# 机械学和物质代写|Mechanics and Matter PX1121 Cardiff University Assignment

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## Instructions:

Mechanics is the branch of physics that deals with the motion of objects under the influence of forces. It is concerned with describing and predicting the behavior of physical systems, including both macroscopic objects like cars and airplanes, as well as microscopic particles like atoms and molecules.

Matter, on the other hand, refers to anything that has mass and takes up space. It is the basic building block of the universe, and includes everything from tiny subatomic particles to massive planets and stars.

Mechanics and matter are intimately connected, as the behavior of matter is governed by the laws of mechanics. For example, the motion of a ball thrown into the air can be described using the principles of mechanics, as can the behavior of atoms and molecules in a gas or liquid.

Understanding the relationship between mechanics and matter is crucial for many areas of science and technology, including engineering, materials science, and many others.

Surface tension: Thermodynamic properties of the interface between two phases are described by a state function called the surface tension $\mathcal{S}$. It is defined in terms of the work required to increase the surface area by an amount $d A$ through $d W=\mathcal{S} d A$.
(a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius $R$ is larger than outside pressure by $2 \mathcal{S} / R$. What is the air pressure inside a soap bubble of radius $R$ ?

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
$$P_{\text {film }}-P_o=P_{\text {interior }}-P_{\text {film }}=\frac{2 \mathcal{S}}{R}$$
$$P_{\text {interior }}-P_o=\frac{4 \mathcal{S}}{R}$$
Hence, the air pressure inside the bubble is larger than atmospheric pressure by $4 \mathcal{S} / R$.

(b) A water droplet condenses on a solid surface. There are three surface tensions involved $\mathcal{S}{a w}, \mathcal{S}{s w}$, and $\mathcal{S}_{s a}$, where $a, s$, and $w$ refer to air, solid and water respectively. Calculate the angle of contact, and find the condition for the appearance of a water film (complete wetting).

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.

(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,
$$S \cdot 4 \pi R^2 \approx m g R=\rho V g R=\frac{4 \pi}{3} R^4 g$$

$$R \approx \sqrt{\frac{3 S}{\rho g}} \approx \sqrt{\frac{3 \cdot 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^3 \mathrm{~kg} / \mathrm{m}^3 \times 10 \mathrm{~m} / \mathrm{s}^2}} \approx 1.5 \times 10^{-3} \mathrm{~m}=1.5 \mathrm{~mm} .$$
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,
$$R \approx\left(\frac{2 \times 7 \times 10^{-2} \mathrm{~N} / \mathrm{m}}{10^5 \mathrm{~N} / \mathrm{m}^2}\right) \approx 10^{-6} \mathrm{~m} .$$