Diffusion of Two Gases

Introduction

            This will analyze the diffusion of two samples of  gases where one gas is initially on the left side of the container and the other is on the right side of the container.

Analysis:

Case 1: Simplest case of heat flow into slab:

            The gas diffusion problem is fairly complex so, as a warm-up, I will start by

Figure 2: Slab whose temperature functions are analyzed below.

analyzing the flow of heat into a large, thin slab of thickness L (see Figure 2). 

            The solid state analog of the atoms that we will discuss later are called phonons which are quantum mechanical entities that represent vibrations of the atomic nuclei in the solid.   These phonon density is higher in the hotter regions of the solid and they carry the heat from the hotter parts of the solids to the cooler parts in the same way that one atomic species migrates from a region where its density is high to a region where its density is low.  The phonons collide with other phonons which retards their progress in the same way that the atoms of a gas collide with other atoms retarding their progress toward lower density. 

 

            For the present heat flow problem, the power input into the left side of the slab will be Q (watts-m^-2).  The conductivity will be K (watt m^-1 K^-1) and the heat capacity will be C (Joules m^-3 K^-1).  Further I will assume, as shown, that the slab is insulated on the sides not in contact with the heat source.  Then we can immediately write that, after transients have died out, the temperature everywhere in the slab rises as

$\frac{\partial T}{\partial t}=\frac{Q}{CL}$                                                                                             (1)

The partial differential equation for heat flow is:

$\frac{K}{C}\frac{{\partial }^{2}T}{\partial x^2}=\frac{\partial T}{\partial t}=\frac{Q}{CL}$                                                                             (2)

where the latter equality applies only to steady state.  Integrating equation 2 with respect to x once we obtain:

$\frac{\partial T}{\partial x}=\frac{Q}{KL}x+A$                                                                                                (3)

where A is a constant that will be determined later by boundary conditions.

Integrating equation 3 with respect to x results in:

$T=\frac{Q}{KL}\frac{x^2}{2}+Ax+B$                                                                           (4)

where B is another constant that we must determine.  Note that, since the boundary at x=L is insulated there is no heat flow there so 

${\frac{\partial T}{\partial x})}_{x=L}=0$                                                                                          (5)

Using equation 5 in equation 3 we get for A:

$A=-\frac{Q}{2K}$                                                                                            (6)

Finally we can set B in equation 4 to the initial temperature, T₀, and obtain

${T(x,t))}_{SteadyState}=\frac{Q}{KL}\left(\frac{x^2}{2}-\frac{xL}{2}\right)+\frac{Q}{C}t$                                                            (7)

 

            Equation 7 needs a complementary solution to specify the transients that occur when Q is turned on.  For that we use a technique call "separation of variables" which I will explain next.  We have the equation:

$\frac{K}{C}\frac{{\partial }^{2}T}{\partial x^2}=\frac{\partial T}{\partial t}$                                                                                       (8)

We make the assumption that T can be written:

$T=X(x){\rm T}(t)$

Then equation 8 can be rewritten:

$\begin{array}{l}\frac{d^{2}X}{d{x}^2}=-\Lambda X\\ \frac{d{\rm T}}{dt}=-\Lambda D{\rm T}\end{array}$                                                                                       (9a,b)

where Λ is a constant that will depend on initial boundary conditions and I have used the symbol D=K/C, which is the diffusivity.

The obvious solution set for equation 9a is

$X=E\cos (\sqrt{\Lambda }x)+F\sin (\sqrt{\Lambda }x)$                                                           (10a)

and the solution for equation 9b is

$T=\exp (-\Lambda Dt)$                                                                                    (10b)

If we are going to match boundary conditions in the interval x=(0,L) then the value of Λ has to be

$\sqrt{\Lambda }=\frac{n\pi }{2L}$

where n is an integer.  Then equations 10a and 10b become:

$X=E\cos \left(\frac{n\pi }{2L}x\right)+F\sin \left(\frac{n\pi }{2L}x\right)$                                                        (10a)

${\rm T}=\exp \left(-{\left(\frac{n\pi }{2L}\right)}^{2}Dt\right)$                                                             (10b)

Finally we apply the conditions

${\frac{dT}{dx})}_{x=0}=-\frac{Q}{K}$ 

${\frac{dT}{dx})}_{x=L}=0$

which implies that

$\left(-E\sin \left(\frac{n\pi }{2L}L\right)+F\cos \left(\frac{n\pi }{2L}L\right)\right)\left(\frac{n\pi }{2L}\right)=0$

$\left(-E\sin \left(\frac{n\pi }{2L}0\right)+F\cos \left(\frac{n\pi }{2L}0\right)\right)\left(\frac{n\pi }{2L}\right)=\frac{Q}{K}$

 

so that

$F=-\frac{Q}{K}\left(\frac{2L}{n\pi }\right)$

and n is odd.  Then the combined transient solution is

$T_{transient}(x,t)=\frac{QL}{2K}-\frac{Q}{K}\frac{2L}{\pi }{\displaystyle \sum _{mn=0}^{\infty }\frac{1}{2m+1}\sin \left(\frac{(2m+1)\pi }{2L}x\right)\exp \left(-{\left(\frac{(2m+1)\pi }{2L}\right)}^{2}Dt\right)}$                  (11)

where we note that

${\displaystyle \sum _{}^{}{(-1)}^m\frac{1}{2m+1}=\frac{\pi }{4}}$

so that T(x,t)=0 at x=L.

 

Case 2: Diffusion of Gases

Figure 2. Showing the initial conditions.  For this picture there are 500 randomly placed Red atoms on the left and 500 randomly placed blue atoms on the right.  They have random velocities in the plane of the Figure.

 

The equation for the density, n, of atoms is exactly similar to that for temperature discussed above:

$D\frac{{\partial }^{2}n(x,t)}{\partial x^2}=\frac{\partial n}{\partial t}$                                                                                 (12)

I will not derive the required solution for n(x,t).  It is

$n(x,t)=\frac{n_0}{2}\left[erfc\left(\frac{x}{2\sqrt{Dt}}\right)\right]$                                                                  (13)

where

$erfc(u)=1-\frac{2}{\sqrt{\pi }}{\displaystyle {\int }_0^u\exp (-v^2)dv}$

where D is the diffusivity and n₀ is the starting density atoms on each side of the mid-line of the container.  The function erfc is defined as

$erfc(u)=1-\frac{2}{\sqrt{\pi }}{\displaystyle {\int }_0^u\exp (-v^2)dv}$

 

 

I will now differentiate both sides of equation 12 to confirm that equation 13 is indeed a solution of equation 12.

 

$\frac{\partial n}{\partial t}=D\frac{{\partial }^{2}n}{\partial x^2}=\frac{n_0}{4}\frac{x}{t}\frac{\exp \left(-\frac{x^2}{4Dt}\right)}{\sqrt{\pi Dt}}$

 

            The diffusivity, D, here deserves to be better explained.  It depends on the average speed <v_x> of the atoms in the horizontal direction as well as the mean free path, l_mfp, between collisions.  The latter depends here on the 2 dimensional density of atoms, n₂, which equals

$n_2=\frac{N}{hw}$

where N is the total number of atoms, h is the height of the container, and w is its width.  The equation to determine l_mfp also depends on the 2 dimensional analog of the collision cross section, σ₂.  For 2 dimensions:

${\sigma }_2=r_R+r_B$

where r_R is the radius of the red atoms and r_B is the radius of the blue atom.

 Then the equation for the mean free path is:

$\begin{array}{l}{n}_2{\sigma }_2{l}_{mfp}=1or\\ l_{mfp}=\frac{1}{n_2{\sigma }_2}\end{array}$ 

The equation to obtain D from these quantities is:

$D=\frac{1}{2}<v_x>l_{mfp}$

Effect of Different Atomic Mass

            So far we have been assuming that the masses of the red and blue atoms are the same so that, upon collision, speed changes of both types of atoms are the same.  Suppose now that the masses, M, of the atoms (since we are just working in two dimensions and therefore the atoms are really discs of the same thickness) is proportional to their radii squared:

$\frac{M_{\mathrm{Re}d}}{M_{Blue}}={\left(\frac{r_R}{r_B}\right)}^2$ 

Similarly we expect the average speeds, since the atomic speeds are generated by temperature, to be in the ratio:

$\frac{<v_{\mathrm{Re}d}>}{<V_{Blue}>}=\sqrt{\frac{M_{Blue}}{M_{\mathrm{Re}d}}}=\frac{r_B}{r_R}$

Note that the product of mass times speed is the momentum,  p, and the ratio of the average momentums is

$\frac{p_{\mathrm{Re}d}}{p_{Blue}}=\frac{M_{\mathrm{Re}d}}{M_{Blue}}\frac{<v_{\mathrm{Re}d}>}{<v_{Blue}>}=\frac{r_{\mathrm{Re}d}}{r_{Blue}}$

If, for example, r_R>r_B this asymmetry results in a net random walk of momentum (Mv) from left to right.  Therefore the equation developed in the previous section will not be valid.