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Fluid - Work done to blow a soap bubble

  1. Dec 18, 2015 #1
    • Member advised to use the homework template!
    Hi everybody, I now encounter some problems when try to solve this:
    Problem statement: Calculate the work done to blow a soap bubble to radius R. Knowing that the process is isotherm, the atmospheric pressure is H, the surface tension is γ.

    Solution:
    $$A = A_{1}+A_{2}$$
    In which A1 is the work done to blow the bubble with 2 layers :
    $$A_{1} = 2(4 \gamma πR^{2}) $$
    A2 is the work done to blow the bubble to pressure p' = H + 4γ/R
    $$A_{2}=p' V ln \dfrac{p'}{H}$$

    The solution above is in my textbook, but I still help some questions
    First, why the soap bubble has 2 layers, which leads to the Laplace pressure is 4γ/R?
    Secondly, in A2 equation, why the ratio is natural logarithm is p' over H? In the isotherm, we have: $$\dfrac{V_{i}}{V_{f}}=\dfrac{p'}{H}$$
    It doesn't match the formular used to calculate work done in the isotherm process.

    Can anyone explain these problems in details for me?
    Thanks a lot.
     
  2. jcsd
  3. Dec 18, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    A soap bubble is made of a thin film of water. The film has an outer surface and an inner surface.
    http://www.webexhibits.org/causesofcolor/15E.html
    It should match. Can you state the formula that you have in mind for the work done in an isothermal process?
     
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