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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 · 2018 · Paper I

Q4(c)(i). Effusion of H₂ and N₂ through a fine glass filter

  • 10 marks
  • Answered
Question

One side of glass cylinder is initially filled with H₂ at a pressure P and other side with N₂ at a pressure 2P. This glass cylinder is closed at both ends and a glass filter in the middle of the cylinder divides it into two compartments of equal volume. Describe what happens, starting at time, t=0 at constant temperature and assuming that the pores of filter have a diameter much smaller than the mean free paths of the gases.

Answer in brief

The gases pass through the filter by effusion, each independently, at a rate proportional to p/√M. H₂ effuses 3.73 times faster per unit pressure, so at first more molecules pass from the H₂ side to the N₂ side: the pressure on the H₂ side first falls below P and that on the N₂ side rises above 2P. Later both gases even out and each side ends at 1.5P (p_(H₂)=P/2, p_(N₂)=P on both sides).

Key points

  • The pores are much narrower than the mean distance between collisions (mean path λ), so molecules pass through them one at a time without colliding in the pore: effusion, not bulk (viscous) flow. There is no pressure-driven…
  • Rate of effusion of each gas through the filter /√M (Graham's law). Per unit pressure H₂ goes √(28.01/2.016)=3.73 times faster than N₂.
  • H₂ flux to the right ∝ P/√(2.016); N₂ flux to the left ∝2P/√(28.01). Ratio =1.86: more molecules leave the H₂ side than enter it.
  • So the pressure on the H₂ side first falls below P and the pressure on the N₂ side rises above 2P: the pressure difference across the filter grows for a while, although the gases move “against” it. In a simple model the low…

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. The pores are much narrower than the mean distance between collisions (mean path λ), so molecules pass through them one at a time without colliding in the pore: effusion, not bulk (viscous) flow. There is no pressure-driven stream; each gas moves independently of the other and of the total pressure.
  2. Rate of effusion of each gas through the filter /√M (Graham's law). Per unit pressure H₂ goes √(28.01/2.016)=3.73 times faster than N₂.
  3. At t=0: H₂ flux to the right ∝ P/√(2.016); N₂ flux to the left ∝2P/√(28.01). Ratio =1.86: more molecules leave the H₂ side than enter it.
  4. So the pressure on the H₂ side first falls below P and the pressure on the N₂ side rises above 2P: the pressure difference across the filter grows for a while, although the gases move “against” it. In a simple model the low side reaches about 0.92P and the high side 2.08P.
  5. H₂ soon becomes nearly uniform; N₂ keeps effusing slowly to the left, so the left pressure then rises and the right pressure falls.
  6. Final state: each gas fills both compartments uniformly. Amounts are conserved and the volumes equal, so on each side p_(H₂)=P/2 and p_(N₂)=P: pₜₒₜₐₗ=1.5P on both sides, same composition
  7. The pressure equalises only because both gases equalise; through such a filter a pressure difference is not removed by flow. At constant temperature on both sides there is no thermal transpiration.

Application. the same faster effusion of the lighter molecule is the basis of isotope separation of ²³⁵UF₆ from ²³⁸UF₆ by gaseous diffusion through porous barriers, stage after stage.

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Model answer to Q4(c)(i), 2018 Paper I: solution page 26
Solution page 26 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 4