I'm a physicist, so apologies if the answer below is in a foreign language; but this was too interesting of a problem to pass up. I'm going to focus on a particular question: If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region? The TL;DR is that thermal effects are going to make this idea basically impossible.

The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$, where $\vec{B}$ is the magnetic field. Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point: $\vec{m} = \alpha \vec{B}$, where $\alpha$ is what we might call the "molecular magnetic susceptibility." Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$. But potential energy is given by $\vec{F} = - \vec{\nabla} U$; which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$.

Now, if we're talking about a sample of gas at a temperature $T$, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor: $$ \rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT} $$ In the limit where $kT \gg \alpha B^2$, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample. To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$; in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.

So how high does this have to be? The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$; and so we have¹ $$ \chi_\text{mol} B^2 \gtrsim \mu_0 RT $$ and so we must have $$ B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}. $$ If you believe Wikipedia, the molar susceptibility of oxygen gas is $4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$; and plugging in the numbers, we get a requirement for a magnetic field of $$ B \gtrsim 258\ \text{T}. $$ This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines. Even at 91 kelvin (just above the boiling point of oxygen), you would need a magnetic field of almost 150 T; still well out of range.

¹ I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules. A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation; and if the gas gets sufficiently dense, then all bets are off.

I'm a physicist, so apologies if the answer below is in a foreign language; but this was too interesting of a problem to pass up. I'm going to focus on a particular question: If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region? The TL;DR is that thermal effects are going to make this idea basically impossible.

The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$, where $\vec{B}$ is the magnetic field. Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point: $\vec{m} = \alpha \vec{B}$, where $\alpha$ is what we might call the "molecular magnetic susceptibility." Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$. But potential energy is given by $\vec{F} = - \vec{\nabla} U$; which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$.

Now, if we're talking about a sample of gas at a temperature $T$, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor: $$ \rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT} $$ In the limit where $kT \gg \alpha B^2$, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample. To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$; in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.

So how high does this have to be? The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$; and so we have¹ $$ \chi_\text{mol} B^2 \gtrsim \mu_0 RT $$ and so we must have $$ B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}. $$ If you believe Wikipedia, the molar susceptibility of oxygen gas is $4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$; and plugging in the numbers, we get a requirement for a magnetic field of $$ B \gtrsim \pu{258 T}. $$ This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines. Even at $\pu{91 Kelvin}$ (just above the boiling point of oxygen), you would need a magnetic field of almost $\pu{150 T}$; still well out of range.

¹ I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules. A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation; and if the gas gets sufficiently dense, then all bets are off.

I'm a physicist, so apologies if the answer below is in a foreign language; but this was too interesting of a problem to pass up. I'm going to focus on a particular question: If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region? The TL;DR is that thermal effects are going to make this idea basically impossible.

The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$, where $\vec{B}$ is the magnetic field. Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point: $\vec{m} = \alpha \vec{B}$, where $\alpha$ is what we might call the "molecular magnetic susceptibility." Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$. But potential energy is given by $\vec{F} = - \vec{\nabla} U$; which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$.

Now, if we're talking about a sample of gas at a temperature $T$, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor: $$ \rho(\vec{r}) \propto e^{-U(\vec{r})/kT} = e^{-\alpha B^2/kT} $$ In the limit where $kT \gg \alpha B^2$, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample. To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$; in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.

So how high does this have to be? The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_A \alpha$; and so we have¹ $$ \chi_\text{mol} B^2 \gtrsim \mu_0 RT $$ and so we must have $$ B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}. $$ If you believe Wikipedia, the molar susceptibility of oxygen gas is $4.3 \times 10^{-8} \text{ m}^3/\text{mol}$; and plugging in the numbers, we get a requirement for a magnetic field of $$ B \gtrsim 258 \text{ T}. $$ This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25-100 times stronger than most MRI machines. Even at 91 Kelvin (just above the boiling point of oxygen), you would need a magnetic field of almost 150 T; still well out of range.

¹ I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules. A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation; and if the gas gets sufficiently dense, then all bets are off.

I'm a physicist, so apologies if the answer below is in a foreign language; but this was too interesting of a problem to pass up. I'm going to focus on a particular question: If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region? The TL;DR is that thermal effects are going to make this idea basically impossible.

The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$, where $\vec{B}$ is the magnetic field. Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point: $\vec{m} = \alpha \vec{B}$, where $\alpha$ is what we might call the "molecular magnetic susceptibility." Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$. But potential energy is given by $\vec{F} = - \vec{\nabla} U$; which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$.

Now, if we're talking about a sample of gas at a temperature $T$, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor: $$ \rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT} $$ In the limit where $kT \gg \alpha B^2$, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample. To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$; in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.

So how high does this have to be? The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$; and so we have¹ $$ \chi_\text{mol} B^2 \gtrsim \mu_0 RT $$ and so we must have $$ B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}. $$ If you believe Wikipedia, the molar susceptibility of oxygen gas is $4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$; and plugging in the numbers, we get a requirement for a magnetic field of $$ B \gtrsim 258\ \text{T}. $$ This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines. Even at 91 kelvin (just above the boiling point of oxygen), you would need a magnetic field of almost 150 T; still well out of range.

¹ I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules. A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation; and if the gas gets sufficiently dense, then all bets are off.