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Pressure Inside a Soap Bubble

  1. Aug 9, 2004 #1
    Show that inside a soap bubble, there must be a pressure [itex]\Delta P[/itex] in excess of that outside equal to [itex]\Delta P = 4 \gamma /r[/itex] where r is the radius of the buble and [itex]\gamma[/itex] is the surface tension.

    My first question is: The pressure on what!? Are they referring to the pressure on the bubble itself or on the surface enclosed by the bubble?

    I already "solved" the problem, but I did it without knowing how and why (I just plugged in what I thought would work and it worked!). For one thing, they seem to be referring to the pressure on the surface enclosed by the bubble. How is the bubble applying pressure here? It seems the surface tension is responsible for keeping the bubble from bursting, but I don't really know why.
     
  2. jcsd
  3. Aug 9, 2004 #2

    Doc Al

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    Think of the bubble as a shell. It has an inside pressure (pushing out) and an outside pressure (pushing in).
     
  4. Aug 10, 2004 #3
    So the pressure is one the bubble? But when I derived the equation, I used the area enclosed by the bubble and not the area of the bubble. If the pressure is on the bubble, why can't I derive the equation.
     
  5. Aug 10, 2004 #4

    Doc Al

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    I'm really not sure what you are asking. The internal pressure within the bubble is greater than the outside pressure. The surface tension is what makes up the difference.
     
  6. Aug 10, 2004 #5

    HallsofIvy

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    "Are they referring to the pressure on the bubble itself or on the surface enclosed by the bubble?"

    What "surface enclosed by the bubble"? If I read this correctly you are thinking of a hemi-spherical bubble floating on a flat water surface (which is the "surface enclosed by the bubble"). But a "bubble" could as easily be a completely spherical bubble in the air. The pressure they are talking about is the pressure on the bubble itself.
     
  7. Aug 11, 2004 #6
    Yes, this exactly what I was thinking. Maybe this is not the purported bubble in the question. I will consider a bubble floating in air. Let F' be the force pushing out on the bubble and let F be the force pushing in. So

    [tex]\Delta P = \frac{F + F'}{A}[/tex]

    where [itex]A = 4\pi r^2[/itex] (surface area of the bubble) and [itex]F = F' = 2\pi r \gamma[/itex]. Simplifying I get [itex]\Delta P = \gamma / r[/itex]. Since this is wrong, I guess the equation for F and F' are wrong and since I don't understand surface tension all that well I don't really know what to do.

    I'm given the relation [itex]\gamma = F/L[/itex] which is the force F per unit length L acting across the surface in question. I chose [itex]L = 2\pi r[/itex], which I think is perfectly reasonable. Nonetheless, I fail to obtain the desired equation.
     
  8. Aug 11, 2004 #7

    Doc Al

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    divide and conquer

    Rather that taking the entire bubble as a whole, just examine a section of bubble and analyze the forces on it. I recommend slicing (mentally) the bubble in half. First identify all the forces acting on that half-bubble: the force on it from the inside pressure, the force on it from the outside pressure, and the surface tension.

    Take another crack at it.

    PS: Remember that force is a vector. For example, the net force on the entire bubble pushing in (due to outside pressure) equals zero! So taking the bubble as a whole doesn't help much. What's the net force pushing in on half a bubble?
     
  9. Aug 12, 2004 #8
    Correct me if I' wrong, but isn't the surface tension vector normal to pressure force vectors? I just don't see how to relate the surface tension with pressure forces. Isn't the net force on any part of the bubble zero anyways (otherwise the bubble would burst or the surface would be moving/wobbling)?
     
  10. Aug 12, 2004 #9

    Doc Al

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    bubble, bubble... toil and trouble?

    Start by considering that half-bubble as I suggested. For simplicity, say you chose the top half:
    (1) What's the force due to internal pressure on that half bubble?
    (2) What's the force due to external pressure on that half bubble?
    (3) What's the force due to surface tension on that half bubble?
    Of course the net force is zero. We make use of that fact to solve this problem.

    In my last post, when I said:
    I was referring only to the force due to the external pressure--not due to external + internal + surface tension. I was pointing out that you won't get far by considering the entire bubble: consider a piece of the bubble.
     
  11. Aug 12, 2004 #10
    (1) What's the force due to internal pressure on that half bubble?
    In terms of the internal pressure P', then it is just F' = P'A (A is the area of half the bubble). I don't know how to express F' in other terms.

    (2) What's the force due to external pressure on that half bubble?
    Similar to the answer in (1).

    (3) What's the force due to surface tension on that half bubble?
    Let T be the force due to surface tension. On any part of the bubble, 0 = F' + F' + T, so T = -(F' + F).
    I don't know how to express T in other terms.

    I just don't know...
     
  12. Aug 12, 2004 #11

    Doc Al

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    half a bubble is better than nothing

    The pressure is outward at all points on the inner surface of the bubble. So what's the net force (due to internal pressure) on that half-bubble? It's an upward force equalling P'A, where A = [itex]\pi r^2[/itex] and r is the radius of the bubble. Think about this until it makes sense.

    Your turn.
     
  13. Aug 12, 2004 #12
    OK. I drew myself a diagram (finally) and I see what you mean now (i.e. the sum of the horizontal components of the force vectors is zero). So essentially the force on the inner surface of the bubble is acting on an area equal to [itex]\pi r^2[/itex]. So what about the surface tension [itex]\gamma[/itex]. Should I leave it as it is?
     
  14. Aug 12, 2004 #13

    Doc Al

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    Now what about the net force from the outer pressure on the half bubble?
    Leave it as what? Calculate it! What surface tension force is pulling on that half bubble? Remember that [itex]F_{s.t.} = \gamma L[/itex]. What's L? Hint: how many surfaces does that bubble have?
     
  15. Aug 12, 2004 #14
    Same deal as with the inner surface, except for its direction (i.e. it is opposite).

    The bubble has two surfaces, so [itex]T = 2\gamma L[/itex] supposedly (I just made this up, I really have no clue). This is the source of my misundertandings. What is L? Should I make it equal to [itex]\pi r[/itex] (i.e. half the circumference)? What determines the value of L?
     
  16. Aug 12, 2004 #15

    Doc Al

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    Right.
    Right!
    Why half the circumference? What determines L is where we drew the boundary for calculating the force. Since we split the bubble in half, the length of the boundary is [itex]2 \pi r[/itex]. (Of course we have to double that, since there are two surfaces.)

    Now put it all together.
     
  17. Aug 12, 2004 #16
    Thank goodness for this forum because my stupid book did not say anything about L being the length of the boundary of the surface.

    So you're saying that [itex]T = 2\gamma \cdot 4\pi r = 8\gamma \pi r[/itex]? So then, do I set [itex]T = A\Delta P[/itex]? I don't understand.
     
  18. Aug 13, 2004 #17

    Doc Al

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    I'm saying that [itex]T = 2\gamma \cdot 2\pi r = 4\gamma \pi r[/itex]. (Where did that [itex]4\pi r[/itex] come from? The circumference is [itex]2\pi r[/itex].)
    Eactly right.
    OK, you've suffered enough. Let's put it all together. The net force on our half-bubble must equal zero. So:
    [tex]P_i \pi r^2 = P_o \pi r^2 + 4\gamma \pi r[/tex]

    thus:
    [tex]\Delta P = \frac {4 \gamma}{r}[/tex]
     
  19. Aug 13, 2004 #18
    You said
    where "that" is the circumference of the bubble (see your penultimate post).

    I was having trouble seeing how the surface tension is acting in the same general direction as the pressure forces, but no more. I have been enlightened. Thanks.
     
  20. Aug 13, 2004 #19

    Doc Al

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    I meant that the force needs to be doubled since there are two surfaces, not that the circumference needs to be doubled. Sorry for the sloppy wording.
    Cool. :approve:
     
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