Surface Tension

A hot air balloon is a sphere of radius $8\text{ m}$. The air inside is at a temperature of $60° \text{C}$. How large a mass can the balloon lift when the outside temperature is $20° \text{C}$? Assume air is an ideal gas, $R=8.314 J mol{e}^{-1}{K}^{-1}$, $1 atm=1.013\times {10}^5 \text{Pa}$ ; the membrane tension is $5 {\text{Nm}}^{-1}$.

Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water$L_v=540{\text{ kcal kg}}^{-1}$, the mechanical equivalent of heat $J=4.2{\text{ J cal}}^{-1}$, density of water ${\rho }_w={10}^3 {\text{kg l}}^{-1}$ , Avagadro’s Number  $N_A=6\times {10}^{26} {\text{k mole}}^{-1}$ and the molecular weight of water $M_A=\text{18 kg for 1 k mole}$.

Calculate the surface tension of water. Consider the intermolecular force between water molecules is $2.05\times {10}^{-11} \mathrm{N}$.
 

 Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water $L_v=540{\text{ kcal kg}}^{-1}$, the mechanical equivalent of heat $J=4.2 {\text{J cal}}^{-1}$, density of water ${\rho }_w={10}^3 {\text{kg/m }}^{-3}$, Avagadro’s number  $N_A=6.0\times {10}^{26} {\text{kmole}}^{-1}$and the molecular weight of water $M_A=18\text{ kg for 1 k mole}$. During vaporisation a molecule overcomes a force $F$, assumed constant, to go from an inter-molecular distance $d$ to $d´$. Estimate the value of $F$. Take the energy is required for one molecule of water to evaporate is  $6.8\times {10}^{-20} \mathrm{J}$.

Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water $L_v=540 {\text{kcal kg}}^{-1}$ , the mechanical equivalent of heat $J=4.2\text{ J} {\text{cal}}^{-1}$ ,density of water ${\rho }_w={10}^3 {\text{kg l}}^{-1}$, Avagadro’s Number $N_A=6.0\times {10}^{26} k mol{e}^{-1}$ and the molecular weight of water $M_A=\text{18 kg for 1 k mole}$ and $1 \text{g}$ of water in the vapour state at $1 \text{atm}$ occupies $1601 {\text{cm}}^3$, estimate the intermolecular distance at boiling point, in the vapour state.

Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water $L_v=540{\text{ kcal kg}}^{-1}$, the mechanical equivalent of heat $J=4.2{\text{ J cal}}^{-1}$ , density of water ${\rho }_w={10}^3 {\text{kg l}}^{-1}$ , Avagadro’s Number $N_A=6.0 \times {10}^{26} {\left(\mathrm{kmole}\right)}^{-1}$  and, the molecular weight of water $M_A=18\text{ kg for 1 k mole.}$  Show that the inter–molecular distance for water is $d={\left[\frac{M_A}{N_A}\times \frac{1}{{\rho }_w}\right]}^{\frac{1}{3}}$ and find its value.

Surface tension is exhibited by liquids due to the force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at the boiling point of the liquid. Given that the latent heat of vaporisation for water $L_v=540 {\text{kcal kg}}^{-1}$, the mechanical equivalent of heat  $J=4.2 {\text{J cal}}^{-1}$, density of water ${\rho }_w=103 {\text{kg/m}}^3$ , Avagadro’s Number $N_A=6.0\times {10}^{26} {\text{kmol}}^{-1}$, and the molecular weight of water $M_A=18 \text{kg for 1 kmol}$, estimate the energy required for one molecule of water to evaporate. 

The surface tension and vapour pressure of water at $20°\text{C}$ is $7.28\times {10}^{-2} {\text{Nm}}^{-1}$ and $2.33\times {10}^3 \text{Pa}$, respectively. What is the radius of the smallest spherical water droplet which can be formed without evaporating at $20° \text{C}$?
 

If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius $R$, break into $N$ small droplets each of radius $r$. Estimate the drop in temperature.
 

Two mercury droplets of radii $0.1\text{ cm}$. and $0.2 \text{cm}$ coalesces into one single drop. What amount of energy is released? The surface tension of mercury $T=435.5 \times {10}^{-3} \frac{\mathrm{N}}{\mathrm{m}}$.
 

The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle $\theta$. If the acceleration is $a\frac{m}{s^2}$, what will be the slope of the free surface?.
 

The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $r=2.5\times {10}^{-5}\text{m}$. The surface tension of sap is $T=7.28\times {10}^{-2} \frac{\mathrm{N}}{\mathrm{m}}$ and the angle of contact is $0°$. Does surface tension alone account for the supply of water to the top of all trees?
 

 Is surface tension a vector quantity?

Which of the following statements are correct for a surface molecule?

The angle of contact at the interface of water-glass is $0°$, Ethyl alcohol-glass is $0°$, Mercury-glass is $140°$ and, Methyl iodide-glass is $30°$. A glass capillary is put in a trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is: