Sphere particle dissolution, volume loss vs time

In summary, the conversation discusses finding the equation of the volume lost during the dissolution of a sphere particle over time. It is mentioned that the dissolution rate is proportional to the surface area of the sphere and a function of time. The conversation also includes equations for area and volume of a sphere, as well as the derivatives of these equations. The final part of the conversation involves finding a volume function of time, which is eventually solved for using the equation r=r0-kt.
  • #1
hammal
3
0
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.
 
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  • #2
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]
 
  • #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.

Cheers
 
  • #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]
Canceling:
[tex]\frac{dr}{dt}=-k[/tex]
or, ##r=r_0-kt##
 
  • #5
Yes, got it, thanks both of you
 

What is sphere particle dissolution?

Sphere particle dissolution is the process by which a spherical particle dissolves in a solvent or medium. This can occur through a variety of mechanisms, such as surface dissolution, diffusion, or erosion.

How does volume loss of a sphere particle change over time?

The volume loss of a sphere particle typically follows an exponential decay curve, with a rapid initial decrease followed by a slower decrease over time. This is due to the limited surface area available for dissolution as the particle shrinks.

What factors affect the dissolution rate of a sphere particle?

The dissolution rate of a sphere particle can be affected by several factors, including the solubility of the particle in the solvent, the surface area to volume ratio of the particle, the concentration of the solvent, and the temperature.

Can the dissolution rate of a sphere particle be predicted?

The dissolution rate of a sphere particle can be predicted through mathematical models and experimental data. However, it is important to note that the dissolution rate may vary depending on the specific conditions and properties of the particle and solvent.

How is the volume loss of a sphere particle measured?

The volume loss of a sphere particle can be measured through various techniques, such as visual observation, gravimetric analysis, or particle size analysis. These methods can provide information about the dissolution rate and the size of the remaining particle over time.

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