Sphere particle dissolution, volume loss vs time

[hammal]

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
t=time
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
d_V/d_t=kA

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.

 

[HallsofIvy]

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, A= 4\pi r^2 and V=n (4/3)\pi r^3.
Differentiating with respect to t,
\frac{dA}{dt}= 8pi r\frac{dr}{dt} and
\frac{dV}{dt}= 4\pi r^2\frac{dr}{dt}

From the first, \frac{dr}{dt}= \frac{1}{8\pi r}\frac{dA}{dt}
so that \frac{dV}{dt}= 4\pi r^2\frac{1}{8\pi r}\frac{dA}{dt}= \frac{r}{4}\frac{dA}{dt}

 

[hammal]

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.

Cheers

 

[Chestermiller]

Hi Hammel. Welcome to Physics Forums.

From HallsofIvy's response,
\frac{dV}{dt}= 4\pi r^2\frac{dr}{dt}=-k (4\pi r^2)
Canceling:
\frac{dr}{dt}=-k
or, $r=r_0-kt$

 

[hammal]

Yes, got it, thanks both of you