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Question
Obtain an expression for the excess of pressure inside a
- liquid drop
- liquid bubble
- air bubble
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Solution
i. Excess of pressure inside air bubble in a liquid: Consider an air bubble of radius R inside a liquid having surface tension T. Let P1 and P2 be the pressures outside and inside the air bubble, respectively.
Now, the excess pressure inside the air bubble is ∆P = P1 – P2
Air bubble
In order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,
- The force due to surface tension acting towards right around the rim of length 2πR is FT = 2πRT
- The force due to outside pressure P, is to the right acting across a cross-sectional area of πR2 is `"P"_1π"R"^2`
- The force due to pressure P2 inside the bubble, acting to the left is `"F"_("P"_2) = "P"_2π"R"^2`
As the air bubble is in equilibrium under the action of these forces, `"F"_("P"_2) = "F"_"T" + "F"_("P"_1)`
Excess pressure is ∆P = P2 – P1 = `(2"T")/"R"`
ii. Excess pressure inside a soap bubble: Consider a soap bubble of radius R and the surface tension of the soap bubble be T. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble. Therefore, the force on the soap bubble due to surface tension is 2 × 2πRT.
Soap bubble
The various forces acting on the soap bubble are,
- Force due to surface tension FT = 4πRT towards right.
- Force due to outside pressure, `"F"_("P"_1) = "P"_1π"R"^2` towards right
- Force due to inside pressure, `"F"_("P"_2) = "P"_2π"R"^2` towards left
As the bubble is in equilibrium, `"F"_("P"_2) = "F"_"T" + "F"_("P"_1)`
P2πR2 = 4πRT + P1πR2 ⇒ (P2 − P1)πR2 = 4πPT
Excess pressure is ∆P = P2 – P1 = `(4"T")/"R"`
iii. Excess pressure inside the liquid drop: Consider a liquid drop of radius R and the surface tension of the liquid is T.
Liquid drop
The various forces acting on the liquid drop are,
- Force due to surface tension FT = 2πRT towards right
- Force due to outside pressure, `"F"_("P"_1) = "P"_1π"R"^2` towards right
- Force due to inside pressure, `"F"_("P"_2) = "P"_2π"R"^2` towards left
As the drop is in equilibrium, `"F"_("P"_2) = "F"_"T" + "F"_("P"_1)`
P2πR2 = 2πRT + P1πR2 ⇒ (P2 − P1)πR2 = 2πPT
Excess pressure is ∆P = P2 – P1 = `(2"T")/"R"`
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