Surface Energy, Drops, and Coalescence
At fixed temperature and composition, the reversible change in surface energy equals surface tension times the change in interfacial area, with each physical interface counted once.
Why this shows up in the exam
Atomisation and sprays · Drop coalescence and emulsions · Estimating work to create bubbles or films
Learn the idea
Creating interface requires energy, so splitting raises area while coalescence releases surface energy. Many small drops expose more total area than one large drop of the same volume. Breaking a drop therefore needs work, while merging drops can release energy as heat or motion.
🧠 Memory hook: Conserve volume first, compare total area second.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Delta U_s = T Delta A — surface-energy change when T is effectively constant and the relevant interfaces are counted
- N r³ = R³ — volume conservation when N equal spherical drops of radius r form one sphere of radius R
- W_required = T(A_final-A_initial) — minimum reversible work; a positive value means new area must be created
How to approach it
- 1Use volume conservation to find the new radius
- 2Write initial and final total interfacial areas
- 3Multiply the signed area change by the correct surface tension
Common slip-ups that cost marks
- •Conserving radius instead of volume
- •Forgetting both surfaces of a soap bubble or film
- •Reporting released energy with the sign of required work
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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