# Deriving the kinetic energy flux in an effusion process

• rogdal
In summary, the energy flux through the opening is proportional to the number of particles with momentum in the range of interest.
rogdal
Homework Statement
Derive the kinetic energy flux for ideal gases in 2 dimensions. Assume there are $$N$$ particles in a 2-d box of area $$A$$. At time $$t=0$$, a hole of length $$l$$ is made in one of the walls (let's say it lies on the x-axis), so a quasi-static process of effusion begins (we also assume that $$l \ll A$$). Also considering that there are no potentials acting on the box and so the energy of a particle is just its kinetic energy $$\epsilon$$, what would be the expected value of the energy flux of all the particles that escape from the box?
Relevant Equations
$$\overline{\Phi_{effusion,\epsilon}} = \overline{\dot{N_{ef}}}\overline{\epsilon_{ef}}=\frac{3Nl}{2A}\sqrt{\frac{(k_BT)^3}{2\pi m}}$$
I could not find any derivations in the litterature, except for the expected value of the energy flux expression itself:

$$\overline{\Phi_{effusion,\epsilon}} = \overline{\dot{N_{ef}}}\overline{\epsilon_{ef}}=\frac{3Nl}{2A}\sqrt{\frac{(k_BT)^3}{2\pi m}}$$

I've started off by calculating the particle flux through the hole, since the final expression is rather similar.

First of all, the particles can only move in the x and y directions. If the hole is placed in a wall in the x axis, the particle density of the effused particles $$n_{ef}$$ is half of the one of all the particles, $$n_{ef}= n/2$$, since the probability of a particle to approach the wall where the hole is, is (on average) half of the one to move away from it. Therefore, we come to the particle flux expression:

$$\Phi_{ef}=ln_{ef}\frac{\vec{p}}{m}·\vec{u_n}=ln_{ef}\frac{p_x}{m}=\frac{Nlp_x}{2Am}$$

Since the particles follow a Maxwell-Boltzmann distribution,

$$\overline{\dot{N_{ef}}}=\overline{\Phi_{ef}}=\int\nolimits_{-\infty}^{+\infty}\int\nolimits_{-\infty}^{+\infty}\Phi_{ef}\rho_{ef}(\vec{p})dp_xdp_y$$

Where $$\rho_{ef}(\vec{p})$$ is the probability distribution. Since we are only considering positive momenta approaching the wall where the hole is, $$\rho_{ef}(\vec{p})=2\rho(\vec{p})\theta(p_x)$$, where the 2 is the normalization constant and $$\rho(\vec{p})= \frac{1}{2\pi mk_BT}\exp\left({-\frac{|\vec{p}|}{2mk_BT}}\right)$$

Solving the double integral the flux of particles per unit of time that escape is:

$$\overline{\dot{N_{ef}}}=\frac{Nl}{A}\sqrt{\frac{k_B T}{2\pi m}}$$

Now, if we apply the same approach to $$\overline{\Phi_{effusion,\epsilon}}=\overline{\Phi_{ef,\epsilon}}$$, we would get

$$\Phi_{ef,\epsilon}=ln_{ef}\frac{|\vec{p}|}{m}\epsilon \rightarrow {|\vec{p}|=\sqrt{2m\epsilon}} \rightarrow \frac{Nl}{A}\frac{\epsilon^{\frac{3}{2}}}{\sqrt{2m}}$$

and, expressing the MB distribution in terms of the energy (not a MB distribution anymore)

$$\rho_\epsilon(\epsilon)=\frac{1}{k_B T}\exp{\left(-\frac{\epsilon}{k_B T}\right)}$$

I'm struggling to find any relation between $$\rho_\epsilon$$ and $$\Phi_{ef,\epsilon}$$ from which I could derive the expression for $$\overline{\Phi_{ef,\epsilon}}$$. Do you have any ideas?

Thanks.

I don't know what you are supposed to assume about this system, but you might want to consider that the MB distribution could be different in 2 dimensions. At least the factor that comes from the density of states should change, I would think.

Philip Koeck said:
I don't know what you are supposed to assume about this system, but you might want to consider that the MB distribution could be different in 2 dimensions. At least the factor that comes from the density of states should change, I would think.

Thanks for your answer. Yes, sorry, the MB distribution in 2 dimensions is

$$\rho(\vec{p})=\frac{1}{2\pi m k_B T}e^{-\frac{p^2}{2mk_B T}}$$

I forgot to square the p above. But the rest of the conclusions were obtained with the correct form of the 2D MB distributions. Do you know how to come up with the energy flux density for this system?

Last edited:
Philip Koeck
rogdal said:
Now, if we apply the same approach to $$\overline{\Phi_{effusion,\epsilon}}=\overline{\Phi_{ef,\epsilon}}$$, we would get$$\Phi_{ef,\epsilon}=ln_{ef}\frac{|\vec{p}|}{m}\epsilon$$
This doesn't look right. The rate of effusion of particles of a certain momentum should be proportional to ##p_x## rather than ##|\vec p|##.

Suppose we define a distribution function ##n(p_x, p_y)## such that the number of particles per unit area which have ##x## and ##y## components of momentum in the range ##p_x## to ##p_x+dp_x## and ##p_y## to ##p_y+dp_y## is $$n(p_x,p_y)dp_xdp_y.$$ You can express ##n(p_x, p_y)## in terms of ##N##, ##A##, and the Maxwell-Boltzmann distribution.

Show that the number of particles with momentum in this range that effuse through the opening per unit time is $$d\Phi_{particles} =l \left(\frac{p_x}{m}\right)\left[n(p_x,p_y)dp_xdp_y\right]$$ Then show that the flux of energy through the opening carried by these particles is $$d\Phi_\epsilon= l \left(\frac{p_x}{m}\right)\epsilon(p_x, p_y)\left[n(p_x,p_y)dp_xdp_y\right]$$ where ##\epsilon(p_x, p_y)## is the energy of a particle with momentum components ##p_x## and ##p_y##.

The total flux of energy through the opening is then ##\Phi_\epsilon = \int d\Phi_\epsilon##.

Last edited:
rogdal
Thank you very much, @TSny, I understand it now. Yes, I've shown both flux differentials and then the final answer is straightforward.

TSny

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