Fluid Mechanics Concept Page - 16

Definition

Types of intermolecular Forces and range of influence

Intermolecular forces are of two types - short range (<3A˚$<3\mathring{A}$) and long range forces (van der waals forces, >3A˚$>3\mathring{A}$).
Short range forces are repulsive whereas the long range forces are attractive. They are responsible for surface tension, friction, etc.

Definition

Cohesive and Adhesive Forces

Cohesion is the property of like molecules (of the same substance) to stick to each other due to mutual attraction. Adhesion is the property of different molecules or surfaces to cling to each other. For example, solids have high cohesive properties so they do not stick to the surfaces they come in contact. On the other hand, gases have weak cohesion. Water has both cohesive and adhesive properties.

Water molecules stick to each other to form a sphere. This is the result of cohesive forces. When contained in a tube, the water molecules touching the surface of the container are at a higher level. This is due to the adhesive force between the water molecules and the molecules of the container.

Example

Examples of cohesive and adhesive forces

Some of the real-life examples of cohesive and adhesive forces are as follows:
1) Cohesion is the term for molecules of a substance sticking together. One of the most common examples is water beading up on a hydrophobic surface.
2) Water molecules are not only attracted to each other, but to any molecule with positive or negative charges. When a molecule attracts a different substance, this is termed adhesion.
Think about what happens when you dip one end of a piece of paper towel into a glass of water. The water will climb up the fibers of the paper, getting it wet above the level of the water in the glass.

Definition

Effect of temperature in surface tension

In general, surface tension decreases when temperature increases because cohesive forces decrease with an increase of molecular thermal activity. The influence of the surrounding environment is due to the adhesive action of  liquid molecules that they have at the interface.

Definition

Variation of surface tension with temperature

Surface tension is the energy that is required to stretch the surface of a liquid one incremental amount of area (Assume the surface of the liquid is in contact with air.). As one might expect this requires an input of energy, that is, the surface tension is positive. It is easier to stretch the surface of a liquid the warmer it gets, because the molecules at the surface are "hopping around" more, the higher the temperature is. So the surface tension always decreases with increasing temperature.

Definition

Detergent action using surface tension

Detergents help the cleaning of clothes by lowering the surface tension of the water so that it more readily soaks into pores and soiled areas.

Definition

Formation of droplets

Spray some mercury and some water on a glass surface. It will be observed that mercury forms spherical drops while water spreads. Then reason for this difference is that the force of cohesion between the molecules of mercury is more than the force of adhesion between the molecules of mercury and glass. Whereas, the force of cohesion between the molecules of water is less than the force of adhesion between the molecules of water and glass.

Example

Surface Energy of a liquid drop and work done in creating it

Surface energy of a liquid drop is given by the product of surface area and surface tension. 
Hence for a spherical liquid drop, surface energy is given by:
S=AT$S=AT$ 
S=4πR2T$S=4\pi R^{2}T$

Work done in creating a drop is equal to the surface energy of the drop.

Example

Change in surface energy of a block with changing total surface area

Example: A spherical liquid drop of radius R$R$ is divided into eight equal
droplets. If the surface tension is T$T$, then find the work done in this
process?

Solution:
Let us say r$r$ is the radius of each eight of the bubble. Since volume of water remains the same, we have
43πR3=8×43πr3⇒r=R2$$\begin{gathered} {\displaystyle \frac{4}{3}}\pi R^3=8\times {\displaystyle \frac{4}{3}}\pi r^3 \\ \Rightarrow r={\displaystyle \frac{R}{2}} \end{gathered}$$
So the work done in this process will be = change in the surface energy i.e.
(8×4πr2×T)−(4πR2×T)=(8×4π(R2)2×T)−(4πR2×T)∵r=R2=8πR2T−4πR2T=4πR2T$$\begin{gathered} (8\times 4\pi r^2\times T)-(4\pi R^2\times T) \\ =(8\times 4\pi {\left(\frac{R}{2}\right)}^2\times T)-(4\pi R^2\times T)\because r=\frac{R}{2} \\ =8\pi R^{2}T-4\pi R^{2}T=4\pi R^{2}T \end{gathered}$$

Shortcut

Finding excess pressure in a liquid drop

For a liquid drop in equilibrium, extra surface energy in expanding the drop from radius r$r$ to Δr$\mathrm{\Delta }r$ is balanced by the energy gain due to pressure difference inside and outside the bubble.
Hence, [4π(r+Δr)2−4πr2]S=8πrΔrS=ΔP4πr2Δr$[4\pi (r+\mathrm{\Delta }r{)}^2-4\pi r^2]S=8\pi r\mathrm{\Delta }rS=\mathrm{\Delta }P4\pi r^2\mathrm{\Delta }r$
Solving, ΔP=2Sr$\mathrm{\Delta }P={\displaystyle \frac{2S}{r}}$

Example: A soap bubble has radius R$R$ and thickness d(<<R)$d(<<R)$ as shown. It collapses into a spherical drop. If the ratio of excess pressure in the drop to the excess pressure inside the bubble is (Rxd)13${\left(\frac{R}{xd}\right)}^{\frac{1}{3}}$. Find x$x$.

Solution:
Let r  be the radius of the water drop formed.$\text{Let}r\text{ be the radius of the water drop formed.}$Since the volume of the water forming bubble and drop is same,$\text{Since the volume of the water forming bubble and drop is same,}$43π(R3−(R−d)3)=43πr3${\displaystyle \frac{4}{3}}\pi (R^3-(R-d{)}^3)={\displaystyle \frac{4}{3}}\pi r^3$
⟹r3≈3R2d (neglecting d2  and d3)$⟹r^3\approx 3R^{2}d(\text{neglecting}{d}^2\text{ and}{d}^3)$Ratio of excess pressure in the drop to the excess pressure inside the bubble is given by,$\text{Ratio of excess pressure in the drop to the excess pressure inside the bubble is given by,}$
Ratio=2σ/r4σ/R$\text{Ratio}={\displaystyle \frac{2\sigma /r}{4\sigma /R}}$
Ratio=12(Rr)$\text{Ratio}={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{R}{r}}\right)$
Substituting the value of r gives: Ratio=(R24d)1/3$\text{Substituting the value of }\text{r}\text{ gives}:\text{Ratio}={\left({\displaystyle \frac{R}{24d}}\right)}^{1/3}$
So, x=24

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