What is the equation for the pressure of a coalesced soap bubble?

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SUMMARY

The equation for the pressure of a coalesced soap bubble is derived from the balance of forces acting on the bubble. When two spherical soap bubbles with radii a[1] and a[2] coalesce, the resulting bubble's radius r satisfies the equation: p*r^3 + 4*γ*r^2 = p(a[1]^3 + a[2]^3) + 4*γ(a[1]^2 + a[2]^2), where p represents ambient pressure and γ denotes surface tension. The derivation involves considering the forces due to surface tension and internal pressure acting on the bubble's cross-section. The final equation reflects the equilibrium state of the coalesced bubble.

PREREQUISITES
  • Understanding of fluid mechanics principles
  • Knowledge of surface tension concepts
  • Familiarity with spherical geometry
  • Basic thermodynamics related to gas behavior
NEXT STEPS
  • Study the derivation of pressure equations in fluid mechanics
  • Explore the effects of surface tension on bubble dynamics
  • Learn about the thermodynamics of gas mixtures
  • Investigate applications of soap bubbles in material science
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Students in physics or engineering, particularly those focusing on fluid dynamics and thermodynamics, will benefit from this discussion. Additionally, researchers interested in the properties of soap films and bubbles will find the insights valuable.

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Homework Statement


Two spherical soap bubbles of radii a[1] and a[2] are made to coalesce. Show that when the temperature of the gas in the resulting soap bubble has returned to its initial values, r of the bubble is given by p*r^3+4*[tex]\gamma[/tex]*r^2=p(a[1]^3+a[2]^3)+4[tex]\gamma[/tex](a[1]^2+a[2]^2) where p is the ambient pressure and [tex]\gamma[/tex] is the surface tension.


Homework Equations





The Attempt at a Solution



I cut the bubble in half and found that the surface tension along the circumference is 2[tex]\gamma[/tex] therefore the total force is 2[tex]\gamma[/tex]*circumference (2[tex]\pi[/tex]*r) we then have the pressure inside the bubble acting over the entire cross section which gives total force of p*[tex]\pi[/tex]*r^2. adding the two together, my final equation is p*r^2+4*[tex]\gamma[/tex]*r=p(a[1]^2+a[2]^2)+4[tex]\gamma[/tex](a[1]+a[2]). I have no idea how I did it but my entire equation has been divided by r somehow. Please help. Thanks.
 
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