The colloidal particles carry an electric charge. The most important property of colloidal solution is that all suspended particles possess either positive or a negative charge. i.e. they carry the same nature of the charge. The mutual forces of repulsion between similarly charged particles prevent them from aggregating and settling under the action of gravity.  This gives stability to the sol.

The dispersion medium carries the opposite charge, hence as a whole, the colloidal solution is electrically neutral.

Origin of the Charge on Colloidal Particles:

A small quantity of electrolyte is always present in the colloidal dispersion. Its presence is necessary for the stability of the sol, as complete removal of the sol causes coagulation of the sol.

Presence of Some Acidic or Basic Groups in Colloidal Solution: 

Colloidal particles may acquire electric charge due to the presence of certain acidic or basic groups in colloidal solution.

For example, protein molecules give rise to the formation of colloidal solutions. Thus the particles of protein sol either have a positive charge or a negative charge depending upon the pH of the medium. A molecule of protein contains a carboxylic acid (COOH) group and also a basic amino (–NH₂) group, it will form a positively charged particle in the acidic medium and a negatively charged particle in the basic or alkaline medium.

In the case of the colloidal solution of proteins, the nature of charge on colloidal particles depends on the pH of the solution. The isoelectric point of a colloid is a pH at which net charge on colloidal particles is zero. Above this pH, the particles are negatively charged and below this pH, particles are positively charged. At isoelectric point, the particles exist in the form of Zwitter ion. Hence they do not migrate under the influence of the electric field.

The isoelectric pH for some proteins is Haemoglobin (pH 4.3-5.3), Casein from human milk (pH 4.1 –  4.7), and Gelatin (pH 4.7).

Due to Self Dissociation:

The dissociation of surface molecules leads to electric charge on colloidal particles of the sol.

For example, Consider an aqueous solution of soap (sodium palmitate) which undergoes dissociation into ions.

The cations (Na⁺) pass into the solvent. Due to the weak attractive forces present in the long hydrocarbon chains, the anions (C₁₅H₃₁ COO^–) have a tendency to form negatively charged aggregates of colloidal dimensions. This type of development of charge is only possible with electrolytes. This is not possible in colloidal solutions of non-electrolytes such as clay, smoke etc.

Electron Capture by Colloidal Particles: 

It is believed that the colloidal solutions prepared by Bredig’s Arc Dispersion Method acquire a charge by electron capture.

Selective or Preferential Adsorption of Ions:

When two or more ions are present in the dispersion medium, then the colloidal particles adsorb preferentially positive or negative ions present in the dispersion medium.

Example: Positively charged Ferric hydroxide sol: 

If FeCl₃ solution is added to the Ferric hydroxide, preferentially Fe³⁺ ions adsorb on Fe(OH)₃ molecules, that is why colloidal particles of ferric hydroxide are positively charged.

FeCl₃ ⇌ Fe³⁺ + 3 Cl^–

Fe(OH)₃ + Fe³⁺ → Fe(OH)₃ / Fe³⁺

Example: Positive charged Siver iodide sol:

When dilute KI is added in excess dilute AgNO₃, the Ag⁺ ions are adsorbed on Agl, and [AgI]Ag⁺ is formed. Thus the colloidal particles have a positive charge.

AgNO₃ + KI  → AgI + KNO₃

AgNO₃ ⇌ Ag⁺ + NO₃^–

AgI + Ag⁺ → AgI / Ag⁺

Example: Negative charged Siver iodide sol: 

When dilute AgNO₃ is added in excess dilute KI, the I^– ions are adsorbed on Agl, and [AgI]I^– is formed. Thus colloidal particles have a negative charge.

AgNO₃ + KI  → AgI + KNO₃

KI ⇌ K⁺ + I^–

AgI + Ag⁺ → AgI / I^–

Example: Negatively charged Arsenious Sulphide Sol: 

It is prepared by passing H₂S gas slowly through the solution of AS₂O₃.

AgNO₃ + KI  → AgI + KNO₃

H2S ⇌ 2H⁺ + S^2-

AS₂S₃ + S^2- → AS₂S₃ / S^2-

Frictional Electrification:

The origin of charge on the colloidal particles may be due to frictional electrification. By mutual rubbing of colloidal particles with molecules of the dispersion medium, the charge is developed on the sol. This view is not satisfactory.

Electrical Double Layer:

Colloidal particles are charged. This charge on the particle is balanced by an opposite charge in the dispersion medium. The charge in the fluid dispersion medium is in the form of free ions. There is a region around each colloidal particle where the charge on particle attracts the free ions from the dispersion medium to form an electrical cloud around it and is called the electrical double layer (Helmholtz electrical double layer).

An electric double layer consists of three regions

• Surface charge – charged ions adsorbed on the particle surface.
• Stern layer – counterions (charged opposite to the surface charge) attracted to the particle surface and closely attached to it by the electrostatic force.
• Diffuse layer – a film of the dispersion medium (solvent) adjacent to the particle. The diffuse layer contains free ions with a higher concentration of the counterions. The ions of the diffuse layer are affected by the electrostatic force of the charged particle. The boundary of this layer is called the slipping plane (shear plane).

The value of the electric potential at the slipping plane is called Zeta potential. The zeta potential is an important parameter for a colloid. Zeta potential depends on the properties of the colloid. For example, adding salt to a colloid shrinks the electrical double layer, and reduces the zeta potential. Zeta potential is given by the relation

Where, ξ = zeta potential

η = coefficient of viscosity

u = velocity of colloidal particles

D = Dielectric constant of the medium = K

Zeta potential and particle size are key indicators of the way colloids behave both in storage and in use. Zeta potential influences the effective size of the particles in the colloid.

Importance of Charge on Colloidal Particles:

• The mutual forces of repulsion between similarly charged colloidal particles prevent them from aggregating and settling under the action of gravity.
• This gives stability to the sol. In the case of lyophobic sols, charge on colloidal particles is fully responsible for its stability.
• In the case of a lyophilic sol, the stability is due to the charge on colloidal particles and solvation.

Electrophoresis or Cataphoresis:

The unidirectional migration of sol particles or dispersed phase particles or colloidal particles towards the oppositely charged electrode under the influence of the applied electric field is called electrophoresis or cataphoresis.

Cause of Electrophoresis:

All sol particle (colloidal particles) carry the same electric charge either positive or negative.  If an electric potential is applied across two platinum electrodes dipping in a sol, the sol particles move towards oppositely charged electrodes.

Illustration:

Consider a sol of As₂S₃ is taken in a ‘U’ shaped glass tube. The sol particles of the sol are negatively charged. Now the dispersion medium with little quantity of electrolyte is introduced over the colloidal solution. There should be a sharp boundary between the sol and the dispersion medium.

Electric potential is applied across the two platinum electrodes dipped in a sol in two limbs, it is observed that the level of sol drops at the negative electrode and rises at the positive electrode side. This shows that sol particles have migrated to the positive electrode, indicating that the particles are negatively charged.

If colloidal particles are allowed to reach the electrode, their charges are neutralised and coagulation takes place.

Applications of Electrophoresis:

• Electrophoresis is used to detect the nature of charge on colloidal particles.
• It is used in the removal of carbon particles from chimney gases.
• It is used in electro-deposition of rubber on metal, wood or cloth surfaces from latex.
• It is used to bring about coagulation of sol.

Electro-osmosis:

The migration of the dispersion medium of a colloidal solution under the influence of the electric field when the movements of colloidal particles are prevented is called as electro-osmosis.

Cause of Electro-osmosis:

Since the sol as a whole is electrically neutral, dispersion medium has an opposite electric charge as compared with that of the sol particles. If the dispersed phase has a positive charge we say that the dispersion medium has a negative charge.

Illustration:

A sol of As₂S₃ is filled in a glass tube. The sol particles of the sol are negatively charged. Hence the dispersion medium (water) is positively charged. The colloidal solution and pure dispersion medium in a glass tube are separated by a semipermeable membrane.

When an electric potential is applied across the platinum electrodes dipping in each arm, sol particles cannot pass through the semipermeable membrane but dispersion medium (water) move to the negative electrode through the semipermeable membrane.  The level of sol drops at the +ve electrode and rises at -ve electrode. This movement of dispersion medium towards -ve electrode shows that the charge on the dispersion medium is positive.

Applications of Electro-osmosis:

• Electro-osmosis is used in dewatering of moist clay
• It is used in the drying of dye-pastes
• It is used in the removal of water from peat.

Applications of Electrical Properties of Colloids

Sewage Precipitation:

Dirty and muddy water from gutters and drainages is called sewage is in colloidal form (colloidal solution).

Sewage water containing colloidal particles of mud, rubbish etc. is collected in a tank fitted with electrodes.

On applying an electric field, colloidal particles are attracted towards oppositely charged electrodes. As their charge gets neutralised, they settle as a precipitate. The precipitated or coagulated matter called sludge is used as manure while clear water is used for irrigation.

Smoke Precipitation:

Smoke is a colloidal solution of negatively charged carbon particles in the air (aerosol)

These carbon particles may condense water vapour on them and thus cities may have a thick cover of smog (smoke + fog). This smog causes air pollution.

Cottrel’s precipitator is a widely used smoke precipitator. Smoke is passed between metal electrodes at high voltage (about 50,000 V) The charged particles are neutralized at the oppositely charged electrode and get deposited there. The gases free from carbon particles are passed to a chimney or for further purification.

Science > Chemistry > Colloids > Charge on Colloidal Particles

The post Electrical Properties of Colloids appeared first on The Fact Factor.

In this article, we shall study the general, mechanical and optical properties of colloids.

General Properties of Colloids:

Heterogenous Character:

The ultramicroscopic examination indicates that colloidal dispersion is a heterogeneous system consisting of a continuous dispersion medium and discontinuous disperse phase.

Visibility:

Colloidal particles cannot be seen through naked eyes or ordinary microscope due to their very small size. The shortest wavelength in visible spectra is about 4000 Å. Hence we cannot see any object less than 200μm and colloidal particles have sizes less than 200μm.

Recently new techniques like Scanning Electron Microscope (SEM), Transmission Electron Microscope (TEM), and Scanning Transmission Electron Microscope (STEM) are used to determine the size and shape of colloidal particles.

Filterability:

The colloidal particles readily pass through ordinary filter paper. The range of particle size of colloidal substance is in between 5 × 10^-9 m to 2 × 10^-7 m. The pore size of ordinary filter paper is bigger i.e. of order 10^-7 m. So Colloidal particles can pass through it and thus filter paper can be used to separate colloidal particles from coarse suspension.

Sols and true solutions pass through filter paper. The colloids cannot pass (diffuse) through parchment membrane but crystalloids can pass through parchment membrane. The process of separating colloids from other dissolved substance using parchment membrane as the filter is called dialysis. This process is used for purification of colloids.

When the impure sol is placed in specially created ultrafilter, with small pores, the sol particles being bigger than the pores remain behind while dispersion medium and dissolved electrolyte pass through. This process is known as ultra-purification.

Surface Tension and Viscosity:

Lyophilic sols have a higher viscosity and lower surface tension than dispersion medium and lyophobic sols have a nearly same viscosity and surface tension as the dispersion medium.

Molecular Mass:

Colloidal particles are of two types i) Multimolecular and ii) Macromolecular

• Multimolecular colloidal particles are aggregates of a number of small molecules or atoms. e.g. Sulphur sol, gold sol.
• Macromolecular particles are very big molecules or polymers. e.g. Starch, proteins.

As the colloidal particles are aggregates of a number of molecules or a large molecule themselves their molecular mass is very high.

Colligative Properties:

Colloidal particles are bigger aggregates. The colligative properties depend on the number of particles. Due to less number of particles compared to true solution colligative properties are lower. Hence the values of colligative properties like osmotic pressure., depression in freezing point and elevation in boiling points are of small order compared to the values shown by true solutions at the same concentration.

Colour:

Many sols are coloured. Sol particles are able to scatter light rays. Colour of the sol depends upon the wavelength of scattered light by the sol particles and which again depends on the size of the sol particles. The colour of colloidal solution also changes with the way the observer receives the light.

Let us consider silver sol (colloidal solution of the same substance) having different types of particles. It is found that the sols show different colours. 

Notes:

• The blue sky is due to the blue light scattered by small dust particles in the atmosphere. The atmosphere is a colloidal system consisting of dust particles suspended in air.
• The red sky is due to red light scattered by larger dust particles in the atmosphere.
• Depending on the size of the dust and water particles, different colours are seen in the cloud.
• Fine gold sol is red but as particle size increases it becomes blue or purple.

Mechanical Properties of Colloids:

Brownian Movement:

The English Botanist Robert Brown, in 1927 observed that colloidal particles exhibit continuous random motion in all directions in a straight line.  He found such movement when pollen grains were suspended in water. The phenomenon of continuous zig-zag movement of colloidal particles in straight line paths in a random direction is known as a Brownian movement.

Explanation:

Colloidal particles are surrounded by a large number of dispersion medium molecules which constantly bombard the colloidal particles. On unequal bombardment, the colloidal particles get pushed in certain directions. Since colloidal particles possess like charge, they repel each other.

Factors Affecting Brownian Movement:

• Brownian movement depends on the viscosity of the dispersion medium. Brownian movement is more in less viscous solution.
• Brownian movement depends on the size of the particle. If the particles are of smaller size. The Brownian movement is more rapid.

Applications Brownian Movement:

• Due to the Brownian movement colloidal particles hardly settle down and prevent aggregation of colloidal particles. Thus colloidal solution becomes stable.
• Avogadro’s number can be calculated by Brownian movement.  

Optical Properties of Colloids:

Tyndall effect:

When an intense beam of light is passed through the colloidal solution (taken in a glass vessel) placed in a dark the path of light through the colloidal solution is clearly visible due to the scattering of light by sol particles.  This effect is known as Tyndall effect.

This fact was first noted in 1857 by Faraday and then studied in  details by Tyndall in 1868. True solutions do not exhibit Tyndall effect.

The emitted light emerges in the form of a bright cone called Tyndall cone. Through ultramicroscope, each colloidal particle appears a bright point against the dark background, due to the scattering. Thus the colloidal particles become self-luminous. As a result, the path of the beam of light through colloidal solution becomes clearly visible. The nature of scattering depends on the size of the sol particle and the refractive indices of sol particle.

Explanation:

Colloidal particles are not large enough like suspension particles to reflect the light nor they are small enough, like true solution particles to allow the light to pass through them. Due to the intermediate size of colloidal particles, they scatter part of the absorbed light, from their surfaces in all directions. Thus the cause of Tyndall effect is a scattering of light by colloidal particles.

Conditions to be Satisfied for Viewing Tyndall Effect:

• The diameter of the dispersed particle is not much smaller than the wavelength of light used.
• There should be a large difference between the magnitudes of refractive indices of the dispersed phase and the dispersion medium.

Applications of Tyndall Effect:

• Tyndall effect is useful to distinguish colloidal solution from the true solution
• To test the purity of gases in the manufacture of H₂SO₄ by the contact process.
• Count the number of colloidal particles in colloidal sols using ultra-microscope.

Science > Chemistry > Colloids > Properties of Colloids

The post Properties of Colloids appeared first on The Fact Factor.

In this article. we shall study numerical problems on the change in surface energy due to breaking of drop or coalescence of number of drops into a one drop.

Example – 01:

Two mercury drops one of radius 2 mm and the other of radius 1 mm coalesce to form a single drop. Find the change in the free surface energy. S.T. of mercury = 0.544 N/m.

Given: radius of first drop = r₁ = 2 mm = 2 × 10^-3 m, radius of second bubble = r₂ = 1 mm = 1 × 10^-3 m, Surface tension = T = 0.544 N/m

To Find: Change in free surface energy = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of two drops = Final volume of single drop

∴ 4/3 π r₁³ +    4/3 π r₂³ =     4/3 π r³

∴   r₁³ +   r₂³ =   r³

∴   2³ +   1³ =   r³

∴   r³ =   9

∴   r =   2.08 mm = 2.08 × 10^-3 m

∴   Change in surface energy = dU = U₁ – U₂

∴    dU = T. A₁ – T.A₂   = T. (A₁ – A₂)

∴    dU = T. [(4πr₁² +4πr₂²)  – 4πr²) = T × 4π [(r₁² +r₂²)  – r²)]

∴    dU = 0.544 × 4π [((2 × 10^-3)² + (1 × 10^-3)²)  – (2.08 × 10^-3)²)]

∴    dU = 0.544 × 4π [4 × 10^-6 + 1 × 10^-6 – 4.3264 × 10^-6]

∴    dU = 0.544 × 4 × 3.142 × 0.6736 × 10^-6

∴    dU = 4.61 × 10^-6  J

Ans: The change in free surface energy is 4.61 × 10^-6 J

Example – 02:

How much energy would be liberated if 10³ water droplets each 10^-8 m in diameter coalesce to produce a single drop? S.T. of water = 0.072 N/m. Assume drops are spherical.

Given: Initial number of drops = n₁ = 10³, initial diameter of each drop = d₁ = 10^-8 m, initial radius of each drop = r₁  = 0.5 × 10^-8 m = = 5 × 10^-9 m Final number of drop = n₂ = 1, Surface tension = T = 0.072 N/m

To Find: Energy liberated = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n₁ drops = Final volume of n₂ drop

∴   n₁ × 4/3 π r₁³   =   n₂ × 4/3 π r³

∴   n₁ ×   r₁³   =   n₂ × r³

∴   10³ ×   (5 × 10^-9)³   =   1× r³

∴   10 × 5 × 10^-9 =     r

∴ r = 5 × 10^-8 m

∴   Energy released = dU = U₁ – U₂

∴    dU = T. A₁ – T.A₂   = T. (A₁ – A₂)

∴    dU = T. ( n₁  × 4πr₁²   – n₂  × 4πr²) = T × 4π (n₁  ×  r₁²  –  n₂  × r²)

∴    dU =  0.072 × 4π [10³ ×  (5 × 10^-9)²  –  1 × (5 × 10^-8)²]

∴    dU =  0.072 × 4 × 3.142 × [10³ × 25 × 10^-18 –  25 ×10^-16]

∴    dU =  0.072 × 4 × 3.142 × [250 × 10^-16 –  25 ×10^-16]

∴    dU =  0.072 × 4 × 3.142 × 225 × 10^-14

∴    dU =  2.036 × 10^-14 J

Ans: The energy liberated is 2.036× 10^-14 J

Example – 03:

A drop of mercury of radius 0.1 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 540 dyne/cm.

Given: Initial number of drops = n₁ = 1, initial radius of each drop = r₁ = 0.1 cm = 0.1 × 10^-2 m = 1 × 10^-3 m, Final number of drop = n₂ = 8, Surface tension = T =540 dyne/cm = 540 × 10^-3 N/m

To Find: Work done = W =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n₁ drops = Final volume of n₂ drop

∴   n₁ × 4/3 π r₁³   =   n₂ × 4/3 π r³

∴   n₁ ×   r₁³   =   n₂ × r³

∴   1 ×   (1 × 10^-3)³   =   8× r³

∴ 1 × 10^-3=     2 r

∴ r = 0.5 × 10^-3 m

∴   Work done = Change in surface energy = dU = U₁ – U₂

∴    dU = T. A₂ – T.A₁   = T. (A₂ – A₁)

∴    dU = T. (n₂  × 4πr²   –   n₁  × 4πr₁² ) = T × 4π (n₂  × r²  –  n₁  ×  r₁² )

∴    dU =  540 × 10^-3 × 4π [8  ×  (0.5 × 10^-3)²  –  1 × (1 × 10^-3)²]

∴    dU =  540 × 10^-3 × 4π [8  ×  0.25 × 10^-6 –  1 × 10^-6]

∴    dU =  540 × 10^-3 × 4π [2 × 10^-6 –  1 × 10^-6]

∴    dU =  540 × 10^-3 × 4 × 3.142  × 1 × 10^-6

∴    dU =  6.786 × 10^-6 J

Ans: The work done is 6.786 × 10^-6 J

Example – 04:

A mercury drop of radius 10^-3 m breaks up into 125 small droplets. Calculate the change in energy assuming that the drops are spherical and S.T. of the mercury is 0.55 N/m.

Given: Initial number of drops = n₁ = 1, initial radius of each drop = r₁ = 10^-3 m, Final number of drop = n₂ = 125, Surface tension = T =0.55 N/m

To Find: Change in area = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n₁ drops = Final volume of n₂ drop

∴   n₁ × 4/3 π r₁³   =   n₂ × 4/3 π r³

∴   n₁ ×   r₁³   =   n₂ × r³

∴   1 ×   (10^-3)³   =   125× r³

∴ 1 × 10^-3=     5 r

∴ r = 0.2 × 10^-3 m

∴   Work done = Change in surface energy = dU = U₁ – U₂

∴    dU = T. A₂ – T.A₁   = T. (A₂ – A₁)

∴    dU = T. (n₂  × 4πr²   –   n₁  × 4πr₁² ) = T × 4π (n₂  × r²  –  n₁  ×  r₁² )

∴    dU =  0.55 × 4π [125  ×  (0.2 × 10^-3)²  –  1 × ( 10^-3)²]

∴    dU =  0.55 × 4π [125  ×  0.04 × 10^-6 –  1 × 10^-6]

∴    dU =  0.55 × 4π [5 × 10^-6 –  1 × 10^-6]

∴    dU =  0.55 × 4 × 3.142  × 4 × 10^-6

∴    dU =  2.76 × 10^-5 J

∴    Change in energy is 2.76 × 10^-5 J

Example – 05:

A mercury drop of radius 0.1 cm breaks up into 27 small droplets. Calculate the work done assuming that the drops are spherical and S.T. of the mercury is 540 dyne/cm.

Given: Initial number of drops = n₁ = 1, initial radius of each drop = r₁ = 0.1 cm = 0.1 × 10^-2 m = 1 × 10^-3 m, Final number of drop = n₂ = 27, Surface tension = T =540 dyne/cm = 540 × 10^-3 N/m

To Find: Work done = W =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n₁ drops = Final volume of n₂ drop

∴   n₁ × 4/3 π r₁³   =   n₂ × 4/3 π r³

∴   n₁ ×   r₁³   =   n₂ × r³

∴   1 ×   (1 × 10^-3)³   =   27× r³

∴ 1 × 10^-3=     3 r

∴ r = 0.33 × 10^-3 m

∴   Work done = Change in surface energy = dU = U₁ – U₂

∴    dU = T. A₂ – T.A₁   = T. (A₂ – A₁)

∴    dU= T. (n₂  × 4πr²   –   n₁  × 4πr₁² ) = T × 4π (n₂  × r²  –  n₁  ×  r₁² )

∴    dU =  540 × 10^-3 × 4π [27  ×  (0.33 × 10^-3)²  –  1 × (1 × 10^-3)²]

∴    dU =  540 × 10^-3 × 4π [27  ×  0.1089 × 10^-6 –  1 × 10^-6]

∴    dU=  540 × 10^-3 × 4π [3 × 10^-6 –  1 × 10^-6]

∴    dU =  540 × 10^-3 × 4 × 3.142  × 2 × 10^-6

∴    dU =  1.357 × 10^-5 J

Ans: The work done is 1.357 × 10^-5 J

Example – 06:

Eight droplets of water each of radius 0.2 mm coalesce together to form a single drop, Find the change in total surface energy, given: surface tension of water = 0.072 N/m.

Given: Initial number of drops = n₁ = 8, initial radius of each drop = r₁ = 0.2 cm = 0.2 × 10^-3 m = 2 × 10^-4 m, Final number of drop = n₂ = 1, Surface tension = T =0.072 N/m

To Find: Change in total surface energy = dU =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n₁ drops = Final volume of n₂ drop

∴   n₁ × 4/3 π r₁³   =   n₂ × 4/3 π r³

∴   n₁ ×   r₁³   =   n₂ × r³

∴   8 ×   (2 × 10^-4)³   =   1× r³

∴ 64× 10^-12=     2 r

∴ r = 4 × 10^-4 m

∴   Work done = Change in surface energy = dU = U₁ – U₂

∴    dU = T. A₂ – T.A₁   = T. (A₁ – A₂)

∴    dU = T. (n₁  × 4πr₁²   –   n₂  × 4πr² ) = T × 4π (n₁  × r²  –  n₂  ×  r² )

∴    dU =  0.072 × 4π [8  ×  (2 × 10^-4)²  –  1 × (4 × 10^-4)²]

∴    dU =  0.072 × 4π [8  ×  4 × 10^-8 –  16 × 10^-8]

∴    dU =  0.072 × 4π [32 × 10^-8 –  16 × 10^-8]

∴    dU =  0.072 × 4 × 3.142  × 16 × 10^-8

∴    dU =  1.448 × 10^-7 J

Ans: The change in surface energy is 1.448 × 10^-7 J

Example – 07:

A mercury drop of radius 0.5 cm falls from a height on glass plate and breaks up into million droplets, all of the same size, Find the height from which the drop must have fallen. Density of mercury 13600 kg/m3. Surface tension of mercury = 0.465 N/m

Given: Initial number of drops = n₁ = 1, initial radius of each drop = r₁ = 0.5 cm = 0.5 × 10^-2 m = 5 × 10^-3 m, Final number of drop = n₂ = 10⁶, Surface tension = T = 0.465 N/m

To Find: Height from which the drop falls = h =?

Solution:

Let ‘r’ be the radius of new drop formed.

The initial volume of n₁ drops = Final volume of n₂ drop

∴   n₁ × 4/3 π r₁³   =   n₂ × 4/3 π r³

∴   n₁ ×   r₁³   =   n₂ × r³

∴   1 × (5 × 10^-3)³   = 10⁶× r³

∴ 5 × 10^-3=     10² r

∴ r = 5 × 10^-5 m

∴   Work done = Change in surface energy = dU = U₁ – U₂

∴    dU = T. A₂ – T.A₁   = T. (A₂ – A₁)

∴    dU= T. (n₂  × 4πr²   –   n₁  × 4πr₁² ) = T × 4π (n₂  × r²  –  n₁  ×  r₁² )

∴    dU=  0.465 × 4π [10⁶ ×  (5 × 10^-5)²  –  1 × (5 × 10^-3)²]

∴    dU =  0.465 × 4π [10⁶ × 25 × 10^-10 –  25 × 10^-6]

∴    dU =  0.465 × 4π [2500 × 10^-6 –  25 × 10^-6]

∴    dU=  0.465 × 4 × 3.142  × 2475 × 10^-6

∴    dU =  0,01446J

∴    Work done is 0.01446J

Now, by work energy principle

Poential energy = work done

∴    mgh = 0.01446

∴    (Volume × density) gh = 0.01446

∴   4/3 π r³ × ρ × g ×h = 0.01446

∴   4/3 × 3.142 × (5 × 10^-3)³ × 13600 × 9.8 ×h = 0.01446

∴   h = (3 × 0.01446) / (4 × 3.142 × (5 × 10^-3)³ × 13600 × 9.8)

∴   h = (3 × 0.01446) / (4 × 3.142 × 125 × 10^-9 × 13600 × 9.8)

∴   h = 0.2072 m

Ans: The height from which the drop may have fallen is 0.2072 m.

Example – 08:

n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4π R²T (n^1/3   – 1)

Solution:

The initial volume of n drops = Final volume of one drop

∴   n×  4/3 π r³   =   1 ×  4/3 π R³

∴ n ×  r³  =   R³

∴ R = r × n^1/3

∴ R × n^{– 1/3} = r

∴   Energy liberated = dU = U₁ – U₂

∴    dU = T. A₁ – T.A₂   = T. (A₁ – A₂)

∴    dU = T. ( n × 4πr²   – 1 × 4πR²)

∴    dU = T. ( n × 4π(R × n^{– 1/3})²   – 1 × 4πR²)

∴    dU = T. ( n × 4πR² × n^{– 2/3}   – 1 × 4πR²)

∴    dU = T. 4πR² ( n × n^{– 2/3}   – 1)

∴    dU = 4π R²T (n^1/3   – 1)

Example – 09:

n droplets of equal size of radius r coalesce to form a bigger drop of radius R. If T is the surface tension of the liquid, then show that the energy liberated is 4πr² T. (n – n^2/3)

Solution:

The initial volume of n drops = Final volume of one drop

∴   n×  4/3 π r³   =   1 ×  4/3 π R³

∴   n ×   r³   =   R³

∴ R = r × n^1/3

∴   Energy liberated = dU = U₁ – U₂

∴    dU = T. A₁ – T.A₂   = T. (A₁ – A₂)

∴    dU = T. ( n × 4πr²   – 1 × 4πR²)

∴    dU = T. ( n × 4πr²   – 1 × 4π(r  × n^1/3)²)

∴    dU = T. ( n × 4πr²   – 1 × 4π r²  × n^2/3)

∴    dU = T. 4πr²( n  –  n^2/3)

∴    dU = 4πr² T. ( n  –  n^2/3)

Note: 

For competitive exams use following formulae directly

Coalescence of n droplets of radius r into a single drop of radius R, then energy liberated

dU = 4πr² T. ( n  –  n^2/3) or dU = 4π R²T (n^1/3   – 1)

Breaking of one drop of radius R into n droplets of radius r, then energy liberated

dU = 4πR² T. ( n  –  n^2/3) or  dU = 4πr²T (n^1/3   – 1)

Previous Topic: The Concept of Surface Energy

Next Topic: Angle of Contact

Science > Physics > Surface Tension >Change in Surface Energy

The post Change in Surface Energy appeared first on The Fact Factor.

Science > Physics > Surface Tension > Numerical Problems on Surface Tension

Example – 1:

A needle 5 cm long can just rest on the surface of the water without wetting. What is its weight? Surface tension of water = 0.07 N/m.

Given: Length of needle = l = 5 cm = 5 × 10^-2 m, Surface tension = T = 0.07 N/m

To Find: Weight of needle = F =?

Solution:

In case of a needle, the water wets it from two sides, Hence, the effective length = 2 l

We have, T = F / effective length

∴   F = T × effective length = T × 2 l

∴   F = 0.07 × 2 × 5 × 10^-2 = 0.007 N

Ans: The weight of needle is 0.007 N

Example – 2:

A light square wireframe each side of which is 10 cm long hangs vertically in the water with one side just touching the water surface. Find the additional force necessary to pull the frame clear of the water (T=0.074 N/m)

Given: Length of frame in contact with water = l = 10 cm = 10 × 10^-2 m, Surface tension = T = 0.074 N/m

To Find: Pull required = F =?

Solution:

In case of wire, the water wets it from two sides,

Hence, the effective length = 2 l

We have, T = F / effective length

∴   F = T × effective length = T × 2 l

∴   F = 0.074 × 2 × 10 × 10^-2 = 0.0148 N

Ans: The pull required is 0.0148 N

Example – 3:

A rectangular film of liquid is formed in a frame of wire and a movable rod of length 4 cm. What force must be applied to the rod to keep it in equilibrium if the surface tension of the liquid is 40 × 10^-3 N/m?

Given: Length of the rod = l = 4 cm = 4 × 10^-2 m, Surface tension = T = 40 × 10^-3 N/m

To Find: Force required = F =?

Solution:

In case of rod, the liquid wets it from two sides,

Hence, the effective length = 2 l = 2 × 4 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 40 × 10^-3 × 2 × 4 × 10^-2 = 3.2 × 10^-3 N

Ans: Force required is 3.2 × 10^-3 N

Example – 4:

A thin and light ring of the material of radius 3 cm is rested flat on the liquid surface. When slowly raised, it is found that the pull required is 0.03 N more before the film breaks than after. Find the surface tension of the liquid.

Given: Radius of the ring = l = 3 cm = 3 × 10^-2 m, Pull required = F = 0.03 N

To Find: Surface tension = T =?

Solution:

In case of ring, the liquid wets it from two sides (inside and outside),

Hence, the effective length = 2 × 2πr

We have, T = F / effective length

∴   T = F / (2 × 2πr) =    0.03 / (2 × 2 × 3.142 × 3 × 10^-2) = 0.08 N/m

Ans: The surface tension of liquid is 0.08 N/m.

Example – 5:

A horizontal circular loop of wire of radius 0.02 m is lowered into crude oil form film. The force due to the surface tension of the liquid is 0.0113 N. Calculate the surface tension of crude oil.

Given: Radius of the loop = l = 0.02 m, Force due to surface tension = F = 0.0113 N

To Find: Surface tension = T =?

Solution:

In case of loop, the liquid wets it from two sides (inside and outside),

Hence, the effective length = 2 × 2πr

We have, T = F / effective length

∴   T = F / (2 × 2πr) =    0.0113 / (2 × 2 × 3.142 × 0.02) =  0.04496 N/m

Ans: The surface tension of liquid is 0.04496 N/m.

Problem – 6:

Calculate the force required to pull away a horizontal circular loop of wire of radius 0.02 m from the surface of the water. The surface tension of water is 0.075 N/m.

Given: Radius of the ring = l = 0.02 m, Surface tension = T = 0.075 N/m.

To Find: Pull required = F =?

Solution:

In case of a ring, the liquid wets it from two sides (inside and outside),

Hence, the effective length = 2 × 2πr

We have, T = F / effective length

∴ F = T × (2 × 2πr) =    0.075 × 4 × 3.142 × 0.02 =  0.0188 N

Ans: The pull required is 0.0188 N

Example – 7:

Calculate the force required to take away a flat circular plate of radius 0.01 m from the surface of the water. The surface tension of water is 0.075 N/m.

Given: Radius of disc = r = 0.01 m, Surface tension = T = 0.075 N/m.

To Find: Force required = F =?

Solution:

In case of a circular plate, the water wets only its outer edges,

Hence, the effective length = circumference = 2 π r = 2 × 3.142 × 0.01 m

We have, T = F / effective length

∴   F = T × effective length

∴   F = 0.075 × 2 × 3.142 × 0.01 = 4.713 × 10^-3 N

Ans: Force required is 4.713 × 10^-3 N

Example – 8:

A thin wire is bent in the form of a rectangle of length 4 cm and breadth 3 cm. What force due to the surface tension does the side experience when a soap film is formed in the frame? S.T. of soap solution = 0.030 N/m.

Solution:

Given: Length of frame = l = 4 cm = 4 × 10^-2 m, Breadth of frame = b = 3 cm = 3 × 10^-2 m, Surface tension = T = 0.030 N/m

To Find: surface tension experienced = F =?

In a case of a rectangular frame, the water wets it from two sides (inside and outside),

Hence, the effective length = 2 × perimeter = 2 × 2 (l + b)

∴ The effective length = 2 × perimeter = 2 × 2 (4 + 3) = 28 cm = 28 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = T × 2 l = 0.030 × 28 × 10^-2 = 8.4 × 10^-3 N

Ans: The surface tension experienced is 8.4 × 10^-3 N

Example – 9:

A rectangular glass plane (height 10 cm, breadth 8 cm, and thickness = 0.2 cm) rests with the largest face flat on the surface of the water. Calculate the additional force required to pull the plane clear of the water. What is the force due to surface tension acting on the plane it is held vertically with the edge of the longest side just touching the water surface? Surface Tension of water = 0.070 N/m.

Part – I

Given: Length of face touching water = l = 10 cm = 10 × 10^-2 m, Breadth of face touching water = b = 8 cm = 8 × 10^-2 m, Surface tension = T = 0.070 N/m

To Find: surface tension experienced = F =?

Solution:

In case of a rectangular plate, the water wets only its outer edges,

Hence, the effective length = perimeter = 2 (l + b)

Effective length =   2 (10 + 8) = 36 cm = 36 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 0.070 × 36 × 10^-2 = 0.0252 N

Ans:  Force due to surface tension is 0.0252 N

Part – II

Given: Length of face touching water = l = 10 cm = 10 × 10^-2 m, Breadth of face touching water = b = 0.2 cm = 0.2 × 10^-2 m, Surface tension = T = 0.070 N/m

To Find: surface tension experienced = F =?

Solution:

In case of a rectangular plate, the water wets only its outer edges,

Hence, the effective length = perimeter = 2 (l + b)

Effective length =   2 (10 + 0.2) = 20.4 cm = 20.4 × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 0.070 × 20.4 × 10^-2 = 0.0143 N

∴ Force due to surface tension is0.0143 N

Example – 10:

A square glass plate 10 cm long and 1 mm in thickness is suspended vertically with the lower edge horizontal from one arm of a balance and counterpoised so that the beam is horizontal. When the glass plate is allowed to touch the surface of a soap solution, an additional mass of 0.72g has to be added to the other pan to keep the beam horizontal. Find the S.T. of the liquid. g = 9.8 m/s².

Given: Length of face touching water = l = 10 cm = 10 × 10^-2 m, Breadth of face touching water = b = 1 mm = 0.1 cm = 0.1 × 10^-2 m, Pull required = F = 0.74 gf = 0.74 × 9.8 × 10^-3 N.

To Find: surface tension = T =?

Solution:

In case of rectangular plate, the water wets only its outer edges,

Hence, the effective length = perimeter = 2 (l + b)

Effective length =   2 (10 + 0.1) = 20.2 cm = 20.2 × 10^-2 m

We have, T = F / effective length

∴   T = (0.74 × 9.8 × 10^-3) / (20.2 × 10^-2) =  0.074 N/m

Ans: The surface tension of liquid is 0.074 N/m

Example – 11:

By dipping a U-shaped wire in a soap solution, a film is formed between it and a light sliding wire resting on it. The sliding wire supports a weight of 0.01 N when its length is 20 cm. Find the surface tension of the liquid.

Given: Length of movable wire = l = 20 cm = 20 × 10^-2 m, Supporting weight = F = 0.01 N

To Find: Surface tension = T =?

Solution:

In case of wire, the liquid wets it from two sides,

Hence, the effective length = 2 l

We have, T = F / effective length

∴   T = F /2 l =    0.01 / (2 × 20 × 10^-2) = 0.025 N/m

Ans: The surface tension of liquid is 0.025 N/m.

Example – 12:

A glass tube of internal diameter 3.5 cm and thickness 0.5 cm is held vertically with its lower end immersed in water. Find the downward pull on the tube due to surface tension. The surface tension of water is 0.074 N/m.

Given: Internal diameter of the tube = 3.5 cm, Internal radius of tube = r_i = 3.5/2 = 1.75 cm  = 1.75 × 10^-2 m, thickness of tube = 0.5 cm, external radius of the tube = r_o =1.75 cm + 0.5 cm = 2.25 cm = 2.25 × 10^-2 m, Surface tension = T = 0.074 N/m.

To Find: Downward pull = F =?

Solution:

In case of a tube, the liquid wets it from two sides (inside and outside),

Hence, the effective length = (2πr_i + 2πr_o) = 2π (r_i + r_o)

∴ The effective length = 2π (1.75 + 2.25) = 2π × 4 = 8 π cm = 8 π × 10^-2 m

We have, T = F / effective length

∴   F = T × effective length = 0.074 × 8 π × 10^-2 = 0.074 × 8 × 3.142 × 10^-2 = 0.0186 N

Ans: Downward pull is 0.0186 N

Example – 13:

A ring of glass is cut from a tube 7.4 internal and 7.8 cm external diameter. This ring with its lower side horizontal is suspended from one arm of a balance so that the lower edge is just immersed in a vessel of water. It is found that an additional weight of 3.62 g must be placed in the other scale pan to compensate for the pull of surface tension on the ring. Calculate the S.T. of water. g = 9.8 m/s².

Given: Internal diameter of the tube = 7.4 cm, Internal radius of tube = r_i = 7.4/2 = 3.7 cm = 3.7 × 10^-2 m, external diameter of the tube = 7.8 cm, external radius of tube = r_o = 7.8/2 = 3.9 cm = 3.9 × 10^-2 m, Pull required = F = 3.62 gf = 3.62 × 9.8 × 10^-3 N.

To Find: Surface tension = T =?

Solution:

In case of tube, the liquid wets it from two sides (inside and outside),

Hence, the effective length = (2πr_i + 2πr_o) = 2π (r_i + r_o)

∴ The effective length = 2π (3.7 + 3.9) = 2π × 7.6 = 15.2 π cm = 15.2 π × 10^-2 m

We have, T = F / effective length

∴   T = (3.62 × 9.8 × 10^-3) / (15.2 π × 10^-2) = 0.074 N/m

Ans: The surface tension of liquid is 0.074 N/m.

Example – 14:

The surface tension of water at 0 °C is 70 dynes/cm. Find the surface tension of water at 25 °C. Given α for water = 0.0027/°C.

Given: Initial temperature θ₁ = 0 °C, Final temperature θ₂ = 25 °C, α for water = 0.0027/°C, Initial surface tension = T₁ = 70 dynes/cm

To Find: Final surface tension = T₂ =?

Solution:

T₂ = T₁ (1 – αθ) = T₁ (1 – α (θ₂ – θ₁))

∴ T₂ = 70 × (1 – 0.0027 × (25 – 0))

∴ T₂ = 70 × (1 – 0.0027 ×25)

∴ T₂ = 70 × (1 – 0.0675)

∴ T₂ = 70 × 0.9325

∴ T₂ = 65.28 dyne/cm

Ans: The surface tension of water at 25 °C is 65.28 dyne/cm

Previous Topic: Everyday Examples of Surface Tension

Next Topic: Concept of Surface Energy

Science > Physics > Surface Tension > Numerical Problems on Surface Tension

The post Numerical Problems on Surface Tension appeared first on The Fact Factor.

In this article, we shall study the concept of surface energy, relation between the surface energy and surface tension.

A molecule on the surface of a liquid experiences a downward pull due to surface tension.  If a molecule deep inside the liquid is to be lifted up towards the fine surface, as soon as it enters the surface film, work will have to be done to lift it further against the unbalanced downward forces of molecular attraction.

This work will be stored in the molecule as potential energy. This is true of every molecule on the surface film and surface film possesses a certain amount of potential energy. Thus the molecules in surface film possess extra energy. This energy is called as surface energy. Its S.I. unit is J and its dimensions are [L²M¹T^-2].

Relation Between Surface Tension and Surface Energy:

Consider a rectangular frame ABCD in which side CD is made of loose write and other three sides are fixed. The frame is immersed in a soap solution and taken out and held horizontally. A film of soap solution will be formed on the frame and it will at once try to shrink and pull the loose wire CD towards AB. If the length of loose wire CD is  ‘l’ and the film is of finite thickness, therefore the film will be in contact with the wire both on the upper surface as well as along its lower surface.  Hence the length of the wire in contact with the film is ‘ 2l ‘.  The force acting on the wire is directed towards AB, per unit length of the contact line is surface tension (T). By definition of surface tension, we have

T = F / 2l

∴    F =  T . 2 l     …  (1)

Imagine an external force is applied on CD which is equal and opposite to force F Let the wire at CD moves to C’D’ through small distance dx.  Then the work done against the force of surface tension is given by

dW = F.dx         …  (2)

From equations (1) and (2),

dW = T.2l. dx

But, 2l . dx = dA = increase in Area of both the surface of the film.

∴ dW = T.dA

This work done is stored inside the films as potential energy dU.

∴   dU =  T.dA

If, initially CD is very close to AB, initial energy and initial area of the film can be taken as zero and dU and dA can be treated as total energy and the total area of the film respectively.

∴ T  =  dU / dA

This expression indicates that surface tension is equal to surface energy per unit area of the surface film.

Numerical Problems:

Example -1:

A water film is formed between two straight parallel wires of 10 cm each with a separation of 0.1 cm. If the distance between the wires is increased by 0.1 cm, how much work is done? T = 0.072 N/m

Given: Length of wire= l = 10 cm = 10 × 10^-2 m, Increase in width = 0.1 cm = 0.1 × 10^-2 m, Surface tension = T = 0.072 N/m

To Find: Work done = W =?

Solution:

In this case, the film has two surfaces

dA = 2 (l × b) = 2 × 10 × 10^-2 × 0.1 × 10^-2 m

dA = 2 × 10^-4 m²

Work done = Change in surface energy = T. dA

Ans: The work done = 0.072 × 2 × 10^-4 = 1.44 × 10^-5  J

Work done is 1.44 × 10^-5 J

Example – 2:

A liquid film is formed in a rectangular frame one side of which is a light and thin movable wire of length 4 cm. The frame is held in a vertical plane with the movable wire in the downward position. Find what weight required to hold the movable wire in equilibrium. How much work is done if the film is stretched by pulling the movable wire downwards by 2 cm? T = 0.025 N/m.

Given: Length of wire= l = 4 cm = 4 × 10^-2 m, displacement = dx = 2 cm = 2 × 10^-2 m, Surface tension = T = 0.025 N/m.

To Find: Weight required = F =?, Work done = W = ?

Solution:

In case of wire, the water wets it from two sides,

Hence, the effective length = 2 l

We have, T = F / effective length

∴   F = T × effective length = T × 2 l

F = 0.025 × 2 × 4 × 10^-2  = 0.002 N

In this case, the film has two surfaces

dA = 2 (l × dx) = 2 × 4 × 10^-2 × 2 × 10^-2 m

dA = 16 × 10^-4 m²

Work done = Change in surface energy = T. dA

Work done = 0.025 × 16 × 10^-4  = 4 × 10^-5  J

Ans: The weight required is 0.002 N and work done is 4 × 10^-5  J

Example – 3:

There is a soap film on a rectangular wire frame of area 4cm × 4 cm. If area of the frame is increased to 4cm x 5 cm, find work done in the process. Surface tension of soap film is 3 × 10^-2 N/m.

Given: Initial area= A₁ = 4cm × 4 cm = 16 cm² = 16 × 10^-4 m², Final area= A₂ = 4cm × 5 cm = 20 cm² = 20 × 10^-4 m², Surface tension = T =3 × 10^-2 N/m.

To Find: Work done = W =?

Solution:

In case of rectangular frame, the water wets it from two sides,

Hence, the change in area = dA = 2 (A₂ – A₁) 

dA =  2 × (20 × 10^-4  – 16 × 10^-4)

∴ dA = 2 (A₂ – A₁)  =  2 × 4 × 10^-4  =  8 × 10^-4  m²

Work done = Change in surface energy = T. dA

Work done = 3 × 10^-2 × 8 × 10^-4  = 2.4 × 10^-5  J

Ans: The work done is 2.4 × 10^-5  J

Example – 4:

Find the work done in blowing a soap bubble of radius 5 cm. S.T. of soap solution = 0.035 N/m.

Given: Radius of bubble = r = 5 cm = 5 × 10^-2 m, Surface tension = T = 0.035 N/m

To Find: Work done = W =?

Solution:

In case of bubble it has two free surfaces

dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T × 8 π r²

∴ Work done = 0.035 × 8  × 3.142 × ( 5 × 10^-2)² 

∴ Work done = 0.035  × 8  × 3.142 × 25 × 10^-4  = 2.2 × 10^-3  J

Ans: The work done is  2.2 × 10^-3  J

Example – 5:

Find the work done in blowing a soap bubble of radius 10 cm. S.T. of soap solution = 30 dyne/cm.

Given: Radius of bubble = r = 10 cm = 10 × 10^-2 m, Surface tension = T =30 dyne/cm = 30 × 10^-3 N/m.

To Find: Work done = W =?

Solution:

In case of bubble it has two free surfaces

dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T × 8 π r²

∴ Work done = 30 × 10^-3 × 8 × 3.142 × (10 × 10^-2)² 

∴ Work done = 30 × 10^-3 × 8 × 3.142 × 100 × 10^-4 = 7.54 × 10^-3 J

Ans: The work done is 7.54 × 10^-3 J

Example – 6:

Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of a soap solution is 30 dyne/cm.

Given: Initial radius of bubble = r₁ = 1 cm = 1 × 10^-2 m, Final radius of bubble = r₂ = 2 cm = 2 × 10^-2 m, Surface tension = T = 30 dyne/cm  = 30 × 10^-3  N/m.

To Find: Work done = W =?

Solution:

In case of bubble it has two free surfaces

dA = A₂ – A₁ = 2 (4 π r₂²) – 2 (4 π r₁²)

∴ dA = 8 π (  r₂²  –   r₁²) = 8 π (  (2 × 10^-2)²  –  (1 × 10^-2)²)

∴ dA =  8 π (  4 × 10^-4  –  1 × 10^-4)

dA =  8 π  × 3 × 10^-4

Work done = Change in surface energy = T. dA

Work done = 30 × 10^-3 × 8 π × 3 × 10^-4

Work done = 30 × 10^-3 × 8 × 3.142 × 3 × 10^-4 = 2.262 × 10^-4 J

Ans: The work done is 2.262 × 10^-4 J

Example – 7:

The total energy of the free surface of a liquid drop is 2π times the surface tension of the liquid. What is the diameter of the drop? Assume all terms in S.I. unit.

Solution:

In case of liquid drop  it has one free surfaces

dA = 4 π r²

Surface energy = T. dA = T  × 4 π r²

∴    2π  T = T  × 4 π r²

∴    2 =  4  r²

∴    0.5 =   r²

∴   r =  0.707 m

Ans: The diameter = d = 2 r =  2  × 0.707 = 1.414 m

Problem – 8:

A soap bubble of radius 12 cm is blown. The surface tension of soap solution is 30 dynes/cm. Calculate the work done in blowing the soap bubble.

Given: Radius of bubble = r = 12 cm = 12 × 10^-2 m, Surface tension = T =30 dyne/cm = 30 × 10^-3 N/m.

To Find: Work done = W = ?

Solution:

In case of bubble it has two free surfaces

dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T  × 8 π r²

∴  Work done = 30 × 10^-3  × 8  × 3.142 × ( 12 × 10^-2)² 

∴  Work done = 30 × 10^-3  × 8  × 3.142 × 144 × 10^-4  = 0.01086J

Ans: The work done is 0.01086J

Problem – 9:

The energy of the free surface of liquid drop is 5π times the surface tension of the liquid. Find the diameter drop in c.g.s. System. 

Solution:

In case of liquid drop, it has one free ssurface

dA = 4 π r²

Surface energy = T. dA = T  × 4 π r²

∴    5π  T = T  × 4 π r²

∴    5 =  4  r²

∴    1.25 =   r²

∴   r =  1.12 cm

Ans:  diameter = d = 2 r =  2  × 1.12 = 2.24 cm

Problem – 10:

The total free surface energy of a liquid drop is π √2  times the surface tension of the liquid. Calculate the diameter of the drop in S.I. unit. 

Solution:

In case of liquid drop, it has one free ssurface

dA = 4 π r²

Surface energy = T. dA = T  × 4 π r²

∴    π √2   T = T  × 4 π r²

∴    5√2 =  4  r²

∴      5√2 / 4 =   r²

∴   r =  1.33 m

Ans:  The diameter = d = 2 r =  2  × 1.33 = 2.66 cm

Problem – 11:

Two soap bubbles have radii in the ratio 4:3. What is the ratio of work done to blow these bubbles?

Given:  r₁ / r₂=4/3, Surface tension = T₁ =T₂ = T

To Find: W₁ / W₂  =?

Solution:

In case of bubble it has two free surfaces

dA = 2 (4 π r²) = 8 π r²

Work done = Change in surface energy = T. dA = T  × 8 π r²

For First Bubble    W₁  = T₁  × 8 π r₁²  …………. (1)

For second Bubble    W₂  = T₂  × 8 π r₂²  …………. (2)

Dividing equation (1) by (2) we get

W₁ / W₂  = (T₁  × 8 π r₁² ) / (T₂  × 8 π r₂² )

W₁ / W₂  = (r₁ / r₂ )²  = = (4/3 )² = 16/9

Ans:  The ratio of work done to blow these bubbles is 16 : 9

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Science > Physics > Surface Tension > Concept of Surface Energy

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In this article, we shall study different phenomena due to surface film formed on the liquid surface due to surface tension.

Examples of Surface Tension (Due to formation of the surface film):

• When a glass rod is dipped in water and taken out it is found that the drop of water is sticking to the end of the rod.  The molecules of water appear to be contained in a bag whose surface is made up of a film of water molecules themselves.  This example shows that the free surface of a liquid behaves like a film which is under tension.

• When water falls slowly in drops from a tap, each drop gradually forms at the tip of the nozzle.  As the drop of water grows in size, it looks like a hanging elastic bag whose surface area is expanding as it fills up.

• Water spiders are able to walk on the surface o the water because their feet produce dimples on the surface film without rupturing the film.  As soon as a foot is lifted the film which is under tension again becomes flat. The feet of the spider produce dimples on the surface film without rupturing the film. Due to which the surface tension acts in an inclined manner. The vertical component of the surface tension supports the weight of the insect. As soon as a foot is lifted the film which is under tension again becomes flat.

• A safety razor blade, when placed gently with its flat surface on water floats on it even though the density of steel, is nearly eight times greater than that of water. The surface film forms due to surface tension support the needle/blade and their thickness is not sufficient to break the film.

• When a paintbrush is immersed in paint, there is no free surface area available for liquid ion the paint and the hair are spread. When the brush is taken out, its bristles are drawn close together due to the surface tension in the film formed between adjacent hair.

• A liquid tries to acquire a minimum surface area on account of its surface tension. For a given volume sphere has the smallest surface area. Hence rainwater falling down acquires spherical shape i.e. raindrops are spherical.

• When irregular shaped camphor piece is dropped on the water surface, it gets dissolved gradually. Thus the surface tension of water around it decreases. Due to irregular shape the unbalanced forces makes it move haphazardly on the surface of the water. Thus the irregularly shaped camphor piece dances on the water surface.
• Oil does not spread on cold water but spreads on hot water because when water is cold the surface tension of oil is less than that of cold water but when water is hot the surface tension of oil is greater than that of hot water.
• High tides in seawater are reduced by sprinkling oil on the water surface.
• Rainwater does not pass through tiny holes in the fabrics of umbrellas, tents, raincoats, etc. because, on account of surface tension, the holes are blocked by fine films of water.
• The surface tension of liquid prevents it from spreading over a solid surface. But lubricating oil must readily spread on the solid surface which is to be lubricated. Therefore, by blending different types of oil, the surface tension of a lubricating oil is made as low as possible.
• When soap dissolves in water, the surface tension of soap solution is greatly reduced and this enables soap solution to quickly spread over large areas. This is one of the reasons why soap is used in washing.
• Soap is used in toothpaste to enables the paste to quickly spread over a large area and give a good mouthwash.
• Soap is used as wetting agent in dying.
• Ice skating is a very enjoyable, because, the tiny droplets of water, formed under the skates, act almost like rigid ball bearings over which the skates glide very smoothly.
• Surface tension is used for diagnosis of jaundice.

Use of Surface Tension in Washing of Clothes:

The surface tension of water is such that water does not spread on oily or greasy clothes. The surface tension of liquid decreases with increase in temperature.  By adding detergent the surface tension of water decreases and it increases the wetting of the cloth by water. The water penetrates into the pores of the cloth and removes the dirt. The dirt is then suspended in a detergent solution. Similarly, the detergent solution suspends the dirt in the solution and does not allow the dirt to settle back on the cloth.

Antiseptics Have Low Surface Tension:

Antiseptic is a medicine used externally on the wounded or injured surface of the body. The antiseptic should spread on body surface easily to get a proper result. Due to lower surface tension the antiseptic can spread easily on body surface.

It is possible to produce a fairly vertical film of soap solution but not of water:

As we add impurity or any soluble substance ion water, the surface tension of water decreases. Thus the surface tension of soap solution is very less compared to water itself.

Oil spreads over the surface of the water while water does not do so:

The surface tension of oil is less than that of water. Hence when oil is dropped on the water it cannot break the surface film of water hence spreads on water. While when water is poured on oil it is capable of breaking the surface film of oil very easily.

Ends of a glass rod become round on heating:

On heating glass melts. The molten glass behaves like a liquid to obtain a minimum surface area on account of its surface tension. For a given volume sphere has the smallest surface area. Thus the molten glass at the end of the tube acquires spherical shape. Thus the end of a glass rod becomes round on heating.

Notes:

• Surface tension is a joint property of the interface separating two substances at least one of which is a fluid. It is not the property of a single fluid alone.
• Surface film is the thin film of liquid near its surface and having the thickness equal to the molecular range of attraction for that liquid. The phenomenon of surface tension is associated intimately with this film.

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Science > Physics > Surface Tension > Surface Tension in Everyday Life

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In this article, we shall study the concept of the surface tension, its characteristics and the factors affecting it.

Intermolecular Forces:

Between any two molecules, there exists a force of attraction. This force is called the intermolecular force. These intermolecular forces of attraction are of two types. a) Cohesive force and b) Adhesive force.

Cohesive Force and Cohesion:

The attractive force between the two molecules of the same substance is called as a cohesive force and the attraction itself is called cohesion. e.g. attraction between water and water molecules.

Adhesive Force and Adhesion:

The attractive force between the two molecules of the different substance is called an adhesive force and the attraction itself is called adhesion. e.g. attraction between water and glass molecules.

Why does water wets glass?

The attractive force between the two molecules of the same substance is called as a cohesive force and the attractive force between the two molecules of the different substance is called an adhesive force. In case of water, the cohesive force between water molecules is very less compared to the adhesive force between glass and water molecules. Thus there is a strong attraction between water and glass molecules. Hence water wets glass.

Why does mercury not wet glass?

The attractive force between the two molecules of the same substance is called as a cohesive force and the attractive force between the two molecules of the different substance is called an adhesive force. In case of mercury, the cohesive force between its molecules is very strong compared to negligible adhesive force between glass and mercury molecules. Thus there is a strong attraction between mercury molecules and almost no attractive force between glass and mercury molecules. Hence mercury does not wet glass.

Range of molecular attraction (r):

The maximum distance between two molecules up to which the intermolecular forces are effective is called the range of molecular attraction. It is of order 10^-9 m.

Sphere of Molecular Influence:

A sphere drawn by taking the radius equal to the range of molecular attraction and centre as the centre of the molecule is called sphere of molecular influence.

Concept of Surface Tension:

Explanation on the Basis of Molecular Theory:

Consider three molecules A, B and C in a liquid, such that molecule A is well inside the liquid, molecule B is close to the free surface and the molecule C is on the free surface.

The sphere of influence of the molecule A is completely inside the liquid, and hence it will be acted upon by equal forces in all directions and these forces will balance one another and the net force acting on it is zero.

For molecule B, a part of the upper half of the sphere of influence is in the air, which contains air molecules.  Air molecules exert very negligible adhesive forces on molecule B.  Therefore, the cohesive forces due to molecules in the liquid remain unbalanced and thus a net force in downward direction acts on the molecule.

For the molecule C, the upper half of the sphere of influence is completely in the air. Due to this, the force of attraction due to the molecules inside the lower half of the sphere will remain unbalanced.  This molecule experiences the maximum possible unbalanced force in the downward direction.

Thus the molecules on the surface and in a surface film of thickness equal to the range of molecular attraction of the liquid molecule experience a net force in the downward direction.  The magnitude of the force depends upon the distance of the molecule from the free surface.  The behaviour of this film is different from that of the rest of the liquid.  It is called the surface film.  This film behaves like a film which is under tension.  This phenomenon is known as surface tension.

If any molecule is brought to the surface from the liquid, the work is to be done against this net downward force. This work increases the potential energy of the surface. But the liquid surface will have the tendency to have minimum potential energy. So a minimum number of molecules remains on the surface of the liquid.

Thus the free surface of a liquid behaves like a stretched elastic membrane and has a tendency to contract so as to minimize its surface area.

Definition of Surface Tension:

Surface tension or force of surface tension is the force per unit length of an imaginary line drawn in any direction on the free surface of a liquid, the line of action of the force being on the surface and at right angles to the length of the imaginary line.

Dimensions and Units of Surface tension:

Surface Tension (T) = Force (F) / Effective Length (L)

∴   [T] = [F] / [L]

∴   [T] = [L¹M¹T^-2] / [L¹]

∴   [T] = [L⁰M¹T^-2]

∴   S.I. unit of T = S.I. unit of F /S.I. unit of L

∴    S.I. unit of T= newton (N) / metre (m)

Thus the dimensions of surface tension are [L⁰M¹T^-2] and its S.I. unit is N/m and c.g.s. unit is dyne/cm.

Factors Affecting Surface Tension:

Temperature:

The surface tension at t °C is given by 

T = T₀ (1 – α t)

Where    T₀ = Surface tension at 0° C and α = Temperature coefficient of surface tension. Its value depends on the nature of the liquid.

• Generally, the surface tension of liquid decreases with an increase in temperature. it is due to the increase in the kinetic energy of liquid molecules.
• The temperature at which surface tension of a liquid becomes zero is called a critical temperature of the liquid.
• The surface tension of a liquid at the boiling point and at critical temperature is zero.
• The surface tension of molten cadmium and copper increases with an increase in temperature.

Impurities:

If the impurity added in a liquid is highly soluble in it, then the surface tension of liquid increases. If the impurity added in a liquid is partly soluble in it, then it decreases.

Contamination:

The presence of dust particles in liquid decreases its surface tension.

Next Topic: Surface Tension in Everyday Life

Science > Physics > Surface Tension > Surface Tension

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When water is taken in a glass vessel, the free surface of the water near the walls is curved concave upward. If mercury is taken in a glass vessel, the free surface of mercury near the walls is convex upwards.

When the liquid is in contact with solid, the angle between the solid surface and the tangent to the free surface of the liquid at the point of contact, measured from inside the liquid is called the angle of contact.

When the liquid surface is curved concave upwards, the angle of contact is acute and when the liquid surface is curved convex upwards, the angle of contact is obtuse.

Characteristics of the Angle of Contact:

• The angle of contact is constant for a given liquid-solid pair.
• When the angle of contact between the liquid and a solid surface is small (acute), the liquid is said to wet the surface. Thus water wets glass.
• If the angle of contact is large the surface is not wetted. Mercury does not wet glass.
• If there are impurities in liquid, then they alter the values of the angle of contact.
• The angle of contact decreases with an increase in temperature.
• For a liquid which completely wets the solid, the angle of contact is equal to zero.

Notes:

• The smearing of glycerine on a glass increases the angle of contact of water with glass from acute to obtuse, hence a glass window is smeared with glycerine.
• Smaller the angle of contact, better is the detergent.
• Teflon coating is done on the surface of the non-stick pan because Teflon increases the angle of contact from acute to obtuse.

The Shape of a Liquid Drop on Solid Surface:

When a small quantity of liquid is poured on a plane solid surface, a force of surface tension acts along a surface separating the two media. There is a formation of the liquid drop.

When the drop is in equilibrium, its shape depends on the forces of interface media. Thus there is surface tension along a surface between a) liquid and air b) solid and air and c) liquid and solid.

Let, θ  = the angle of contact of given solid-liquid pair.

T1 = Surface tension at the liquid-solid interface

T2 = Surface tension at the air-solid interface

T3 = Surface tension at the air-liquid interface

For equilibrium of the drop

T₂ = T₁ + T₃ cos θ

Case -1:

If T₂ > T₁, and T₂ -T₁ < T₃, then cos θ is positive and the angle of contact is acute. e.g. kerosene on the glass surface

Case – 2:

If T₂ < T₁, and T₁ -T₂ < T₃, then cos θ is negative and the angle of contact is obtuse. e.g. mercury on the glass surface Mercury does not wet the glass and thus it forms a drop on the glass.

Case – 3:

If T₂ -T₁ = T₃, then cos θ is 1 and the angle of contact is zero. For pure water, cos θ is almost 1 i.e. water wets the glass and thus water spreads on the glass.

Case – 4:

If T₂ -T₁ > T₃, then cos  θ > 1 which is not possible. Thus liquid will spread over the solid and drop shall not be formed.

Notes:

For the small drop, gravitational potential energy is very small and hence can be neglected. Hence for stability, the surface tension of the drop should be least i.e. the surface area should be least. It can be achieved by forming a spherical drop because the surface area of a sphere is least for the given volume.

For the large drop, gravitational potential energy dominates to make the potential energy minimum. It does so by bringing the centre of gravity as low as possible, due to which the drop flattens.

The Formation of Concave Surface of Liquid (Explanation of Acute Angle of Contact):

When impure water or kerosene is taken in a glass vessel, it is found that the surface near the walls is curved concave upwards.

Consider a molecule of water M on the free surface very close to the wall of the glass vessel. The force of cohesion C due to other water molecules is as shown in the figure. In addition to this, a force of adhesion A acts due to the glass molecule as shown in the figure. The net adhesive force between water molecules and air molecules is negligible. The gravitational force on the molecule is also negligible.

The magnitude of A is greater than the magnitude of C and resultant of the two molecular forces of attraction R is directed towards the glass or outside the liquid. Hence the molecule A is attracted towards the walls of the glass vessel.

The free surface of water adjusts itself at right angles to the resultant R. Therefore molecules like M creep upward on the solid surface. Thus the water surface is curved concave upwards and the angle of contact is acute.

The Formation of Convex Surface of Liquid (Explanation of Obtuse Angle of Contact):

When mercury is taken in a glass vessel, it is found that the surface near the walls is curved convex upwards.

Consider a molecule of mercury M on the free surface very close to the wall of the glass vessel. The force of cohesion C due to other mercury molecules is as shown in the figure. In addition to this, a force of adhesion A acts due to the glass molecule as shown in the figure. The net adhesive force between mercury molecules and air molecules is negligible. The gravitational force on the molecule is also negligible.

The magnitude of A is very less than the magnitude of C and resultant of the two molecular forces of attraction R is directed inside the liquid. Hence the molecule A is attracted towards other molecules of mercury.

The free surface adjusts itself at right angles to the resultant R. The molecule A creeps downwards on the glass surface. Thus the surface of the mercury in the glass is curved convex upwards and the angle of contact is obtuse.

The Formation of Flat Surface of Liquid:

When pure is taken in a glass vessel, it is found that the surface of the water is perfectly flat.

Consider a molecule of water M on the free surface very close to the wall of the glass vessel. The force of cohesion C due to other water molecules is negligible. The force of adhesion A acts due to the glass molecule as shown in the figure. The gravitational force on the molecule is also negligible.

The magnitude of A is large and resultant of the two molecular forces of attraction R is along the direction of the adhesive force. Hence the molecule A is attracted towards glass molecules.

The free surface adjusts itself at right angles to the wall. Thus the surface of the pure water is perfectly flat.

Mercury Drops Coalesce Together When Brought Together:

The attractive force between the two molecules of the same substance is called as a cohesive force and the attractive force between the two molecules of the different substance is called an adhesive force.

In the case of mercury, the cohesive force between its molecules is very strong compared to negligible adhesive force between any surface and mercury molecules. The net adhesive force between mercury molecules and air molecules is negligible. The gravitational force on the molecule is also negligible. Thus there is a strong attraction between mercury molecules and almost no attractive force between surface and mercury molecules. Hence mercury drops coalesce together when brought together.

Water on a clean glass surface tends to spread out, while mercury on the same surface tends to form a drop. 

The shape of the drop depends on the angle of contact of the liquid with glass. In the case of a glass water interface, the angle of contact is acute. Thus to achieve this angle water spreads on the glass. In the case of a glass mercury interface, the angle of contact is obtuse. Thus to achieve this angle mercury forms a drop as shown.

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Science > Physics > Surface Tension > Angle of Contact

The post Angle of Contact appeared first on The Fact Factor.

In this article, we shall study Laplace’s Law about the excess of pressure inside a drop or a bubble of a liquid.

Excess of Pressure Inside a Drop of a Liquid and Soap Bubble:

Due to surface tension, free liquid drops and bubbles are spherical, if the effect of gravity and air resistance are negligible. Due to the spherical shape, the inside pressure P_i is always greater than the outside pressure P_o. The excess of pressure is P_i– P_o.

Laplace’s Law of a Spherical Membrane for a Liquid Drop:

Due to the spherical shape, the inside pressure P_i is always greater than the outside pressure P_o. The excess of pressure is P_i– P_o.

Let the radius of the drop increases from r to r + Δr, where Δr is very very small, hence the inside pressure is assumed to be constant.

Initial surface area = A₁ = 4 π r²

Final surface area = A₂ = 4 π (r + Δr)²

Final surface area = 4 π (r² + 2r.Δr + Δr²)

Final surface area =  4 πr² + 8 πr.Δr + 4 πΔr²

Δr is very very small, hence Δr² still smsll hence the term 4 πΔr² can be neglected.

Final surface area = A₂ = 4 πr² + 8 πr.Δr

Hence Change in area = A₂ – A₁ = 4 πr² + 8 πr.Δr  –  4 πr²

Change in area = dA = 8 πr.Δr

Now, work done in increasing the surface area is given by

dW = T. dA = T.  8 πr.Δr    …………… (1)

By definition of work in mechanics we have

dW = Force ∴ displacement = F .Δr  …………… (2)

But P = F /A, Hence F = Excess pressure × Area

F = (P_i– P_o) × 4 πr²

Substituting in equation (2) we have

dW = (P_i– P_o) × 4 πr².Δr  …………… (3)

From equations (3) and (4) we have

(P_i– P_o) × 4 πr².Δr = T.  8 πr.Δr

(P_i– P_o) = 2T / r

This relation is known as Laplace’s law for the spherical membrane for a liquid drop.

Laplace’s Law of a Spherical Membrane for a Liquid Bubble:

Due to the spherical shape, the inside pressure P_i is always greater than the outside pressure P_o. The excess of pressure is P_i– P_o.

Let the radius of the bubble increases from r to r + Δr, where Δr is very very small, hence the inside pressure is assumed to be constant.

Bubble has two free surfaces, one inside and another outside

Initial surface area = A₁ = 2 × 4 π r² = 8 π r²

Final surface area = A₂ = 2 × 4 π (r + Δr)²

Final surface area = 8 π (r² + 2r.Δr + Δr²)

Final surface area =  8 πr² + 16 πr.Δr + 8 πΔr²

Δr is very very small, hence Δr² still smsll hence the term 8 πΔr² can be neglected.

Final surface area = A₂ = 8 πr² + 16 πr.Δr

Hence Change in area = A₂ – A₁ =  8 πr² + 16 πr.Δr  –  8 πr²

Change in area = dA = 16 πr.Δr

Now, work done in increasing the surface area is given by

dW = T. dA = T.  16 πr.Δr    …………… (1)

By definition of work in mechanics we have

dW = Force × displacement = F .Δr  …………… (2)

But P = F /A, Hence F = Excess pressure × Area

F = (P_i– P_o) × 4 πr²

Substituting in equation (2) we have

dW = (P_i– P_o) × 4 πr².Δr  …………… (3)

From equations (3) and (4) we have

(P_i– P_o) × 4 πr².Δr = T.  16 πr.Δr

(P_i– P_o) = 4T / r

This relation is known as Laplace’s law for spherical membrane for a liquid bubble.

Difference in Pressure Across the Surface Film:

Rise and fall of liquid in a capillary tube can be explained by knowing the fact that a pressure difference exists across a curved free surface of the liquid.

When the liquid surface is flat:

In this case, the surface tension is horizontal and hence has no normal component to the horizontal surface. Due to which there is no extra pressure on outside or inside. Thus the pressure on the liquid side is equal to the pressure on the vapour side.

When the liquid surface is concave:

In this case, the component of surface tension acts vertically upward. Due to which pressure inside liquid decreases. Thus for the equilibrium of concave surface, the pressure on the vapour side should be more than that on the liquid side. Hence liquid rises in the capillary tube. In this case, the adhesion is greater than the cohesion.

When the liquid surface is convex:

In this case, the component of surface tension acts vertically downward. Due to which pressure inside liquid increases. Thus for the equilibrium of convex surface, the pressure on the vapour side should be less than that on the liquid side. Hence liquid dips in the capillary tube. In this case, the cohesion is greater than the adhesion.

Conclusion:

In case 2 and 3 we can see that, when liquid surface is curved, the surface tension give rises to a pressure which is directed towards the centre of curvature of the surface, to balance this there is excess pressure acting on the surface. Thus, there is always excess of pressure on the concave side of a curved liquid surface over the pressure on its convex side due to surface tension.

Numerical Problems on Excess of Pressure:

Example – 1:

A raindrop of diameter 4 mm is about to fall on the ground. Calculate the pressure inside the raindrop. Surface tension of water = 0.072 N/m. Atmospheric pressure = 1.013 x 10⁵ N/m².

Given: Diameter of soap bubble = 4 mm, Radius of raindrop = r = 4/2 = 2 mm = 2 × 10^-3 m, Surface tension = T = 0.072 N/m, Outside pressure = P_o = Atmospheric pressure = 1.013 × 10⁵ N/m².

To Find: Inside pressure = P_i =?

Solution:

By Laplace’s law of spherical membrane for a drop

(P_i– P_o) = 2T / r

∴    P_i   = (2T / r) + P_o

∴    P_i   = (2 × 0.072 / 2 × 10^-3) + 1.013 × 10⁵

∴    P_i   =72 + 1.013 × 10⁵

∴    P_i   = 1.01372 × 10⁵ N/m²

Ans: The pressure inside the raindrop is 1.01372 × 10⁵ N/m²

Example – 2:

Find the excess pressure inside the soap bubble of diameter 3 cm. The surface tension of the soap solution is 3 × 10^-2 N/m. 

Given: Diameter of soap bubble = 3 cm, Radius of soap bubble = r = 3/2 =1.5 cm = 1.5 × 10^-2 m, Surface tension = T = 3 × 10^-2 N/m

To Find: excess pressure = (P_i– P_o) =?

Solution:

By Laplace’s law of spherical membrane for a bubble

(P_i– P_o) = 4T / r

∴   (P_i– P_o) = 4 × 3 × 10^-2 / 1.5 × 10^-2

∴   (P_i– P_o) = 8 N/m²

The excess pressure inside the soap bubble is 8N/m²

Example – 3:

What should be the diameter of a soap bubble, in order that the excess pressure inside it is 51.2 N/m². The surface tension of the soap solution is 3.2 × 10^-2 N/m. 

Given: excess pressure = (P_i– P_o) = 51.2 N/m², Surface tension = T = 3.2 × 10^-2 N/m

To Find: Diameter of soap bubble =?

Solution:

By Laplace’s law of spherical membrane for a bubble

(P_i– P_o) = 4T / r

∴ r = 4T / (P_i – P_o)

∴ r = 4 × 3.2 × 10^-2 / 51.2

∴ r = 2.5 × 10^-3 m = 2.5 mm

Diameter of soap bubble = 2 × 2.5 mm = 5 mm

Ans: The diameter of soap bubble is 5 mm

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Science > Physics > Surface Tension > Laplace’s Law of Spherical Membrane

The post Laplace’s Law of Spherical Membrane appeared first on The Fact Factor.

In this article, we shall study to solve problems on capillary action.

Important Formulae:

and Jurin’s law, hr = constant

where h  =  height of liquid level in the capillary

T  =  Surface tension

ρ =  Density of liquid

r  =  Radius of the bore of the capillary tube

g  =  Acceleration due to gravity

θ =  Angle of contract

Example – 1:

A capillary tube of a uniform bore is dipped vertically in water which rises by 7 cm in the tube. Find the radius of the capillary if the surface tension is 70 dynes/cm g= 9.8 m/s².

Given: Rise in the tube = h = 7 cm = 7× 10^-2 m, Surface tension = T = 70 dynes/cm = 70 × 10^-3 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density of water = ρ = 1 ×10³ kg/m³, Angle of contact = θ  = 0°

To Find: Radius of the tube = r =?

Solution:

The rise in the tube is given by

∴ r = (2 × 70 × 10^-3 × cos 0°) / (7× 10^-2 × 1 ×10³ × 9.8)

∴ r = 2 × 10^-4 m = 0.2 × 10^-3 m = 0.2 mm

Ans: The radius of the capillary is 0.2 mm

Example – 2:

Water rises to a height of 4.5 cm in a capillary tube of radius r. Find r assuming the surface tension of water is 72 dynes/cm. Take the angle of contact of water in the glass as 15°. Density of water=1000 kg/m³, g = 9.8 m/s².

Given: Rise in the tube = h =4.5 cm = 4.5 × 10^-2 m, Surface tension = T = 72 dynes/cm = 72 × 10^-3 N/m, Acceleration due to gravity = g = 9.8 m/s², Density of water = ρ = 10³ kg/m³, Angle of contact = θ = 15°

To Find: Radius of the tube = r =?

Solution:

The rise in the tube is given by

∴ r = (2 × 72 × 10^-3 × cos 15°) / (4.5× 10^-2× 1 ×10³ × 9.8)

∴ r = (2 × 72 × 10^-3 × 0.9659) / (4.5× 10^-2× 1 ×10³ × 9.8)

∴ r = 3.15 × 10^-4 m = 0.315 × 10^-3 m = 0.315 mm

Ans: The radius of the tube is 0.315 mm

Example – 3:

A liquid rises to a height of 9 cm in a glass capillary of radius 0.02 cm. What will be the height of the liquid column in a similar glass capillary of radius 0.03 cm?

Given: Rise in first tube = h₁ = 9 cm, Radius of first tube = r₁ = 0.02 cm, Radius of second tube = r₂ = 0.03 cm,

To Find: Radius in second tube = h₂ =?

Solution:

For a particular liquid-solid interface by Jurin’s law, hr = constant

∴    h₁ r₁ = h₂ r₂

∴    h₂   = h₁ r₁ / r₂

∴    h₂ = 9 × 0.02 / 0.03 = 6 cm

Ans: The height of the liquid column in a similar glass capillary is 6 cm

Example – 4:

A liquid rises to a height of 4.5 cm in a glass capillary of radius 0.01 cm. What will be the height of the liquid column in a similar glass capillary of radius 0.02 cm?

Given: Rise in first capillary tube = h₁ = 4.5 cm, Radius of first capillary tube = r₁ = 0.01 cm, Radius of second capillary tube = r₂ = 0.02 cm,

To Find: Radius in second capillary tube = h₂ =?

For a particular liquid-solid interface by Jurin’s law, hr = constant

∴ h₁ r₁ = h₂ r₂

∴ h₂   = h₁ r₁ / r₂

∴ h₂ = 4.5 × 0.01 / 0.02 = 2.25 cm

Ans: The height of the liquid column in a similar glass capillary is 2.25 cm

Example – 5:

Water rises to height 3.2 cm in glass capillary tube. Find height to which same water rises in another capillary having half area of cross-section.

Given: Rise in first capillary tube = h₁ = 3.2 cm, Area of cross-section of first capillary tube = A₁,  Area of cross-section of second capillary tube = A₂ , A₂ = ½ A₁,

To Find: Rise in second capillary tube = h₂ =?

Solution:

Given A₂ = ½ A₁

∴    π r₂²  = ½ π r₁²

∴    r₂² = ½ r₁²

∴    r₂ = 0.707 r₁

For a particular liquid-solid interface by Jurin’s law hr = constant

∴    h₁ r₁ = h₂ r₂

∴    h₂   = h₁ r₁ / r₂

∴    h₂ = 3.2.5 × r₁ / 0.707  r₁  = 4.525 cm

Ans: The height of the liquid column in the second glass tube is 4.525 cm

Example – 6:

A liquid of density 900 kg/m³ rises to a height of 7 mm in a capillary tube of 2 mm internal diameter. If the angle of contact of the liquid in the glass is 25°, find the surface tension of the liquid. g=9.8 m/s².

Given: Rise in capillary tube = h = 7 mm = 7× 10^-3 m, Internal diameter = 2 mm, Radius of capillary tube = r = 2/2 = 1 mm = 1 × 10^-3 m, Acceleration due to gravity = g = 9.8 m/s², Density of liquid = ρ =900 kg/m³, Angle of contact = θ = 25°

To Find: Surface tension = T =?

Solution:

The rise in the capillary tube is given by

∴ T = (7× 10^-3 × 1 × 10^-3 × 900 × 9.8) / (2 × cos 25°)

∴ T = (7× 10^-3 × 1 × 10^-3 × 900 × 9.8) / (2 × 0.9063)

∴ T = 0.064 N/m

Ans: The surface tension of the liquid is 0.064 N/m

Example – 7:

When a capillary tube of radius 0.45 mm is dipped into water, the level inside the capillary tube rises to height 3 cm above the surface of the water. Calculate the surface tension of water, if its angle of contact is zero and its density is 1 g/cm³, g = 9.8 m/s².

Given: Rise in capillary tube = h =3 cm = 3 × 10^-2 m, Radius of capillary tube = r = 0.45 mm = 0.45 × 10^-3 m, Acceleration due to gravity = g = 9.8 m/s², Density of water =  1 g/cm³ = 1 × 10³ kg/m³, Angle of contact = θ  = 0°

To Find: Surface tension = T =?

Solution:

The rise in the capillary tube is given by

∴ T = (3× 10^-3 × 0.45 × 10^-3 × 1 × 10³ × 9.8) / (2 × cos 0°)

∴ T = (3× 10^-3 × 0.45 × 10^-3 × 1 × 10³ × 9.8) / (2 ×1)

∴ T = 0.006615 N/m

Ans: The surface tension of the liquid is 0.066615 N/m

Problem – 8:

The capillary rise when a glass capillary tube of diameter 0.4 mm is dipped vertically in olive oil is 3.82 mm. If the angle of contact of olive oil with glass is 0°and the density of olive oil is 920 kg/m³, find the surface tension of olive oil.

Given: Rise in capillary tube = h =3.82 mm = 3.82 × 10^-3 m, Radius of capillary tube = r = 0.4 mm = 0.4 × 10^-3 m, Acceleration due to gravity = g  = 9.8 m/s², Density of olive oil = 920kg/m³, Angle of contact = θ  = 0°

To Find: Surface tension = T =?

Solution:

The rise in the tube is given by

∴ T =(3.82× 10^-3  × 0.4 × 10^-3 × 920 × 9.8) / (2 × cos 0°)

∴ T =(3.82 × 10^-3  × 0.4 × 10^-3 × 920 × 9.8) / (2 ×1)

∴ T = 6.889 × 10^-3 N/m

Ans: The surface tension of olive oil is 6.889 × 10^-3 N/m

Example – 9:

A mercury barometer tube is 2.5 mm in internal radius. Find the error introduced in the observed reading because of surface tension. g = 9.8 m/s², T=0.540 N/m. Density of mercury = 13600 kg /m³, angle of contact of mercury in glass = 135°.

Given: Radius of capillary tube = r =2.5 mm = 2.5 × 10^-3 m, Surface tension = T = 0.540N/m, Acceleration due to gravity = g = 9.8 m/s², Density of mercury = ρ = 13600 kg/m³, Angle of contact = θ  = 135°

To Find: Error introduced in the observation = h =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 0.540× cos 135°) / (2.5× 10^-3× 13600 × 9.8)

∴ h= (2 × 0.540× (-0.707)) / (2.5× 10^-3× 13600 × 9.8)

∴ h= – 2.29 × 10^-3 m = – 2.29 mm

Ans: The error introduced in the observation is – 2.29 mm

Hence 2.29 mm should be added to height of mercury measurement.

Example – 10:

The tube of a mercury barometer is 3 mm in diameter. What error is introduced in the reading because of surface tension? Angle of contact of mercury in glass = 135° and S.T. of mercury = 0.460 N/m. Density of mercury = 13.6 g/cc, g=9.8 m/s²

Given: Diameter of capillary tube = 3 mm, Radius of capillary tube = r =3/2 = 1.5 mm = 1.5 × 10^-3 m, Surface tension = T = 0.460N/m, Acceleration due to gravity = g  = 9.8 m/s², Density of mercury = ρ = 13.6 g/cc= 13.6 × 10³ kg/m³, Angle of contact = θ  = 135°

To Find: Radius of capillary tube = r =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 0.460× cos 135°) / (1.5× 10^-3× 13.6 × 10³ × 9.8)

∴ h= (2 × 0.460× (-0.707)) / (1.5× 10^-3× 13.6 × 10³ × 9.8)

∴ h= – 3.25 × 10^-3 m = – 3.25 mm

Ans: The error introduced in the observation is – 3.25 mm

Hence 3.25 mm should be added to height of mercury measurement.

Example – 11:

A mercury barometer tube is 1 cm in diameter. Find the error introduced in the observed reading because of surface tension. g = 9.8 m/s², T=435.5 dyne/cm. The density of mercury = 13600 kg /m³, angle of contact of mercury in glass = 140°.

Given: Diameter of capillary tube = 1 cm, Radius of capillary tube = r =1/2 = 0.5 cm = 0.5 × 10^-2 m, Surface tension = T =435.5 dyne/cm = 435.5 × 10^-3 N/m, Acceleration due to gravity = g = 9.8 m/s², Density of mercury = ρ = 13600 kg/m³, Angle of contact = θ  = 140°

To Find: Correction due to capillary action = h =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 435.5 × 10^-3 × cos 140°) / (0.5× 10^-2× 13600 × 9.8)

∴ h= (- 2 × 435.5 × 10^-3 × sin 140°) / (0.5× 10^-2× 13600 × 9.8)

∴ h= (- 2 × 435.5 × 10^-3 × 0.766) / (2.5× 10^-3× 13600 × 9.8)

∴ h= – 1.01 × 10^-3 m = – 1.01 mm

Ans: The correction due to capillary action is – 1.01 mm

Hence 1.01 mm should be added to height of mercury measurement.

Example – 12:

A capillary tube of radius 0.5 mm is dipped vertically in a liquid of surface tension 0.04 N/m and density 0.8 g/cc. Calculate the height of capillary rise, if the angle of contact is 10°.

Given: Radius of capillary tube = r =0.5 mm = 0.5 × 10^-3 m, Surface tension = T =0.04 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density = ρ = 0.8 g/cc = 0.8  × 10³  kg/m³, Angle of contact = θ  = 10°

To Find: Height of capillary rise = h =?

Solution:

The rise in capillary tube is given by

∴ h= (2 × 0.04× cos 10°) / (0.5× 10^-3× 0.8 × 10³ × 9.8)

∴ h= (2 × 0.04× 0.9848) / (0.5× 10^-3× 0.8 × 10³ × 9.8)

∴ h= 2.01 × 10^-2 m = 2.01 cm

Ans: The height of capillary rise is 2.01 cm

Problem – 13:

A capillary tube 0.12 mm in diameter has its lower end immersed in a liquid of surface tension 0.054 N/m. If the density of the liquid is 860 kg/m³, find the height to which the liquid rises in the tube. Given the angle of contact 30° and g = 9.8 m/s².

Given: Radius of capillary tube = r =0.12 mm = 0.12 × 10^-3 m, Surface tension = T =0.054 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density = ρ = 860 kg/m³, Angle of contact = θ  = 30°

To Find: Height of capillary rise = h =?

Solution:

The rise in the tube is given by

∴ h= (2 × 0.054× cos 30°) / (0.12× 10^-3×860 × 9.8)

∴ h= (2 × 0.054× 0.5) / (0.12× 10^-3 × 860 × 9.8)

∴ h= 5.34 × 10^-2 m = 5.34 cm

Ans: The height of capillary rise is 5.34 cm

Example – 14:

A capillary tube of radius 0.5 mm is immersed in a beaker of mercury. The mercury level inside the tube is found to be 0.80 cm below the level of the reservoir. Determine the angle of contact between mercury and glass. Surface tension of mercury = 0.465 N/m and density is 13.6 × 10³ kg/m³, g = 9.8 m/s^2.

Given: Radius of capillary tube = r =0.5 mm = 0.5 × 10^-3 m, Height of capillary rise = h = – 0.80 cm = – 0.80 × 10^-2 m, Surface tension = T =0.465 N/m, Acceleration due to gravity = g = 9.8 m/s², Density = ρ =13.6  × 10³  kg/m³,

To Find:  Angle of contact = θ =?

Solution:

The rise in capillary tube is given by

∴ cos θ =   (- 0.80 × 10^-2 × 0.5 × 10^-3 × 13.6  × 10³ × 9.8)/ (2× 0.465)

∴ cos θ =   – 0.5732

∴ cos (π – θ) =  0.5732

∴ (π – θ) = cos^-1 (0.5732)

∴   (180° – θ) =55°2′

∴ θ = 180° – 55°2′ = 124°58′

Ans: The angle of contact is 124°58′

Example – 15:

Calculate the density of paraffin oil, if glass capillary of diameter 0.25 mm dipped in a paraffin oil of the surface tension 0.0245 N/m rises to a height of 4 cm. Angle of contact of paraffin oil with glass is 28°, g  = 9.8 m/s²,

Given: Diameter of capillary tube = 0.25 mm, Radius of capillary tube = r = 0.25/2 = 0.125 mm = 0.125 × 10^-3 m, Height of capillary rise = h = 4 cm = 4 × 10^-2 m , Surface tension = T =0.0245 N/m, Acceleration due to gravity = g  = 9.8 m/s², Angle of contact = θ  = 28°

To Find:   Density = ρ =?

Solution:

The rise in capillary tube is given by

∴ ρ = (2 × 0.0245 × cos 28°) / (0.125 × 10^-3 × 4 × 10^-2 × 9.8)

∴ ρ = (2 × 0.0245 × 0.8829) / (0.125 × 10^-3 × 4 × 10^-2 × 9.8)

∴ ρ = 882.9 kg/m³

Ans: The density of paraffin oil is 882.9 kg/m³

Example – 16:

Water rises to a height of 4 cm in a certain capillary tube. If the same capillary tube is dipped in mercury, the level of mercury decreases to 3 cm. Compare the surface tension of water and mercury, if densities of mercury and water are 13.6 × 10³ kg/m³ and 10³ kg/m³ respectively. The angle of contact for water and mercury are 0° and 135° respectively.

Given:   

For water: The rise in tube = h_w =4 cm, Density = ρ_w = 10³ kg/m³, Angle of contact = θ_w =0°, Radius of capillary = r_w= r.

For mercury: The rise in tube = h_m = – 3 cm, Density = ρ_m = 13.6 × 10³ kg/m³, Angle of contact = θ_m =135°, Radius of capillary = r_m = r

To Find: T_w / T_m =?

Solution:

The rise in the tube is given by

Ans: The ratio of the surface tension of water to the surface tension of mercury 0.06932:1

Example – 17:

A U tube has one limb of radius 0.5 cm and the other limb of radius 0.1 cm. Water is poured in the tube. Find the difference in the level of water in the two tubes. S.T. of water = 0.070 N/m. g = 9.8 m/s².

Given: Radius of first limb = r₁ =0.5 cm =  0.5 × 10^-2 m ,  Radius of second limb = r₂ =0.1 cm =  0.1 × 10^-2 m , Surface tension = T = 0.070 N/m, Acceleration due to gravity = g  = 9.8 m/s², Density = ρ = 1  × 10³  kg/m³, Angle of contact = θ  = 0°

To Find: Differene in levels   = |h₂ – h₁| =?

Solution:

The rise in the capillary tube is given by

Hence h ∝ 1/r

Now r₁ > r₂, hence  h₂ > h₁

Ans: The difference in the level of water in the two tubes is 1.142 cm

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