The work done by a water droplet on the outside world, needed to increase the radius from $R$ to $R+\Delta R$ is
$$
\Delta W=\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R
$$
where $P$ is the pressure inside the drop and $P_o$ is the atmospheric pressure. In equilibrium, this should be equal to the increase in the surface energy $\mathcal{S} \Delta A=\mathcal{S} \cdot 8 \pi R \cdot \Delta R$, where $\mathcal{S}$ is the surface tension, and
$$
\Delta W_{\text {total }}=0, \Longrightarrow \Delta W_{\text {pressure }}=-\Delta W_{\text {surface }},
$$
resulting in
$$
\left(P-P_o\right) \cdot 4 \pi R^2 \cdot \Delta R=\mathcal{S} \cdot 8 \pi R \cdot \Delta R, \quad \Longrightarrow \quad\left(P-P_o\right)=\frac{2 \mathcal{S}}{R}
$$
In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures, and
$$
P_{\text {film }}-P_o=P_{\text {interior }}-P_{\text {film }}=\frac{2 \mathcal{S}}{R}
$$
leading to
$$
P_{\text {interior }}-P_o=\frac{4 \mathcal{S}}{R}
$$
Hence, the air pressure inside the bubble is larger than atmospheric pressure by $4 \mathcal{S} / R$.

When steam condenses on a solid surface, water either forms a droplet, or spreads on the surface. There are two ways to consider this problem:
Method 1: Energy associated with the interfaces
In equilibrium, the total energy associated with the three interfaces should be minimum, and therefore
$$
d E=S_{a w} d A_{a w}+S_{a s} d A_{a s}+S_{w s} d A_{w s}=0 .
$$

Since the total surface area of the solid is constant,
$$
d A_{a s}+d A_{w s}=0 .
$$
From geometrical considerations (see proof below), we obtain
$$
d A_{w s} \cos \theta=d A_{a w} .
$$
From these equations, we obtain
$$
d E=\left(S_{a w} \cos \theta-S_{a s}+S_{w s}\right) d A_{w s}=0, \quad \Longrightarrow \quad \cos \theta=\frac{S_{a s}-S_{w s}}{S_{a w}} .
$$
Proof of $d A_{w s} \cos \theta=d A_{a w}$ : Consider a droplet which is part of a sphere of radius $R$, which is cut by the substrate at an angle $\theta$. The areas of the involved surfaces are
$$
A_{w s}=\pi(R \sin \theta)^2, \quad \text { and } \quad A_{a w}=2 \pi R^2(1-\cos \theta) .
$$
Let us consider a small change in shape, accompanied by changes in $R$ and $\theta$. These variations should preserve the volume of water, i.e. constrained by
$$
V=\frac{\pi R^3}{3}\left(\cos ^3 \theta-3 \cos \theta+2\right)
$$
Introducing $x=\cos \theta$, we can re-write the above results as
$$
\left{\begin{aligned}
A_{w s} & =\pi R^2\left(1-x^2\right), \
A_{a w} & =2 \pi R^2(1-x), \
V & =\frac{\pi R^3}{3}\left(x^3-3 x+2\right) .
\end{aligned}\right.
$$

The variations of these quantities are then obtained from
$$
\left{\begin{aligned}
d A_{w s} & =2 \pi R\left[\frac{d R}{d x}\left(1-x^2\right)-R x\right] d x \
d A_{a w} & =2 \pi R\left[2 \frac{d R}{d x}(1-x)-R\right] d x \
d V & =\pi R^2\left[\frac{d R}{d x}\left(x^3-3 x+2\right)+R\left(x^2-x\right)\right] d x=0 .
\end{aligned}\right.
$$
From the last equation, we conclude
$$
\frac{1}{R} \frac{d R}{d x}=-\frac{x^2-x}{x^3-3 x+2}=-\frac{x+1}{(x-1)(x+2)}
$$

Substituting for $d R / d x$ gives,
$$
d A_{w s}=2 \pi R^2 \frac{d x}{x+2}, \quad \text { and } \quad d A_{a w}=2 \pi R^2 \frac{x \cdot d x}{x+2},
$$
resulting in the required result of
$$
d A_{a w}=x \cdot d A_{w s}=d A_{w s} \cos \theta .
$$
Method 2: Balancing forces on the contact line
Another way to interpret the result is to consider the force balance of the equilibrium surface tension on the contact line. There are four forces acting on the line: (1) the surface tension at the water-gas interface, (2) the surface tension at the solid-water interface, (3) the surface tension at the gas-solid interface, and (4) the force downward by solid-contact line interaction. The last force ensures that the contact line stays on the solid surface, and is downward since the contact line is allowed to move only horizontally without friction. These forces should cancel along both the $y$-direction $x$-directions. The latter gives the condition for the contact angle known as Young’s equation,
$$
\mathcal{S}{a s}=\mathcal{S}{a w} \cdot \cos \theta+\mathcal{S}{w s}, \Longrightarrow \cos \theta=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}} .
$$
The critical condition for the complete wetting occurs when $\theta=0$, or $\cos \theta=1$, i.e. for
$$
\cos \theta_C=\frac{\mathcal{S}{a s}-\mathcal{S}{w s}}{\mathcal{S}{a w}}=1 $$ Complete wetting of the substrate thus occurs whenever $$ \mathcal{S}{a w} \leq \mathcal{S}{a s}-\mathcal{S}{w s} .
$$

(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances surface tension effects are all important. At room temperature, the surface tension of water is $\mathcal{S}_o \approx 7 \times 10^{-2} \mathrm{Nm}^{-1}$. Estimate the typical length-scale that separates “large” and “small” behaviors. Give a couple of examples for where this length-scale is important.

• Typical length scales at which the surface tension effects become significant are given by the condition that the forces exerted by surface tension and relevant pressures become comparable, or by the condition that the surface energy is comparable to the other energy changes of interest.

Example 1: Size of water drops not much deformed on a non-wetting surface. This is given by equalizing the surface energy and the gravitational energy,
$$
S \cdot 4 \pi R^2 \approx m g R=\rho V g R=\frac{4 \pi}{3} R^4 g
$$

leading to
$$
R \approx \sqrt{\frac{3 S}{\rho g}} \approx \sqrt{\frac{3 \cdot 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^3 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2}} \approx 1.5 \times 10^{-3} \mathrm{~m}=1.5 \mathrm{~mm} .
$$
Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel (about $1 \mathrm{~atm})=$ surface tension of water, gives
$$
\pi_{g e l} \approx \frac{N}{V} k_B T \approx \frac{2 S}{R}
$$
where $N$ is the number of counter ions within the gel. Thus,
$$
R \approx\left(\frac{2 \times 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^5 \mathrm{~N} / \mathrm{m}^2}\right) \approx 10^{-6} \mathrm{~m} .
$$