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Sphere particle dissolution, volume loss vs time

  1. Jun 13, 2013 #1
    Hi all,
    it might be a silly question, but my math is "a bit" rusty.

    I want to find the equation of the volume lost during the dissolution of a sphere particle as function of time:

    A=area of sphere
    V=volume of sphere
    k=dissolution rate as volume dissolved over area time

    A=(4/pi) r^2
    V=(4/3)pi r^3
    Volume change from dissolution Vloss=kAt

    what I know is:
    d_V/d_r=4 pi r^2
    d_A/d_r=(8/pi) r

    but from here I don't know how to find V(t), so data I can plot the volume against time or integrate it to get the amount of volume lost against time.

    Please any help is much appreciated.
  2. jcsd
  3. Jun 13, 2013 #2


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    Science Advisor

    I presume that you are talking about a sphere, such as a moth ball, that is "evaporates" at a rate proportional to the surface area of the sphere. That is your dV/dt= kA.
    Now, as you say, [itex]A= 4\pi r^2[/itex] and [itex]V=n (4/3)\pi r^3[/itex].
    Differentiating with respect to t,
    [tex]\frac{dA}{dt}= 8pi r\frac{dr}{dt}[/tex] and
    [tex]\frac{dV}{dt}= 4\pi r^2\frac{dr}{dt}[/tex]

    From the first, [itex]\frac{dr}{dt}= \frac{1}{8\pi r}\frac{dA}{dt}[/itex]
    so that [itex]\frac{dV}{dt}= 4\pi r^2\frac{1}{8\pi r}\frac{dA}{dt}= \frac{r}{4}\frac{dA}{dt}[/itex]
  4. Jun 13, 2013 #3
    Yes, that is exactly what I meant. Thanks for your answer, but it is not clear how to find a volume function of time:
    I need something like d_V/d_t= whatever function of t

    I'm not able to do it from your last part.

  5. Jun 13, 2013 #4
    Hi Hammel. Welcome to Physics Forums.

    From HallsofIvy's response,
    [tex]\frac{dV}{dt}= 4\pi r^2\frac{dr}{dt}=-k (4\pi r^2)[/tex]
    or, ##r=r_0-kt##
  6. Jun 13, 2013 #5
    Yes, got it, thanks both of you
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