Q. 1.69 - An Introduction to Thermal Physics

Question

Consider a narrow pipe filled with fluid, where the concentration of a specific type of molecule varies only along its length (in the x direction). Fick's second law is derived by considering the flux of these particles from both directions into a short segment$∆x$

$\frac{\partial n}{\partial t}=D\frac{{\partial }^{2}n}{\partial x^2}$

Step-by-Step Solution

Answer

Consider a narrow pipe filled with fluid, where the concentration of a specific type of molecule varies only along its length (in the x direction).  The Fick's second law is$\frac{\mathrm{\partial }n}{\mathrm{\partial }t}=D\frac{{\mathrm{\partial }}^{2}n}{\mathrm{\partial }{t}^2}$

Step1:Explain type of molecule varies only along the length of the pipe

The flux describes the rate at which molecules diffuse per unit area and per unit time in diffusion. $J_x$. Assume we have a narrow pipe filled with fluid or gas, and the molecular concentration varies only along its length. Take two adjacent narrow slices of pipe. each of width$\Delta x$. The first slice is bounded by$x_1$ and $x_2$ and the second slice by$x_2$ and $x_3$.

Step2:derive Fick's second law

$J_x=-D\frac{dn}{dx}$$\left(1\right)$

The number of molecules entering slice$N1$ 2 from slice in time$∆t$ is equal to the flux multiplied by the slice's cross sectional area and time interval, so:

$N_1=J_{x,1}A\mathrm{\Delta }t$

substitute from$J_x=-D\frac{dn}{dx}$ , so:

$N_1=-\left[D\frac{n_2-n_1}{\mathrm{\Delta }x}\right]A\mathrm{\Delta }t$$\left(2\right)$

Similarly, the number of molecules exiting the second slice on the opposite side, $N_2$, is:

$N_2=-\left[D\frac{n_3-n_2}{\mathrm{\Delta }x}\right]A\mathrm{\Delta }t$$\left(3\right)$

The difference $N_1-N_2$ is the net change in the number of molecules in the second slice, so:

$\mathrm{\Delta }N=N_1-N_2$$\left(4\right)$

subtract equation  from$N_1=-\left[D\frac{n_2-n_1}{\mathrm{\Delta }x}\right]A\mathrm{\Delta }t$ and$N_2=-\left[D\frac{n_3-n_2}{\mathrm{\Delta }x}\right]A\mathrm{\Delta }t$ substitute from$J_x=-D\frac{dn}{dx}$ 

$N_1-N_2=\left[D\frac{n_3-n_2}{\mathrm{\Delta }x}\right]A\mathrm{\Delta }t-\left[D\frac{n_2-n_1}{\mathrm{\Delta }x}\right]A\mathrm{\Delta }t$

Step3:Solution

$\begin{array}{r}\frac{\mathrm{\Delta }N}{\mathrm{\Delta }t}=DA\left[\frac{n_3-n_2}{\mathrm{\Delta }x}-\frac{n_2-n_1}{\mathrm{\Delta }x}\right]\\ \frac{\mathrm{\Delta }N}{\mathrm{\Delta }t}=DA\left[\frac{n_3-n_2-n_2+n_1}{\mathrm{\Delta }x}\right]\\ \to \frac{\mathrm{\Delta }N}{\mathrm{\Delta }t}=DA\left[\frac{n_3-2n_2+n_1}{\mathrm{\Delta }x}\right]\end{array}$

dividing both sides by the slice's volume$V=A\Delta x$, so:

$\frac{\mathrm{\Delta }\frac{N}{V}}{\mathrm{\Delta }t}=\frac{DA}{A\mathrm{\Delta }x}\left[\frac{n_3-2n_2+n_1}{\mathrm{\Delta }x}\right]$

utilizing molecular concentration$n=\frac{N}{V}$ on the LHS we get:

$\frac{\mathrm{\Delta }n}{\mathrm{\Delta }t}=D\left[\frac{n_3-2n_2+n_1}{(\mathrm{\Delta }x{)}^2}\right]$$\left(5\right)$

In the limit of,

$\mathrm{\Delta }t,\mathrm{\Delta }x\to 0$

In the following relation:

$\frac{{\mathrm{\partial }}^{2}y}{\mathrm{\partial }{x}^2}=\frac{y_3-2y_2+y_1}{(\mathrm{\Delta }x{)}^2}$

 will become

$\frac{\mathrm{\partial }n}{\mathrm{\partial }t}=D\frac{{\mathrm{\partial }}^{2}n}{\mathrm{\partial }{t}^2}$$\left(6\right)$

The solutions to this equation are the same as the solutions to the heat equation because they are formally equivalent. Starting with any concentration distribution, it will gradually spread out over time until the concentration is the same everywhere.