91影视

$N2$Consider a thin gas slab with cross-sectional area A. We cut the box in half so that the number of molecules on one side of the partition is$N1$ and the number of molecules on the other side is. We'll assume that the entire box is at a constant temperature $N1$and that Schroeder's approximations for mean free path and average speed are correct:

$\mathrm{鈩�}鈮�\frac{1}{4蟺r^2}\frac{V}{N}$$\left(1\right)$

$\stackrel{炉}{v}鈮�\sqrt{\frac{3KT}{m}}$$\left(2\right)$

where $r$is the molecular radius and m the mass of one molecule.

Those molecules within a distance$\mathrm{鈩�}$of the midpoint of the slab can cross the midpoint if they are travelling towards the midpoint. Since only half the molecules will, on average, be moving towards the partition (statistically), the net number$螖N$of molecules that cross the partition in a time$螖t$, which is the time it takes a molecule to move distance $\mathrm{鈩�}$, is:

$\mathrm{螖}N=\frac{1}{2}\left(N_1鈭�N_2\right)$$\left(3\right)$

If the molecule number gradient is$\frac{dN}{dx}$then we have:

$\frac{N_1鈭�N_2}{\mathrm{鈩�}}=\frac{dN}{dx}$$\left(4\right)$

where $\mathrm{鈩�}$is the distance that a molecule travels before colliding, substituting from into$\frac{N_1鈭�N_2}{\mathrm{鈩�}}=\frac{dN}{dx}$$\mathrm{螖}N=\frac{1}{2}\left(N_1鈭�N_2\right)$so:

$\mathrm{螖}N=\frac{1}{2}\mathrm{鈩�}\frac{dN}{dx}$$\left(5\right)$

The flux is the net rate at which molecules cross the partition per unit area $J_x$, its given by:

$\left|J_x\right|=\frac{\mathrm{螖}N}{A\mathrm{螖}t}$$\left(6\right)$

substitute from equation$\mathrm{螖}N=\frac{1}{2}\mathrm{鈩�}\frac{dN}{dx}$into equation $\left|J_x\right|=\frac{\mathrm{螖}N}{A\mathrm{螖}t}$with $螖N$, so:

$\left|J_x\right|=\frac{1}{A\mathrm{螖}t}\left[\frac{1}{2}\mathrm{鈩�}\frac{dN}{dx}\right]$

multiply with $\frac{\mathrm{鈩�}}{\mathrm{鈩�}}$,so:

$\left|J_x\right|=\frac{{\mathrm{鈩�}}^2}{2\mathrm{鈩�}A\mathrm{螖}t}\left[\frac{dN}{dx}\right]$$\left(7\right)$

However, the volume is equal to the free mean path multiplied by the cross sectional area, and the time it takes the molecule to move one free mean path is equal to the time it takes the molecule to move one free mean path$\mathrm{鈩�}$, or:

$\mathrm{鈩�}A=V\frac{\mathrm{鈩�}}{\mathrm{螖}t}=\stackrel{炉}{v}$

substitute, so equation$\left|J_x\right|=\frac{{\mathrm{鈩�}}^2}{2\mathrm{鈩�}A\mathrm{螖}t}\left[\frac{dN}{dx}\right]$ will become:

$\left|J_x\right|=\frac{\mathrm{鈩�}\stackrel{炉}{v}}{2V}\left[\frac{dN}{dx}\right]$

but $\frac{N}{V}=n$, so:

This is the magnitude of the flux, as indicated by the absolute value. Because the flux is moving in the opposite direction of the gradient, we will have:

$J_x=鈭�\frac{1}{2}\mathrm{鈩�}\stackrel{炉}{v}\frac{dn}{dx}$$\left(8\right)$

The quantity $\frac{1}{2}\mathrm{鈩�}\stackrel{炉}{v}$is a close approximation to the diffusion constant D for an ideal gas, with units of ${\mathrm{m}}^2路{\mathrm{s}}^{-1}$. The following is a more general form of this equation:

$J_x=鈭�D\frac{dn}{dx}$

Using the values from Schroeder's book for air at room temperature, $\mathrm{鈩�}=$$1.5脳{10}^{-7}\mathrm{m}$and $\stackrel{炉}{v}=500\mathrm{m}路{\mathrm{s}}^{-1}$, so:

$D=\frac{1}{2}\mathrm{鈩�}\stackrel{炉}{v}=\frac{1}{2}脳1.5脳{10}^{鈭�7}脳500=3.75脳{10}^{鈭�5}{\mathrm{m}}^2{\mathrm{s}}^{鈭�1}$

$D_{\text{air}}=3.75脳{10}^{鈭�5}{\mathrm{m}}^2鈰�{\mathrm{s}}^{鈭�1}$

The measured value is around $2脳{10}^{-5}{\mathrm{m}}^2路{\mathrm{s}}^{-1}$so this isn't too far off for a rough estimate

From the values for $\mathrm{鈩�}$and$\stackrel{炉}{v}$in equations $\mathrm{鈩�}鈮�\frac{1}{4蟺r^2}\frac{V}{N}$and $\stackrel{炉}{v}鈮�\sqrt{\frac{3KT}{m}}$We can calculate the relationship between the diffusion constant and temperature at constant pressure as follows:

$D=\frac{1}{2}\left[\mathrm{鈩�}\right]\left[\stackrel{炉}{v}\right]=\frac{1}{2}\left[\frac{1}{4蟺r^2}\frac{V}{N}\right]\left[\sqrt{\frac{3KT}{m}}\right]$

based on the ideal gas law,$\frac{V}{N}=\frac{kT}{P}$, so:

$D=\frac{1}{2}\left[\frac{1}{4蟺r^2}\frac{kT}{P}\right]\left[\sqrt{\frac{3KT}{m}}\right]$

$鈫�D=\frac{1}{8蟺r^2}\frac{(KT{)}^{\frac{3}{2}}}{P}\sqrt{\frac{3}{m}}$

So, at constant pressure,

$D鈭�(KT{)}^{\frac{3}{2}}$