VKC ScienceChemistry for written exams Telegram

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

Q4(b). Capillary action, contact angle and γ=12hρ gr

  • 10 marks
  • Answered
Question

Explain the phenomenon of capillary action with showing relation between the angle of contact and the nature of liquid and derive the expression, γ=1/2hρ gr (Symbols have usual meanings).

Answer in brief

A liquid that wets the glass (θ<90°) rises in a narrow tube, one that does not (θ>90°) is depressed. At equilibrium the upward surface-tension force 2π rγθ balances the weight π r²hρ g; for θ≈0, γ=12hρ gr.

Key points

  • the rise or fall of a liquid in a narrow tube relative to the outside level. It comes from the competition between adhesion (liquid–glass) and cohesion (liquid–liquid), expressed through the contact angle θ between the liquid…
  • θ<90° (adhesion > cohesion; water, alcohol on clean glass): the liquid wets the wall, the meniscus is concave, and the liquid rises. For water on clean glass θ≈0.
  • θ>90° (cohesion > adhesion; mercury on glass): the liquid does not wet, the meniscus is convex, and the liquid is depressed (θ<0, h<0). At θ=90° the surface is flat and there is no change.
  • The liquid surface meets the wall along a circle of length 2π r. Surface tension acts along the surface, at angle θ to the wall; its vertical component pulls the column up: Fᵤₚ=2π r γθ

The full model answer

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

Capillary action and the contact angle

  1. Capillary action: the rise or fall of a liquid in a narrow tube relative to the outside level. It comes from the competition between adhesion (liquid–glass) and cohesion (liquid–liquid), expressed through the contact angle θ between the liquid surface and the wall, measured through the liquid.
  2. θ<90° (adhesion > cohesion; water, alcohol on clean glass): the liquid wets the wall, the meniscus is concave, and the liquid rises. For water on clean glass θ≈0.
  3. θ>90° (cohesion > adhesion; mercury on glass): the liquid does not wet, the meniscus is convex, and the liquid is depressed (θ<0, h<0). At θ=90° the surface is flat and there is no change.

Derivation

  1. The liquid surface meets the wall along a circle of length 2π r. Surface tension acts along the surface, at angle θ to the wall; its vertical component pulls the column up: Fᵤₚ=2π r γθ
  2. The liquid rises until the weight of the raised column (meniscus volume neglected) balances this pull: W=mg=π r²h ρ g
  3. At equilibrium: 2π rγθ=π r²hρ g, so γ=(hρ gr)/(2θ)
  4. For a liquid that wets the glass completely, θ=0, θ=1: γ=12hρ gr
  5. Same result from pressure: the concave surface of radius r has a pressure deficit 2γ/r beneath it (Laplace); the column rises until hρ g=2γ/r.

Example: water (γ=72.75 mN m⁻¹, ρ=998 kg m⁻³) in a tube of radius 0.10 mm rises h=2γ/ρ gr=14.9 cm; h∝1/r (Jurin's law).

Application. the capillary-rise method is a standard way to measure surface tension; the same action draws water up soil pores and plant xylem, ink into a fountain-pen nib and solvent up a TLC plate.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q4(b), 2020 Paper I: solution page 18
Solution page 18 of 45