Excess pressure inside a Liquid drop

  • #1

Summary:

How does the surface tension acting along the base of hemisphere holds the curved surface against pressure?

Main Question or Discussion Point

While studying about the effects of surface tension i came across the excess pressure inside a liquid drop.
Here they considered a hemisphere ABCDE from the drop and listed out the conditions for it to be in equilibrium.
IMG_20191217_192915.jpg

The forces acting on them are taken as
IMG_20191217_193416.jpg


F1= 2πRS
F2= P1×(Projection of hemispherical surface on ABCD)
=>F2=P1×πR²
F3=P2×πR²


For equilibrium we take
F1+F2=F3

But what role does the surface tension (T) has in maintaining the equilibrium for a hemisphere.
My doubts are:
1. Isn't the pressure due to Atmosphere and the pressure inside the hemispherical drop enough to balance each other. Why do we need a surface tension here?
2. When i referred some other sources, it said that the surface tension holds the drop from bursting. If it is so, the force due to T we calculates here acts only along the base periphery of the hemisphere. What effect would it have on the curved surface?


Hope this won't come under Homework help
 

Answers and Replies

  • #2
gleem
Science Advisor
Education Advisor
1,667
1,021
The surface tension of the bubble is due to the attraction of molecules with their neighbors thus pulling them closer and causing the surface area to shrink. This naturally will increase the pressure inside until an equilibrium situation is produced.

In the analysis three forces are identified. That due to the outside pressure pushing on the surface (P0πr2) that due to the surface tension acting in the surface on the circumference (2π rσ) which act together and that due to the inside pressure (P πr2) countering the other two to produce the equilibrium situation. where r is the radius and σ is the surface tension

The pressure in the bubble is constant throughout . The splitting of the bubble into two hemispheres simplifies the analysis.

thus

P πr2 = P0πr2 + 2π rσ

or P = P0+ 2σ/r
 
  • #3
The surface tension of the bubble is due to the attraction of molecules with their neighbors thus pulling them closer and causing the surface area to shrink. This naturally will increase the pressure inside until an equilibrium situation is produced.

In the analysis three forces are identified. That due to the outside pressure pushing on the surface (P0πr2) that due to the surface tension acting in the surface on the circumference (2π rσ) which act together and that due to the inside pressure (P πr2) countering the other two to produce the equilibrium situation. where r is the radius and σ is the surface tension

The pressure in the bubble is constant throughout . The splitting of the bubble into two hemispheres simplifies the analysis.

thus

P πr2 = P0πr2 + 2π rσ

or P = P0+ 2σ/r
What if the surface tension is zero?
How would the hemispherical part of the drop behave?
 
  • #6
Lord Jestocost
Gold Member
597
406
See comment #2 by @.Scott in the thread:

Zero surface tension: "It would signify that the molecules within the fluid have the same attraction to each other as to the atmosphere at the surface."
 
Last edited:

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