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Properties of matter

  1. Mar 13, 2006 #1
    When two soap bubbles of different radii coalesce, will the newly formed
    bubble be spherical in shape? Also will there be a change in volume due to this process?Assume that temperature is constant.
  2. jcsd
  3. Mar 14, 2006 #2

    Andrew Mason

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    If the two volumes of air within the bubbles combine into an undivided volume surrounded by a soap membrane, then I would say that the bubble has to be spherical (since there is no preferred direction of motion of the molecules in the gas and since the soap bubble assumes the shape which has the smallest area - a sphere).

    The volume question is a bit tricky. Since they are at the same temperature, the question is really: how do the pressures of the two separate bubbles compare to the pressure in the combined bubble?

    I am not sure but here is my take on it: The pressure inside the bubble is a function of the surface tension of the bubble and the temperature. If temperature is constant, the higher the surface tension, the higher the pressure within the bubble. So if the two bubbles combine to form one bubble, the surface area of the combined bubble is smaller, meaning the membrane is thicker so the soap molecules are not stretched as much as before. Therefore, the surface tension is less than before and the pressure is less, so the volume increases.

  4. Mar 14, 2006 #3
    thats a great thought problem, and good answer A Mason, what class/book is that from Amith?
  5. Mar 15, 2006 #4


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    Andrew, I think that the pressure inside a soap bubble can be written as a function of the Surface Tension and the radius of the bubble.

    A soap bubble has some definite (but very small) thickness and therefore has two layers, one towards the outside air and one towards the enclosed air, and between these layers is the soap solution.

    So if the pressure of the air outside is [itex] P_{atm} [/itex] and the pressure in the soap solution between these layers is [itex] P_1 [/itex] and the pressure of the enclosed air is [itex] P_2 [/itex]

    [tex] P_1 - P_{atm} = \frac{2S}{R} [/tex]
    where R is the radius of the soap bubble and S is the surface tension of the solution. I'm assuming the thickness of the soap bubble is negligible to the radius. So, again,

    [tex] P_2- P_1 = \frac{2S}{R} [/tex]

    [tex] P_2 - P_{atm} = \frac{4S}{R} [/tex].

    And the pressure inside is greater than the pressure outside by an amount

    [tex] \frac{4S}{R} [/tex]

    Also, for the OP's question, my assumption would be the mass of the air inside the two smaller soap bubbles will be conserved, and so the volume of the new, larger bubble will be the sum of the volumes of the two smaller bubbles. Since the radii of the smaller bubbles are known, the radius of the larger one can be found, and from this the pressure inside the new bubble can also be found.

    What do you reckon?
    Last edited: Mar 15, 2006
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