User:Niel Malan/Effusion

This is a draft for a new page on effusion.

Consider a gas at scale where the collisions between gas molecules are negligible. For any given surface inside the gas, the flux of molecules is given by $$\Phi = \frac{a}{b}$$

The number of collisions with the walls of a container per unit are per unit time is given by


 * $$J_\text{collision} = \frac{1}{4}n \bar v = \frac{n}{4} \sqrt{\frac{8 k_{B} T}{\pi m}}$$

where $$n$$ is the the number of molecules, $$k_B$$ is the Boltzmann constant, $$T$$ is the temperature, and $$m$$ is the mass of the molecule. Note that this rate is independent of the pressure. This equation will hold as long as the assumption of no collisions between molecules is valid: this might be a low pressure, but it might also be at small scale.

If the container is one compartment of a larger container divided into two parts by a thin membrane that, and aperture with area $$A$$ is punched through the membrane, then the flux of molecules from one compartment to the other will be given by


 * $$J_\text{collision} A = {A n} \sqrt{\frac{8 k_{B} T}{2 \pi m}}$$

If a given gas with at a given temperature is assumed, and if the gas behaviour is close to that of an ideal gas, then $$n$$ can be estimated from the ideal gas law $$P V = n k_B T$$, and the equation becomes


 * $$J_\text{collision} A = \frac{P A}{\sqrt{2 \pi m k_{B} T}}$$

For calculating experimental values, the molar flow rate can be be calculated,


 * $$Q = \frac{P A}{\sqrt{2 \pi M R T}}$$

where $$M$$ is the molar mass, $$N_A$$ is the Avogadro constant, and $$R = N_A k_B$$ is the gas constant.

Because the movement of molecules through the aperture in the membrane is independent of conditions on the other side of the membrane, the total flow is simply the sums of the flows on either side. Because the flow can be either into or out of one of the compartments, we need to assign a sign to the flow. Conventionally, flow out of container is negative. So, using Compartment 1 as reference, the total flow through the aperture is given by


 * $$Q_{total} = Q_2 - Q_1 = \frac{A}{\sqrt{2 \pi M R}}\left(\frac{P_2}{\sqrt{T_2}} - \frac{P_1}{\sqrt{T_1}}\right)$$

where $$Q_i$$ is the flow due to temperature $$T_i$$ and pressure $$P_i$$ in Compartment $$i$$.

This expression shows that under isothermal conditions (i.e. $$T_1 = T_2$$), $$Q_{total}$$ will be negative, indicating that the gas will flow from compartment with the higher pressure into the compartment with the lower pressure. Under isobaric conditions (i.e. $$P_1 = P_2$$), flow will be from the compartment with the lower temperature into the compartment with the higher temperature.

The expression also shows that when


 * $$\frac{P_2}{P_1} = \frac{\sqrt{T_2}}{\sqrt{T_1}}$$

a stationary state is achieved and flow ceases ($$Q_{total} = 0$$).

These results are useful if the Knudsen number is large ($$\mathrm{Kn} > 1 $$). If $$\mathrm{Kn} < 0.01 $$, the flow trough a round pipe is given by the Hagen–Poiseuille equation derived from continuum mechanics:


 * $$Q = \frac{(P_2 - P_1) A^2}{8 \pi \mu L} $$

where $$\mu$$ is the kinematic viscosity and $$L$$ is the length of the pipe. Note that in this formula the flow is independent of the temperature (except so far it affects the viscosity), is proportional to the square of the area, and inversely proportional to the length of the pipe.

If the two compartments are linked so that a free flow of gas is possible without a disturbance of the temperatures of the two compartments, the pressures will be equal ($$P_1 = P_2 = P$$), so that the flow formula can be written


 * $$Q = \frac{P A}{\sqrt{2 \pi M R}}\left(\frac{\sqrt{T_1} - \sqrt{T_2}}{\sqrt{T_2 T_1}}\right)$$

For a given flow, and typical temperature and pressures, what is the area of the aperture?


 * $$A = \frac{Q \sqrt{2 \pi M R}}{P}\left(\frac{\sqrt{T_1 T_2}}{\sqrt{T_1} - \sqrt{T_2}} \right)$$

The flow in the link between the compartments can be conveniently measured in units of volume per unit time. If we take the equation $$P V = N R T$$ and divide each side by a time interval $$t$$ and rearrange, we get


 * $$Q = \frac{N}{t} = \frac{P}{R T}\frac{V}{t}= \frac{P}{R T}Q_V$$

Substitute


 * $$A = \frac{Q_V \sqrt{2 \pi M R}}{R T}\left(\frac{\sqrt{T_1 T_2}}{\sqrt{T_1} - \sqrt{T_2}} \right)$$

Here $$T$$ is the temperature measured in the link between the two compartments, where the flow is measured.

In chemistry the volume flow rate is conveniently measured in millilitres per minute, so if $$Q_v$$ is given in ml/min, the equation including a conversion factor


 * $$A = \frac{Q_V \sqrt{2 \pi M R}}{R T}\left(\frac{\sqrt{T_1 T_2}}{\sqrt{T_1} - \sqrt{T_2}} \right) \times \frac{1}{6 \times 10^7}$$

will give an area in square metres.

Now we can calculate $$A$$ for a typical set of conditions. We can take the environmental temperature $$T$$ and the temperature at the cold side of the membrane ($$T_1$$) as 25 °C (298 K), and the temperature at the hot side of the membrane ($$T_2$$) as 30 °C (303 K). We can use nitrogen gas as the working gas, for which $$M = 28.0134 \mathrm{\ g\ mol^{-1}}$$.