximate numbers): H₂ must be pre-cooled below 205 K before a J–T expansion will liquefy it.

Model answer to Q2, 2026 Mains Paper II: solution page 12
Solution page 12 of 22
Read the whole 2026 solution2026 question paper

Coversvan der Waals constantsCriticalBoyle and inversion temperatures of a van der Waals gas

All parts of Question 2

  • 2 Critical, Boyle and inversion temperatures of a van der Waals gas10 marks