Surface tension — origin, Young-Laplace equation: Exercises

Throughout, remember the two headline results built in the parent note: $Δ P_{drop} = \frac{2 γ}{r}, Δ P_{soap bubble} = \frac{4 γ}{r}, Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) .$ Here $γ$ (surface tension) is force-per-length or energy-per-area, $r$ is the sphere radius, and $R_{1}, R_{2}$ are the two principal radii (how sharply the surface bends in two perpendicular directions). Unless told otherwise, take $γ_{water} = 0.072 N/m$ and atmospheric pressure $P_{0} = 1.013 \times 1 0^{5} Pa$ .

Level 1 — Recognition

Recall Solution — L1·Q1

WHAT to spot: the only difference is the number of liquid–air interfaces.

A drop has one interface (liquid inside, air outside) → $Δ P = \frac{2 γ}{r}$ .
A soap bubble is a thin film with air on both sides → two interfaces → $Δ P = \frac{4 γ}{r}$ .

With $r = 2 \times 1 0^{- 3} m$ : $Δ P_{drop} = \frac{2 ( 0.072 )}{2 \times 1 0 ^{- 3}} = 72 Pa, Δ P_{bubble} = \frac{4 ( 0.072 )}{2 \times 1 0 ^{- 3}} = 144 Pa .$ The bubble’s excess pressure is exactly twice the drop’s — because it has twice as many skins squeezing.

Recall Solution — L1·Q2

Force per length: $\frac{N}{m}$ . Energy per area: $\frac{J}{m ^{2}} = \frac{N \cdot m}{m ^{2}} = \frac{N}{m}$ . Same unit $\Rightarrow$ the two definitions describe one physical number. That is why the parent note could write $γ = F / L = energy / area$ .

Level 2 — Application

Recall Solution — L2·Q1

WHY energy, not pressure: the question asks for work done against the surface, which is $γ \times (area created)$ . A soap bubble has two surfaces (inner + outer), each of area $4 π R^{2}$ : $A_{total} = 2 \times 4 π R^{2} = 8 π R^{2} .$ $W = γ A_{total} = 0.030 \times 8 π (0.03)^{2} = 0.030 \times 8 π \times 9 \times 1 0^{- 4} .$ $W = 6.786 \times 1 0^{- 4} J \approx 6.79 \times 1 0^{- 4} J .$ What it means: less than a milli-joule — surfaces are cheap, which is why a gentle breath makes a big bubble.

Recall Solution — L2·Q2

A film has two faces, so both pull down on the wire: $F = 2 γ L = 2 (0.025) (0.08) = 4.0 \times 1 0^{- 3} N = 4.0 mN .$ WHY the 2: forgetting it halves the answer — the classic $F = 2 γ L$ frame result.

Level 3 — Analysis

Recall Solution — L3·Q1

WHY compare $1/ r$ : each bubble's excess pressure is $Δ P = \frac{4 γ}{r} \propto \frac{1}{r}$ . Smaller $r$ ⇒ larger excess pressure.

Bubble B ( $r = 2 cm$ ) has the higher internal pressure.
Air therefore flows from B (small) to A (big) — the small bubble shrinks, the big one grows.

(b) Both bubbles sit in the same atmosphere $P_{0}$ , so the driving difference is the difference of excess pressures: $Δ P_{A} = \frac{4 ( 0.030 )}{0.04} = 3.0 Pa, Δ P_{B} = \frac{4 ( 0.030 )}{0.02} = 6.0 Pa .$ $Δ P_{A} - Δ P_{B} = 3.0 - 6.0 = - 3.0 Pa .$ The negative sign confirms B's pressure exceeds A's by $3.0 Pa$ , pushing air toward A. See the figure: the tighter (redder) skin squeezes harder.

Figure — Surface tension — origin, Young-Laplace equation

Recall Solution — L3·Q2

WHAT the middle film feels: on one side is bubble B's inside (higher pressure $P_{B}$ ), on the other side is bubble A's inside (lower pressure $P_{A}$ ). The film curves toward the low-pressure side (bulges into A), and its own Young–Laplace jump must equal the pressure difference across it. A single dividing film still has two faces, so: $jump across mid film \frac{4 γ}{r _{mid}} = P_{B} - P_{A} = \frac{4 γ}{r _{B}} - \frac{4 γ}{r _{A}} .$ Cancel $4 γ$ : $\frac{1}{r _{mid}} = \frac{1}{r _{B}} - \frac{1}{r _{A}} = \frac{1}{0.02} - \frac{1}{0.04} = 50 - 25 = 25 m^{- 1} .$ $r_{mid} = \frac{1}{25} = 0.04 m = 4 cm = r_{A} .$ Neat result: $r_{mid} = \frac{r _{A} r _{B}}{r _{A} - r _{B}}$ ; here it equals $r_{A}$ because $r_{A} = 2 r_{B}$ .

Level 4 — Synthesis

Recall Solution — L4·Q1

(a) Volume is conserved (liquid is nearly incompressible): $8 \times \frac{4}{3} π a^{3} = \frac{4}{3} π R^{3} \Rightarrow R^{3} = 8 a^{3} \Rightarrow R = 2 a = 2 mm .$ (b) Compare surface areas (a drop has one interface): $\frac{A _{big}}{A _{small,total}} = \frac{4 π R ^{2}}{8 \times 4 π a ^{2}} = \frac{R ^{2}}{8 a ^{2}} = \frac{( 2 a ) ^{2}}{8 a ^{2}} = \frac{4}{8} = \frac{1}{2} .$ Surface energy $= γ A$ , so the energy halves — the merged drop stores half the surface energy of the eight droplets. (c) Energy went down, so surface energy is released (appears as heat / slight warming). This is why small droplets spontaneously coalesce: it lowers the system's energy.

Recall Solution — L4·Q2

(a) A hemispherical meniscus is one interface with $R_{1} = R_{2} = r$ : $Δ P = γ (\frac{1}{r} + \frac{1}{r}) = \frac{2 γ}{r} = \frac{2 ( 0.072 )}{2.0 \times 1 0 ^{- 4}} = 720 Pa .$ The pressure just below the curved meniscus is $720 Pa$ lower than atmospheric (the surface curves away from the liquid). (b) That deficit is balanced by the weight of the raised column, $ρ g h$ : $ρ g h = \frac{2 γ}{r} \Rightarrow h = \frac{2 γ}{ρ g r} = \frac{2 ( 0.072 )}{1000 \times 9.8 \times 2.0 \times 1 0 ^{- 4}} .$ $h = \frac{0.144}{1.96} = 0.07347 m \approx 7.35 cm .$ What it means: the same Young–Laplace jump that pressurises a drop pulls water up a thin tube — the physics is identical, just curved the other way.

Level 5 — Mastery

Recall Solution — L5·Q1

WHY $P V$ conservation: temperature is fixed, so for the trapped air Boyle's law $P V = const$ holds for the total enclosed air. Each bubble's internal absolute pressure is $P = P_{0} + \frac{4 γ}{r}$ (excess pressure added to atmosphere), and its volume is $\frac{4}{3} π r^{3}$ .

Conserve total $P V$ : $(P_{0} + \frac{4 γ}{r _{1}}) \frac{4}{3} π r_{1}^{3} + (P_{0} + \frac{4 γ}{r _{2}}) \frac{4}{3} π r_{2}^{3} = (P_{0} + \frac{4 γ}{r}) \frac{4}{3} π r^{3} .$ Cancel $\frac{4}{3} π$ and expand: $P_{0} r^{3} + 4 γ r^{2} = P_{0} (r_{1}^{3} + r_{2}^{3}) + 4 γ (r_{1}^{2} + r_{2}^{2}) .$ Plug numbers ( $r_{1} = 0.03, r_{2} = 0.04 m, 4 γ = 0.12$ ):

$P_{0} (r_{1}^{3} + r_{2}^{3}) = 1.013 \times 1 0^{5} (2.7 \times 1 0^{- 5} + 6.4 \times 1 0^{- 5}) = 1.013 \times 1 0^{5} \times 9.1 \times 1 0^{- 5} = 9.2183 J-ish (Pa\cdotm^{3}) .$
$4 γ (r_{1}^{2} + r_{2}^{2}) = 0.12 (9 \times 1 0^{- 4} + 1.6 \times 1 0^{- 3}) = 0.12 \times 2.5 \times 1 0^{- 3} = 3.0 \times 1 0^{- 4} .$

So the equation is: $1.013 \times 1 0^{5} r^{3} + 0.12 r^{2} = 9.2183 + 3.0 \times 1 0^{- 4} = 9.21860.$ The surface terms are tiny next to $P_{0} r^{3}$ , so to good accuracy $r^{3} \approx r_{1}^{3} + r_{2}^{3} = 9.1 \times 1 0^{- 5}$ : $r \approx (9.1 \times 1 0^{- 5})^{1/3} = 0.04497 m \approx 4.50 cm .$ Solving the full cubic numerically shifts $r$ by well under $0.01 mm$ , so $r \approx 4.50 cm$ . Insight: at ordinary atmospheric pressure the surface-energy terms barely matter — the merged volume is essentially $r_{1}^{3} + r_{2}^{3}$ . The Young–Laplace correction only dominates in vacuum, where $P_{0} \to 0$ .

Recall Solution — L5·Q2

With $P_{0} = 0$ the $P V$ equation loses its $P_{0} r^{3}$ term. Using $P = \frac{4 γ}{r}$ : $\frac{4 γ}{r _{1}} \cdot \frac{4}{3} π r_{1}^{3} + \frac{4 γ}{r _{2}} \cdot \frac{4}{3} π r_{2}^{3} = \frac{4 γ}{r} \cdot \frac{4}{3} π r^{3} .$ Cancel $\frac{4}{3} π \cdot 4 γ$ : $r_{1}^{2} + r_{2}^{2} = r^{2} \Rightarrow r = r_{1}^{2} + r_{2}^{2} = 0.0 3^{2} + 0.0 4^{2} = 0.0025 = 0.05 m = 5 cm .$ WHY it changed: with no external pressure, the only pressure comes from surface tension, so the conserved quantity becomes surface area ( $4 γ / r \times r^{3} \propto r^{2}$ ) rather than volume. The merged bubble is a 3-4-5 right triangle in radii — a clean signature that areas add, not volumes. In atmosphere, the huge $P_{0}$ makes volume the near-conserved quantity instead (Q1). This is the same "which quantity is conserved" theme as the coalescing drops, now sharpened.

Recall wrap-up

Recall Cover-and-check

Drop vs soap-bubble excess pressure? ::: $2 γ / r$ vs $4 γ / r$ (one surface vs two). Work to blow a bubble of radius $R$ ? ::: $γ \times 8 π R^{2}$ (two surfaces). Eight drops of radius $a$ merge — big radius and energy factor? ::: $R = 2 a$ ; surface energy halves. Coalescing bubbles in air — what is nearly conserved? ::: total volume, $r^{3} \approx r_{1}^{3} + r_{2}^{3}$ . Coalescing bubbles in vacuum — what is conserved? ::: total area, $r^{2} = r_{1}^{2} + r_{2}^{2}$ .