Physics Topics can also be used to explain the behavior of complex systems, such as the stock market or the dynamics of traffic flow.
Pressure difference between the two sides of a curved liquid surface
i) Suppose the free surface of a liquid is plane [Fig.(a)]. A molecule lying on its surface is attracted by other surface molecules equally in all directions. So the resultant tangential force on the molecule is zero.
ii) If the free surface of the liquid is concave, then every molecule on the surface experiences an upward resultant force due to attraction by other surface moleculers [Fig.(b)].
iii) If the liquid surface is convex, then the resultant force on a molecule on the surface due to attraction by other surface molecules will be directed downwards [Fig.(c)].
Obviously there must be a difference of pressure between the two sides of a curved surface for equilibrium of it. This difference of pressure i.e., the excess pressure force will balance the resultant force due to surface tension. The pressure on the concave side must be greater than the pressure on the convex side.
We shall now calculate the excess pressure on the concave side of spherical surfaces in case of a liquid drop, an air bubble in a liquid and a soap bubble.
Excess pressure Inside a liquid drop: Let us consider a liquid drop of radius R of a liquid of surface tension T [Fig.]. Every molecule on its surface experiences a resultant pull normally inwards due to surface
tension. So the internal pressure of the drop becomes greater than the pressure outside it. The internal excess pressure of the drop produces a force acting outwards which balances the force due to surface tension and maintains the equilibrium of the drop.
Suppose, external pressure on the drop = P, internal pressure of the drop = (P + p).
So, the excess pressure inside the drop = p.
Suppose this internal excess pressure acting normally outwards increases the radius of the drop from R to R + ΔR
i.e., it increases the surface area of the drop. Here ΔR is taken to be so small that the pressure inside the drop may be taken as unchanged.
Work done by the excess pressure,
W = Excess pressure × area × displacement
= p.4πR2 ᐧ ΔR ……….. (1)
Increase of surface area of the liquid drop,
ΔA = 4π(R + ΔR)2 – 4πR2
= 4π{R2 + 2R ᐧ ΔR + (ΔR)2 – R2}
= 8πR ᐧ ΔR; [neglecting the term (ΔR)2 which is very small]
∴ Increase in surface energy,
E = increase in surface area × surface tension
= 8πrR ᐧ ΔR ᐧ T ……… (2)
This increase in surface energy of the liquid drop takes place at the cost of work done by the excess pressure i.e., E = W.
So, from equation (1) and (2) we have,
p ᐧ 4πr2 ᐧ ΔR = 8πR ᐧ ΔR. T
or, p = \(\frac{2 T}{R}\) ……… (3)
Excess pressure inside an air bubble in a liquid. Let us consider an air bubble of radius R formed in a liquid of surface tension T [Fig.]. Like a liquid drop
the air bubble has also one surface in contact with the liquid. So proceeding similarly as in the case of a liquid drop we can prove that the excess pressure inside the air bubble in a liquid is given by p = \(\frac{2 T}{R}\).
Excess pressure inside a soap bubble: Let us consider a thin soap bubble of radius R formed from a soap solution of surface tension T [Fig.].
Suppose, the pressure outside the bubble = P, internal pressure = (P + p)
So, the excess pressure inside the bubble = p. Suppose, the radius of the bubble increases from R to R + ΔR due to this internal excess pressure acting normally outwards, i.e., the surface area of the bubble increases. Here ΔR is taken to be so small that the pressure inside the bubble may be taken as unchanged.
Work done by the excess pressure
W = excess pressure × area × displacement
= p ᐧ 4πR2 ᐧ ΔR
The soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble.
So, increase of surface area of the soap bubble,
ΔA = 2(4π(R + ΔR)2 – 4πR2]
= 8π[R2 + 2R ᐧ ΔR + (ΔR)2 – R2]
= 16πR ᐧ ΔR; [neglecting the term (ΔR)2 ………. (2)
This increase in surface energy of the soap bubble takes place at the cost of work done by the excess pressure i.e., E = W.
So, from equations (1) and (2) we have,
p ᐧ 4πR2 ᐧ R = 16πR ᐧ ΔR ᐧ T
or, p = \(\frac{4 T}{R}\) ……….. (3)
Numerical Examples
Example 1.
Find the excess pressure Inside a rain water drop of diameter 0.02 cm. The surface tension of water = 0.072 N ᐧ m-1.
Solution:
Water drop has only one curved surface.
So, excess pressure of a water drop, p = \(\frac{2 T}{r}\)
where, T = surface tension of water
r = radius of a water drop = \(\frac{0.02}{2}\) = 0.01 cm
= 0.01 × 10-2m
∴ p = \(\frac{2 \times 0.072}{0.01 \times 10^{-2}}\) = 1440 N ᐧ m-2.
Example 2.
Surface tension of soap solution = 27 dyn ᐧ cm-1. Calculate the excess pressure (in N ᐧ m-2) inside a soap bubble of radius 3cm.
Solution:
The excess pressure inside a soap bubble,
p = \(\frac{4 T}{r}\) = \(\frac{4 \times 27}{3}\) = 36 dyn ᐧ cm-2 = 3.6 N ᐧ m-2.
Example 3.
Find the pressure inside an air bubble of radius 0.1 mm just inside the surface of water. Surface tension of water = 72 dyn ᐧ cm-1.
Solution:
Excess pressure inside an air bubble
= \(\frac{2 T}{r}\) = \(\frac{2 \times 72}{0.01}\)
[∵ T = 72 dyn ᐧ cm-1 and r = 0.1 mm = 0.01 cm]
= 14400 dyn ᐧ cm-2.
Atmospheric pressure = 76 × 13.6 × 980 dyn ᐧ cm-2
∴ Total pressure inside an air bubble
= (76 × 13.6 × 980 + 14400)
= 1.0274 × 106 dyn ᐧ cm-2.
Example 4.
The excess pressure inside a soap bubble of radius 8 mm raises the height of an oil column by 2 mm. Find the surface tension of the soap solution. Den-sity of the oil = 0.8g ᐧ cm-3.
Solution:
Excess pressure in a soap bubble (p) = \(\frac{4 T}{r}\)
Again, p = hρg, h = height of the oil column.
Now, \(\frac{4 T}{r}\) = hρg or, T = \(\frac{h \rho g \times r}{4}\) = \(\frac{0.2 \times 0.8 \times 980 \times 0.8}{4}\)
[∵ h = 2mm = 0.2 cm; ρ = 0.8 g ᐧ cm-3, g = 980 cm ᐧ s-2 and r = 8mm = 0.8cm].
∴ T = 31.36 dyn ᐧ cm-1
Example 5.
In an isothermal process, two soap bubbles of radii a and b combine and form a bubble of radius c. If the external pressure is p, then prove that the surface tension of the soap solution is T = \(\frac{p\left(c^3-a^3-b^3\right)}{4\left(a^2+b^2-c^2\right)}\)
Solution:
We know, the excess pressure inside the soap bubble = internal pressure — external pressure.
∴ For the bubble of radius a, excess pressure,
\(\frac{4 T}{a}\) = pa – p
∴ pa = (p + \(\frac{4 T}{a}\))
Similarly, for the bubble of radius b,
pb = (p + \(\frac{4 T}{b}\))
For the bubble of radius c,
Pc = (p + \(\frac{4 T}{c}\))
Boyle’s law is applicable in isothermal process. According to this law,
Example 6.
Two soap bubbles of radii 0.04 m and 0.03 m are: combined in such a way that a common surface is formed between the two bubbles. What is the radius of curvature of the common surface?
Solution:
Let, the radii of the two soap bubbles are r1 and r2 and the internal pressures are p1 and p2 respectively
The radius of the common surface = r, atmospheric pressure = p0
For the first bubble,
p1 – p0 = \(\frac{4 T}{r_1}\) ……. (1)
and for the second bubble,
p2 – p0 = \(\frac{4 T}{r_2}\) ……. (2)
where, T = surface tension of soap solution.
Subtracting (1) from (2) we get,
p2 – p1 = 4T(\(\frac{1}{r_2}\) – \(\frac{1}{r_1}\)) = 4T(\(\frac{1}{0.03}\) – \(\frac{1}{0.04}\))
= \(\frac{4 \times 100}{12}\) × T ………… (3)
But, for the common surface,
p2 – p1 = \(\frac{4 T}{r}\) …… (4)
Comparing (3) and (4) we get,
\(\frac{4 T}{r}\) = \(\frac{4 \times 100}{12}\) × T
∴r = 0.12 m.