# How can I convert volume as a function of time to mass or mole as a function of time for a dissolving drug particle?

I have been given a spherical drug particle that dissolves by surface erosion into the bloodstream. The rate of dissolution is proportional to the exposed surface area of the particle. I have to model the volume of the dissolving particle as a function of time, which I did, and I got

$$V(t) = \left(\dfrac{3V_0^{1/3} - kt}{3}\right)^3$$

where $V_0 = V(t=0)$, the initial volume of particle. When I plot using Excel with $V_0=100$, I get a reasonable downward sloping curve which reaches 0 in about 3 sec.

Now I have to derive the rate of drug release as a function of time (g/s or mol/s) from the volume function. There is no other information given. How do I do this? Please help!

• Well, you know the rate of volume release, and if you know the density, then you know the mass? So, what's the density of this drug? – Jonathan R Swift Sep 20 '18 at 17:04
• The density is not given numerically. But we need to take a constant rho and derive the function of rate of drug release from this volume function. How do I incorporate rho? – kazi Sep 23 '18 at 6:42

This is actually more of a calculus question than anything.

That function gives you the volume over time. You are looking for the rate of change in volume over time.

That is the definition of the derivative of that function. So all we need to do is get $$\dfrac{\partial V}{\partial t}$$.

For that, we define $$V(x) = f(g(x))$$, where $$g(x) = 3V_0^{1/3}−kt$$ and $$f(x) = \left(\dfrac{g(x)}{3}\right)^3 = \dfrac{1}{27}g(x)^3$$. Via the chain rule, we know that $$\dfrac{\partial V}{\partial t} = f'(g(x))\cdot g'(x)$$. Therefore, we have that

\begin{alignat}{4} \dfrac{\partial V}{\partial t} &= &&\dfrac{1}{9}(3V_0^{1/3}−kt)^2\cdot(-k) \\ &= -&&\dfrac{k}{9}(3V_0^{1/3}−kt)^2 \end{alignat}

The result here is negative (assuming a positive $$k$$ and that the result between parentheses is also positive), which indicates that the volume is dropping over time. Obviously, the volume lost is precisely what's released into the bloodstream. So if that latter interpretation (rate of release) is most relevant, just remove the negative sign.

• The unit of rate of drug release is g/s or mol/s, not L/s. So just removing the negative sign from volume function won't do. I also don't have the density, so I don't know how to convert. – kazi Sep 21 '18 at 18:11
• @kazi if you only know the volume without knowing the density, you can't convert to grams or mol, by definition. Density is the conversion factor between them. – Wasabi Sep 21 '18 at 18:30
• The density is not given numerically. But we need to take a constant rho and derive the function of rate of drug release from this volume function. How do I incorporate rho? – kazi Sep 23 '18 at 6:43
• @kazi I'm having trouble understanding what you're asking for (I'm a civil engineer, so maybe there's a basic knowledge gap we're not aware of). Do you mean how to convert this equation into a rate of mass loss by adding in a variable $\rho$ which represents the (unknown) density? If so, just get the equation and multiply all of it by $\rho$. Note that if the density can be expected to change in the blood stream (hell if I know!), then $\rho$ will be a function of time as well. – Wasabi Sep 23 '18 at 7:24