Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2020 / Paper 1 / Section B / 10

A soap film has two air-water interfaces. The Young–Laplace equation therefore givesThe radius is half the specified diameter; using the single-interface droplet expression would miss a factor of two.

For the same ideal gas composition and temperature, $\rho =pM/(RT)$, hence $\Delta \rho /{\rho }_0=\Delta p/p_0=10^{-5}$. The bubble volume is $V=4\pi r^3/3=4.189\times 10^{-3}{\mathrm{m}}^3$. The gas's mass in excess of displaced ambient air isThis is a positive buoyant mass, meaning a net downward weight after subtracting ordinary air buoyancy. It is not an upward lifting capacity. Temperature, humidity and gas-composition differences have been excluded.

Treat the two film surfaces as an effective interfacial tension $\Gamma =2\gamma$. The relevant Bond number for the gas-density difference is $\mathrm{Bo}=\Delta \rho g{r}^2/\Gamma$. Substituting the Young–Laplace equation and the preceding density relation givesThus the formal crossover $\mathrm{Bo}>1$ gives $r>h_0/2\simeq 4.25\mathrm{km}$. This compares gravity on the excess gas with capillary restoration and neglects film weight. It does not say gravity-driven drainage of a real water film is negligible below kilometre scales. At this enormous radius, treating ambient pressure and density as uniform is itself no longer quantitatively reliable.

Approximating the skin as a uniform thin water shell with density $1000\mathrm{kg}{\mathrm{m}}^{-3}$,This is about 2500 times the $5.03\times 10^{-8}\mathrm{kg}$ buoyant mass of the enclosed gas. The film weight therefore dominates the bubble's downward gravitational force under the stated assumptions. Equal-temperature air inside does not provide a net lift to balance it: an apparently floating bubble can be transiently supported by air currents or have different internal temperature/composition. Film thickness is generally nonuniform and changes with drainage and evaporation; the uniform $1\mu \mathrm{m}$ thickness is an estimate, not a stable equilibrium condition.

For a spherical liquid droplet, the Kelvin equation isHere $V_m=0.018/957=1.881\times 10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, soThe formal answer is about $1.0\mathrm{nm}$. At molecular dimensions, bulk surface tension, incompressible-liquid and continuum curvature assumptions are only approximate. There is one liquid-vapour interface here, unlike a soap bubble's two interfaces.

Under uniform geometric scaling, write ${\mathbf{r}}_i^{\prime }=\lambda {\mathbf{r}}_i$. Each volume element contributes a factor ${\lambda }^3$, while the inverse-sixth-power separation contributes ${\lambda }^{-6}$. The double integral therefore scales as ${\lambda }^3{\lambda }^3{\lambda }^{-6}=1$, andThis assumes unchanged material Hamaker constant, pairwise-additive nonretarded interactions and the same intervening medium. It is a simultaneous scaling of both body sizes and their separation, not merely moving fixed bodies apart.

For nonretarded attraction between planar half-spaces, the interaction free energy per area is $W(h)=-A_H/(12\pi h^2)$. The Derjaguin approximation gives $F=2\pi R_{\mathrm{eff}}W(h)$, with $R_{\mathrm{eff}}=R_1{R}_2/(R_1+R_2)=a/2$ for equal spheres. ThusThe force is $13.0\mu \mathrm{N}$, attractive. Scaling both radii and the gap by two gives $F^{\prime }=F/2$, or $6.50\mu \mathrm{N}$ attractive. This agrees with the invariant energy $V\simeq -A_{H}a/(12h)$: differentiating with respect to a gap that doubles makes the force halve. The approximation requires $h\ll a$ and neglects roughness and retardation.