Can anyone help me write a differential equation for a certain situation? I'm having trouble remembering how to do this.

Question

kkutie7 · Answer

so i have a light intensity that changes due to absorption in a vial of dissolved chlorophyll is proportional to:
1. the distance increment traversed in the vial dz
2. the light intensity at a particular location in the vial I(z)
3. number of molecules per cm3, N
4. absorption crosssection (incm2), *sigma*

kainui · Answer

Is light intensity unitless? I'm sorta trying to figure this out but I know we can cheat for this one and look up the derivation of like the Beer-Lambert law or w/e it is

kkutie7 · Answer

I have no idea what the unit light intensity would have. This is all i was giving and im confused

kainui · Answer

I think best is to draw a picture, start labelling all that stuff, and writing it out. I'm a bit distracted but I'll focus more on this I guess since I'm kinda curious to figure it out.

kkutie7 · Answer

I dont understand how im supposed to start. i know i need a function of I=? and i need a constant in there k but thats it

kkutie7 · Answer

from my understanding "proportional"
$$dI=dz,dI=I(z), dI=N , dI=\sigma$$

kainui · Answer

My guess is this so far, that at some place z, the intensity will decrease by a little bit dz further by some amount proportional to the number of molecules in that area which is n.

$$I(z+dz)=I(z)- k n$$

Now n is infinitesimally small, since it's the number of molecules in an infinitesimally small volume:

$$n = N dV$$

But we know that infinitesimal volume is really the cross section $\sigma$ times dz so:

$$dV = \sigma dz$$

Plugging this all in we get:

$$I(z+dz)=I(z) - k N\sigma dz$$

(Mind you this has all been mostly just my guess idk if this is 100% right but feels good to me)

Now we can see that a little rearrangement gets us:

$$\frac{I(z+dz)-I(z)}{dz} = -k N \sigma$$

which since dz is infinitesimally small, the derivative...

$$\frac{dI}{dz} = -kN \sigma$$

kkutie7 · Answer

i understand how you plugged everything in but i fuzzy on the set up in the begging:
$$I(z+dz)=I(z)- k n$$

kainui · Answer

I sorta just made it up, so it could be wrong.

kainui · Answer

My idea was that $I(z+dz)$ is the intensity a little bit further ahead of $I(z)$. Since it's ahead of it, it should be lower in intensity. 

$$I(z+dz) < I(z)$$

in other words just some random positive number A:

$$I(z+dz) = I(z) - A$$

I realized ok well what's decreasing the intensity? The number of molecules, which should be a constant for the entire beaker cause the concentration is the same of chlorophyll everywhere in there I think. But I don't know by how much it decreases the intensity per number of molecules $n$ so I put a $k$ on it.
$$I(z+dz)=I(z)-kn$$

Really though $n=N dV$ since N is molecules/volume so maybe I should use $dn$ instead, but I felt like that made it look confusing since chlorophyll molecules are finite and not infinitesimally small it's probably more reasonable to try to derive this from a limit or something... The main thing is that $N$ is like the density.

Hopefully that clears up my reasoning a bit

kainui · Answer

Looks like I'm wrong cause it gets the wrong answer so idk what exactly I just found a derivation of it here that I'm trying to decrypt haha http://www.pci.tu-bs.de/aggericke/PC4e/Kap_I/beerslaw.htm

kkutie7 · Answer

i think i understand, but i think i'm getting confused because this was all my teacher gave me:
$$R\rightarrow P$$
find a diff eq for the concentration as it changes with time:
$$C_{R}k=\frac{dC_{R}}{dt} $$
I feel lik the problem i was given is too much of a jump in complexity =(

kainui · Answer

Oh you're probably told or it's in your book that for any sorta thing you have that the change in it is proportional to the rate in minus the rate out like this:

$$\frac{dQ}{dt} = rate_{in} - rate_{out}$$

Not sure, I guess this is a ordinary differential equations class? It shouldn't be so crazy I think I definitely was taking a more bare bones approach of trying to create the model out of nothing but I think you're allowed to assume some stuff from what you're given.

kkutie7 · Answer

its mathematics for physical chemistry
yes ive definitely heard the whole rate in and out thing before

caozeyuan · Answer

so you have change in I as a function of the 4 variables given. Lets take an infinitesimal change in I is dI

caozeyuan · Answer

since it says proportional to, you just make the independent varibales time toghether

kkutie7 · Answer

i found this dont know if its right though. let my type it up

caozeyuan · Answer

so that would be sigma*N*I*dz

kkutie7 · Answer

$$-\frac{dI}{I(z)}=Ndz\rightarrow \frac{dI}{I(z)}=-\sigma Ndz\rightarrow \frac{dI}{dz}=-\sigma N I(z)$$

caozeyuan · Answer

yes, and thats exactly whwat Im going to say

caozeyuan · Answer

My apporaoch is to simplhy put dI on the LHS, and on the RHS you have all your independent variables, which gives sigma*N*I*dz, the minus sign is necessay because lignt intensity decereases as distance increases.

caozeyuan · Answer

now you divide by dz and thats the same formula as you just wrote

kkutie7 · Answer

sweet thanks