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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 · 2020 · Paper I

Q5(b). van der Waals constants of ethylene from T_c and p_c

  • 10 marks
  • Answered
Question

Calculate van der Waals constants for ethylene.
Given: critical temperature (T_c) = 282 K and critical pressure (P_c) = 50 atm.

Answer in brief

a=27R²T_c²64p_c=4.51 L² atm mol⁻² (0.457 Pa m⁶ mol⁻²); b=(RT_c)/(8p_c)=0.0578 L mol⁻¹ (5.78×10⁻⁵ m³ mol⁻¹).

Key points

  • the critical point is the point of inflexion of the T_c isotherm. For p=(RT)/(Vₘ-b)-aVₘ², setting the first and second derivatives to zero gives V_c=3b, p_c=a27b², T_c=(8a)/(27Rb)
  • Eliminating V_c (which is hard to measure): (T_c)/(p_c)=8b/R, so b=(RT_c)/(8p_c); and a=27b²p_c=27R²T_c²64p_c.
  • Given: T_c=282 K, p_c=50 atm, R=0.082 L atm K⁻¹ mol⁻¹.
  • b=((0.082)(282))/(8×50)=(23.12)/(400), b=0.0578 L mol⁻¹=5.78×10⁻⁵ m³ mol⁻¹

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. Critical constants: the critical point is the point of inflexion of the T_c isotherm. For p=(RT)/(Vₘ-b)-aVₘ², setting the first and second derivatives to zero gives V_c=3b, p_c=a27b², T_c=(8a)/(27Rb)
  2. Eliminating V_c (which is hard to measure): (T_c)/(p_c)=8b/R, so b=(RT_c)/(8p_c); and a=27b²p_c=27R²T_c²64p_c.
  3. Given: T_c=282 K, p_c=50 atm, R=0.082 L atm K⁻¹ mol⁻¹.
  4. b: b=((0.082)(282))/(8×50)=(23.12)/(400), b=0.0578 L mol⁻¹=5.78×10⁻⁵ m³ mol⁻¹
  5. a: a=27(0.082)²(282)²64×50=(14437)/(3200), a=4.51 L² atm mol⁻²=0.457 Pa m⁶ mol⁻²
  6. Check: V_c=3b=0.173 L mol⁻¹; Z_c=p_cV_c/RT_c=3/8=0.375 for every van der Waals gas (real gases give about 0.27–0.30, so b from T_c, p_c is only approximate).

Meaning: a measures the attraction between molecules (ethylene, a polarisable π system, has a fairly large a); b is about four times the volume of the molecules in a mole: from b=23π N_Ad³ the molecular diameter is about 358 pm.

Application. with a and b the Boyle temperature (T_B=a/Rb=952 K) and the inversion temperature of the Joule–Thomson effect (2a/Rb) can be estimated; ethylene, with T_c=282 K just below room temperature, cannot be liquefied by pressure alone at 25°C.

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Model answer to Q5(b), 2020 Paper I: solution page 24
Solution page 24 of 45
Read the whole 2020 solution2020 question paper

All parts of Question 5