- #1

rogdal

- 10

- 2

- 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.

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

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.