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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 · 2019 · Paper I

Q1(e). Maxwell distribution of molecular speeds and the effect of temperature

  • 10 marks
  • Answered
Question

Write Maxwell's distribution of molecular velocities for gases. What is the effect of temperature on the distribution? Represent graphically.

Answer in brief

(dN_c)/(N)=4π((M)/(2π RT))^(3/2)c² e^-Mc²/2RT dc. On heating, the curve becomes lower and broader and its maximum moves to higher speed; the area stays 1.

Key points

  • the fraction of molecules with x-velocity between vₓ and vₓ+dvₓ is f(vₓ) dvₓ=((M)/(2π RT))^(1/2)e^-Mvₓ²/2RTdvₓ, a Gaussian centred on zero.
  • combining three components over a spherical shell of radius c (volume 4π c²dc) gives the formula above. f(c)=1/N(dN_c)/(dc) starts at zero (c² factor), rises to a maximum and falls in an exponential tail.
  • cₘₚ=√((2RT)/(M)), c=√((8RT)/(π M)), cᵣₘₛ=√((3RT)/(M)); ratio 1: 1.128: 1.225. For N₂ at 300 K: 422, 476 and 517 m s⁻¹.
  • cₘₚ grows as √T, so the maximum moves to the right (N₂: 422 m s⁻¹ at 300 K, 771 m s⁻¹ at 1000 K).

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. One component: the fraction of molecules with x-velocity between vₓ and vₓ+dvₓ is f(vₓ) dvₓ=((M)/(2π RT))^(1/2)e^-Mvₓ²/2RTdvₓ, a Gaussian centred on zero.
  2. Speeds: combining three components over a spherical shell of radius c (volume 4π c²dc) gives the formula above. f(c)=1/N(dN_c)/(dc) starts at zero (c² factor), rises to a maximum and falls in an exponential tail.
  3. Three speeds: cₘₚ=√((2RT)/(M)), c=√((8RT)/(π M)), cᵣₘₛ=√((3RT)/(M)); ratio 1: 1.128: 1.225. For N₂ at 300 K: 422, 476 and 517 m s⁻¹.
  4. Effect of temperature: cₘₚ grows as √T, so the maximum moves to the right (N₂: 422 m s⁻¹ at 300 K, 771 m s⁻¹ at 1000 K).
  5. The peak height fₘₐₓ=(4)/(π)((M)/(2RT))^(1/2)e⁻¹ falls as 1/√T (N₂: 1.97×10⁻³ to 1.08×10⁻³ s m⁻¹).
  6. The area is always 1, so the lower peak must spread out: more molecules at high speeds, fewer slow ones.

Application. the growth of the high-speed tail with temperature explains why reaction rates rise steeply on heating: only molecules in the tail carry energy above Eₐ.

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Model answer to Q1(e), 2019 Paper I: solution page 7
Solution page 7 of 40
Read the whole 2019 solution2019 question paper

All parts of Question 1