Q2(b). Two mercury drops merge: change of area and energy
Find the change in surface area and energy when two identical mercury droplets of diameter 2 mm merged isothermally to one drop. Given that surface tension of mercury is 490 dyne cm⁻¹ at experimental temperature.
Answer in brief
The area falls by 5.18 mm² (5.18×10⁻⁶ m², 20.6 %); the surface energy falls by 2.54×10⁻⁶ J (25.4 erg), released as heat.
Key points
- Radius of the new drop. Volume is conserved: 43π R³=2×43π r³, so R=2^(1/3)r=1.260 mm.
- Initial area: A₁=2×4π r²=8(3.14)(1.0)²=25.12 mm²
- Final area: A₂=4π R²=4(3.14)(1.260)²=19.94 mm²
- Δ A=A₂-A₁=-5.18mm²=-5.18×10⁻⁶m² (-0.0518cm²) a fall of 20.6 %.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Given: r=1 mm =1.0×10⁻³ m (diameter 2 mm); γ=490 dyn cm⁻¹=0.490 N m⁻¹; π=3.14 (as printed).
- Radius of the new drop. Volume is conserved: 43π R³=2×43π r³, so R=2^(1/3)r=1.260 mm.
- Initial area: A₁=2×4π r²=8(3.14)(1.0)²=25.12 mm²
- Final area: A₂=4π R²=4(3.14)(1.260)²=19.94 mm²
- Change in area: Δ A=A₂-A₁=-5.18mm²=-5.18×10⁻⁶m² (-0.0518cm²) a fall of 20.6 %.
- Change in surface energy. γ is the work to create unit area isothermally, so Δ E=γ Δ A=(0.490N m⁻¹)(-5.18×10⁻⁶m²) Δ E=-2.54×10⁻⁶ J=-25.4 erg
- The surface energy decreases: 2.54×10⁻⁶ J is released (as heat, to keep the process isothermal). This is why small drops coalesce spontaneously: merging lowers the area and the surface Gibbs energy.
Application. the same loss of surface energy drives the coalescence of droplets in emulsions and the growth of large crystals at the cost of small ones (Ostwald ripening); emulsifiers lower γ to slow coalescence.
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All parts of Question 2
- 2(a)(i) Viscosity coefficient: dimension, SI unit, and its rise with T in a gas10 marks
- 2(a)(ii) Boyle temperature of a van der Waals gas15 marks
- 2(b) Two mercury drops merge: change of area and energy10 marks
- 2(c) Equilibrium constant for disproportionation of Cu⁺15 marks