Understanding Surface Tension: Laplace's Law and Balloon Experiment Explained

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SUMMARY

The discussion centers on the application of Laplace's Law to understand the behavior of two connected balloons with differing sizes and equal surface tension. When the valve between the balloons is opened, air flows from the smaller balloon to the larger one, as predicted by the equation T(surface tension) = P(pressure) x R(radius). The confusion arises from the assumption that a larger balloon contains more air and thus should have higher pressure; however, the material properties of the balloons, such as stiffness and stretchiness, influence the pressure dynamics. Ultimately, the larger balloon has a lower internal pressure due to the increased surface tension from its stretched rubber material.

PREREQUISITES
  • Understanding of Laplace's Law in fluid mechanics
  • Knowledge of surface tension and its effects on pressure
  • Familiarity with the properties of different balloon materials
  • Basic principles of gas behavior in connected systems
NEXT STEPS
  • Study the mathematical derivation of Laplace's Law in fluid dynamics
  • Explore the relationship between surface tension and pressure in soap bubbles
  • Investigate the mechanical properties of various rubber materials used in balloons
  • Learn about gas laws and their application in connected systems
USEFUL FOR

Students of physics, educators explaining fluid dynamics, and anyone interested in the principles of surface tension and material properties in practical applications.

seraphodiabolus
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Having trouble to understand a classical example of surface tension:
Two balloons are connected to each other with a valve. If the surface tension of the two balloons is the same but one balloon is bigger than the other, when the valve is lifted open so the air in the two balloons is now connected, what will happen to the two balloons?
The correct answer is the small balloon will get smaller and the big balloon gets bigger. This answer follows the application of Laplace's law where it says (something like the following) T(surface tension) = P(pressure) x R (radius). So the bigger the radius, the smaller the pressure. Thus when the valve is open, air will flow from high pressure (small balloon) to low pressure (big balloon), as always.
The confusion I have is, I assume a bigger balloon will have MORE air inside. If two balloons are made of the same material (so as to have the same surface tension) and I pump air into the balloons, shouldn't the bigger balloon contain more air, which means higher the pressure. Then the logical conclusion will be opposite to what Laplace's law predict: if I open the valve, air will flow from big balloon (more air) to small balloon (less air). What's wrong with my assumption?
 
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... it is possible to have the same tension and different materials.
Some rubber is stiffer than others so it requires a smaller expansion to build the same tension. You should know this from blowing up different brand/shape party balloons. In order that the tension is the same, the pressure in the big one must be smaller. Maybe there is less air in it and the material is very stretchy.

See: http://hyperphysics.phy-astr.gsu.edu/hbase/ptens.html
 
I think this question relates to soap bubbles not balloons.
The surface tension in a soap bubble is constant and this is why the pressure in a small soap bubble is greater than in a large bubble.
A balloon is made of rubber and the tension increases as the rubber is stretched as it gets bigger.
 

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