# Is it possible to split up air into its parts by something like centrifuging? [duplicate]

Is it possible to centrifuge cases to split them up into their compounds (at least temporarily)?

For instance, the air is composed of 78% nitrogen, 21% oxygen, 0.9% argon, and other gases. Under what conditions might we be able to centrifuge it to split it up?

• U238 is separated this way, as are air plants that extract noble gasses from the atmosphere. Commented Feb 3, 2023 at 15:10

The kind of separation that you describe occurs in a device that is called gas centrifuge.

For every G-load of a gas there will be a corresponding bias in density of the components.

At 1 G of acceleration the bias is negligable. At high rpm there will be a high G-load, and the higher density material will be present in higher concentration at the perimeter of the rotation, and in lower concentration close to the center of rotation.

It is standard to implement this type of separation in a continuous and cascaded setup.

Quote from the wikipedia article:

Each centrifuge receives one input and produces two output lines, corresponding to light and heavy fractions. The input of each centrifuge is the output (light) of the previous centrifuge and the output (heavy) of the following stage.

It is not possible to achieve splitting in one go. Hence the cascade setup.

The reason that complete splitting does not even occur at high G-load arises from probability. This is an example of a situation where potential energy and probability are acting in opposite direction. In terms of energy it is favorable for the higher density material to migrate down the potential. But a more mixed state of the gas is more probable than a less mixed state.

For instance, the air is composed of 78% nitrogen, 21% oxygen, 0.9% argon, and other gases. Under what conditions might we be able to centrifuge it to split it up?

This is a good opportunity to review scaling strategies and some fundamental thermodynamics relations. We seek an equilibrium state where heavier and lighter gases are enriched to a certain degree by centrifugation.

First, what are we working with—what's the equilibrium pressure distribution of gas in a gravity well? We know that Nature prefers low potentials (low enthalpy) but many possibilities (high entropy). These tendencies can both be traced back to the Second Law.

Lower enthalpy involves more stratification of gas molecules; higher entropy involves a better distribution of gas molecules. The equilibrium balance—the happy medium—comes from the surrounding thermal energy (broadly, $$kT$$, where $$k$$ is Boltzmann's constant and $$T$$ is the temperature) driving some atmospheric distribution while maintaining a density gradient arising from the gravitational field. This is, of course, what we observe in our atmosphere: a fairly smooth density gradient from the surface up toward the void of space.

Approach 1: A scaling relation

We know that the gravitational potential energy of an system (say, a gas molecule) is $$Mgh$$, with mass $$M$$, gravitational acceleration $$g$$, and height $$h$$; let's just set this equal to $$kT$$ to achieve a scaling relation. In other words, we seek a characteristic height $$h$$ of natural stratification (with the aim to alter it by centrifugation). The results may be off from the exact solution by a factor of 2 or 3, say, but they'll generally be within an order of magnitude.

So we write $$kT\sim Mgh$$, or perhaps more conveniently $$RT\sim mgh$$, where $$R$$ is Boltzmann's constant per mole and $$m$$ is the molar mass. This gives, at room temperature,

$$h \sim \frac{RT}{mg}=\frac{(8.314\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1})(293\,\mathrm{K})}{(0.029\,\mathrm{kg}\,\mathrm{mol}^{-1})(9.81\,\mathrm{m}\,\mathrm{s}^{-2})}\approx 9\,\mathrm{km}.$$

Thus, within a few multiples of 9 km, we should expect an atmosphere at 20°C to fade to nothing, as a reflection of the counteracting tendencies of potential energy minimization and entropy maximization for a given temperature.

What about the acceleration needed to enrich a molecule of a higher weight over a certain length scale? Here, the length scale would correspond to the radius or arm length of the centrifuge. We have, for a 3 m arm length, say,

$$g \sim \frac{RT}{\Delta mh}=\frac{(8.314\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1})(293\,\mathrm{K})}{(0.015\,\mathrm{kg}\,\mathrm{mol}^{-1})(3\,\mathrm{m})}\approx 54\mathrm{,}000\,\mathrm{m}\,\mathrm{s}^{-2},$$

where I've used the molar mass difference between air and the heavier carbon dioxide as an example. This corresponds to an acceleration of >5000g, and it's not as if we end up with a nearly pure band of carbon dioxide to skim off—perhaps the concentration has doubled.

What if we seek a more exact calculation than a scaling relation? I'll first review a first-principles derivation that starts with the Second Law; then, I'll apply a familiar equation for the gas pressure in a gravity field. The results will be the same, as the former is actually used to obtain the latter.

Approach 2: Uniformity of the chemical potential

The thermodynamic potential that incorporates both enthalpy minimization and entropy maximization is the Gibbs free energy, which is minimized under familiar conditions involving ambient temperatures and pressures. Furthermore, the partial molar Gibbs free energy is minimized for each chemical component; this parameter is called the chemical potential, which is the true arbiter of how matter will move. (In the case of near-ideal gases such as atmospheric gases, the chemical potential is related in a fairly simple way to the partial pressure, and we could start our problem by considering the latter. However, the former is the more fundamental description, useful for broader adaptability, so I'll review that approach first before covering the partial-pressure solution. Note that the chemical potential doesn't necessarily involve chemical reactions; the name is somewhat a historical artifact.)

Consider a component A (here, the gas we desire to enrich) in a mixture. The chemical potential of A in the mixture is generally defined as $$\mu_{A\text{,mixture}}=\mu_A^\circ+RT\ln a,\tag{1}$$

where $$\mu_A^\circ$$ is the molar Gibbs free energy of pure A in a reference state and $$a$$ is the activity, which simplifies to the concentration $$C_A$$ for an ideal gas mixture.

In a gravity well, the chemical potential must satisfy

$$\frac{\partial (\mu_{A\text{,mixture}}+m_Agh)}{\partial h}=0.\tag{2}$$

At equilibrium, in other words, the chemical potential for each component must be constant at different heights, or that component would shift. Without gravity, this reduces to the familiar result that a gas expands to equilibrate its concentration/partial pressure.

The chemical potential $$\mu_{A\text{,mixture}}$$ is a function of the temperature $$T$$, pressure $$P$$, and concentration $$C_A$$, so we expand Eq. (2) in these variables to obtain

$$\frac{\partial \mu_{A\text{,mixture}}}{\partial T}\frac{\partial T}{\partial h}+\frac{\partial \mu_{A\text{,mixture}}}{\partial P}\frac{\partial P}{\partial h}+\frac{\partial \mu_{A\text{,mixture}}}{\partial C_A}\frac{\partial C_A}{\partial h}+m_Ag=0,$$

where we seek $$\frac{\partial C_A}{\partial h}$$: the height-dependent concentration gradient that we wish to maximize by centrifugation.

Now,

• For height-independent temperature, $$\frac{\partial T}{\partial h}=0$$, so the first term disappears;
• We always have $$\frac{\partial \mu}{\partial P}=V$$ (or $$\frac{RT}{P}$$ for an ideal gas);
• $$\frac{\partial P}{\partial h}=-\rho g=-\frac{m_\text{mixture}}{V}g$$ (hydrostatic equilibrium, where $$m_\text{mixture}$$ is the total molar mass); and
• $$\frac{\partial \mu_{A\text{,mixture}}}{\partial C_A}=\frac{RT}{C_A}$$ (from Eq. (1)), giving

$$0-m_\text{mixture}g+\frac{RT}{C_A}\frac{\partial C_A}{\partial h}+m_Ag=0$$

or

$$RT\frac{\partial \ln C_A}{\partial h}=m_\text{mixture}g-m_Ag.$$

Integrating, we have

$$\frac{C_A}{C_{A\text{,0}}}=\exp\left(\frac{(m_\text{mixture}-m_A)gh}{RT}\right)\tag{3},$$

where $$C_{A\text{,0}}$$ is the concentration at $$h=0$$. With centrifugation, the goal is to drive most of A toward $$h=0$$ (or $$h=h_\text{max}$$, when separating a lighter gas) and therefore to make $$\frac{|m_\text{mixture}-m_A|g}{RT}$$ large compared to the centrifuge radius or arm length. Note that this is exactly what the scaling relation $$g \sim \frac{RT}{\Delta mh}$$ above indicates, with the exponential function now specifying the ultimate concentration distribution.

Approach 3: The barometric equation

As noted above, the result from this derivation strategy matches that from simply using partial pressures as a surrogate for concentrations and concentrations as a surrogate for activities. The well-known barometric equation at constant temperature is

$$P_A=P_0 \exp\left(\frac{-m_Agh}{RT}\right);$$

which is just a single-component version of the relation in Eq. (3) derived above. Normalizing the partial pressure of component A to the total pressure gives

$$\frac{C_A}{C_{A\text{,0}}}=\frac{\frac{P_{A}}{P}}{\frac{P_{A,0}}{P_0}}=\frac{P_{A}P_0}{P_{A,0}P}=\exp\left(\frac{(m_\text{mixture}-m_A)gh}{RT}\right),$$

which is a perfect match to Eq. (3).

Since we’re assuming ideal gases, applying the barometric equation is a perfectly valid solution strategy, albeit less extensible to other materials than the chemical potential approach outlined above (which can be adapted to incorporate chemical reactions, for instance).